Open Ultrasound Lab

Interview preparation

368 interview and qualifying-exam questions from all 33 modules, each with a model answer and the follow-up an interviewer would ask next. Tags mark the level: screen, onsite or research. Click a question to reveal the answer.

Modules

01Complex numbers and phasors

11 questions · open in the module

I1What does j mean geometrically, and why is j² = −1 not mysterious? · screen

j is the operator "rotate by +90°". Multiplying by −1 is a 180° rotation, so \(j^2 = -1\) says two quarter turns make a half turn. The plane is forced because no arrow on the real line can turn 90°. Follow-up: "So what is 1/j?" A quarter turn back, −j, since \(j(-j) = 1\).

I2Compute (3 + 4j)(1 + 2j) and explain the result in polar form. · screen

\(3 + 6j + 4j + 8j^2 = -5 + 10j\). Polar: \(\lvert z\rvert = 5\), \(\lvert w\rvert = 2.236\), product 11.18; angles 53.13° + 63.43° = 116.57°, and atan2(10, −5) = 116.57°. Multiplication multiplies lengths and adds angles because \(wz = c\,z + d\,(jz)\) is a rotation matrix times \(\lvert w\rvert\). Follow-up: "And the quotient?" \((3 + 4j)(1 - 2j)/5 = 2.2 - 0.4j\), length 2.236, angle −10.30°.

I3Why does the FFT of a real cosine show two peaks? What is a negative frequency? · screen

\(\cos\theta = (e^{j\theta} + e^{-j\theta})/2\): two half-amplitude phasors turning in opposite directions, whose imaginary parts cancel at every instant. A negative frequency is clockwise rotation, not time reversal. For any real signal \(X(-f) = X^*(f)\), so the negative half is redundant. Follow-up: "Amplitude of each peak for A = 1, N points, no normalization?" N/2 each, when the tone is exactly on a bin.

I4What is the physical meaning of the Q channel, and how is it measured? · onsite

\(A\cos(\omega t + \varphi) = I\cos\omega t + Q\cos(\omega t + 90^\circ)\), with \(I + jQ = A e^{j\varphi}\). Q is the amount of the 90°-shifted reference in the signal. It is measured with a second mixer whose oscillator is shifted by a quarter period (\(Q = -2\operatorname{LPF}\{x\sin\omega_d t\}\)), or, for a narrowband signal, by sampling the RF a quarter period earlier: \(x(t_0 - T/4) = Q\). Follow-up: "When does the quarter-period trick fail?" When the envelope changes within T/4: about 20 % rms error at 60 % bandwidth for a 5 MHz Gaussian echo.

I5Derive Euler's formula without the Taylor series. · onsite

Let \(f(\theta) = e^{j\theta}\), so \(f' = jf\), \(f(0) = 1\). Then \(d\lvert f\rvert^2/d\theta = f'f^* + f f'^* = j\lvert f\rvert^2 - j\lvert f\rvert^2 = 0\), so \(\lvert f\rvert = 1\); the speed \(\lvert f'\rvert = 1\), so the point covers arc length θ on the unit circle starting at 1, counterclockwise because the velocity at 1 is j. Its coordinates are (cos θ, sin θ). Alternatively, \((1 + j\theta/n)^n\) is n rotations by atan(θ/n) with total stretch \(e^{\theta^2/(2n)} \to 1\). Follow-up: "What does the limit tell you about forward Euler on an oscillator?" It spirals outward by \(\sqrt{1 + \omega^2\Delta t^2}\) per step for any Δt.

I6You demodulate at fd = 4.5 MHz but the echo is centred at 5 MHz. What changes in I, Q and the envelope? · onsite

\(\mathrm{IQ}(t) = A(t - T_e)\,e^{j[2\pi(f_0 - f_d)t - 2\pi f_0 T_e]}\): the envelope \(\lvert\mathrm{IQ}\rvert\) is unchanged (as long as the band stays inside the low-pass), but the arrow keeps turning at +0.5 MHz, one turn per 2 µs. I and Q oscillate under the envelope; phase-based estimators see a frequency offset. Follow-up: "Why does depth matter?" Attenuation lowers the echo centre frequency with depth, so a fixed \(f_d\) creates a depth-dependent residual rotation that biases phase-to-delay conversions.

I7Two equal point scatterers lie in one resolution cell, separated axially by Δz. When do they cancel at 5 MHz? · onsite

Round-trip phase difference \(4\pi\Delta z/\lambda\); \(\lvert\text{sum}\rvert = 2\lvert\cos(2\pi\Delta z/\lambda)\rvert\), zero at Δz = λ/4 = 77 µm (λ = 0.308 mm), maximum at multiples of λ/2 = 154 µm. Follow-up: "So what is speckle?" The sum of many such arrows with random phases: a 2-D random walk whose length is Rayleigh distributed (Module 07).

I8In phase-based displacement estimation, what limits the displacement per frame, and how do you extend it? · research

The phase step must stay below π, i.e. \(\lvert\Delta z\rvert < \lambda/4\) per frame (77 µm at 5 MHz), otherwise unwrapping chooses the wrong turn and the error propagates. Extensions: higher frame rate; lower \(f_0\) for the coarse estimate; coarse delay from cross-correlation of envelopes or RF and fine delay from phase; spatial or temporal continuity priors for 2-D/3-D unwrapping; or using the phase of a lower difference frequency (two-frequency methods). Follow-up: "What does noise do?" A phase noise σ near 1 rad produces steps beyond π by itself, so unwrap fails even for slow motion.

I9Your I and Q channels have a 1 % gain mismatch and a 1° phase error. What artefact appears in spectral Doppler, and how large is it? · research

\(z = \cos\theta + j g\sin(\theta + \delta)\) \(= e^{j\theta}\frac{1 + g e^{j\delta}}{2} + e^{-j\theta}\frac{1 - g e^{-j\delta}}{2}\). The second term is a mirror image at \(-f_D\) with relative level \(\lvert 1 - g e^{-j\delta}\rvert / \lvert 1 + g e^{j\delta}\rvert\) = −40.0 dB for g = 1.01, δ = 1° (−26 dB for 5 % and 5°). Geometrically the counter-rotating halves are unequal in the wrong direction, so the IQ circle becomes an ellipse. Follow-up: "How do you correct it?" Estimate g and δ from a calibration tone and apply the inverse 2×2 transform to (I, Q).

I10Why do differentiation-based estimates (velocity from displacement, strain, instantaneous frequency) amplify noise, and by how much? · research

d/dt multiplies each component by \(j2\pi f\), so the noise power spectrum is weighted by \((2\pi f)^2\). For white noise of density \(N_0\) band-limited to ±B, the derivative variance is \(\int_{-B}^{B} N_0 (2\pi f)^2\,df = N_0 (2\pi)^2\,\frac{2B^3}{3}\), growing as B³ while the signal's derivative grows only with its own bandwidth. Strain estimators therefore fit a least-squares slope over a kernel (a low-pass differentiator), trading resolution for SNR. Follow-up: "Discrete first difference?" \(\lvert H\rvert = 2 f_s\lvert\sin(\pi f/f_s)\rvert\) with a half-sample delay.

I11How does a time delay appear on IQ data, and why must you rotate after interpolating IQ samples? · onsite

RF delayed by τ has analytic signal \(A(t - \tau)e^{j2\pi f_0(t - \tau)}\). After demodulation at \(f_d\), \(\mathrm{IQ}_{\text{delayed}}(t) = A(t - \tau)e^{j2\pi f_0(t - \tau)}e^{-j2\pi f_d t}\) \(= \mathrm{IQ}(t - \tau)\,e^{-j2\pi f_d\tau}\). A beamformer undoes the delay: interpolating a channel at \(t + \tau\) realigns the envelope but not the carrier phase, which still carries the factor \(e^{-j2\pi f_d\tau}\). So the interpolated sample must be multiplied by \(e^{+j2\pi f_d\tau}\) (the sign is for demodulation with \(e^{-j2\pi f_d t}\); it flips with the opposite convention). Without it, channels add with phase errors of \(2\pi f_d\tau\) and the beam is destroyed. Follow-up: "Which frequency, \(f_0\) or \(f_d\)?" \(f_d\), the frequency actually removed (Module 08).

02The DFT and FFT by hand

10 questions · open in the module

I1What does the imaginary part of an FFT bin physically mean? · screen

It is minus the correlation of the signal with a sine at that bin's frequency, i.e. the quadrature content. For real x, x[n] contains \(\frac{2}{N}[\mathrm{Re}\,X[k]\cos(2\pi kn/N)\) \(- \mathrm{Im}\,X[k]\sin(2\pi kn/N)]\), so \(-\frac{2}{N}\mathrm{Im}\,X[k]\) is the sine amplitude (02.10). A pure sine at bin 1 of an 8-point FFT gives \(X[1] = -4j\). It is not "imaginary" in any physical sense: it is the 90°-shifted projection, and it changes if you move the time origin. Follow-up: what is \(\mathrm{Im}\,X[0]\)? (Zero for real x; there is no sine at DC.)

I2A 1 V amplitude sinusoid sits exactly on bin 100 of a 1024-point FFT. What does abs(X(101)) read, and how do you convert to volts? · screen

\(AN/2 = 512\). Divide by N/2 for the one-sided amplitude (1 V). With a Hann window it reads 256 (coherent gain 0.5). If the tone sits half a bin off, the rectangular peak drops to \(512 \times 2/\pi \approx 326\) (−3.92 dB). Follow-up: where is the other half? (Bin 924, the conjugate.)

I3Why do the upper N/2 bins mirror the lower ones for real data, and when do they not? · screen

\(e^{-j2\pi(N-k)n/N} = e^{+j2\pi kn/N}\), so bin N − k is frequency −k, and for real x it equals \(X[k]^{*}\). A real cosine is two counter-rotating phasors, so each appears twice with half amplitude. Complex I/Q data have no such symmetry: after demodulation the upper bins are genuine negative frequencies, which encode flow direction in Doppler.

I4Derive the radix-2 FFT butterfly. · onsite

Split n into even and odd: \(X[k] = \sum x[2m]W_N^{2mk}\) \(+ W_N^k\sum x[2m+1]W_N^{2mk}\). Use \(W_N^2 = W_{N/2}\), so the sums are N/2-point DFTs \(E[k]\), \(O[k]\). They have period N/2 and \(W_N^{k+N/2} = -W_N^k\), so \(X[k] = E + W^kO\) and \(X[k + N/2] = E - W^kO\). Recursion gives \(\log_2 N\) stages of N/2 butterflies: \((N/2)\log_2 N\) multiplies. Follow-up: why is the input bit-reversed? (Each split sorts by the lowest remaining bit.)

I5Does zero padding improve frequency resolution? · screen

No. It samples the same DTFT more finely; the main-lobe width is set by the record length, ≈ \(1/T = f_s/N\). Padding a 2 µs gate (80 samples at 40 MHz) to 256 gives a 156 kHz grid, but two components closer than about 500 kHz cannot be reliably separated. It does improve peak location by interpolation and prevents wrap-around in FFT convolution. Follow-up: what does improve resolution? (A longer gate, or model-based methods such as AR/Burg with their own bias, Module 27.)

I6What is the scalloping loss, and how would you remove it for amplitude measurements? · onsite

The peak-bin amplitude deficit for a tone between bins: \(\lvert D_N(\tfrac12)\rvert/N \approx 2/\pi\), −3.92 dB, rectangular window; −1.42 dB with Hann. Remedies: a flat-top window (wide main lobe, small amplitude error), zero padding and peak interpolation (parabolic on log magnitude), or fitting the known kernel. Follow-up: what does that cost in noise bandwidth? (Wider windows raise ENBW, 1.5 bins for Hann, more for flat-top.)

I7A tone is 20 dB below the per-sample noise. How long an FFT do you need to see it at 10 dB SNR in its bin? · onsite

Bin SNR = \((N/2)\) × per-sample SNR, so \(N/2 = 10^{(10 + 20)/10} = 1000\), N ≈ 2000; take 2048 (gain 30.1 dB). That assumes the tone is coherent over the whole record, sits on a bin, and the noise is white. Follow-up: with Hann? (ENBW 1.5 costs 1.76 dB, so go to 4096 or accept 8.3 dB.)

I8How do you build the frequency axis for fftshift(fft(x)) with N odd? · screen

f = (-floor(N/2):ceil(N/2)-1)*fs/N, which for N = 7 is \((-3 \dots 3)\cdot f_s/7\): no Nyquist bin. For N even it is −N/2 … N/2 − 1, with \(-f_s/2\) first. Undo with ifftshift, not fftshift, which differ by one sample for odd N. Follow-up: where is DC after the shift? (Index floor(N/2) + 1 in MATLAB.)

I9Show that the DFT's inverse is \(\frac{1}{N}\mathbf{F}^{H}\) without inverting anything. · research

\((\mathbf{F}^{H}\mathbf{F})_{lk} = \sum_n e^{j2\pi(l-k)n/N} = N\,\delta[l-k]\) by the geometric series with \(r^N = 1\). So \(\mathbf{F}^{H}\mathbf{F} = N\mathbf{I}\), \(\mathbf{F}\) is \(\sqrt N\) times a unitary matrix, and Parseval follows. Follow-up: what are the eigenvalues of \(\mathbf{F}/\sqrt N\)? (Only ±1 and ±j: \(\mathbf{F}^4 = N^2\mathbf{I}\).)

I10You need the spectrum of 128 RF lines of 2048 samples each, real-time. What FFT tricks do you use? · research

Pack pairs of lines as \(x + jy\) and split with the conjugate-symmetry formulas (02.22) (64 complex FFTs instead of 128); keep only N/2 + 1 bins; batch the transforms (GPU or FFTW plans reused across lines); if only a band is needed, demodulate and decimate first (Module 08), so N shrinks by the decimation factor. Cost per line ≈ \((N/2)\log_2 N\) = 11 264 complex multiplies vs 4.2 million direct. Follow-up: how does the 2D FFT of the whole frame relate? (Rows then columns: k-space, Module 04.)

03Convolution, LTI systems and filters

11 questions · open in the module

I1Is time-gain compensation an LTI operation? Is envelope detection? · screen

TGC, \(y[n] = g[n]\,x[n]\), is linear (scaling and adding inputs scales and adds outputs) but time-varying: an echo at a later time gets a larger gain, so delaying the input does not just delay the output. Envelope detection is time-invariant but nonlinear (the envelope of \(-x\) equals the envelope of \(x\)). Neither has a transfer function. Follow-up: "So where may you filter?" Anywhere before the envelope; and TGC commutes with an LTI filter only approximately, when \(g\) varies slowly over the filter length.

I2Convolve [1, 2, 3] with [2, 1, −1] by hand and give two checks. · screen

\([2, 5, 7, 1, -3]\). Checks: length \(3 + 3 - 1 = 5\); \(\sum y = 12 = \sum x \cdot \sum h = 6 \cdot 2\); and the alternating sum (\(\omega = \pi\)): \(2 - 5 + 7 - 1 - 3 = 0\) \(= (1 - 2 + 3)(2 - 1 - 1) = 2 \cdot 0\). Follow-up: "Do it with a 4-point FFT." You get \([-1, 5, 7, 1]\) because the −3 wraps onto \(n = 0\); pad to 5 or more.

I3Why are complex exponentials the eigenfunctions of LTI systems? · screen

\(y[n] = \sum_k h[k]\,e^{j\omega(n-k)}\) \(= e^{j\omega n}\sum_k h[k]\,e^{-j\omega k}\) \(= H(e^{j\omega})\,e^{j\omega n}\). The shift-invariance turns a delay into multiplication by \(e^{-j\omega k}\), and exponentials are the functions whose shifts are scalar multiples of themselves. Matrix view: convolution matrices are Toeplitz (circulant with padding), and the DFT vectors diagonalize every circulant. Follow-up: "Are sines eigenfunctions?" Not individually: a sine comes out as a scaled sine plus a cosine unless \(H\) is real at that \(\omega\).

I4What does the imaginary part of a frequency response mean physically? · onsite

For a cosine input, \(y = \mathrm{Re}\,H\cos\omega n - \mathrm{Im}\,H\sin\omega n\). \(\mathrm{Re}\,H\) is the gain of the in-phase copy, \(-\mathrm{Im}\,H\) the gain of a copy shifted 90° (a quarter period late if \(\mathrm{Im}\,H < 0\)). Example: \(h = [\tfrac12, \tfrac12]\) at \(\omega = \pi/2\) has \(H = 0.5 - 0.5j\), so the output is half in-phase plus half lagging quadrature, net 0.707 at −45°, a half-sample delay. \(\mathrm{Im}\,H\) comes from the odd part of \(h\) about \(n = 0\). Follow-up: "Which filter has \(\mathrm{Re}\,H = 0\) everywhere?" Any odd \(h\), e.g. the Hilbert transformer (\(-j\,\mathrm{sgn}\,\omega\)) or the centered difference (\(j\sin\omega\)).

I5Phase delay versus group delay: which one tells you where the echo envelope lands? · onsite

Group delay \(\tau_g = -d\varphi/d\omega\) at the pulse center frequency gives the envelope delay; phase delay \(-\varphi/\omega\) gives the carrier delay. For linear phase they coincide; for a 41-tap symmetric FIR at 40 MHz both are 20 samples = 0.5 µs, 0.385 mm of depth. Follow-up: "Why does this matter for beamforming?" A phase-delay mismatch between channel filters looks like a time-delay error: 10° at 5 MHz is 5.6 ns.

I6Derive the convolution theorem and say what changes for the DFT. · onsite

\(Y = \sum_n\sum_k x[k]\,h[n-k]\,e^{-j\omega n}\); write \(e^{-j\omega n} = e^{-j\omega k}\,e^{-j\omega(n-k)}\) and set \(m = n - k\): \(Y = X H\). For the N-point DFT the indices are mod \(N\), so DFT products give circular convolution; linear convolution needs \(N \ge L + M - 1\). Follow-up: "And the dual?" Multiplication in time is circular convolution of the DFTs divided by \(N\): windowing smears lines (leakage).

I7A one-pole smoother has a = 0.9. Give its impulse response, DC gain, −3 dB frequency, delay and stability. · onsite

\(h[n] = 0.1 \cdot 0.9^n\), \(\sum h = 1\) (DC gain 1). \(\lvert H\rvert^2 = 0.01/(1 - 1.8\cos\omega + 0.81) = \tfrac12\) at \(\cos\omega = 1 - 0.01/1.8\), \(\omega = 0.105\) rad/sample (\(\approx 1 - a\)). Time constant \(-1/\ln 0.9 = 9.5\) samples; group delay at DC \(a/(1-a) = 9\) samples. Pole at 0.9, inside the circle: stable. Noise variance gain \(0.1/1.9 = 1/19\). Follow-up: "Frame rate 50 Hz?" Time constant 0.19 s, cutoff \(0.105 \times 50/(2\pi) = 0.84\) Hz.

I8Why can't you deconvolve an RF line by dividing spectra, and what do you do instead? · research

\(\hat S = Y/P = S + W/P\): outside the transducer band \(\lvert P\rvert\) is tiny (for a 5 MHz, 60 % pulse, −67 dB at 10 MHz), so noise is amplified enormously. Use Wiener \(G = P^{*}/(\lvert P\rvert^2 + \varepsilon)\) with \(\varepsilon\) the noise-to-signal ratio, whose gain is capped at \(1/(2\sqrt\varepsilon)\), or sparse or Bayesian inversion with priors on \(\rho\). Follow-up: "Limits?" No method restores frequencies with no signal, \(P\) must be known (it changes with depth through attenuation, so blind or depth-adaptive estimates are used), and phase errors in \(P\) bias the result.

I9Is the pulse-echo RF line really an LTI convolution of a pulse with reflectivity? · research

Only locally. Frequency-dependent attenuation low-passes the pulse progressively with depth (center frequency downshift), diffraction changes the pulse off focus, TGC adds a time-varying gain, and multiple scattering and nonlinear propagation add non-convolutional terms. Practice: assume shift invariance within short windows (a few pulse lengths), estimate a local pulse, and compensate attenuation. Follow-up: "How does this affect sound-speed estimation?" Methods that compare phase or delays between windows or channels assume the same local pulse on both sides; depth-dependent spectral change biases them unless both are measured over matched windows.

I10Can a causal filter have a purely real frequency response? · research

Real \(H\) for all \(\omega\) requires \(h\) even about \(n = 0\) (for real \(h\)); a causal \(h\) is zero for \(n < 0\), so evenness forces \(h = c\,\delta[n]\). Any useful causal filter has phase. The best you can do causally is linear phase (symmetric FIR, pure delay). Offline, filter forward and then backward: \(H H^{*} = \lvert H\rvert^2\), zero phase, magnitude squared. Follow-up: "Price of zero phase?" Doubled dB figures, non-causal transients at both ends, and it cannot run in real time.

I11An IIR filter has a pole at z = 1.02. What do you see, and how do you fix it? · onsite

The impulse response contains \(1.02^n\), doubling every \(\ln 2/\ln 1.02 = 35\) samples: the output eventually overflows whatever the input. Fix: redesign so all poles are inside the circle, or if the pole moved because of coefficient quantization, implement as cascaded biquads (second-order sections) where each pole pair depends on two coefficients. Follow-up: "Pole exactly at 1?" An integrator: a bounded input of constant sign (DC) makes the output grow linearly without bound, so it is not BIBO stable; zero-mean noise makes it wander like a random walk.

04Sampling, Fourier & LTI systems

12 questions · open in the module

I1Why is the beam pattern the Fourier transform of the aperture? · onsite

Each aperture point radiates a spherical wave; in the paraxial approximation the path length from \(\xi\) to (x, z) is \(z + (x - \xi )^{2}/(2z)\), so the field is \(\int a(\xi ) e^{j\pi \xi ^{2}/(\lambda z)} e^{-j2\pi \xi x/(\lambda z)} d\xi\) times a phase. When the quadratic term is negligible (\(z \ge 2D^{2}/\lambda\), the Fraunhofer region) it is the FT of \(a(\xi )\) at \(x/(\lambda z) = \sin \theta /\lambda\). Focusing applies \(e^{-j\pi \xi ^{2}/(\lambda F)}\), which cancels the quadratic phase at \(z = F\), so the focal plane is a Fraunhofer plane for any F. Rect aperture → sinc with first null at \(\lambda F\#\).

Follow-up: what happens away from focus? The quadratic phase returns (defocus), nulls fill and the mainlobe widens; the depth of field is ≈ \(7 \lambda F\#^{2}\).

I2Why do two-way sidelobes double in dB, and when is that not true? · screen

By reciprocity the receive sensitivity pattern equals the transmit pattern of the same aperture, and the pulse-echo response is their product. With identical apertures and focus the amplitude is squared, so 20 log of the square doubles every level: −13.26 → −26.52 dB for rect, −31.5 → −63 dB for Hann. It fails when Tx and Rx differ (fixed transmit focus with dynamic receive, different apodizations or aperture sizes): levels add in dB angle by angle, and a rect receive with Hann transmit has its first sidelobe at −24.8 dB. It also assumes narrowband; a broadband pulse fills the nulls.

Follow-up: what is the two-way −6 dB width? \(0.886 \lambda F\#\), equal to the one-way −3 dB width.

I3Where do grating lobes appear for pitch = \(\lambda\) at 20° steering? · screen

\(\sin \theta _{g} = \sin \theta _{0} \pm \lambda /d\) \(= 0.342 \pm 1\). The + sign gives 1.342 (invisible); the − sign gives \(-0.658, \theta _{g}\) = −41.1°, at full height for point elements. With element width \(0.9\lambda\), the element factor puts it at −4.4 dB one-way, −8.7 dB two-way relative to the main lobe. To remove it for \(\pm 20^{\circ}: d\) < λ/(1 + sin 20°) = \(0.745\lambda\) (0.230 mm at 5 MHz).

Follow-up: what does a broadband pulse do to it? The grating-lobe angle depends on \(\lambda\), so the lobe spreads over angles and its peak drops, but its energy stays.

I4What sets axial versus lateral resolution? · screen

Axial: the pulse length, i.e. bandwidth: \(\Delta z = c\tau /2\) with \(\tau \approx 0.883/B\) (Gaussian, −6 dB). 5 MHz, 60 %: \(B = 3\) MHz, \(\Delta z = 0.227\) mm, independent of \(f_{0}\). Lateral: diffraction, i.e. aperture and wavelength: −6 dB two-way width \(0.886 \lambda F\# = 0.546\) mm at 5 MHz, F# 2. Raising \(f_{0}\) at fixed bandwidth improves lateral but not axial; widening the band improves axial only; elevation is set by the lens and the elevation aperture.

Follow-up: why does ULM beat both? It localizes isolated sources, which is estimation (precision scales with SNR and PSF width), not resolution.

I5A 7 MHz tone is sampled at 10 MHz. What do you see? · screen

\(k = \operatorname{round} (7/10)\) \(= 1, f_{0} - k f_{s} = -3\) MHz, so a 3 MHz tone. The negative sign means it comes from the −7 MHz image: the spectrum is mirrored and the phase conjugated, \(\cos (2\pi \cdot 7t + \varphi )\) \(\equiv \cos (2\pi \cdot 3t - \varphi )\) at the sample instants. The samples cannot distinguish them; only an anti-alias filter before the ADC can.

Follow-up: the Doppler analogue? \(v_{\mathrm{Nyq}} = c\cdot \mathrm{PRF}/(4f_{0})\); flow faster than that wraps to the opposite sign.

I6Why does zero padding not improve frequency resolution? · screen

Resolution is the width of the kernel the spectrum is convolved with, set by the record length T (\(\approx 1/T\) for rect, 1.44/T −3 dB for Hann). Zero padding appends samples that carry no information, so the DTFT of the padded record is identical; the DFT just samples it more densely (\(f_{s}/(PN)\) instead of \(f_{s}/N\)). Two equal, in-phase tones 0.6/T apart stay one lobe at any padding (other relative phases can split them, Demo 04.4); a 4× longer record separates them.

Follow-up: what does padding buy? Interpolated peak location, less scalloping in display, and the length needed to avoid circular wrap in FFT convolution.

I7What bandpass sampling rate works for a 3–7 MHz band? · onsite

\(B = 4\) MHz, \(\lfloor f_{H}/B\rfloor = \lfloor 7/4\rfloor = 1\), so only the ordinary Nyquist zone exists: \(f_{s} \ge 14\) MHz. Undersampling needs \(f_{L} \ge B\) (fractional bandwidth ≤ 2/3); this band is 80 %. Alternatives: demodulate to baseband IQ and sample I and Q at ≥ 4 MHz each, or narrow the band with a filter first. For [3.5, 6.5] MHz, \(6.5 \le f_{s} \le 7.0\) MHz works, spectrally inverted \((n = 2)\), with only 0.5 MHz of margin.

Follow-up: why \(4f_{0}?\) Consecutive samples fall at carrier phases \(0, \pi /2, \pi ,\ 3\pi /2\) and give I, −Q, −I, Q directly.

I8You filter RF data by multiplying FFTs and see artifacts at the start of each line. Why? · screen

The DFT diagonalizes circular convolution. With an N-point FFT, the output is the linear result folded with period N; if \(N < L + M - 1\), the filter tail past sample N wraps onto the first samples, so late echoes contaminate shallow depths. Fix: \(N \ge L + M - 1\) \((2048 + 63 - 1 = 2110 \to 4096)\), or overlap-add/save. In 2D image filtering the same thing makes edges bleed across the image; pad both dimensions.

Follow-up: and the 63-tap filter's delay? 31 samples = 0.775 µs = 0.60 mm at 40 MHz; compensate it identically on every channel.

I9How does F-number trade lateral resolution against depth of field? · onsite

Lateral −6 dB width (two-way) ≈ \(0.886 \lambda F\#\), linear in F#; depth of field ≈ \(7 \lambda F\#^{2}\), quadratic. At 5 MHz: F# 2 gives 0.55 mm and 8.7 mm; F# 4 gives 1.09 mm and 35 mm. Halving F# doubles resolution but cuts depth of field by 4, which is why systems use multiple transmit foci or dynamic receive focusing with an expanding aperture (constant receive F#).

Follow-up: what limits the smallest F#? Element directivity (acceptance angle) and the number of channels; below \(F\# \approx 1\) the paraxial formulas also degrade.

I10Sketch the 2D spectrum of an RF image and explain how plane-wave compounding fills it. · research

Two lobes at \(k_{z} = \pm 2f/c\) (\(\pm 6.49\) cycles/mm at 5 MHz, width set by the pulse band) forming a bow-tie: at each frequency \(|k_{x}| \le D/(\lambda z) = 1/(\lambda F\#)\), the convolution of Tx and Rx aperture spectra. A single plane wave has no transmit aperture, so the lateral extent comes from receive only; each angle \(\theta\) shifts it by \(\sin \theta /\lambda\). Compounding angles lays these regions side by side and synthesizes a transmit aperture. A uniform step \(\Delta (\sin \theta )\) samples that synthetic aperture, so the focus is replicated every \(\lambda /\Delta (\sin \theta )\) laterally (17.6 mm for 1° at 5 MHz).

Follow-up: how does a sound-speed error show up? The k-space mapping \(k_{z} = 2f/c\) is scaled and the regions from different angles no longer add coherently: defocus, which is what coherence- and phase-based sound-speed estimators detect.

I11For a QUS power-spectrum estimate from a 3 mm gate, which window and which corrections? · onsite

The gate is 3.9 µs, so spectral resolution is ≈ 0.26 MHz with rect and ≈ 0.37 MHz (−3 dB) with Hann. Backscatter spectra are smooth and span tens of dB, so Hann is the usual choice: its −31.5 dB sidelobes and −18 dB/octave roll-off limit leakage from the passband peak into the band edges. Corrections: divide by \(\mathrm{CG}^{2}\) (Hann 0.25) for coherent levels, or by \(\sum w^{2} (\mathrm{ENBW})\) for noise-like power; the same window on the sample and the reference phantom cancels most of it in the ratio. Average over independent gates to reduce the variance of the estimate.

Follow-up: longer gate? Better spectral resolution and lower variance, worse depth resolution, and attenuation within the gate biases the estimate.

I12Your simulated narrowband beam shows deep nulls, the measured beam does not. Why? · research

A real pulse contains many frequencies, and each puts its nulls at \(x \propto 1/f\), so the energy pattern has none: with a 60 % band the rect two-way first null fills to −24 dB while the sidelobe level drops from −26.5 to −28.8 dB. Other contributors: element-to-element gain and delay errors (random phase errors of variance \(\sigma ^{2} \mathrm{rad}^{2}\) add a diffuse sidelobe floor of about \(\sigma ^{2}/N\) relative to the peak, one-way, uniform weights), quantized delays, noise, and sound-speed or aberration-induced defocus. Compare against a broadband simulation before blaming hardware.

Follow-up: how would you measure the beam? Scan a wire or small sphere laterally in a water tank in the focal plane and record the peak or energy of the echo versus offset.

05Matrices as operators

10 questions · open in the module

I1In MATLAB, what is the difference between ' and .', and which do you use on IQ data? · screen

' is the conjugate (Hermitian) transpose, .' the plain transpose; on real data they agree. For z = (1, j), z.'*z = 1 + \(j^{2} = 0\) while z'*z = 2: only \((\cdot)^{H}\) gives an energy. Every energy, correlation, covariance (\(R = X^{H}X/K\)), projection and adjoint on IQ data uses '. A pure layout change (samples × elements to elements × samples) must use .', otherwise every phase flips sign and Doppler or phase-aberration estimates change sign.

Follow-up: "numpy?" x.T does not conjugate, x.conj().T does, and np.dot does not conjugate while np.vdot conjugates its first argument.

I2Why is DAS not the least-squares inverse of the forward model? · onsite

With \(\mathbf{d} = H\mathbf{\gamma} + \mathbf{n}\), the least-squares estimate is \((H^{H}H)^{-1}H^{H}\mathbf{d}\). DAS applies only \(H^{H}\) (up to apodization and the pulse matched filter): it sums each pixel's data along that pixel's hyperbola, the adjoint, so the image is \(H^{H}H\mathbf{\gamma}\), reflectivity blurred by the PSF matrix. \(H^{H}H\) is far from the identity (finite aperture and bandwidth) and is badly conditioned, so the exact inverse would amplify noise along weak singular vectors; in practice one uses \((H^{H}H + \lambda I)^{-1}H^{H}\), solved iteratively with B and \(B^{H}\) (CG/LSQR), never by forming \(H^{H}H\).

Follow-up: "Then why is DAS so widely used?" It is the matched filter: for white noise and a single scatterer it maximizes SNR at that pixel, it is cheap (one sparse product), and it is robust to model error, which the inverse is not.

I3What does a large condition number mean for pulse-echo sound-speed tomography? · research

In \(A \Delta \mathbf{s} = \Delta \mathbf{t}, \kappa (A)\) is the worst-case amplification of relative delay error into relative slowness error, and \(\log_{10}\kappa\) is the number of digits lost. Limited-angle pulse-echo geometry makes axially adjacent pixels see nearly the same steep paths, so columns are nearly parallel and κ is huge (effectively a null space for axial variations). With delays known to 1% and \(\kappa = 10^{3}\), the unregularized map can be 10× wrong in its weak directions. The cure is regularization (Tikhonov, smoothness or TV priors) and more angular diversity (more transmit/receive angles), with bias as the price: the reconstruction blurs or biases exactly the patterns the geometry cannot see.

Follow-up: "How do you choose λ?" L-curve corner, discrepancy principle (residual ≈ expected delay noise), or GCV; and validate on a phantom with known layers, because the bias is structured.

I4K snapshots across N elements: what is the rank of the sample covariance, and what does that imply for minimum-variance beamforming? · onsite

\(\hat{R} = (1/K)X^{H}X\) with \(X \in \mathbb{C}^{K\times N}\), so \(\operatorname{rank} \hat{R} \le \min (K,N)\). For K < N it is singular and \(\hat{R}^{-1}\) in the MV weights \(w = \hat{R}^{-1}\mathbf{a}/(\mathbf{a}^{H}\hat{R}^{-1}\mathbf{a})\) does not exist. Ultrasound MV therefore averages over subarrays of length L < N (and over a few depth samples) to get enough snapshots, and adds diagonal loading \(\hat{R} + \varepsilon I\), which is Tikhonov. The Reed–Mallett–Brennan result says about K = 2N independent snapshots keep the expected SINR loss near 3 dB when the snapshots do not contain the desired signal; in ultrasound they do, so convergence is slower and loading is needed even at K ≥ 2N.

Follow-up: "What does the eigen-spectrum tell you?" The number of large eigenvalues is the number of coherent sources; the rest is the noise floor, the basis of subspace methods (MUSIC, eigenspace MV) and of the clutter-rank discussion in Module 06.

I5How do you compute a pseudo-inverse stably? · onsite

Through the SVD: \(A^{+} = V\Sigma^{+}U^{H}\) with \(1/\sigma_{i}\) only for \(\sigma_{i}\) above a tolerance (MATLAB pinv uses about \(\max (m,n)\cdot \sigma_{\max}\cdot \mathrm{eps}\); pinv(A,tol) raises it, which is truncated-SVD regularization). Never form \((A^{T}A)^{-1}A^{T}\) for an ill-conditioned A: it squares κ. For full column rank, QR (A\b) gives the same least-squares solution more cheaply; for rank-deficient problems lsqminnorm or a complete orthogonal decomposition gives the minimum-norm solution. For large sparse operators, LSQR with early stopping acts as a regularizer.

Follow-up: "How do you pick the tolerance?" From the noise: drop \(\sigma_{i}\) whose \(1/\sigma_{i}\) amplification pushes the noise above the signal along \(v_{i}\), i.e. the discrete Picard condition (\(\lvert u_{i}^{H}b\rvert\) should decay faster than \(\sigma_{i}\)).

I6Size, nonzeros and memory of the DAS matrix for 128 elements, 2048 samples and a 256 × 256 image? Would you store it? · screen

B is 65 536 × 262 144. With linear interpolation each pixel row has 2 taps per element: 65 536 · 256 = 16.8 M nonzeros, density 0.1%. Sparse double storage ≈ 16 B per nonzero ≈ 270 MB; dense would be 137 GB. A product \(B\mathbf{d}\) is 16.8 M multiply-adds per frame (complex data), the same work as a DAS loop. Store it when it is reused many times (fixed geometry over thousands of frames on a GPU, iterative solvers needing B and \(B^{H}\)); compute on the fly when geometry or sound speed changes per frame.

Follow-up: "And with 8 plane-wave angles compounded?" Stack the 8 operators side by side: 8× the columns and nonzeros (134 M, ≈ 2 GB double), or apply them one at a time and sum.

I7Show that the DFT diagonalizes circular convolution. What goes wrong if you FFT-filter an RF line without zero padding? · screen

A circulant is \(C = \sum c_{k}S^{k}\) with S the cyclic shift. The DFT vectors are eigenvectors of S (a cyclic shift multiplies DFT bin k by the unit-modulus phase \(e^{\pm i2\pi k/N}\), sign set by the DFT convention), hence of every polynomial in S, with eigenvalues fft(c): \(C = F^{H} \operatorname{diag} (F\mathbf{c}) F/N, F\) the unnormalized DFT matrix. Linear convolution of N samples with L taps is circular only if both are zero-padded to \(N_{c} \ge N + L - 1\); with \(N_{c} = N\) the last L − 1 output samples wrap into the first L − 1 (error 0.363 for a 31-tap 5 MHz pulse in m05_convolution_matrix.m).

Follow-up: "When is direct convolution faster?" For short kernels: for a 2048-sample line the crossover is near L ≈ 100 taps, lower with real-input FFTs and overlap-add (that count already assumes a precomputed filter spectrum).

I8Adding a quadratic term to your attenuation fit with raw frequency columns makes the coefficients jump around. Why, and what is the fix? · onsite

The columns 1, \(f, f^{2}\) over 2–8 MHz are nearly parallel (a Vandermonde matrix on an off-center interval), so κ grows fast: cond([1 f]) = 16.2 but order 3, 5, 7 give \(6.4\times 10^{3}, 3.8\times 10^{6}, 2.6\times 10^{9}\). Center and scale, t = (f − 5)/3, and the same orders give 7.6, 39 and 215. If you also form \(A^{T}A\) you square these numbers. Fix: centered/scaled (or orthogonal-polynomial) columns, QR solve, and report the covariance \(\sigma^{2}(A^{T}A)^{-1}\) so the reader sees the coefficient uncertainty.

Follow-up: "Does centering change the fitted curve?" No, the column space is the same, so P and the fitted values are identical; only the coefficients and their conditioning change.

I9Write a phase-screen correction as a matrix. When does it fail? · research

For narrowband IQ with element delays \(\delta \tau_{e}\): \(\mathbf{d}_{\mathrm{ab}} = (D \otimes I_{t})\mathbf{d}\), \(D = \operatorname{diag} (e^{-i\omega_{0}\delta \tau_{e}})\); correction is \(D^{H}\) before DAS, N complex multiplies per sample, exact because D is unitary. It fails when the aberrator is not a thin screen at the array: a distributed or deep aberrator gives a field-dependent correction (isoplanatic patches of a few mm), so no single diagonal D works across the image. Broadband data need a delay (phase linear in frequency), not one phase.

Follow-up: "How would you estimate D?" Cross-correlate neighbouring channels on a bright target or speckle (Module 11), or find the phases that maximize coherence or image sharpness; the distortion-matrix approach estimates it per isoplanatic patch.

I10Is \(P = I - \mathbf{1}\mathbf{1}^{T}/N\) a good clutter filter for Doppler ensembles? What are its eigenvalues? · onsite

P is the orthogonal projector onto the complement of the constant vector: eigenvalues 1 (N − 1 times) and 0, \(\operatorname{rank} N - 1, P^{2} = P = P^{T}\). It removes exactly the slow-time mean, i.e. perfectly stationary tissue. Its frequency response is a narrow notch at 0 Hz of width about PRF/N, so moving tissue (breathing, cardiac motion) leaks through and slow flow near the notch is attenuated. Polynomial regression filters project out 1, \(t, t^{2}, \dots\); the SVD filter \(I - U_{k}U_{k}^{H}\) projects out the dominant spatiotemporal components, which handles moving tissue.

Follow-up: "Why does a projector, not a general high-pass, matter?" A projector has no transient for a finite ensemble and leaves the orthogonal complement untouched; IIR filters on short ensembles need initialization tricks.

06Eigendecomposition & SVD

11 questions · open in the module

I1What is the relation between PCA, the eigendecomposition of a covariance and the SVD? Which one do you compute, and why? · screen

For centered data \(\mathbf{X} (N\times p), \mathbf{X} = \mathbf{U}\Sigma \mathbf{V}^{H}\) gives \(\mathbf{C}\) \(= \mathbf{X}^{H}\mathbf{X}/(N-1)\) \(= \mathbf{V}[\Sigma^{2}/(N-1)]\mathbf{V}^{H}\): principal axes = right singular vectors, variances = \(\sigma^{2}/(N-1)\), scores = \(\mathbf{U}\Sigma\). Compute the SVD of \(\mathbf{X}\) when accuracy matters, because forming \(\mathbf{C}\) squares κ; use the eigendecomposition of the small Gram/covariance matrix when speed matters and the needed components are well above \(\sqrt{\varepsilon}\cdot \sigma_{1}\). Follow-up: "Does centering matter for a Casorati matrix?" Yes: removing each pixel's temporal mean is a degree-0 regression wall filter, so \(\sigma_{1}\) then describes tissue motion rather than static brightness.

I2Walk me through an SVD clutter filter, and explain why it beats a high-pass filter when tissue moves. · screen

Reshape the n frames into the Casorati matrix \(\mathbf{S}\) (pixels × frames), take its SVD (in practice the eigendecomposition of \(\mathbf{S}^{H}\mathbf{S}\)), keep the band (06.15) \([k_{\mathrm{lo}}, k_{\mathrm{hi}}]\): \(\mathbf{S}_{f} = \mathbf{SV}_{b}\mathbf{V}_{b}^{H}\), and PD = \((1/n)\sum_{\mathrm{band}}\sigma_{k}^{2}\lvert u_{k}\rvert^{2}\). A high-pass is a projection onto a fixed frequency band, so tissue at \(f_{D} = 2v/\lambda\) (52 Hz at 8 mm/s, 5 MHz) that overlaps flow Doppler cannot be removed without removing flow. The SVD projection is data-derived: coherent tissue occupies its own temporal vectors even at the same \(f_{D}\). Numbers from the MATLAB lab: at 8 mm/s, SVD 15.2 dB vessel contrast vs −0.4 dB for the best IIR; at rest, 24.7 vs 23.1 dB. Follow-up: "When does the SVD lose?" Large vessels (coherent, climb into the tissue ranks), non-rigid or out-of-plane motion (tissue spreads into the blood ranks), noise-dominated depth.

I3How do you choose the SVD thresholds automatically? · onsite

\(k_{\mathrm{hi}}\): from the noise model. Estimate \(\sigma_{n}\) (no-transmit frames, or match the median singular value to the Marchenko–Pastur median) and keep \(\sigma_{k} > \sigma_{n}(\sqrt{m}+ \sqrt{n})\), or use the Gavish–Donoho threshold \((4/\sqrt{3})\sqrt{n}\sigma_{n}\) for a square matrix. \(k_{\mathrm{lo}}\): from what distinguishes tissue: (1) the spatial-similarity matrix of \(\lvert \mathbf{u}_{k}\rvert\) (Baranger et al. 2018), where tissue ranks form a correlated block; (2) the Doppler content of the temporal vectors (mean or RMS frequency above a tissue threshold); (3) the knee of \(\sigma_{k}\) in dB; (4) energy-based rules (cut when the cumulative tissue energy reaches a fraction). Per block if the field is non-stationary. The best \(k_{\mathrm{lo}}\) moves with tissue motion (2 → 15 for 0 → 24 mm/s in the MATLAB sweep), which is why it must be adaptive. Follow-up: "How do you validate a threshold rule without ground truth?" Phantoms with known flow, in vivo reproducibility, residual tissue at bright moving interfaces, slow-flow preservation in known vessels, and for ULM the localization count and velocity distribution vs \(k_{\mathrm{lo}}\).

I4What happens to SVD filtering under large tissue motion, and what do you do about it? · onsite

Rank-1 requires separability. A uniform phase rotation stays rank 1, but speckle that slides across the PSF needs more ranks, roughly one extra per \(\sigma_{z}\) of displacement over the ensemble (0.1 mm → 3 ranks, 1 mm → 13 ranks to −50 dB in the module's simulation). Tissue then spreads into the ranks where blood lives, the optimal \(k_{\mathrm{lo}}\) rises and contrast drops (16.5 → 6.4 dB from 4 to 24 mm/s in the MATLAB sweep). Remedies: shorter ensembles or sliding windows (less displacement per window), motion estimation and registration before the SVD, block-wise SVD with local cutoffs, rejecting frames with large motion, and higher-order methods that model motion explicitly. Follow-up: "What does a shorter ensemble cost?" Doppler resolution PRF/N, fewer blood ranks, higher PD variance.

I5What is the computational cost of an SVD clutter filter, and how would you make it real-time? · onsite

Thin SVD of an M×N Casorati matrix (M ≫ N) is \(O(MN^{2})\). The fast route: \(G = \mathbf{S}^{H}\mathbf{S}\) (\(MN^{2}\), a GEMM that GPUs do well), eig of \(G (N^{3})\), then \(\mathbf{SV}_{b}\) (MNb). For 128×128 pixels and 300 frames: \(1.5\times 10^{9}\) complex MACs for \(G, 2.7\times 10^{7}\) for the eig. Further: compute only the leading \(k_{\mathrm{hi}}\) vectors (randomized SVD or subspace iteration, O(MNk)), update across overlapping windows instead of recomputing, use blocks in parallel. Follow-up: "Any catch with single precision?" Yes: forming G squares κ, so \(\sigma_{k}\) below about \(\sqrt{\varepsilon_{\mathrm{single}}}\cdot \sigma_{1}\) (−69 dB) are corrupted; measured on the simulated ensemble the first > 1 % error appears at −66.8 dB, right at the blood/noise boundary. Either keep G in double, or run the SVD on \(\mathbf{S}\) directly.

I6Why are singular vectors unique only up to sign or phase, and when does that bite you? · onsite

\(\sigma \mathbf{uv}^{H}\) \(= \sigma (\mathbf{u}e^{j\varphi})(\mathbf{v}e^{j\varphi})^{H}\), so any common phase is valid, and for a repeated σ any unitary rotation within the singular subspace is valid. Only the projectors onto groups of singular vectors separated by a gap are unique, with perturbation sensitivity of order \(\lVert \mathbf{E}\rVert /\text{gap}\). It bites when you average or track vectors across blocks, windows or acquisitions (align phases first, or compare \(\lvert u_{k}\rvert\) or projectors), when you compare MATLAB with another library (signs differ), and inside the noise bulk where gaps are tiny. The filter \(\mathbf{SV}_{b}\mathbf{V}_{b}^{H}\) and PD are invariant. Follow-up: "How do you align two sets of singular vectors?" Orthogonal Procrustes: the SVD of \(\mathbf{V}_{1}^{H}\mathbf{V}_{2}\) gives the best unitary alignment.

I7Prove that the truncated SVD is the best rank-k approximation in the spectral norm. Does the same hold in Frobenius norm? · research

\(\lVert \mathbf{A} - \mathbf{A}_{k}\rVert_{2} = \sigma_{k+1}\) by unitary invariance (06.9). For any \(\mathbf{B}\) with rank ≤ k, \(\operatorname{null} \mathbf{B}\) has dimension ≥ n − k and therefore intersects the (k+1)-dimensional \(\operatorname{span} (\mathbf{v}_{1}, \dots, \mathbf{v}_{k+1})\) in a unit vector \(\mathbf{w}\). Then \(\lVert (\mathbf{A} - \mathbf{B})\mathbf{w}\rVert\) \(= \lVert \mathbf{Aw}\rVert \ge \sigma_{k+1}\). Frobenius: Weyl's inequality \(\sigma_{i+j-1}(\mathbf{X} + \mathbf{Y}) \le \sigma_{i}(\mathbf{X}) + \sigma_{j}(\mathbf{Y})\) with \(\mathbf{X} = \mathbf{A} - \mathbf{B}\), \(\mathbf{Y} = \mathbf{B}\), \(j = k+1\) gives \(\sigma_{k+i}(\mathbf{A}) \le \sigma_{i}(\mathbf{A} - \mathbf{B})\), so \(\lVert \mathbf{A} - \mathbf{B}\rVert_{F}^{2} \ge \sum_{i>k}\sigma_{i}^{2}\). It holds for every unitarily invariant norm (Mirsky). Follow-up: "Is the minimizer unique?" In Frobenius norm yes if \(\sigma_{k} > \sigma_{k+1}\); in the spectral norm no, many matrices achieve \(\sigma_{k+1}\).

I8What does the dominant eigenvector of the channel (spatial) covariance matrix mean physically? Connect it to DORT and aberration correction. · research

\(\mathbf{R} = E[\mathbf{xx}^{H}]\) across the aperture. For a single dominant scatterer \(\mathbf{x} = a\mathbf{g}\) + noise, so \(\mathbf{R}\) \(= E\lvert a\rvert^{2}\mathbf{gg}^{H} + \sigma_{n}^{2}\mathbf{I}\); the dominant eigenvector is the Green's function \(\mathbf{g}\) (including any aberration), and its phase conjugate is the transmit law that refocuses on the scatterer. \(\lambda_{1}/\sum \lambda_{i}\) measures coherence (a coherence-factor analogue). DORT applies the same idea to the inter-element response matrix \(\mathbf{K}\): one significant singular value per resolved point scatterer, each singular vector a selective focusing law (Prada & Fink 1994); iterative time reversal is power iteration on \(\mathbf{K}^{H}\mathbf{K}\). In speckle there is no single scatterer, so the distortion-matrix method averages over an isoplanatic patch to recover a rank-1 structure whose first singular vector is the aberration law. Follow-up: "What breaks it?" Scatterers within one resolution cell (vectors mix), patches larger than the isoplanatic area (rank grows), low SNR (\(\sigma_{1}\) near the noise bulk).

I9A static tissue ensemble is rank 1. Is tissue moving with a uniform axial Doppler shift also rank 1? What actually raises the rank? · research

If the scene only changes phase, \(\mathbf{S}\) \(= \mathbf{s}_{0}[e^{j2\pi f_{D}t_{1}} \dots e^{j2\pi f_{D}t_{n}}]\) is an outer product: rank 1 at any \(f_{D}\), which a high-pass filter cannot remove if \(f_{D}\) is in the passband but the SVD removes in one term. A real displacement also moves the envelope relative to the pixel grid; each pixel then sees a different, time-varying speckle amplitude, and the pattern is no longer separable. The number of ranks grows with displacement over the PSF size: about 3 ranks for \(1\sigma_{z}\), 6 for \(3\sigma_{z}\), 13 for \(10\sigma_{z}\) at a −50 dB residual. Follow-up: "So how would you make moving tissue low rank again?" Register the frames (undo the displacement) before the SVD, or work in blocks small enough that motion is locally a pure translation over a short window.

I10Where is the noise floor in a singular-value plot, and can a component below that floor still be detected? · research

For i.i.d. noise of standard deviation σ per entry, the noise singular values fill the Marchenko–Pastur bulk \([\sigma (\sqrt{m}- \sqrt{n}), \sigma (\sqrt{m}+ \sqrt{n})]\); for 1600 pixels × 80 frames that is 31.1σ to 48.9σ, only 4.0 dB wide (3.8 dB measured), so the tail of a Casorati spectrum looks like a gently sloping plateau. A rank-1 component of strength θ separates from the bulk only above \(\theta_{c} = \sigma (mn)^{1/4} = 18.9\sigma\) (the BBP transition); just above it \(\sigma_{1}\) barely clears the edge (\(\theta = 1.2\theta_{c}\) gives \(\sigma_{1} = 49.4\sigma\) vs edge 48.9σ) and its singular vectors are only partly aligned with the truth (\(\lvert \langle \mathbf{v}_{1}, \mathbf{v}\rangle \rvert^{2} = 0.45\)). Below \(\theta_{c}\) it is invisible to any SVD threshold. Follow-up: "Why are the temporal vectors better estimated than the spatial ones?" The overlap formulas weight by the aspect ratio: with \(m \gg n\), \(\lvert \langle \mathbf{v}_{1}\), \(\mathbf{v}\rangle \rvert^{2} = 0.89\) but \(\lvert \langle \mathbf{u}_{1}, \mathbf{u}\rangle \rvert^{2} = 0.44\) at \(\theta = 2\theta_{c}\): the n-dimensional temporal vector is estimated from m ≫ n pixels of evidence, while the m-dimensional spatial vector has only n frames.

I11Your ensemble is 128 frames at 1 kHz with a 5 MHz probe. What Doppler and velocity resolution do the temporal singular vectors have, and what would doubling the ensemble change? · screen

Δf = PRF/N = 7.8 Hz, i.e. (λ/2)Δf = 0.154 mm × 7.8 Hz = 1.20 mm/s axial; Nyquist ±500 Hz = ±77 mm/s. Doubling to 256 frames halves Δf to 3.9 Hz (0.60 mm/s), adds blood ranks and lowers PD variance, but doubles the time over which tissue moves (more tissue ranks, possibly non-stationary motion), quadruples the Gram cost \(MN^{2}\) and halves the PD frame rate if windows do not overlap. Follow-up: "Can you measure a 0.5 mm/s flow with 1.2 mm/s resolution?" Its mean frequency can be estimated below Δf at high SNR; resolving it from tissue at a similar Doppler needs the spatial separation the SVD provides.

07Random vectors, covariance & correlation

12 questions · open in the module

I1Derive the van Cittert–Zernike result for pulse-echo speckle intuitively. What does the coherence curve of a uniform focused aperture look like? · onsite

Diffuse scatterers are independent sources, so their contributions add in power: the focal plane acts as an incoherent source whose brightness is the transmit intensity \(|P_{t}(\xi )|^{2}\). After receive focusing, a source at lateral position \(\xi\) reaches element i with phase \(-kx_{i}\xi /z\), a plane wave across the aperture. The covariance of elements i and j is the sum over sources of \(e^{-jk(x_{i} - x_{j})\xi /z}\) weighted by \(|P_{t}|^{2}\): a Fourier transform of the source intensity evaluated at the separation. Since \(|P_{t}|^{2}\) is the transform of the transmit-aperture autocorrelation, \(C(m) = (t \star t)(m)/(t \star t)(0)\). Uniform: a triangle \(1 - m/N_{t}\), zero at the transmit width; \(N_{t} = 64\) gives \(C(32) = 0.5\).

Follow-up: 32-element transmit, 64-element receive? Zero at lag 32; receive apodization does not change the normalized curve. Away from the focus? The insonified width grows to ≈ \(D|1 - z/z_{f}|\) and the first null moves to a separation ≈ \(\lambda z/w\): at 27 mm with a 30 mm focus, C(16) drops from 0.73 to 0.18.

I2Why does short-lag spatial coherence suppress noise and clutter while keeping speckle? · screen

Channel noise is uncorrelated between elements, so it adds only to the diagonal of \(\mathbf{R}\): it raises lag 0 and scales every lag ≥ 1 by \(\mathrm{SNR}/(1 + \mathrm{SNR})\). SLSC never uses lag 0, so noise does not add to it; at 0 dB SNR the lag-1 value halves (0.98 → 0.49) instead of being swamped. Much reverberation clutter is also nearly incoherent beyond lag 0. Mainlobe speckle, in contrast, keeps \(C(m) \approx 1 - m/N\) at short lags regardless of its brightness, so a clutter-filled anechoic region turns dark. Short lags are used because long lags are both small (triangle) and noisy, and are the first to be destroyed by aberration.

Follow-up: why the real part? The magnitude of a normalized correlation of pure noise is biased up to ≈ \(\sqrt{\pi /(4K)}\) for K snapshots; the real part has zero mean.

I3You sweep the assumed sound speed and plot coherence (CF or SLSC). What do the position, height and width of the peak tell you? · research

Position: in a homogeneous medium the residual focusing phase vanishes only at \(\hat{c} = c\), so the argmax estimates c (the MATLAB lab recovers 1510 m/s to within 1 m/s). In layered media it is a path-averaged speed. Height: compared with the VCZ ideal (SLSC 5.67 for N = 64, M = 6; CF 0.667), a low peak means noise (constant factor at lags ≥ 1) or aberration (lag-dependent loss). Width: set by the edge defocus phase \((\pi D^{2}/(2\lambda z))(\Delta c/c):\ 1.6\) rad at 40 m/s for F/1.6, giving −4% SLSC but about −30% CF. A broad, flat peak means low sensitivity (large F-number, short lags only) or aberration. A shifted, flattened peak with a screen present (+25 m/s for SLSC in the lab) is the signature of aberration bias; noise alone lowers the peak without moving it.

Follow-up: how to make it local? Estimate per region and invert along paths (tomographic or differentiable beamforming), because the focusing delay integrates slowness along the path.

I4How many snapshots do you need for MVDR on a 64-element array, and what do people actually do in ultrasound? · onsite

\(\hat{R} = XX^{H}/K\) has rank ≤ K, so \(K \ge N = 64\) just to invert it. Reed–Mallett–Brennan: with i.i.d. signal-free snapshots the SINR ratio is \(\mathrm{Beta}(K - N + 2, N - 1)\), mean \((K - N + 2)/(K + 1)\), so \(K = 2N - 3 = 125\) for −3 dB and \(K = N\) gives \(2/(N + 1)\) (−15.1 dB for \(N = 64\)). A pixel offers perhaps 1–20 axial samples, so ultrasound MVDR uses sub-aperture averaging (\(L \le N/2\), multiplying snapshots by \(N - L + 1\)) plus diagonal loading \(\delta = \Delta \cdot \operatorname{tr} (\hat{R})/L\) with \(\Delta \approx 10^{-2}\).

Follow-up: why is it worse than RMB in practice? The snapshots contain the desired echo, so errors in steering or delays make the adaptive weights cancel signal; loading bounds the white-noise gain and restores robustness.

I5Distinguish correlation coefficient, correlation function, covariance matrix and coherence. · screen

The correlation coefficient is one number, \(\operatorname{cov} (x, y)/(\sigma _{x}\sigma _{y})\), mean and scale removed, \(|\rho | \le 1\) by Cauchy–Schwarz (NCC peak in speckle tracking). The correlation function \(R_{xy}(\tau )\) is the same inner product as a function of lag (delay estimation, Kasai's arg R(1)). The covariance matrix collects all pairwise covariances of a vector; it is Hermitian PSD, and \(w^{H}Rw\) is beamformer output power. Spatial coherence C(m) is the per-pair normalized covariance averaged along a diagonal of that matrix, a function of element lag. Spectral coherence \(\gamma ^{2}(f)\) normalizes the cross-spectrum per frequency. All coherences are averages: from one segment or snapshot they are identically 1.

Follow-up: is "correlation matrix" the normalized one? It depends on the field: statistics normalizes to unit diagonal; array processing means \(\mathrm{E}[xx^{H}]\), identical to R for zero-mean IQ.

I6State and sketch the Wiener–Khinchin theorem. Why does it matter for speckle and QUS? · onsite

For a WSS process the PSD is the Fourier transform of the autocorrelation, \(S(f) = \int R(\tau )e^{-j2\pi f\tau}d\tau\). Sketch: \(\mathrm{E}|X_{T}(f)|^{2}/T\) for a record of length T is the transform of the triangularly weighted \(R(\tau )(1 - |\tau |/T)\), which tends to S(f) as \(T \to \infty\). Finite-record version: the periodogram is exactly the DFT of the biased \(\hat{R}(k)\). For speckle, white reflectivity through the pulse gives \(S = N_{0}|P(f)|^{2}\) and \(R = N_{0}R_{pp}\): the speckle autocorrelation is the pulse autocorrelation (FWHM 1.25/B for a Gaussian pulse), which is why spectral QUS divides by a reference-phantom spectrum.

Follow-up: why average periodograms? Each bin of a raw periodogram is exponential with std/mean = 1 whatever the record length; Welch averaging of K segments cuts that by about \(\sqrt{K}\) at the cost of resolution.

I7What is the coherence factor for a point target, for diffuse speckle and for noise, with a uniform 64-element aperture? · onsite

\(\mathrm{CF} = |\sum s_{i}|^{2}/(N\sum |s_{i}|^{2})\). Point at focus: identical channels, \(\mathrm{CF} = 1\). Diffuse speckle (ratio of expectations): \(\sum _{m}(N - |m|)C(m)/N^{2}\) with C the triangle, = \(2/3 + 1/(3N^{2}) = 0.667\). Noise: only lag 0 contributes, \(\mathrm{CF} = 1/N = 0.016\). Speckle at 0 dB channel SNR: (N·0.667 + 1)/(2N) = 0.34.

Follow-up: the per-pixel average of CF in speckle is ≈ 0.57, not 0.667, because \(\mathrm{E}[X/Y] \neq \mathrm{E}[X]/\mathrm{E}[Y]\); check which a paper reports, and whether CF was computed per sample or over an axial kernel.

I8Two speckle kernels have true correlation 0.95, each with 10 dB SNR of independent noise. What correlation do you measure, and how would you get channel SNR from coherence? · screen

Noise adds to the variances but not the covariance: \(\rho _{\mathrm{meas}} = \rho /\sqrt{(1 + 1/\mathrm{SNR}_{1})(1 + 1/\mathrm{SNR}_{2})}\) \(= 0.95 \times 10/11\) = 0.864. The same factor scales spatial coherence at lags ≥ 1, so with the VCZ value known, \(\mathrm{SNR} = C(1)/(C_{\mathrm{VCZ}}(1) - C(1)):\) \(N = 128, C(1) = 0.80\) gives \(\mathrm{SNR} = 4.2\) (6.2 dB). This is lag-one coherence (Long, Bottenus and Trahey 2018).

Follow-up: what inflates C(1) artificially? Crosstalk or common-mode electronic noise, which is correlated between neighbouring channels.

I9A QUS parameter correlates with fat fraction at \(\hat{\rho } = 0.6\) in 12 patients. Give a 95% interval. What changes if each patient contributes 50 correlated ROIs? · screen

Fisher \(z: \operatorname{atanh} 0.6 = 0.693\), se = \(1/\sqrt{N - 3} = 1/3\), interval 0.693 ± 0.653 = [0.040, 1.347] in \(z, \rho\) ∈ [0.04, 0.87]. Barely excludes zero. Adding 50 ROIs per patient does not give \(N = 600\): the ROIs within a patient share the same fat fraction and are strongly correlated, so the effective N stays near 12 (use a mixed model or per-patient averages).

Follow-up: what is \(\operatorname{std} (\hat{\rho })\) at \(\rho = 0.9\) with \(N = 16\) independent pairs? \((1 - 0.81)/\sqrt{15} = 0.049\).

I10Why is speckle IQ circular complex Gaussian, and what are the envelope statistics? Are I and |z| independent? · screen

Each IQ sample sums many phasors with independent, uniform phases: the central limit theorem gives a Gaussian, uniform phases give rotation invariance, so \(\mathrm{E}[z^{2}] = 0, I\) and Q uncorrelated with equal variance, and (jointly Gaussian) independent. Then |z| is Rayleigh with mean/std = \(\sqrt{\pi /(4 - \pi )} = 1.91, |z|^{2}\) is exponential with mean/std = 1, the phase is uniform. I and |z| are uncorrelated (symmetry under I → −I) but dependent \((|z| \ge |I|);\) \(\operatorname{corr} (I^{2}, |z|^{2}) = 1/\sqrt{2}\).

Follow-up: what breaks circularity, and would the envelope show it? I/Q gain or phase imbalance gives \(|P|/R \approx 0.42\) for 20% gain error and 10° skew, but the envelope SNR only moves from 1.91 to 1.82: the envelope is a poor detector.

I11Your measured coherence curve falls below the VCZ triangle. How do you tell noise from aberration? · research

Divide by the known triangle. Noise gives a ratio that is flat for all \(m \ge 1\) (equal to \(\mathrm{SNR}/(1 + \mathrm{SNR})\)) with a discontinuity at lag 0. Aberration gives a ratio that decays with lag, ≈ \(e^{-D_{\varphi}(m)}\) two-way, where \(D_{\varphi}\) is the phase structure function; the lag where it falls gives the screen correlation length and the depth gives the rms phase (one screen realization in the MATLAB lab, \(\sigma = 0.7\) rad, \(\ell = 4\) elements: \(C(4) = 0.57\) against 0.94; the ensemble value \(e^{-D_{\varphi}(4)}\cdot 0.94\) is 0.64). A wrong sound speed gives a similar lag-dependent loss, so sweep \(\hat{c}\) first. A linear phase error gives no loss at all.

Follow-up: what figure of merit would you report for an aberration-correction method? The restored coherence curve (or SLSC, CF) relative to the triangle, at matched SNR, plus point-target resolution.

I12An RF kernel is 5 mm long at 5 MHz, 60% bandwidth, 50 MHz sampling. How many independent samples does it contain? · research

It holds 325 samples but about 16 speckle cells of length c·1.25/(2B) = 0.32 mm. The effective N depends on the statistic: \(N_{\mathrm{eff}} = T/\int \rho _{q}(\tau )d\tau\) with \(\rho _{q}\) the correlation of the averaged quantity. For an intensity mean, \(\int |r_{\mathrm{env}}|^{2} = 0.94/B\) gives 20.8; for a correlation coefficient between two RF kernels, \(\int r^{2} \approx 0.47/B\) gives 41.5 (Monte Carlo 40). So \(\operatorname{std} (\hat{\rho })\) at \(\rho = 0.9\) is ≈ 0.03, between the cell-count estimate 0.05 and the naive sample-count estimate 0.011.

Follow-up: why does the RF correlation coefficient get twice the intensity count? The carrier makes \(r(\tau )^{2}\) oscillate as \(\cos ^{2}\), halving its integral.

08Analytic signal & IQ demodulation

12 questions · open in the module

I1What is \(H(f)\) for the Hilbert transform, what is \(\mathcal{H}\{\cos\}\) and \(\mathcal{H}\{\sin\}\), and what is \(\mathcal{H}\) applied twice? · screen

\(H(f) = -j\operatorname{sgn}(f)\): −90° on positive frequencies, +90° on negative, unit magnitude. \(\mathcal{H}\{\cos\omega_0 t\} = \sin\omega_0 t\), \(\mathcal{H}\{\sin\omega_0 t\} = -\cos\omega_0 t\). Twice: \(H^2 = (-j\operatorname{sgn} f)^2 = -1\) for \(f \ne 0\), so \(\mathcal{H}\{\mathcal{H}\{x\}\} = -x\) (for \(x\) with no DC). The analytic signal \(x + j\mathcal{H}\{x\}\) has spectrum \(2u(f)X(f)\). Follow-up: why is the time-domain kernel \(1/(\pi t)\) a problem in practice? It decays slowly, so FIR Hilbert transformers need many taps and FFT implementations are circular.

I2Why can complex IQ be sampled at a rate set by bandwidth while RF needs twice its highest frequency? Give numbers for a 5 MHz, 80 % probe. · screen

A real signal's spectrum has a mirror at negative frequencies; sampling must keep the positive band from overlapping the mirror's replicas, hence \(f_s > 2f_{\max}\). IQ has only one band, so \(f_{s,\mathrm{IQ}} > B\) (plus \(2\lvert\Delta\rvert\) for a demodulation offset). \(\sigma_f = 4/2.355 = 1.70\) MHz; the −20 dB width is 7.3 MHz, so 8 to 10 MHz complex; RF needs ≥ 17.3 MHz at the −20 dB edge, 30 to 40 MHz in practice. Follow-up: is IQ cheaper in bits? At \(f_s/4\) complex it carries the same number of reals as RF at \(f_s/2\); the gain is against oversampled RF and in interpolation accuracy, not against Nyquist-rate RF.

I3Three ways to get a B-mode envelope from RF. Which are exact? · screen

(1) \(\lvert x_a\rvert\) via Hilbert/FFT: exact by definition (subject to edge effects). (2) IQ demodulation then \(\lvert\mathrm{IQ}\rvert\): equal to (1) for an ideal LPF that passes the band and rejects the image; a real FIR adds sub-percent error (0.6 % in the MATLAB lab). (3) Rectify and low-pass: an approximation; \(\lvert\cos\rvert\) has mean \(2/\pi\) and harmonics at \(2f_0\), so the result depends on the LPF and is scaled by \(2/\pi\). Follow-up: do (1) and (2) differ in phase? Yes, by \(2\pi f_d t\).

I4A colleague's IQ beamformer gives a PSF that is 13 dB weak and smeared over several millimetres at 20 mm, but fine B-mode speckle texture. Diagnose. · onsite

Missing (or wrong-sign) phase rotation. Each channel carries \(e^{-j2\pi f_d\tau_n}\) after envelope alignment; summing without \(e^{+j2\pi f_d\tau_n}\) gives a Fresnel-integral loss, −13.4 dB for 32 elements, 0.3 mm pitch, 5 MHz at 20 mm, oscillating with depth. Speckle still looks like speckle because random-phase sums have the same statistics. Test: beamform a point target at several depths; a loss that oscillates with depth and tracks \(\bigl\lvert\frac{1}{N}\sum e^{-j2\pi f_d\tau_n}\bigr\rvert\) confirms it. A wrong-sign rotation doubles the phase error instead of cancelling it. Follow-up: which \(f\) to use in the rotation? The stored \(f_d\), not the nominal \(f_0\).

I5With 5 MHz IQ, what axial displacement corresponds to 1° of phase, and what is the largest inter-frame motion you can track by phase? · onsite

\(d\) per radian \(= \lambda/(4\pi)\) = 24.5 µm, so 1° = 0.43 µm. Phase wraps at \(\pm\pi\), i.e. \(\pm\lambda/4\) = ±77 µm between frames; at PRF 5 kHz that is 0.385 m/s (Kasai Nyquist \(c\cdot\mathrm{PRF}/(4f_0)\)). Follow-up: what limits precision? Phase noise ~ 1/SNR per sample, reduced by averaging over a window; and the conversion frequency, which must be the local \(\hat{f}_c\).

I6Derive the attenuation-induced downshift for a Gaussian pulse. Which \(\sigma\), which \(z\), which units? · onsite

Amplitude spectrum \(G(f) = \exp\bigl(-(f-f_0)^2/2\sigma_f^2\bigr)\); round trip to one-way depth \(z\) multiplies by \(\exp(-2\alpha_0 f z)\) with \(\alpha_0\) the amplitude coefficient in Np/cm/MHz. Completing the square: center \(f_0 - 2\alpha_0\sigma_f^2 z\), same \(\sigma_f\). With the power-spectrum \(\sigma_p = \sigma_f/\sqrt{2}\) the same result reads \(f_0 - 4\alpha_0\sigma_p^2 z\). 5 MHz, 60 %, 0.5 dB/cm/MHz (0.0576 Np): −0.187 MHz/cm. Follow-up: turn this into an estimator: \(\hat{\alpha}_0 = -\text{slope}/(2\sigma_f^2)\), with \(\sigma_f\) measured from a reference; then list the confounders (diffraction, non-Gaussian spectra, frequency-dependent backscatter, LPF clipping of the shifted band).

I7For a CUTE-type sound-speed method that measures phase differences between steered IQ images, what biases the delay maps and how do you fix each? · research

(1) Conversion frequency: \(\Delta\tau = -\Delta\varphi/(2\pi f)\), and \(f\) must be the local echo frequency, which decreases with depth (−0.19 MHz/cm in the example above); using \(f_d\) or nominal \(f_0\) scales delays by \(\hat{f}_c/f\). Fix: estimate \(\hat{f}_c\) from fast-time \(R(1)\). (2) Phase wrapping when \(\lvert\Delta\tau\rvert > 1/(2f_c)\): limit the angle step or unwrap. (3) Low-coherence pixels (low envelope, decorrelation between angles) give random phase: weight by \(\lvert R\rvert\) or coherence. (4) Wrong or inconsistent IQ bookkeeping (sign, \(f_d\) in the rotation) between beamforming and phase estimation. Follow-up: why does a constant \(\Delta\) not matter for the phase difference itself? Both images carry the same residual rotation at the same pixel, so it cancels in the product.

I8You are designing a ULM acquisition: 5 MHz probe, 128 channels, 40 mm depth, 10 kHz plane-wave rate. RF or IQ channel data? Quantify. · research

Round trip to 40 mm: 2 × 40/1.54 = 52 µs. RF at 40 MHz, 16-bit: 2078 samples × 128 × 2 B = 532 kB per transmit, 5.3 GB/s at 10 kHz. IQ at 10 MHz: 520 complex = 1040 reals, 266 kB per transmit, 2.7 GB/s, the same as RF at 20 MHz (4 samples per wavelength). IQ is the natural choice: downstream SVD clutter filtering and Doppler need IQ anyway, and IQ DAS with rotation is accurate at 10 MHz (−0.4 dB with linear interpolation). Follow-up: what do you lose? The band outside the LPF; if you want harmonics or full-band spectral analysis, keep RF.

I9How would you estimate the local center frequency robustly from IQ, and why not average the instantaneous frequency? · onsite

Average \(R(1) = \sum \mathrm{IQ}^*[k]\,\mathrm{IQ}[k+1]\) over a depth window and over lines, then \(\hat{f}_c = f_d + (f_{s,\mathrm{IQ}}/2\pi)\arg R(1)\) (08.20). This is equivalent to the \(A^2\)-weighted mean of \(f_i\), which equals the spectral centroid; unweighted \(f_i\) is dominated by spikes at envelope nulls. In the MATLAB lab, 64 lines and 4 mm windows give 0.02 MHz rms scatter and recover \(\alpha_0\) within 3 %. Follow-up: what aliasing limit? \(\lvert f_c - f_d\rvert < f_{s,\mathrm{IQ}}/2\).

I10Your system samples at exactly \(4f_0\) and forms IQ by taking I and Q from neighboring samples. When is that acceptable? · onsite

When the echo is narrowband relative to the \(1/(4f_0)\) offset between the I and Q samples (50 ns at 5 MHz). For a 60 % Gaussian echo the pairwise IQ has 18 % rms complex error averaged over echo positions; averaging the two Q neighbors ([¼ ½ ¼] after the 1, −j, −1, +j mixer) brings it to about 4 %, and a longer interpolating LPF lower still. Follow-up: what does the 3-tap filter do spectrally? \(\cos^2(\pi f/f_s)\): a null at \(f_s/2 = 2f_0\), where the image sits, and 17 % droop at the 2.7 MHz band edge.

I11Why is IQ delay-and-sum exact even though it uses a single frequency \(f_d\) for the rotation? · research

Because the rotation is not a model of the echo; it undoes the demodulator. \(\mathrm{IQ}_n(t) = x_{a,n}(t)\,e^{-j2\pi f_d t}\) exactly, so \(\mathrm{IQ}_n(\tau)\,e^{j2\pi f_d\tau} = x_{a,n}(\tau)\) for any \(\tau\) and all frequencies in the band. The single frequency is the one that was removed, not an approximation of the echo spectrum. Error sources are only interpolation of the baseband and LPF imperfections. Follow-up: what if \(f_d\) varied with depth during demodulation? Rotate by the accumulated demodulator phase \(\theta_d(\tau) = 2\pi\int_0^\tau f_d\,dt\), not \(f_d(\tau)\cdot\tau\).

I12Your phase-based displacement estimates are 15 % low at 4 cm depth in liver and correct near the surface. What is the likely cause? · research

The phase-to-displacement conversion uses the nominal \(f_0\), but the echo is downshifted: at 5 MHz, 60 %, 0.5 dB/cm/MHz, \(f_c(4\ \mathrm{cm}) = 5 - 0.75 = 4.25\) MHz. The phase is \(-2\pi(4.25\ \mathrm{MHz})\tau\), so \(\hat{\tau} = \tau \times 4.25/5 = 0.85\tau\): 15 % low. Fix: convert with the local \(\hat{f}_c\) from fast-time \(R(1)\) (Loupas-style 2-D autocorrelation does this jointly). Follow-up: could the demodulation LPF make it worse? Yes: if the shifted band is clipped, \(\hat{f}_c\) is biased toward \(f_d\).

09Digital filter design

10 questions · open in the module

I1Why are RF and IQ filters in a beamformer almost always FIR? · screen

Linear phase: a constant, known delay \(M/2\) that is identical on every channel, so filtering never looks like an aberration and the delay is subtracted once. Polyphase decimation divides the cost by D, which erases most of an IIR's per-output advantage (for the 8/9 MHz spec, 53 folded multiplies vs about 20 for an 8th-order elliptic, before decimation). No stability, limit-cycle or coefficient-sensitivity problems in fixed point. Follow-up: how many taps for 3–7 MHz at 40 MHz with 1 MHz transitions and 40 dB? The low-pass estimate says about 61; two transitions push the true equiripple minimum to 72.

I2A 129-tap linear-phase FIR runs at 50 MHz on RF. What delay do you remove, in samples, µs and mm? · screen

Order 128, delay 64 samples = 1.28 µs. In pulse-echo depth that is \(c\,t/2\) = 770 m/s × 1.28 µs = 0.99 mm. Follow-up: what if the filter has an even number of taps? Then the delay is a half sample; either pad to odd length or account for the half-sample shift in the delay law, otherwise IQ phase is biased by \(\pi f_0/f_s\).

I3Derive the bandwidth and the peak group delay of a pole at \(r e^{j\theta}\). · onsite

\(\lvert 1 - r e^{j\theta}e^{-j\omega}\rvert^2\) \(= 1 - 2r\cos(\omega - \theta) + r^2\) \(\approx (1 - r)^2 + r(\omega - \theta)^2\) near θ, so half power at \(\lvert\omega - \theta\rvert \approx 1 - r\): full width \(2(1 - r)\) rad/sample. The phase of \(1/(1 - re^{-j\Delta})\) is \(-\operatorname{atan}\left(r\sin\Delta/(1 - r\cos\Delta)\right)\); differentiating at Δ = 0 gives the group delay \(r/(1 - r)\). r = 0.95: 0.1 rad/sample and 19 samples. Follow-up: convert to a color-flow PRF of 4 kHz: 64 Hz wide, 4.75 ms delay, longer than a 12-pulse ensemble (3 ms).

I4Size a Kaiser-window low-pass: 50 dB, 0.5 MHz transition at \(f_s\) = 20 MHz. Then estimate equiripple with \(\delta_p = 0.01\). · onsite

\(\Delta f = 0.025\), \(\Delta\omega = 0.157\) rad/sample. \(\beta = 0.5842 \times 29^{0.4} + 0.07886 \times 29 = 4.53\). \(M = 42/(2.285 \times 0.157) = 117\), round to 118 for an integer delay. Equiripple: \(\delta_s = 0.00316\), \(-20\log_{10}\sqrt{0.01 \times 0.00316} = 45.0\) dB, \(M \approx (45.0 - 13)/(14.6 \times 0.025) = 88\), so budget about 90 after the few-percent underestimate. Follow-up: why can the window design not take advantage of the 0.01 passband? Because its \(\delta_p\) equals \(\delta_s\) by construction.

I5Why does the bilinear transform need prewarping, and why not use impulse invariance instead? · onsite

The bilinear map sends Ω to \(\omega = 2\operatorname{atan}(\Omega T/2)\), so analog edges land at lower digital frequencies (15 MHz lands at 11.0 MHz for \(f_s\) = 40 MHz). Prewarping designs the prototype at \((2/T)\tan(\omega_p/2)\) so the edges land where specified. Impulse invariance samples \(h_a(t)\), which aliases its spectrum: fine for steep low-pass prototypes, impossible for high-pass and band-stop, whose response does not fall off. Follow-up: what does the bilinear transform do to an analog zero at infinity? It maps to z = −1, a null at Nyquist.

I6Your 8th-order elliptic low-pass works in MATLAB but oscillates on the FPGA. Diagnose. · onsite

Almost certainly direct-form coefficients: the \(a_k\) of a narrowband 8th-order filter span tens, so after quantization the clustered poles move outside the unit circle (16-bit direct form moves an 8th-order Butterworth pole at \(0.02\,f_s\) to radius 1.30). Re-implement as four biquads with poles paired to nearest zeros, scale each section against overflow, put the highest-Q section last. If it still hums with zero input, it is a limit cycle: add more state bits or use magnitude truncation / error feedback. Follow-up: why are FIRs immune? No feedback: quantization moves zeros, which changes the stopband floor but cannot destabilize.

I7Color flow shows strong clutter bleeding into the velocity map near vessel walls. The wall filter is a 2nd-order Butterworth high-pass on 10 pulses. What do you check first? · research

The initialization. With zero initial state, the clutter step at the first pulse rings through the ensemble: on N = 10 at \(f_c\) = 0.1 PRF the DC suppression is only 12.6 dB instead of infinite. Step initialization nulls DC exactly but gives only −32 dB at 0.01 PRF; projection initialization gives −43 dB with all 10 outputs kept; a degree-2 regression filter −63 dB. Follow-up: what does projection initialization compute? \(\mathbf{s}_0 = -\mathbf{G}^{+}\mathbf{F}\mathbf{x}\), minimizing output energy over the initial state, so \(\mathbf{A} = (\mathbf{I} - \mathbf{G}\mathbf{G}^{+})\mathbf{F}\).

I8When is a polynomial regression wall filter enough, and when do you need an SVD filter? · research

Regression assumes clutter is a low-order polynomial in slow time at every pixel independently: fine for slowly moving tissue and short ensembles where its cutoff (\(\approx (0.44 + 0.41p)\,\mathrm{PRF}/N\)) sits below the slowest flow of interest. When tissue moves at velocities overlapping blood (cardiac, breathing, handheld probe) or when slow microvascular flow must be kept, frequency separation fails; an SVD filter on a spatiotemporal Casorati matrix separates by spatial coherence (Module 06, Demené et al. 2015) and needs long ensembles (ultrafast Doppler, ULM). Follow-up: what is the regression filter's matrix? \(\mathbf{I} - \mathbf{Q}\mathbf{Q}^H\) with \(\mathbf{Q}\) orthonormal polynomials; an SVD filter is the same with \(\mathbf{Q}\) from the data.

I9You smooth shear-wave displacement vs time with filtfilt and a 4th-order Butterworth at 1 kHz. What did you actually apply? · onsite

\(\lvert H\rvert^2\): zero phase (so arrival times are not biased, which matters for time-of-flight SWE), but −6 dB at 1 kHz, the −3 dB point at 0.896 kHz, and an 8th-order magnitude roll-off. The first and last few periods of the cutoff are affected by edge transients unless the record is reflection-padded and the filter started from steady state. Follow-up: could you do this in real time? Only block-wise with overlap and a latency of at least the block length.

I10You need envelopes from RF on an FPGA at 40 MHz. FFT Hilbert or FIR Hilbert, and how big? · research

FIR: streaming, no block edges, no circular wrap. A 31-tap Hamming-windowed type III Hilbert is within −0.19/+0.05 dB from 2 to 18 MHz; only 16 taps are nonzero and they are antisymmetric, so 8 multiplies per output. Delay the in-phase path by 15 samples to match. Better still, demodulate to IQ and decimate (Module 08), which also cuts the data rate. Follow-up: why not an even-length Hilbert? Type IV has a half-sample delay, awkward to match in the real path.

10Matched filtering & pulse compression

11 questions · open in the module

I1Derive the matched filter and its output SNR. · screen

Signal output at \(t_0\) is \(\int HS\,e^{j2\pi f t_0}df\), noise variance \((N_0/2)\int\lvert H\rvert^2 df\) for real white noise of two-sided PSD \(N_0/2\). Cauchy–Schwarz bounds the ratio by \(\int\lvert S\rvert^2/(N_0/2) = 2E/N_0\), with equality for \(H = \kappa S^* e^{-j2\pi f t_0}\), i.e. \(h(t) = \kappa s^*(t_0 - t)\). The output is the cross-correlation of the record with the template; only \(E\) appears, not the shape.

Follow-up: what if the noise is colored? Whiten and match: \(H = S^* e^{-j2\pi f t_0}/N(f)\), \(\mathrm{SNR} = \int\lvert S\rvert^2/N(f)\,df\).

I2Your simulation shows matched-filter SNR exactly 3 dB below 2E/N₀. Bug? · screen

Probably a convention. If the data are analytic or IQ and SNR is defined as \(\lvert y\rvert^2/\mathrm{E}\lvert n_{\mathrm{out}}\rvert^2\), both noise quadratures are counted and the result is \(E/N_0\). Taking \(\mathrm{Re}\{y\}\) at the peak restores \(2E/N_0\) (the script measures 6.95 vs 9.94 dB). Other usual causes: one-sided \(N_0\), or the analytic signal's energy \(2E\) paired with the real signal's noise.

Follow-up: does envelope detection really lose 3 dB? Not at high SNR, where the quadrature noise adds only a second-order term to the envelope; the loss appears as a detection-threshold effect at low SNR.

I3Why use a chirp instead of simply driving the probe harder? · onsite

Peak rarefactional pressure is capped by the MI limit (1.9, Track 3), by pulser voltage and by nonlinear distortion at high pressure. SNR depends on energy, so at a fixed peak the only lever is duration; compression restores the \(1/B\) resolution. A TB-50 chirp gives 18.8 dB against a Gaussian pulse of the same band, about 3.8 cm more depth at 5 MHz and 0.5 dB/cm/MHz.

Follow-up: what limits \(T\)? Blind zone \(cT/2\), \(I_{\mathrm{SPTA}}\)/TI (energy scales with \(T\)), frame rate in multi-line or plane-wave sequences, attenuation mismatch, and transmit/receive switching.

I4Your chirp shows −13 dB range sidelobes. How do you get to −50 dB, and what does it cost? · onsite

Taper. On receive alone, Hann gives −31.3 dB for 1.76 dB and a 1.65× wider mainlobe, but Hamming and Blackman stall near −36 dB at TB = 50 because of Fresnel ripple. On transmit the PSL drops to −46 (Hann) or −81 dB (Blackman) at a 4–5 dB energy cost at fixed peak; split tapering sits between. Alternatives: nonlinear FM (spectrum shaped by sweep rate, no mismatch loss), least-squares or Wiener mismatched filters, and larger \(TB\).

Follow-up: why does the transmit taper beat the ideal-window number for Hann? Because the output spectrum is then closer to the window squared, and the smooth envelope suppresses the band-edge ripple.

I5Golay pair or chirp for deep plane-wave ULM at 3 MHz? · research

Golay gives exactly zero static sidelobes but needs two firings per angle, halving the compounded frame rate. With \(N\) = 16 and 5 cm/s axial motion at a 5 kHz PRF, \(d\) = 10 µm, \(\varphi = 4\pi \cdot 3\ \mathrm{MHz} \cdot 10\ \mu\mathrm{m}/1540\ \mathrm{m/s}\) = 0.245 rad and the residual is −28.3 dB; tissue and bubble motion at higher speed degrade it further, and the nonlinear bubble echo differs between firings. A tapered chirp compresses in one firing, survives motion (about 12 ns of Doppler bias per m/s at 5 MHz, \(B\) = 5 MHz, \(T\) = 10 µs), and its harmonic remains compressible. For ULM I would choose a tapered chirp and budget its sidelobes against the bubble-to-tissue ratio after SVD filtering.

Follow-up: how would you rescue Golay? Estimate the inter-firing displacement (e.g. Kasai phase, Module 11) and rotate echo B before summing.

I6What does frequency-dependent attenuation do to a compressed chirp, and how do you fix it? · research

It tapers the received spectrum toward low frequencies: centre 5.00 → 3.68 MHz at 6 cm for 0.5 dB/cm/MHz with an 80 % transducer. With the surface filter the peak does not move (zero-phase model), but the mainlobe widens (338 → 374 ns), the PSL rises (−29.8 → −26.1 dB) and the mismatch loss is 2.0 dB. A depth-adapted filter recovers the 2 dB but widens to 475 ns; an equalizer keeps 376 ns and costs 5.8 dB. Depth-dependent filter banks interpolate between them.

Follow-up: why does the raw chirp envelope move by −2.5 µs while the compressed peak stays? The LFM maps frequency to time, so a spectral tilt shifts the envelope centroid by \((f_c - f_0)T/B\); the matched filter removes the quadratic phase first.

I7Is range–Doppler coupling of an LFM a concern in ultrasound? · research

Not for SNR, sometimes for position. \(\Delta\tau \approx \nu T/B\) with \(\nu = 2vf_0/c\): at 1 m/s, 5 MHz, \(T\) = 10 µs, \(B\) = 5 MHz the ridge gives 13.0 ns and the exact peak 11.8 ns (finite-\(TB\) correction \(3/(\pi^2B^2T)\)), i.e. 9 µm of apparent depth for 0.01 dB of loss. That is small against the 0.19 mm resolution but not against ULM localization precision or µm-scale elastography displacements, and it depends on velocity. Alternate up- and down-chirps, or average them, to cancel it.

Follow-up: when does the narrowband ambiguity function fail? When \((2v/c)\,TB\) approaches 1, i.e. when the echo's time compression over the code length is comparable to the resolution: never at tissue speeds with clinical codes.

I8You add a coded B-mode to a cleared system. What changes in the 510(k) acoustic-output section? · onsite

It must be measured in the new mode at worst-case settings. At the same peak pressure MI is unchanged, but PII, \(I_{\mathrm{SPTA.3}}\) and power scale by the energy ratio, so a TB-50 chirp at MI 1.9 and 200 pulses/s at a point goes from 27 to 2003 mW/cm², over the 720 mW/cm² Track 3 limit; TIs rise by the same factor. You would cap MI (×0.60 to 1.14 here) or the pulse rate for the coded mode, define \(f_{\mathrm{awf}}\) for a swept waveform, and check derated values with nonlinear propagation in water.

Follow-up: can you claim the SNR gain if \(I_{\mathrm{SPTA}}\) already binds? No: at a binding energy limit the gain against white noise is zero; coding helps where MI, not energy, was the binding constraint.

I9How much depth does 15 dB of compression gain buy? · screen

\(\Delta z = G/(2\alpha f_0)\): 3.0 cm at 0.5 dB/cm/MHz and 5 MHz, 5.0 cm at 3 MHz. For a broadband pulse with an adapted filter it is more, 4.9 cm at 5 MHz with a 60 % Gaussian band cut at its −24 dB points, because the deep spectrum downshifts to where attenuation per cm is lower.

Follow-up: and in a clutter-limited region? Nothing: clutter scales with the transmit.

I10Will coded excitation improve the contrast of a liver cyst at 6 cm in an obese patient? · onsite

Only to the extent the cyst is noise-limited. Reverberation from the abdominal wall and off-axis clutter are copies of the transmitted code and compress with the same gain, and range sidelobes of bright interfaces add axial fill-in. If the noise floor sits above the clutter at that depth, coding lowers it by up to \(G\) and contrast improves; otherwise harmonic imaging or better aberration and clutter suppression matter more.

Follow-up: how would you tell which regime you are in? Turn the transmit off (or receive with transmit muted) and compare the noise floor with the signal level inside the cyst at the same gain.

I11What is the PSL of Barker-13, and how would you improve it without a longer code? · screen

Sidelobes are 0 or 1 against a peak of 13: −22.28 dB, with an 11.14 dB gain. A least-squares mismatched filter of length 39 gives −38.5 dB for 0.20 dB of SNR loss (65 taps: −58.6 dB). Concatenating or nesting codes, or switching to Golay pairs, are the other options.

Follow-up: why not design the filter on the ±1 chip sequence directly? Because the received chips are band-limited by the transducer; the mismatched filter must be designed on the received waveform.

11Correlation & time-delay estimation

12 questions · open in the module

I1Why is the cross-correlation peak the maximum-likelihood delay estimate, and when is it not? · screen

For a known template in white Gaussian noise, the log-likelihood is \(-\sum (y - s_D)^2/2\sigma^2\); the energy of the shifted template does not depend on \(D\), so the likelihood is maximized by maximizing \(\sum y\,s_D\), the correlation. With both records noisy and the signal unknown, the ML processor is Knapp & Carter's generalized cross-correlation: weight the cross-spectrum by \(\lvert\gamma\rvert^2/[\lvert G_{xy}\rvert(1 - \lvert\gamma\rvert^2)]\). For white noise and high in-band SNR that is flat in band and small outside, so band-pass filtering plus plain correlation is close to ML. It fails for coloured noise or clutter (pre-whiten), for shape changes between records (strain, decorrelation), and below threshold, where the ML estimate is dominated by false peaks. Follow-up: what does PHAT weighting (\(1/\lvert G_{xy}\rvert\)) do? It whitens to phase only: sharp peaks and robustness to reverberation, but it amplifies low-SNR bands, so it is ML only when the coherence is constant across the band.

I2Compare parabolic, cosine and phase-zero subsample estimators: bias, variance, cost, failure modes. · onsite

All three start from the integer NCC peak. Parabolic: model-free and cheap, but the wrong shape for an RF peak; bias 0.012 sample at \(f_s/f_0\) = 8 and 0.052 at \(f_s/f_0\) = 4, toward the integer sample (pixel locking). Cosine: fits \(R \propto \cos(\omega(k - \delta))\), estimates \(\omega\) from the data, bias ≈ 0.0014 sample, residual from the envelope; same cost. Phase-zero: \(\delta = -\angle C(k_0)/(2\pi f_0 T_s)\) on the analytic signals, no interpolation bias if \(f_0\) equals the spectral centroid; biased by \((k_0 - d)\,e/(1+e)\) for an \(f_0\) error \(e\). Above threshold all three have jitter near the CRLB (Demo 11.3); bias, not jitter, separates them. Every symmetric fit is exact at \(d\) = 0 and ½. Follow-up: why is the Gaussian fit the worst on RF and the best on envelope data? Because it models a Gaussian peak: right for \(\lvert\mathrm{IQ}\rvert\) correlation, too sharp-shouldered for a cosine, so on RF it overshoots toward half-samples with 0.025 sample bias.

I3Derive the Kasai estimator and its Nyquist velocity. · screen

Slow-time IQ at a fixed gate: \(z[m] = A\,e^{j(\varphi_0 + m\Delta\varphi)} + n\), with the two-way path change per pulse \(2v/\mathrm{PRF}\), so \(\Delta\varphi = 2\pi f_0 (2v/c)/\mathrm{PRF}\) \(= 4\pi f_0 v/(c\cdot\mathrm{PRF})\). The lag-one autocorrelation \(R(1) = \sum z[m+1]\,z^*[m]\) \(\approx \lvert A\rvert^2 (M-1)\, e^{j\Delta\varphi}\), so \(\hat{v} = c\cdot\mathrm{PRF}\cdot\angle R(1)/(4\pi f_0)\). \(\angle R(1) \in (-\pi, \pi]\), so \(\lvert v\rvert < c\cdot\mathrm{PRF}/(4f_0)\): 0.385 m/s at 5 MHz, 5 kHz, i.e. \(\lambda/4\) = 77 µm per pulse. Equivalently \(\angle R(1)/(2\pi T_{\mathrm{PRF}})\) is the mean Doppler frequency of the slow-time spectrum. Follow-up: is it optimal? Not for a tone at high SNR: the sum telescopes to the first and last pulses, \(\sigma \propto 1/(M-1)\), variance 1.8× the CRLB at \(M\) = 8. Loupas adds the fast-time lag to estimate the local \(\bar{f}\) and remove downshift bias.

I4Why do false peaks appear at multiples of the RF period, and how do you suppress them? · onsite

The expected RF correlation is the inverse transform of the power spectrum: an envelope of width ~\(1/B\) times \(\cos(2\pi f_0\tau)\). Each cosine maximum near \(\tau = m/f_0\) is a local peak with height equal to the envelope there; for a Gaussian pulse, \(\exp(-\pi^2 B^2/(8\ln 2))\) at \(m = 1\): 0.53 at 60 % bandwidth, 0.93 at 20 %. Noise, decorrelation and the finite kernel can push one above the main peak, giving errors of exactly one period (200 ns, 154 µm at 5 MHz). Suppression: wider bandwidth; longer kernels (at least ~5 periods); a coarse search on envelope or \(\lvert\mathrm{IQ}\rvert\) correlation, then RF or phase refinement; small search ranges guided by neighbours or a previous frame; spatial regularization or median filtering of the displacement field; rejecting low-NCC kernels. Follow-up: why do hops persist at high SNR with a 2.5-period kernel? A short window of speckle is nearly a sinusoid, so the window one period away matches it almost as well; the finite-kernel fluctuation, not the noise, decides.

I5State the Walker–Trahey bound and give its scalings. What jitter do you expect at 5 MHz, 60 % bandwidth, 2 µs kernel, 30 dB? · onsite

\(\sigma_D \ge \sqrt{\frac{3}{2\pi^2 f_0^3 T(B^3+12B)}\bigl[\frac{(1+1/\mathrm{SNR}^2)^2}{\rho^2} - 1\bigr]}\) (11.13), the Carter coherence CRLB for a rectangular band with flat coherence \(\rho\cdot\mathrm{SNR}_p/(1+\mathrm{SNR}_p)\). Scalings: \(f_0^{-3/2}\), \(T^{-1/2}\), \(\approx B^{-1/2}\) for \(B < 1\), and a decorrelation floor \(\propto \sqrt{1/\rho^2 - 1}\). Numbers: prefactor \(8.2\times10^{-17}\) s², bracket 0.002 → 0.41 ns (0.31 µm); with \(\rho\) = 0.99, 1.35 ns. Follow-up: your simulation reaches 0.3 ns at 40 dB but the bound says 0.13 ns. Why? Interpolator bias and window-edge decorrelation (\(W - \lvert D\rvert\) overlap) set a floor; the bound assumes an unbiased estimator and an infinite stationary record.

I6How do the CRLB and the Ziv–Zakai bound relate, and what is the threshold? · research

The CRLB is local: it uses the Fisher information, the curvature of the log-likelihood at the true delay, so it only describes small errors around the right peak. The Ziv–Zakai bound converts estimation into binary detection: for a delay with a uniform prior of width \(A\), \(\mathrm{MSE} \ge \frac{1}{A}\int_0^A h\,(A - h)\,P_{\min}(h)\,dh\), where \(P_{\min}(h)\) is the minimum error probability of deciding between delays \(D\) and \(D + h\) (a valley-filling refinement tightens it). At high SNR only small \(h\) contribute and it converges to the CRLB. At low SNR \(P_{\min}(h)\) stays near ½ even for large \(h\), notably near \(h = m/f_0\) where the correlation side peaks are high, and the bound rises toward the prior variance \(A^2/12\) (a standard deviation of 346 ns for ±24 lags at 40 MHz). The SNR where the side-peak terms switch on is the threshold. Follow-up: how would you design around it? Increase \(T\cdot B\) or SNR until the side-peak error probability is negligible, shrink \(A\) with prior information (tracking, coarse-to-fine), and report outlier rates separately.

I7How do you choose the kernel length for quasi-static elastography or speckle tracking? · onsite

Three constraints. (1) Jitter: \(\sigma \propto T^{-1/2}\) through the bound. (2) Decorrelation: under strain \(\varepsilon\) the window ends drift by \(\varepsilon T/2\), and \(1 - \rho^2\) grows as \((\varepsilon T f_0)^2\); the optimum is where the noise and decorrelation terms balance, about \(\varepsilon T f_0 \approx 0.075\) at 20 dB (\(T \approx\) 1.5 µs at 1 %, 5 MHz, \(\sigma \approx\) 2.1 ns; Monte Carlo 2.5 ns). (3) Resolution and false peaks: axial resolution ≈ \(cT/2\) (1.16 mm for 1.5 µs), and kernels below ~5 periods hop. In practice: 5–15 periods with 50–80 % overlap, companding or multi-step (coarse-to-fine) tracking for strains above 2–3 %, and a final displacement-gradient estimator over several points. Follow-up: what is the strain filter? Varghese & Ophir's curve of elastographic SNR versus applied strain: bounded at low strain by jitter and at high strain by decorrelation.

I8How would you estimate a near-field phase screen from channel data, and what limits it? · research

Model a thin delay screen \(\tau_i\) at the aperture. Cross-correlate each element with its neighbour (Flax & O'Donnell) on speckle or a point target, integrate the \(\Delta\hat{\tau}\), apply the correction, re-transmit and iterate, since a better transmit focus raises the inter-element coherence. Limits: (1) \(\rho\) between neighbours is set by van Cittert–Zernike (\(1 - 1/N\) at focus for a uniform \(N\)-element transmit), so jitter has a decorrelation floor; (2) integration is a random walk, \(\sqrt{n}\,\sigma_\Delta\) (\(\sqrt{63} \times 2\) ns ≈ 16 ns at element 64 for \(\sigma_\Delta\) = 2 ns), and one false peak adds a \(1/f_0\) step to every later element; (3) mean and linear terms are unobservable from diffuse speckle; (4) the thin-screen, single-isoplanatic-patch assumption fails for distributed aberrators, where the correction is only valid near the region used. Better: correlate against a beamsum reference, multi-lag least squares with NCC weights, or estimate from the full set of pairs (distortion or reflection matrix methods). Follow-up: why does multi-lag LS sometimes not help? If pair errors are dominated by Markov-like speckle decorrelation, a lag-\(l\) pair is as noisy as \(l\) adjacent pairs and carries little new information; longer lags also hop more.

I9A displacement map shows a faint sawtooth texture with a period of one axial sample. What is it, and how do you remove it? · screen

Pixel (peak) locking: the subsample interpolator is biased toward integer lags, with zero bias at 0 and ½ sample and extremes in between, so a smoothly varying true displacement acquires a periodic error of one-sample period. At \(f_s/f_0\) = 8 the parabolic amplitude is ~0.012 sample (0.23 µm); after differentiation it becomes a strain artefact. Fixes: cosine fit or band-limited upsampling of the correlation, phase-zero estimation on IQ with the local \(\bar{f}\) (Loupas), or a higher sampling rate. Averaging does not help because the error is deterministic in \(d\). Follow-up: can you calibrate and subtract the bias curve? Only approximately: each kernel's interpolation error varies with its own local spectrum (±0.011 sample in m11_ncc_subsample_bias.m), comparable to the mean bias.

I10Your ARFI phase tracker underestimates displacement by about 15 % at 4 cm depth in liver, but not near the surface. Why? · onsite

Frequency-dependent attenuation shifts the received centre frequency down with depth (\(f_c = f_0 - 2\alpha_0\sigma_f^2 z\) for a Gaussian spectrum: 0.75 MHz at 4 cm for 0.5 dB/cm/MHz, 5 MHz, 60 %). The phase measures \(2\pi f_c D\), and dividing by the nominal \(f_0\) gives \(\hat{D} = D\cdot f_c/f_0 = 0.85\,D\). Fix: Loupas' 2D autocorrelator, which estimates the local mean frequency from the fast-time lag-one correlation, or a depth-dependent \(f_c(z)\) from the spectral centroid. Follow-up: does NCC with parabolic interpolation have the same problem? No scale error, because it measures time directly; its bias depends on \(f_s/f_c\) and is small at 40 MHz.

I11Why is lateral speckle tracking so much worse than axial, and what would you do about it? · onsite

Jitter scales as \(1/\sqrt{-R''(0)}\). Axially the RF correlation carries the \(f_0\) carrier, so its curvature is \((2\pi f_{z0})^2 + 1/(2\sigma_z^2)\); laterally there is only the beam envelope, \(1/(2\sigma_x^2)\). For 5 MHz and a 0.6 mm lateral FWHM the predicted ratio is about 15. Lateral sampling is also coarse (line spacing comparable to the beam), and on RF a separable 2D fit through the integer axial row couples axial subsample offsets into the lateral estimate (15.5 µm vs 2.7 µm with axial-first fitting in m11_block_matching_2d.m). Remedies: fit axially first or fit a 2D model with a cross term; denser lateral sampling or interpolated beams; transverse-oscillation beamforming or angle compounding to create a lateral carrier; combine axial estimates from several steering angles. Follow-up: why does the envelope help laterally but hurt axially? It removes the carrier: the axial curvature drops (axial error 3–4× worse) and the coupling disappears.

I12How would you use time-delay estimation to measure local sound speed? · research

Beamform the same region from several transmit (or receive) steering angles with an assumed \(c\). A local sound-speed error makes the echo from a given point arrive with angle-dependent delay errors, a line integral of the slowness error along each path. Measure the phase shift between complex images from neighbouring angles with a Kasai-type lag-one product (CUTE, Jaeger et al. 2015), stack these into a linear system relating phase shifts to the slowness map, and solve it with regularization. At 5 MHz, 1 ns of delay is 1.8° of phase, so the \(\pm\lambda/4\) wrap and the phase jitter fix the usable angle step and the averaging region; the Walker–Trahey bound with the inter-angle coherence predicts the noise. Follow-up: what limits it? Decorrelation between angles (larger steps lower \(\rho\)), the ill-conditioning of limited-angle tomography (poor axial resolution of \(c\)), and the \(f_0\) used to convert phase to delay.

12Estimation theory & the Cramér–Rao bound

11 questions · open in the module

I1What is the Cramér–Rao bound, and what does it assume? · screen

A lower bound on the variance of any unbiased estimator: \(\operatorname{var}(\hat\theta) \ge 1/I(\theta)\), \(I = \mathrm{E}[(\partial\ln p/\partial\theta)^2]\). It assumes the support of \(p(y;\theta)\) does not depend on \(\theta\) and the log-likelihood is differentiable; it is local (curvature at the true value) and says nothing about bias. It is reached exactly in the linear Gaussian model and asymptotically by the MLE. Follow-up: "And for biased estimators?" \((1 + b')^2/I\), and the MSE can be lower than \(1/I\).

I2A paper claims λ/20 resolution with ULM. What do you ask? · screen

Whether λ/20 is localization precision (std of an isolated bubble's position) or the separability of two bubbles. Then the ENR or SNR definition and noise model (\(\sigma_x = \mathrm{FWHM}/(2.355\sqrt{\mathrm{ENR}})\): λ/20 with FWHM ≈ λ needs only ENR ≈ 18.6 dB), whether outliers were excluded, how motion and tracking errors were handled, and how resolution was measured on the final map (for example a Fourier-ring-correlation split of the localizations, or the smallest separated vessel pair). Follow-up: "Why is precision so cheap?" Because the bound scales as \(1/\sqrt{\mathrm{ENR}}\) and bubbles are bright.

I3Derive the localization CRLB for a Gaussian PSF in complex white noise. · onsite

\(s = a\,g(x - x_0)\,g(z - z_0)\), noise density \(N_0\). \(I_{x_0x_0} = (2/N_0)\lvert a\rvert^2\iint\lvert\partial g/\partial x\rvert^2\) \(= (2\lvert a\rvert^2/N_0)\cdot\pi w_x w_z/(2w_x^2)\) \(= \mathrm{ENR}/w_x^2\). Cross terms with \(\mathrm{Re}\,a\), \(\mathrm{Im}\,a\) and \(z_0\) vanish (odd × even), so \(\sigma_x \ge w_x/\sqrt{\mathrm{ENR}}\). Example: FWHM 1λ at 5 MHz, ENR 30 dB: 308/(2.355·31.6) = 4.1 µm. Follow-up: "Real RF instead?" \(\sqrt{2}\,w/\sqrt{\mathrm{ENR}_r}\) with \(\mathrm{ENR}_r = \sum s^2/\sigma^2\), and the axial bound changes if the carrier phase is known.

I4Where does the factor 2 between real and complex noise come from? · onsite

With \(\mathrm{E}\lvert w\rvert^2 = \sigma^2\), each quadrature has \(\sigma^2/2\); summing the real-data information of both quadratures gives \(I = (2/\sigma^2)\,\mathrm{Re}\sum\lvert\partial s\rvert^2\). Phase-known amplitude from IQ: \(\sigma^2/(2\lVert s\rVert^2)\); real pulse in real noise \(\sigma^2\): \(\sigma^2/\lVert s\rVert^2\). Same for frequency: complex tone \(6/(\mathrm{SNR}\cdot N(N^2-1))\) vs Kay's real \(12/(\eta N(N^2-1))\). Follow-up: "Does demodulating RF to IQ change the bound?" No; it is an invertible map of band-limited data. Only the bookkeeping of \(\sigma^2\) changes.

I5Why can an intensity-weighted centroid be nearly efficient while an amplitude-weighted centroid is not? · onsite

Linearize at high ENR: \(\lvert y\rvert^2 \approx \lvert a\rvert^2 g^2 + 2\,\mathrm{Re}(a^* g\, n)\), so the intensity centroid's error is a noise projection onto \(x\cdot g\), which is proportional to \(g'\), the optimal (matched) template: variance \(w^2/\mathrm{ENR}\), the bound. With \(\lvert y\rvert\) weights the error is a projection onto \(x\cdot 1\) over the box: equal-weight noise from every pixel, so variance grows with box area. Follow-up: "What breaks the intensity centroid?" Background (pulls toward the box centre), truncation, neighbours, and the \(\lvert n\rvert^2\) term at low ENR.

I6How does Doppler frequency precision scale with ensemble length, and where is the threshold? · onsite

\(\operatorname{var}(\hat\omega) \ge 6/(\mathrm{SNR}\cdot N(N^2 - 1))\), so \(\sigma \propto N^{-3/2}\); N = 16 → 32 cuts σ by 2.83. The periodogram MLE follows it down to an integrated \(N\cdot\mathrm{SNR}\) of about 13 dB (N = 32); 2 dB lower it is two orders of magnitude worse because noise bins win. Lag-one is 5.7× in variance above the bound at N = 32 even at high SNR. Follow-up: "Convert to velocity": \(\sigma_v = c\cdot\mathrm{PRF}\cdot\sigma_\omega/(4\pi f_0)\): 1.49 mm/s for N = 16, 10 dB, PRF 5 kHz, 5 MHz.

I7How would you bound sound-speed estimation from delay data, and what dominates the bound? · research

Write the delay model \(\tau_e(c, z, x_s, t_0)\), its Jacobian, \(\mathbf{I} = \mathbf{J}^{T}\mathbf{C}_\tau^{-1}\mathbf{J}\) with the delay covariance from the Walker–Trahey/Carter bound (Module 11), and take \([\mathbf{I}^{-1}]_{cc}\). The nuisances dominate: for one target at 30 mm, 128 elements, 1 ns jitter, \(\sigma_c\) goes from 0.065 m/s (\(z\) unknown) to 2.8 m/s (\(z\) and \(t_0\) unknown), because \(t_0\), \(z\) and \(c\) all produce nearly the same bowl-shaped moveout. Break it with several depths and angles, a calibrated lens delay, or a prior. Follow-up: "Correlated delay errors?" Use the full \(\mathbf{C}_\tau\); aberration-like errors are low-order and project onto the same bowl, so they hurt more than white jitter.

I8A learned localizer beats the CRLB on your test set. Is that suspicious? · research

Not necessarily: the CRLB bounds unbiased estimators. A network trained on a distribution of PSFs and positions acts as a Bayesian estimator with that prior, and the relevant bound is the Bayesian CRB \(1/(\mathrm{E}[I] + I_{\mathrm{prior}})\). Check the bias map against true sub-pixel position (pull toward pixel centres or training positions), test out of distribution (other PSF widths, depths, neighbours), and compare the MSE against the Bayesian bound for the same prior. Follow-up: "Can any estimator beat the Bayesian bound?" Not in Bayes MSE under regularity conditions.

I9Why is the MLE of echo power from IQ biased, and how do you correct it? · onsite

Invariance makes \(\lvert\hat a\rvert^2\) the MLE of \(\lvert a\rvert^2\), but \(\mathrm{E}\lvert\hat a\rvert^2 = \lvert a\rvert^2 + \sigma^2/\lVert s\rVert^2\): the estimator variance adds. At ENR = 10 dB the power is 10 % (0.41 dB) high. Subtract an estimate of \(\sigma^2/\lVert s\rVert^2\) from a noise-only region or noise frames. Follow-up: "Where else?" Power Doppler, coherence factor and correlation coefficients at low SNR, all biased by the noise term.

I10Two bubbles are 0.5 FWHM apart. What happens to the precision, and what does it mean for high-concentration ULM? · research

With both positions and amplitudes unknown, the Fisher matrix couples them: std inflation 3.96 for in-phase echoes (1.81 in quadrature), rising as \(d^{-2}\) (\(d^{-1}\)) as they approach. Before that, a single-bubble localizer returns a biased "mid-point". High concentration therefore needs model-order selection (one bubble or two), sparse recovery (Module 13), or temporal separation (separable bubble dynamics). Follow-up: "Does amplitude ratio matter?" Not in the Fisher inflation (only the relative phase), but a weak neighbour is harder to detect, which fails first.

I11To estimate mean backscattered power from speckle, do you average intensity or dB values? · screen

Intensity. The sample mean of exponential intensities is the efficient MLE. Averaging dB values gives a biased result (−2.51 dB, Euler's \(\gamma\) in \(10\log_{10}\)) with variance \(\pi^2/6 = 1.64\) times the bound for \(\ln\mu\): 61 % efficiency, i.e. the same precision needs 64 % more independent samples. Follow-up: "How many independent samples in a gate?" Count speckle cells, not pixels (Module 07).

13Inverse problems and regularization

11 questions · open in the module

I1Why is deconvolution of an RF line ill-posed, and what is in its null space? · screen

Convolution with the pulse is diagonal in frequency with singular values \(\lvert H(f)\rvert\). A transducer passes a band around \(f_0\) and nothing at DC or far above, so those frequencies of the reflectivity are (numerically) in the null space: for a 5 MHz, 60 % pulse, \(\lvert H\rvert < 0.01\) outside about 1.1–8.9 MHz. Dividing by H amplifies noise there by \(1/\lvert H\rvert\), over 100 dB at 12 MHz. Tikhonov returns the minimum-norm, band-limited estimate; anything outside the band must come from a prior.

Follow-up: "Then how do sparse deconvolution methods claim higher resolution?" They assume few reflectors; the prior extrapolates the spectrum, and it fails when the assumption does (for example two same-sign reflectors λ/4 apart, which nearly cancel in band).

I2Write the Tikhonov filter factor. A colleague says "we used λ = 0.01": what do you ask? · screen

\(f_i = \sigma_i^2/(\sigma_i^2 + \lambda^2)\) for the penalty \(\lambda^2\lVert\mathbf{x}\rVert^2\). I ask whether the penalty was \(\lambda\lVert\mathbf{x}\rVert^2\) or \(\lambda^2\lVert\mathbf{x}\rVert^2\) (a factor 100 apart at 0.01), whether A was normalized (λ is meaningful only relative to \(\sigma_1\)), what L was, and how λ was chosen. With λ relative to \(\sigma_1\) in the squared convention, components below 1 % of \(\sigma_1\) are suppressed and no noise component is amplified more than \(1/(2\lambda)\) = 50 times.

Follow-up: "Which components are amplified most?" Those with σ = λ, not the smallest ones.

I3How would you choose λ on in-vivo data, and how can each method fail? · onsite

Discrepancy principle if I can measure the noise level (repeat acquisitions, or a noise-only region): \(\lVert\mathbf{r}\rVert = \tau\sqrt{m}\,\sigma_n\); it is robust but over-smooths a little and is wrong if \(\sigma_n\) is wrong. GCV needs no \(\sigma_n\) but assumes white noise: with correlated noise (delay estimates correlated over the kernel, speckle-induced errors) it picks a tiny λ; in the RF demo with AR(1) noise its error is 31 times the oracle's. The L-curve corner is a heuristic that can over-smooth for smooth solutions and has no corner at very low noise. I would cross-check two rules, test stability to a new noise draw, and in simulation report the oracle too.

Follow-up: "Can you fix GCV for correlated noise?" Whiten with an estimate of the noise covariance first, then apply GCV to the whitened problem.

I4Your CG reconstruction gets worse after 40 iterations. Is that a bug? · onsite

No, it is semiconvergence. CGLS first fits the large singular components (signal) and then the small ones (noise), so the error has a minimum. The iteration count plays the role of λ: for Landweber with \(\omega = 1/\sigma_1^2\), \(k \approx \ln 2\,(\sigma_1/\lambda)^2\); CGLS gets there in roughly the square root of that (35 versus 1823 in the RF example). Stop with the discrepancy principle on \(\lVert\mathbf{r}_k\rVert\), or add an explicit penalty and iterate to convergence.

Follow-up: "Which is better, early stopping or explicit Tikhonov?" Equivalent in error in this example; explicit penalties make the result independent of the solver and let you compute R, early stopping is cheaper for very large problems.

I5Compare L2 and L1 regularization for point reflectors. Why soft thresholding, why FISTA, and what about the amplitudes? · onsite

L2 spreads energy over all components and returns band-limited blobs; L1 prefers few nonzeros and can resolve reflectors closer than the pulse length if their echoes stay distinguishable in band. Soft thresholding is the proximal map of \(\mu\lvert x\rvert\), so ISTA alternates a gradient step on the data term with thresholding; FISTA adds Nesterov momentum and improves the worst-case gap from \(O(1/k)\) to \(O(1/k^2)\). L1 shrinks surviving amplitudes by about \(\mu/\lVert\mathbf{a}\rVert^2\), so I would debias by least squares on the recovered support before quantifying intensities, and start μ at \(\sigma_n\lVert\mathbf{a}\rVert\sqrt{2\ln N}\).

Follow-up: "What breaks L1 recovery?" Highly coherent dictionaries (fine grids, band-limited pulses), off-grid sources, and pairs that cancel in band.

I6When would you use TV for a sound-speed map, and what artifacts would you warn a reviewer about? · onsite

When the tissue is approximately piecewise constant (abdominal wall layers, a lesion in a homogeneous organ) and edges matter. TV preserves edges but lowers small plateaus by about \(2\mu/w\) (1D denoising), turns gradients into staircases, and with the anisotropic form favours axis-aligned edges. In the layered example it roughly halves the RMSE compared with the best quadratic penalty; on a ramp it ties \(\mathbf{D}_2\). I would show a ramp phantom and a small-inclusion phantom to expose both artifacts and quote values from plateau interiors.

Follow-up: "How do you solve it?" ADMM with \(\mathbf{z} = \mathbf{D}\mathbf{x}\): one factorization of \(\mathbf{A}^{T}\mathbf{A} + \rho\mathbf{D}^{T}\mathbf{D}\), then soft thresholding; ρ affects speed, not the minimizer.

I7Give the Bayesian interpretation of Tikhonov and of L1. What does it tell you about λ? · research

With Gaussian noise of variance \(\sigma_n^2\), MAP minimizes \(\lVert\mathbf{A}\mathbf{x} - \mathbf{y}\rVert^2/(2\sigma_n^2) - \log p(\mathbf{x})\). A zero-mean Gaussian prior with variance \(\sigma_x^2\) gives Tikhonov with \(\lambda^2 = \sigma_n^2/\sigma_x^2\), the noise-to-signal ratio, which is the Wiener filter in the circulant case; a Laplace prior of scale β gives L1 with \(\mu = \sigma_n^2/\beta\). So λ should rise when the noise rises and fall when the expected contrast rises. MAP gives a point estimate; the sparsity comes from the mode, not from the prior, whose samples are not sparse.

Follow-up: "How would you get error bars?" From the posterior: for the Gaussian case its covariance is \(\sigma_n^2(\mathbf{A}^{T}\mathbf{A} + \lambda^2\mathbf{L}^{T}\mathbf{L})^{-1}\); for non-Gaussian priors, sampling (MCMC) or approximations.

I8Why do echo-mode sound-speed images smear inclusions in depth, and what would you change in the acquisition? · research

Straight rays from each pixel to a linear array lie within \(\pm\theta_{\max}\) of vertical. By the Fourier-slice theorem, their integrals determine only spatial frequencies in a wedge of half-angle \(\theta_{\max}\) around \(k_x\); the cone around \(k_z\) is missing, so the resolution kernel is elongated axially, roughly by \(1/\sin\theta_{\max}\) (9.4 versus 1.2 mm at 10°, 3.3 versus 1.3 mm at 30° in the toy). Differences between angle pairs also make layered structures nearly invisible (sensitivity \(2(\sec\theta - 1)\)). I would widen the angle span as far as the element directivity allows, add angle pairs with large mid-angle differences, combine with absolute-delay information such as depth-resolved autofocus for layers, and use a prior that matches the anatomy.

Follow-up: "What limits \(\theta_{\max}\)?" Element directivity, grating lobes at large steering, aperture coverage of the oblique rays at depth (fewer valid rows), and speckle decorrelation between widely different angles.

I9A reviewer asks for the resolution of your sound-speed map. What do you report? · research

For a linear (Tikhonov-type) reconstruction, rows of the model-resolution matrix \(\mathbf{R} = (\mathbf{A}^{T}\mathbf{A} + \lambda^2\mathbf{L}^{T}\mathbf{L})^{-1}\mathbf{A}^{T}\mathbf{A}\) at representative pixels: their axial and lateral FWHM, how they vary with depth, and \(\operatorname{trace}\mathbf{R}\) as the effective number of resolved parameters. I would add the noise standard deviation map from the posterior covariance, and for nonlinear regularizers (TV, L1) point-target and edge-response simulations at the same λ. Inclusion-mean accuracy is not resolution.

Follow-up: "Can R be computed for a 3D problem?" Not densely; compute selected rows by solving with unit vectors (one solve each), or estimate the diagonal with randomized probing.

I10How do you validate a regularized reconstruction method without fooling yourself? · research

Avoid the inverse crime: generate data on a finer grid with a different (preferably full-wave) model and realistic, correlated noise. Report the oracle parameter and the parameter from a practical rule. Test priors against phantoms that violate them (gradients for TV, extended targets for L1). Show stability to new noise draws and to model error (wrong c₀, refraction). Then phantoms with independently known sound speed, then in vivo with a reference such as a transmission measurement or a known anatomical value.

Follow-up: "What is the signature of the inverse crime?" Near-perfect reconstructions at unrealistically small λ, and errors that fall without bound as noise goes to zero.

I11Why invert for slowness rather than sound speed, and what does the linearization ignore? · screen

Travel time is the path integral of slowness, so the forward model is linear in \(s = 1/c\) and the Gaussian delay-noise model maps to a quadratic data term. Near 1540 m/s, 1 ns/mm of slowness equals 2.37 m/s (\(\Delta c \approx -c_0^2\,\Delta s\)). The linear model assumes straight rays (no refraction), small contrast and that the misfocused echo stays at the same pixel; larger contrasts call for iterating with ray bending or a wave-based model.

Follow-up: "How big is the delay error from a 50 m/s inclusion?" 20.4 ns/mm, so 204 ns over 10 mm, a full period at 5 MHz.

14Optimization & automatic differentiation

10 questions · open in the module

I1Your gradient-descent reconstruction converges painfully slowly. Diagnose it and give the rate. · screen

Slow linear convergence means a large condition number of the Hessian. With the optimal fixed step \(2/(L + \mu)\) the error contracts by \((\kappa - 1)/(\kappa + 1)\) per iteration, about \((\kappa/2)\ln(1/\varepsilon)\) iterations: 691 for \(\kappa\) = 100 and \(\varepsilon = 10^{-6}\). First check the scaling (parameters in mixed units inflate \(\kappa\) by orders of magnitude; Jacobi scaling fixed \(1.2 \times 10^6 \to 1.17\) in the lab), then add momentum or use conjugate gradients (\(\sqrt\kappa\) dependence), or switch to Gauss–Newton if it is a least-squares problem. Follow-up: why does momentum help? Per mode it is a second-order recursion; tuned, its roots have modulus \((\sqrt\kappa-1)/(\sqrt\kappa+1)\), so iterations scale with \(\sqrt\kappa\).

I2How do you choose the step size when you do not know L? · screen

Backtracking with the Armijo condition: try \(\alpha\) (or 2× the last accepted \(\alpha\)), halve until \(f(\mathbf{x} + \alpha\mathbf{d}) \le f(\mathbf{x}) + 10^{-4}\alpha\nabla f^{T}\mathbf{d}\). Alternatively estimate \(L\) by a few power iterations on Hessian–vector products and use \(1/L\). For least squares, Gauss–Newton or LM chooses the step from curvature and needs no \(\alpha\). Follow-up: what can go wrong? Too short steps (add the Wolfe curvature condition) and, on nonconvex costs, long trial steps that jump basins; cap the trial step at a physically meaningful size (40 m/s in the autofocus demo).

I3You fit a backscatter or attenuation model with 3–5 parameters. GD, Gauss–Newton, LM or Adam? · onsite

LM: it behaves like Gauss–Newton (quadratic convergence for small residuals, about 10 iterations in the lab's echo fit) near the answer and like scaled gradient descent far from it, and never accepts an uphill step. Solve the subproblem with QR or with the scaled normal equations at this size, and scale parameters. At convergence \(\sigma^2(\mathbf{J}^{T}\mathbf{J})^{-1}\) gives the CRLB-type error bars (0.69 ns on \(t_0\)). Adam would take thousands of iterations and give no covariance. Follow-up: when does Gauss–Newton degrade? Large residuals (model error) or strong nonlinearity make the neglected \(\sum r_i\nabla^2 r_i\) term matter; convergence becomes linear.

I4Explain cycle skipping and quantify it for a 5 MHz pulse. · onsite

The L2 misfit between an echo and a delayed model is \(2E - 2R(\tau)\); for an RF pulse \(R(\tau) \approx E\,e^{-\tau^2/4\sigma^2}\cos 2\pi f_0\tau\), so local minima appear near every period (194, 388 ns for \(f_0\) = 5 MHz, \(\sigma\) = 0.125 µs) and the basin of the true minimum ends near \(\pm T_0/2\) = ±97 ns. A start beyond that converges to the wrong cycle. In travel-time form the starting model must satisfy \(\lvert\Delta c\rvert/c < \lambda/(2L)\). Remedies: envelope or correlation-based costs, low frequencies, multiscale schedules, better starting models. Follow-up: why not just low-pass the data? The energy is not there: a 1.5 MHz Gaussian low-pass keeps −29 dB of a 5 MHz, 60 % pulse, so noise dominates; you must transmit low frequencies.

I5Design a gradient-based global sound-speed estimator for a linear array. · research

Acquire several plane-wave angles; beamform IQ data on an ROI fixed in two-way time (pixel depth \(\hat ct/2\)) so that \(\hat c\) only changes focusing; maximize the coherently compounded energy or a coherence metric over ROIs with bright, diffuse echoes; get \(dM/d\hat c\) exactly (chain rule through delays and the interpolator, or AD), check it once by central differences; climb with a line search and a box constraint, from a coarse sweep or several starts. Report bias (calibrate on phantoms: the demo shows +2 to +10 m/s geometric bias) and precision across ROIs. Follow-up: why not maximize the energy of a single plane-wave image, or the RF value at the target depth? The first is nearly invariant to focusing (receive migration preserves energy); the second cycle-skips every \(\hat c^2/(2zf_0) \approx\) 8 m/s.

I6Forward or reverse mode for a differentiable beamformer with a 100 × 100 slowness map and a scalar loss? What limits it? · onsite

Reverse mode: \(10^4\) inputs, 1 output, so one backward pass at a small constant multiple of the forward cost, against \(10^4\) forward-mode passes. The limit is memory: the tape stores delays, interpolation indices and weights per pixel, element and transmit; for 256 × 256 pixels, 128 elements and 11 angles that is \(9.2 \times 10^7\) values, 0.74 GB per float64 quantity. Use checkpointing (recompute delays on the backward pass), float32 and patch-wise losses. Follow-up: when is forward mode better? Few parameters (a single \(\hat c\), a handful of layer speeds): one dual-number pass per parameter, no tape.

I7What is the complex-step derivative, why can h be 10⁻²⁰, and when does it fail? · screen

\(f'(x) \approx \mathrm{Im}\,f(x + ih)/h\). The Taylor series gives \(\mathrm{Im}\,f(x + ih) = hf'(x) - h^3f'''/6 + \dots\), with no subtraction of nearly equal numbers, so round-off does not grow as \(h\) shrinks and \(h = 10^{-20}\) yields machine precision (\(2.2 \times 10^{-16}\) in the lab) against ~\(10^{-8}\) for forward differences. It fails for non-analytic operations (abs, conj, MATLAB's ', max/min branches) and for code that already uses complex numbers, such as an IQ beamformer. Follow-up: what would you use for the IQ beamformer? Dual numbers or reverse-mode AD, or the hand-derived chain rule, verified once by central differences.

I8Derive the gradient of ‖y − Aw‖² with respect to complex w, and say how you would test it. · research

\(\mathcal{L} = (\mathbf{y} - \mathbf{A}\mathbf{w})^{H}(\mathbf{y} - \mathbf{A}\mathbf{w})\). Treating \(\mathbf{w}\) and \(\mathbf{w}^*\) as independent, \(\partial\mathcal{L}/\partial\mathbf{w}^* = -\mathbf{A}^{H}(\mathbf{y} - \mathbf{A}\mathbf{w})\); the steepest-descent update is \(\mathbf{w} \leftarrow \mathbf{w} + \mu\mathbf{A}^{H}(\mathbf{y} - \mathbf{A}\mathbf{w})\), which is the LMS/Landweber step. Test: perturb \(\mathrm{Re}\,w_k\) and \(\mathrm{Im}\,w_k\) separately by \(\pm h\); \(d\mathcal{L}/d\mathrm{Re} + i\,d\mathcal{L}/d\mathrm{Im}\) must equal \(2\,\partial\mathcal{L}/\partial w_k^*\) (agreement \(2 \times 10^{-10}\) in the lab). Follow-up: what happens with the conjugate direction? The imaginary parts of all steps flip sign; the lab's run diverges to \(10^{62}\).

I9Adam converged to a slightly different answer each run and never quite settles. Why? · onsite

With a constant learning rate \(\eta\), Adam moves each parameter by about \(\eta\) per step (\(\hat m/\sqrt{\hat v} \approx \mathrm{sign}\,g\)), so it hovers in a band of order \(\eta\) around the minimum; with stochastic gradients there is also a noise floor proportional to the step (\(\alpha\sigma^2/(2\lambda)\) for SGD). Decay \(\eta\), increase the batch, or finish with a deterministic second-order method on the full cost. Follow-up: is Adam scale invariant? To a diagonal rescaling of the gradient yes (though a step of η still means different physical changes in different parameter units), to rotations no: a rotated narrow valley still slows it (Rosenbrock: 3034 iterations).

I10You maximize coherent energy over element phases on speckle. The optimizer converges, the energy exceeds that of the true correction, but the estimated screen is off by 0.14 rad rms. Explain. · research

The cost is the sample coherent energy \(\mathbf{w}^{H}\hat{\mathbf{R}}\mathbf{w}\) with \(\hat{\mathbf{R}}\) from \(K\) = 64 realizations. Speckle has no unique focal point: \(\hat{\mathbf{R}}\) is far from rank one (\(\lambda_1/\lambda_2 \approx 2.4\) in the demo, against about 80 with a point 10 dB above the speckle) and its top eigenvector fits the sampling fluctuations of \(\hat{\mathbf{R}}\), so the optimizer finds a law that beats the true one on this data set yet differs from it. A point-like target (guide star) makes \(\hat{\mathbf{R}}\) nearly rank one: at 10 dB target-to-speckle ratio the error drops to about 0.02 rad. More independent realizations, isoplanatic averaging or the distortion-matrix approach (Module 06) also help. Follow-up: why remove tilt before scoring? A linear phase only steers the beam (Module 07); it is not identifiable from coherence.

15Wave physics, nonlinear propagation and harmonic imaging

11 questions · open in the module

I1Derive the linear acoustic wave equation. Which terms does linearization drop, and how big are they? · screen

Linearized mass (\(\partial\rho'/\partial t + \rho_0 \nabla \cdot \mathbf{u} = 0\)), momentum (\(\rho_0\, \partial\mathbf{u}/\partial t = -\nabla p\)) and state (\(p = c_0^2 \rho'\)); take \(\partial/\partial t\) of the first, the divergence of the second and eliminate: \(\nabla^2 p = c_0^{-2} \partial^2 p/\partial t^2\). Dropped: \(\nabla \cdot (\rho' \mathbf{u})\), \(\rho_0 (\mathbf{u} \cdot \nabla) \mathbf{u}\), \(\rho'\, \partial \mathbf{u}/\partial t\) and \((B/2) s^2\), all of relative order \(\varepsilon = p_0/(\rho_0 c_0^2)\), about \(4 \times 10^{-4}\) at 1 MPa. Follow-up: "Then why do harmonics matter?" Because the error accumulates with distance: the relevant parameter is \(\sigma = \beta\varepsilon k x\), which reaches 0.5 in 2–3 cm at diagnostic pressures.

I2Why is soft-tissue contrast weak in B-mode, and what amplitude does a fat–muscle boundary return? · screen

Soft-tissue impedances are all within about 10 % of 1.6 MRayl. Fat → muscle (1.38 → 1.66 MRayl): R = 0.093, −20.7 dB; liver → muscle about −48 dB. Most of the image is diffuse scattering from sub-resolution structure, 40–60 dB below a specular reflector, displayed after log compression. Follow-up: "And air?" R = −0.9995, the polarity inverts and 0.1 % of the energy crosses: gel, and acoustic shadows behind gas.

I3A beam hits the skull at 45°. What happens to its energy? · onsite

The longitudinal critical angle is \(\arcsin(1540/2820)\) = 33.1°, so no longitudinal wave enters. The boundary still transmits by mode conversion: solving the three boundary conditions gives about 93 % of the energy as a shear wave and 7 % reflected (\(c_S\) = 1500 m/s, ρ = 1900 kg/m³). The shear wave is refracted toward the normal (\(c_S < c_1\)) and attenuates strongly in bone; at the inner table it converts back to longitudinal. Follow-up: "Why do transcranial studies avoid this regime?" Shear attenuation and the extra mode conversions; normal incidence keeps 71 % in the longitudinal wave.

I4How does a subcutaneous fat layer bias an image and a sound-speed estimate, and how would you correct it? · research

Axially, \(d(c_0/c_f - 1)\): 2 cm of fat at 1450 m/s puts everything 1.24 mm deeper. Laterally, oblique fat–muscle boundaries refract (a 20° wedge deviates 1.9°, 2 mm at 6 cm) and duplicate structures. Across the aperture, path lengths in fat differ, so delays are wrong and the focus loses coherence. A global sound-speed estimator returns a path-averaged speed biased by the layer; local estimators (CUTE-style phase-shift tomography, layered inversion) model the layer explicitly, and aberration correction applies the residual delays. Follow-up: "What does the FDTD/k-Wave check add?" Reflections and mode conversion that ray models miss.

I5Rayleigh scattering goes as \(f^4\). Why are measured backscatter exponents in tissue usually below 4? · onsite

\(f^4\) holds for \(ka \ll 1\). Real scatterers have finite size and correlated structure; the form factor rolls the BSC off. With the Gaussian form factor \(\exp(-0.827 k^2 a_{\mathrm{eff}}^2)\) the local exponent is \(4 - 1.654 (k a_{\mathrm{eff}})^2\), equal to 2 at \(ka = 1.10\) (54 µm at 5 MHz). Fitting that exponent is how QUS estimates effective scatterer size (Module 27). Follow-up: "What else biases the slope?" Uncorrected attenuation and diffraction, and gating, which is why reference-phantom normalization is used.

I6Why must frequency-dependent attenuation come with dispersion? Estimate it for liver. · research

Causality makes the complex wavenumber analytic in the upper half-plane, so its real (phase) and imaginary (loss) parts are a Kramers–Kronig pair. For \(\alpha = \alpha_1 \lvert \omega \rvert\) the nearly local result is \(\Delta(1/c) = -(2/\pi)\alpha_1 \ln(\omega/\omega_0)\). With 0.5 dB/cm/MHz, c rises 1.38 m/s per neper of ln f, 3.2 m/s from 1 to 10 MHz (3.6 m/s for y = 1.1). Follow-up: "Where does this matter?" Comparing sound speeds measured at different frequencies, and time-domain simulators (fractional Laplacian terms in k-Wave) that must include it to be causal.

I7Define B/A and β, and compute the shock distance in liver at 2 MHz, 1 MPa. · onsite

\(B/A = 2\rho_0 c_0 (\partial c/\partial p)_s\), the second-order coefficient of the equation of state relative to the first; \(\beta = 1 + B/2A\) adds convection. Liver: B/A ≈ 7, β = 4.5, ρ = 1060 kg/m³, c = 1578 m/s: \(\bar{x} = \rho c^3/(\beta\omega p_0)\) \(= 1060 \times 1578^3/(4.5 \times 2\pi \times 2 \times 10^6 \times 10^6)\) = 73.7 mm (29.5 mm at 5 MHz). Follow-up: "Will it shock?" No: α at 2 MHz ≈ 12.4 Np/m gives \(\Gamma \approx 1.1\), so absorption balances nonlinearity and the harmonic saturates near \(\sigma \approx 0.5\).

I8Why is the tissue-harmonic beam narrower, with lower sidelobes? Give numbers. · onsite

The harmonic source density is \(\propto p_1^2\), so the transmit profile is squared: for a line aperture the −6 dB width goes from 1.207 to 0.886 \(\lambda_0 F/D\) and the first sidelobe from −13.3 to −26.5 dB. Received at \(2f_0\) the two-way pattern is \(\operatorname{sinc}^2(u) \operatorname{sinc}(2u)\): 0.511 \(\lambda_0 F/D\) (0.58 × the fundamental two-way) with −27.9 dB first sidelobe. Grating lobes of the \(2f_0\) receive pattern are suppressed by the squared transmit. Follow-up: "Why not just transmit at \(2f_0\)?" Attenuation doubles in dB on the way down, and the \(f_0\)-designed array would have transmit grating lobes at \(2f_0\).

I9Why does THI reduce clutter in difficult patients? · screen

\(p^2\) dependence: clutter at −20 dB (reverberation, sidelobes, off-axis scatter) produces harmonic at −40 dB. The harmonic is absent at the skin and builds up with depth, so near-field reverberations in the body wall contribute little, and the harmonic is generated after part of the aberrating path. Follow-up: "Cost?" 15–25 dB less signal, twice the receive attenuation in dB, narrower usable bandwidth with filtering, or half the frame rate with PI.

I10Compare pulse inversion and amplitude modulation. What limits each in vivo? · onsite

With \(e(a) = \sum a^k e_k\): PI, \(e(a) + e(-a)\), keeps even orders (+6 dB on \(2f_0\)) and removes the fundamental at full bandwidth; AM, \(e(a) - 2e(a/2)\), removes the linear term but keeps ½ of the quadratic and ¾ of the cubic terms, including the nonlinear fundamental, useful for bubbles. Both need two transmits (half frame rate) and identical pulses; motion leaves \(2 \lvert \sin(\pi f_0 \delta) \rvert\): −26 dB at 10 mm/s, 5 kHz, 3 MHz. Follow-up: "How do you reduce the motion residual?" Higher PRF, motion compensation, or three-pulse sequences that cancel the linear residual to higher order.

I11How would you simulate harmonic imaging for a new probe, and what would you check first? · research

Linear codes (Field II) cannot make harmonics. For a homogeneous focused beam, a KZK solver (axisymmetric or 3D) with tissue power-law absorption and B/A; for heterogeneous media and aberration, a nonlinear full-wave code (k-Wave with β and power-law absorption). First checks: plane-wave limit against Fubini (\(\sigma \le 1\)) and the Gol'dberg behaviour, the on-axis fundamental against O'Neil, grid convergence with at least the third or fourth harmonic resolved (PPW counted at the highest harmonic). Then compare the simulated \(2f_0\) focal width with the \(p^2\) prediction. Follow-up: "Measurement?" Hydrophone with bandwidth to 3–5 \(f_0\) (Module 21).

16Piezoelectric transduction and equivalent circuits

11 questions · open in the module

I1What does \(k_t^2 = 0.26\) mean physically, and why is it not "26 % efficiency"? · screen

In a quasi-static cycle, 26 % of the energy stored on one side can be extracted on the other; equivalently \(c^E = 0.74 c^D\) and \(\varepsilon^S = 0.74 \varepsilon^T\) for that mode. It is a coupling, not an efficiency: a lossless transducer converts all the absorbed electrical power to acoustic power at matched conditions; \(k^2\) decides how wide the band over which that can be done is (the \(f_r \text{–}f_a\) gap) and how hard you can load the plate before sensitivity collapses. Follow-up: "So what limits efficiency?" Mismatch at the electrical port, backing absorption, and material losses (\(Q_m\), \(\tan \delta\)).

I2Why is \(k_t\) of PZT-5H 0.51 when \(k_{33}\) is 0.75? · screen

Boundary conditions. \(k_{33}\) is for a bar with free lateral faces \((T_1 = T_2 = 0)\), \(k_t\) for a plate clamped laterally by its own extent \((S_1 = S_2 = 0)\). In the plate the \(d_{31}\) lateral contraction is blocked, so the stiffness and permittivity that enter are \(c_{33}^D\) and \(\varepsilon_{33}^S\), and the coupling drops. A sliver frees one direction \((k_{33}^{\prime} \approx 0.70)\), a composite pillar both. Follow-up: "What width-to-thickness ratio does a sliver need?" ≲ 0.6, to push the width mode well above the thickness mode.

I3An air-loaded PZT-5H plate shows \(\lvert Z\rvert\) minimum at 4.41 MHz and maximum at 5.00 MHz. What are \(k_t\) and the thickness? · onsite

\(f_r /f_a = 0.882\), \(x_r = (\pi /2)(0.882) = 1.385\), \(k_t^2 = x \cot x = 0.260\), \(k_t = 0.51\). Thickness from the antiresonance: \(t = v^D /(2f_a) = 4600/(10^{7}) = 0.46\) mm. Small-gap shortcut would give \(k_t = 0.54\). Follow-up: "And if the plate were already bonded to a backing?" Then the extrema shift and broaden; the IEEE formula no longer applies, fit a KLM model instead.

I4Sketch the KLM model and give its element values. Why do transducer designers prefer it to Mason? · onsite

Electrical port: \(C_0 = \varepsilon^S A/t\) in series with \(jX_1\), \(X_1 = (k_t^2 /\omega C_0)\operatorname{sinc} (f/f_a)\), then a transformer \(\varphi = k_t \sqrt{\pi /(\omega_0 C_{0}Z_{0})}\operatorname{sinc} (f/2f_a)\) feeding the centre of an acoustic line of length t, impedance \(Z_0 = \rho v^D A\), split into two halves to the back and front ports. Preferred because the acoustic path stays a continuous line: backing, bond lines, matching layers and the load attach as more line sections or as ABCD factors, and the frequency dependence of the coupling is explicit. Follow-up: "Show me it agrees with Mason." Free plate: \(X_1 + \varphi^2 (Z_0 /2)\tan (kt/2) = (k_t^2 /\omega C_0)\tan x/x\).

I5Why does a thickness transducer have a passband near \(3f_0\) but none near \(2f_0\)? · onsite

The electric field is uniform across the thickness, so it drives the plate only through the face terms h D; an even mode moves both faces in the same direction and its strain integrates to zero, giving no net drive. In KLM, \(\varphi \propto \sin (kt/2)\) is zero at \(2f_a\), and \(\varphi^2\) at \(3f_a\) is sinc²(1.5)/sinc²(0.5) = 1/9 of its value at \(f_a\). Follow-up: "How would you build a dual-band probe?" Stack two plates or use the third harmonic deliberately, or a CMUT; mention superharmonic contrast imaging.

I6An array element shows 750 Ω capacitive at 5 MHz and the system is 50 Ω. What do you do, and what do you gain? · onsite

Cancel \(C_0\) with a series inductor \(L = 1/(\omega^2 C_0)\) (29.5 µH at 4.5 MHz for 42.4 pF), leaving \(R \approx 346\) Ω, then transform the real part toward 50 Ω (1 : 2.5 turns). In the reference design that is −21.6 → −10.1 → −2.6 dB two-way. A shunt inductor alone does nothing useful against a 50 Ω source. Follow-up: "Cost?" Inductor Q and size in the connector, narrower band if the tuning Q exceeds the acoustic Q, and element-to-element tolerance.

I7Why does a heavier backing broaden the band and lower sensitivity? Give a formula. · screen

The plate is a half-wave resonator losing ≈ \(4(Z_B + Z_F)/Z_p\) of its energy per period, so \(Q \approx (\pi /2)Z_p /(Z_B + Z_F)\): 36 air-backed into water, 8.3 at 5 MRayl. The energy that leaves through the back is absorbed, so the forward fraction \(Z_F /(Z_B + Z_F)\) drops and the two-way loss grows about as its square. Follow-up: "Then how do you get 70 % bandwidth at good sensitivity?" Matching layers raise the effective front load; use light backing.

I8Your measured impedance of a kerfed array element shows an extra resonance a little above the thickness resonance that KLM does not predict. What is it, and what do you do? · research

A lateral (width) mode of the element, or of the kerf-filler/element structure, coupling through \(d_{31}\) when the width-to-thickness ratio is not small enough (≈ 0.6 or less separates them). KLM is 1D and cannot show it. Sub-dice the element, change the kerf filler, or move to a composite; verify with 2D/3D finite elements (Module 17) and a measured impedance sweep of every element. Follow-up: "Why does it hurt imaging?" It narrows and distorts the pulse spectrum, adds ringing, and increases crosstalk.

I9Why do single-crystal and composite probes have wider bandwidth than PZT probes with the same matching? · research

Higher \(k_t\) (0.6–0.85 versus 0.5) widens the \(f_r \text{–}f_a\) gap, i.e. the intrinsic electromechanical bandwidth, so the plate can be loaded more heavily for the same sensitivity; composites also lower Z (10–20 MRayl), which makes the transformation to 1.5 MRayl easier with fewer layers. Follow-up: "Any downside?" Crystal: \(T_{\mathrm{rt}}\) and \(E_c\) limits, cost, property spread; composite: polymer loss, lateral modes if not fine enough, and lower \(\varepsilon\) raising the electrical impedance.

I10How would you build a KLM model of a real, lossy element from measurements? · research

Measure the bare plate: \(C_0\) well above resonance gives \(\varepsilon^S\); \(f_a\) and t give \(v^D\); \(f_s\), \(f_p\) give \(k_t\) by IEEE 176; peak widths give \(Q_m\) and the off-resonance phase gives \(\tan \delta\) (or fit complex constants to the whole curve, as in Sherrit's method). Then measure each layer's Z and attenuation separately, build the ABCD chain, and compare the predicted electrical impedance of the finished element and the pulse-echo response against measurements, adjusting bond-line thickness last. Follow-up: "What will not fit?" Lateral modes, kerf crosstalk and lens effects: move to FE.

I11You designed a 5 MHz half-wave plate and the pulse-echo centre is 4.3 MHz. Name the causes and how you would separate them. · onsite

(1) Low-impedance drive and receive put the response near \(f_r = 0.88 f_a\), not \(f_a\). (2) Matching layers designed below \(f_a\) and their mass load the face. (3) The tuning inductor resonates with \(C_0\). (4) Lens and tissue attenuation downshift further in practice. Separate them with the model: switch off tuning, vary layer design frequency, change the receiver impedance, and compare with the measured impedance curve. Follow-up: "How do you hit 5 MHz?" Thin the plate (raise \(f_a\) to ≈ 5.7 MHz) and re-centre the layers.

17Matching layers, backing, lenses and finite-element design

10 questions · open in the module

I1Why does a transducer need a matching layer, and what impedance would you pick for one layer on PZT into tissue? · screen

PZT at ≈ 34 MRayl against tissue at 1.5 MRayl transmits only 16 % of the power per pass and reflects 92 % of the pressure, so the plate rings. A \(\lambda /4\) layer transforms the load to \(Z_1^2 /Z_L\); \(\sqrt{Z_p Z_L } \approx 7.2\) MRayl makes that equal to \(Z_p\) at \(f_0\). For a short pulse I would start lower, at the Desilets value \((Z_p Z_L^2)^{1/3} \approx 4.3\) MRayl, and refine with a KLM model. Follow-up: "Why lower?" Because the plate is already resonant; under-matching at \(f_0\) flattens the band.

I2Two matching layers: give the impedances and say where the numbers come from. · onsite

Desilets, Fraser and Kino (1978): \(Z_1 = Z_p^{4/7} Z_L^{3/7} \approx 9\) MRayl next to the plate and \(Z_2 = Z_p^{1/7} Z_L^{6/7} \approx 2.3\) MRayl outside, from a KLM analysis of the resonant plate. The microwave binomial transformer gives 15.7 / 3.3 MRayl, which maximizes flat transmission from a semi-infinite source and in a transducer gives a wider −6 dB band but a ringier pulse. Follow-up: "What would you actually build?" Whatever filled epoxy or ceramic gets near 9 MRayl with acceptable loss, then re-optimize the second layer in the model with the real first-layer properties.

I3What does the backing do, and why not always use a backing matched to the ceramic? · screen

It absorbs the rearward wave and damps the plate: \(Q \approx (\pi /2)Z_p /(Z_B + Z_L)\). A matched backing gives the shortest pulse but sends at least half of the energy backwards on transmit and loses as much on receive (half each way only if the layers present \(Z_p\) to the plate; a bare PZT face into tissue sends about 96 % backwards). Without matching layers, going from air to a matched backing adds about 28 dB of two-way loss in the reference design (−3.6 to −31.2 dB) for 16 % → 127 % bandwidth. With two good matching layers the band is already about 63 % on air backing, so designers use moderate backing (a few MRayl) and let the layers do the broadbanding. Follow-up: the backing must also be thick and lossy enough that its rear echo is below the dynamic range.

I4Silicone lens: convex or concave, and what radius for a 40 mm elevation focus? · screen

Convex, because silicone (≈ 1000 m/s) is slower than tissue: the thick centre delays the central rays. \(F = R/(c_t /c_l - 1)\), so \(R = 40 \times 0.54 = 21.6\) mm. Follow-up: "What happens in a patient with a thick fat layer?" With 1450 m/s the same lens focuses at 48 mm; the elevation focus is set by the lens and the tissue, not by the scanner.

I5What does lens attenuation do to the image besides costing sensitivity? · onsite

Two things. It low-pass filters the pulse (centre frequency down by about \(\beta \sigma^2\), 0.2–0.3 MHz is typical at 5 MHz), which lowers axial resolution and biases spectral QUS unless the reference has the same filter. And it weights the elevation aperture towards the edges because the convex lens is thickest at the centre: an inverse apodization that narrows the elevation main lobe slightly and raises sidelobes, more so at high frequency and wide aperture. Follow-up: "How would you reduce it?" Lower-loss lens materials, thinner lens with a shorter radius at the cost of more aberration, or elevation focusing in a 1.5D/1.75D array instead.

I6Your measured centre frequency is 10 % below the KLM design. List the suspects in order. · research

(1) The lens: it downshifts the spectrum and is often left out of the model. (2) Element geometry: a slender element resonates at the bar frequency, about 12 % below the plate value for \(\nu \approx 0.3\) (FDTD: 0.440 against 0.5). (3) Layer thickness or speed: a layer 10 % thick moves its band 10 % down. (4) Material constants: \(v^D\), \(k_t\) and \(\varepsilon^S\) differ from the datasheet after dicing and poling. (5) Bondlines at high frequency. The test that separates them: measure the electrical impedance of a bare element (\(f_r\), \(f_a\) give \(k_{\mathrm{eff}}\) and v), then of the stack without lens, then with lens, and fit each step.

I7What are lateral modes and how do you avoid them in an array? · onsite

Width resonances of the element bar, near \(f \approx c_p /(2w)\), that couple to the thickness mode through Poisson's ratio when w is comparable with t. They split the spectrum and add ringing and spurious beam lobes. The guideline keeps w/t below about 0.6; wider elements are sub-diced into two or more slender pieces wired in parallel. In composites the pillar aspect ratio and the lateral period play the same role. Follow-up: "How would you check?" A 2D/3D FE model of the element with kerf and filler, then laser vibrometry or the impedance spectrum of a diced sample.

I8When is a KLM model enough, and when do you need finite elements? · onsite

KLM/ABCD is exact for a laterally infinite plate with plane layers: it is the right tool to choose layer impedances and thicknesses, backing and electrical tuning, and it fits measured impedance well for large single elements. FE is needed for anything with edges: array elements with w/t near 1, kerf and sub-dicing, crosstalk through filler, backing and lens, curved lenses, element directivity, composites with non-trivial pillar geometry. Follow-up: "Explicit or implicit?" Explicit time-domain for broadband transient and large models; implicit frequency-domain for exact modes, admittance at a few frequencies and strongly coupled multiphysics.

I9You build an explicit FE model of a 7.5 MHz probe with a silicone lens. How do you choose the mesh and time step? · research

Find the shortest wavelength anywhere at the highest frequency of interest, including shear waves in the epoxies (treat the soft silicone lens as a lossy, nearly fluid medium) and the upper band edge (say 12 MHz). Use 15–20 elements per that wavelength for second-order schemes (−0.3 to −0.55 % phase speed error at \(S = 0.5\)), more if the waves travel many wavelengths. The time step follows from the fastest wave (ceramic longitudinal) and the smallest element through the CFL limit, \(c_{\mathrm{max}} \Delta t/\Delta x \le 1/\sqrt{d}\) for this stencil (codes add safety factors). Then check convergence by refining once and comparing impedance and pulse-echo. Follow-up: cost grows as \(\mathrm{ppw}^4\) in 3D, so exploit symmetry and model one element plus neighbours.

I10Design a matching stack for a 15 MHz probe on a 1-3 composite at 15 MRayl. What changes compared with PZT? · research

The mismatch is smaller (\(T_I = 0.33\) bare), so lower layer impedances: Desilets one layer \((15 \times 2.25)^{1/3} = 3.2\) MRayl, two layers \(15^{4/7}\,1.5^{3/7} = 5.6\) and \(15^{1/7}\,1.5^{6/7} = 2.1\) MRayl, all reachable with filled epoxies and parylene. At 15 MHz a 2500 m/s layer is 42 µm thick, so lapping tolerance and bondlines (keep them to 1–2 µm) dominate, and layer loss in dB stays the same as at 5 MHz for \(\alpha \propto f\). The composite's lower Z and higher \(k_t\) give bandwidth with lighter backing. Follow-up: check the composite's lateral (inter-pillar) resonances against 2 × 15 MHz if harmonic imaging is planned.

18CMUT, PMUT and new transducer materials

11 questions · open in the module

I1What sets a CMUT's pull-in voltage, and how does it scale with membrane radius, thickness and gap? · screen

Balance of the plate's bending stiffness against the electrostatic pressure: \(V_{\mathrm{PI}}^{2} = c\cdot Dg_{\mathrm{eff}}^{3}/(\varepsilon_0 a^{4})\) with \(c\) = 27.8 for a clamped plate (56.9 for the piston idealization) (18.4), so \(V_{\mathrm{PI}} \propto h^{3/2}g_{\mathrm{eff}}^{3/2}/a^{2}\), and \(g_{\mathrm{eff}}\) includes \(t_i/\varepsilon_r\) of the insulator. A 1 µm Si membrane, 20 µm radius, 0.2 µm gap and 0.1 µm nitride: 53 V. Follow-up: "how would residual tensile stress change it?" It adds a membrane stiffness that raises \(V_{\mathrm{PI}}\) and the resonance; for thin, highly stressed membranes it dominates the bending term.

I2Why does a CMUT not need a matching layer, and what limits its bandwidth? · screen

Its moving mass per area is a few g/m², so \(\omega m\) is a fraction of water's 1.5 MRayl: the water load over-damps the mass–spring (\(Q \approx 0.2\)) and couples directly. The band runs from \(\approx \eta k_{\mathrm{eff}}/(2\pi Z_w)\) (spring) to \(\approx Z_w/(2\pi\eta m)\) (mass, including water added mass) (18.7). In practice higher membrane modes, parasitic capacitance, crosstalk and diffraction bring measured immersion bandwidths to about 100 % (Oralkan et al. 2002: 96 % uncompensated, 105 % compensated). Follow-up: "and what limits the low end?" Spring softening helps (\(k_{\mathrm{eff}}\) falls with bias), but element size limits radiation at low frequency.

I3Explain collapse-mode operation. Why would you use it, and what are the risks? · onsite

Biasing past \(V_{\mathrm{PI}}\) brings the membrane centre onto the insulator; the free annulus then vibrates at a higher, bias-tunable frequency (\(f/f_0 \approx 8.7\) at \(b/a\) = 0.5), with higher coupling (FEM: \(k_T^{2} \approx 0.7\)) and larger average stroke. The state is hysteretic: it releases only below \(V_{\mathrm{SB}}\), set by the insulator's electrical thickness. Risks: dielectric charging under very high contact fields, wear and stiction at the contact, temperature and bias drift of the operating point. Follow-up: "how do you get into collapse reliably?" Overdrive briefly above \(V_{\mathrm{PI}}\), then hold a bias inside the hysteresis window.

I4How do you define and compute the coupling coefficient of a CMUT, and why is the piston value optimistic? · onsite

\(k^{2} = 1 - K_V/K_Q\) (or \(1 - C^{S}/C^{T}\) electrically): the fraction of stiffness that is electrical. For a plate, linearize about the biased shape, form \(\mathbf{K}_V = \mathbf{K}\) minus the nodal softening and \(\mathbf{K}_Q = \mathbf{K}_V + V^{2}\mathbf{c}\mathbf{c}^{T}/C\), and compare the lowest eigenvalues (18.5)–(18.6). The piston puts all softening into one degree of freedom and gets \(2\xi/(1 - \xi) = 0.43\) at 0.9 \(V_{\mathrm{PI}}\); the plate gets 0.26 because only the centre softens strongly. Follow-up: "which one do you measure?" From \(f_r/f_a\) of the impedance or from \(C\) at constant bias, as for a piezo element (Module 16).

I5What is the main crosstalk mechanism in CMUT arrays and how do you suppress it? · onsite

A dispersive guided wave at the fluid–membrane interface, slower than \(c_w\) and cut off near the membrane resonance; Bayram et al. (2007) measured −17 dB on the first neighbour near 2.1 MHz in conventional mode. It is a property of the compliant surface, not of the substrate (the substrate Lamb modes A0 and S0 were about 23 and 48 dB lower). Suppression: operate in collapse mode (cutoff rises, average −39 dB), add a lossy coating such as PDMS, shape the cell layout or trench between elements, or compensate in the beamformer. Follow-up: "why does it get worse with bias?" Softening lowers the cutoff into the band.

I6Sacrificial release or wafer bonding for a new CMUT array: how do you choose? · onsite

Wafer bonding gives single-crystal silicon membranes with excellent thickness control (device layer under 1.5 µm, < 10 nm variation), any cell shape and fill factor, and gap control by oxidation (LOCOS: 40 nm ± 2 nm), but needs SOI wafers, a thicker insulator for the conductive membrane, and is harder to do on top of CMOS. Sacrificial release uses deposited nitride membranes, works at lower temperature and is one natural route for CMUT-on-CMOS (low-temperature bonding onto the CMOS wafer is the other), at the cost of membrane stress and thickness variation and etch-hole sealing. Follow-up: "which would you use for a monolithic 2D array?" Sacrificial release (or low-temperature bonding variants), because the CMOS limits the thermal budget.

I7Sketch a 2D CMUT-on-CMOS array for 3 MHz cardiac imaging. What dominates the design? · research

\(\lambda/2\) = 257 µm pitch; a 20 × 20 mm aperture is about 80 × 80 = 6400 elements, impossible to cable, so each element gets a pulser and LNA under it and micro-beamforming reduces the channel count to the system (Module 19). Dominant constraints: power density (a few mW per channel already gives watts on a few cm2 at the patient), element capacitance versus LNA noise, bias distribution (tens of volts on a CMOS die, isolated per element or shared), acoustic crosstalk through the interface wave, and thermal limits on the probe face (Module 22). Follow-up: "how do you test it?" Per-element impulse response and C–V maps on wafer before dicing, then pulse-echo uniformity.

I8Give the PMUT resonance formula. Why is the electrode only about 70 % of the radius? · screen

\(f = 10.22/(2\pi a^{2})\cdot\sqrt{D/\bar\rho h}\) for a clamped bilayer plate with laminated stiffness \(D\), divided by \(\sqrt{1 + 0.67\rho_w a/\bar\rho h}\) in water (18.9)–(18.11). The electrode applies a uniform moment; the mode curvature changes sign at \(r = a/\sqrt{2}\), so the work done is \(\propto \beta_e^{2}(1 - \beta_e^{2})\): zero for a full electrode, maximal at 0.707. Example: AlN 1 µm on Si 2 µm, \(a\) = 50 µm: 5.6 MHz in vacuum, 2.4 MHz in water. Follow-up: "what does residual stress do?" Tension raises \(f\) and pushes toward membrane behaviour, compression lowers it and can buckle the plate.

I9AlN or PZT film for a PMUT receiver? For a transmitter? · onsite

Receiver: open-circuit voltage per stress \(\propto e_{31,f}/(\varepsilon_0\varepsilon_{33})\): AlN 11.3 GV/m and Sc0.15AlN 15 GV/m beat sol-gel PZT (0.7–1.8 as published) because their permittivity is 10–12 rather than hundreds; AlN also has a lower loss tangent. Transmitter: force per volt \(\propto e_{31,f}\), 8–14 C/m² for PZT versus 1.05 for AlN, and energy FOM \(e^{2}/\varepsilon\) up to 110 GPa for epitaxial PZT. ScAlN doubles AlN's energy FOM while staying CMOS-friendly. Follow-up: "what about one die doing both?" Use separate receive and transmit electrodes, or a ScAlN compromise.

I10Would you put single-crystal composite in a 2D matrix TEE probe? What would worry you? · research

The gain: \(k_t \approx\) 0.8 (composite) versus ≈ 0.5–0.7, which means wider bandwidth and higher sensitivity, valuable for harmonic imaging through a small aperture. The worries: PMN-PT depoles near \(T_{\mathrm{rt}}\) 86–100 °C and has \(E_c \approx\) 2 kV/cm, so bonding, sterilization and self-heating at the probe face, and bipolar drive at high voltage, all eat the margin; PIN-PMN-PT (\(T_{\mathrm{rt}}\) 105–130 °C, \(E_c\) 4.5–6 kV/cm) is the usual answer. Also cost, dicing yield at small pitch, and the higher permittivity, which helps small elements drive the in-probe electronics. Follow-up: "Mn doping?" Raises \(Q_m\) (≈ 500), useful for therapy, less for imaging bandwidth.

I11What is AC poling and why did it attract attention? · research

Poling a relaxor-PT crystal with an alternating field instead of DC. In rhombohedral PMN-PT it changes the domain configuration; Qiu et al. (Nature 2020) removed the light-scattering 71° domain walls and obtained near-perfect optical transparency with \(d_{33}\) above 2100 pC/N and \(k_{33} \approx 0.94\). For ultrasound it offers higher \(d_{33}\) from the same crystal and transparent transducers through which light can be delivered, which simplifies photoacoustic and optical–acoustic probes. Follow-up: "what limits it?" Temperature stability and the same \(T_{\mathrm{rt}}\) and \(E_c\) constraints as any PMN-PT; poling-condition reproducibility in production.

19Array architectures

13 questions · open in the module

I1Why is a phased array pitched at \(\lambda /2\) and a linear array at about \(\lambda\)? · screen

Grating lobes sit at \(\sin \theta_0 \pm \lambda /d\). A phased array steers the whole aperture over ±45° or more, so it needs \(d \le \lambda /(1 + \sin \theta_{\mathrm{max}})\): \(0.59\lambda\) for 45°, \(\lambda /2\) for a full sector. A linear array translates an unsteered sub-aperture; with \(d = \lambda\) the first replica is at ±90°, where the element factor is zero, so it is harmless. Follow-up: "what happens when you steer the linear array 20° for compounding?" The lobe enters at −41°, about 6 dB below the main lobe narrowband; broadband pulses and the transmit/receive product lower it further, but that is why linear-array steering stays modest.

I2Define 1.25D, 1.5D and 1.75D arrays precisely. · screen

All have several rows in elevation. 1.25D: the elevation aperture is switched (outer rows added with depth), focus fixed by the lens. 1.5D: aperture, apodization and focus are dynamic, but rows are paired symmetrically about the centreline, so no elevation steering; channels N(R + 1)/2. 1.75D: rows independent, so the delay profile can be asymmetric and the beam can steer a few degrees; channels NR. 2D: steering in elevation comparable to azimuth. Follow-up: "why does 1.5D help QUS?" Thinner and more uniform slice over depth means less partial-volume and a diffraction term that changes less with depth, but switching steps must be matched in the reference data.

I3What is the element factor, and how does it limit the receive F-number? · screen

The far-field directivity of one element: \(\operatorname{sinc} (w \sin \theta /\lambda)\) for a rigid baffle, times \(\cos \theta\) for a soft baffle, which matches kerfed arrays better. Edge elements of an aperture at F# see the pixel at atan(1/(2F#)); if that exceeds the −6 dB angle they add noise and clutter faster than signal. \(F\#_{\min} = 1/(2 \tan \theta_{-6\,\mathrm{dB}})\): 0.69 for 0.27 mm elements at 5 MHz, 1.05 at 7.5 MHz. Follow-up: "so what F# should the beamformer use for a broadband pulse?" A frequency-dependent one, or apodization based on element directivity, because the acceptance cone is twice as wide at the low band edge as at the high one.

I4You must design a 3D TEE probe with a few thousand elements, but the cable carries about 100 conductors. Walk through the options. · onsite

Required reduction ≈ 30×. Fully wired is impossible through a gastroscope. Sparse: keeps the aperture with \(K \approx 100\) elements but the floor is ≈ −20 dB one-way and sensitivity drops by K/N; unacceptable for cardiac contrast. RCA: 2N lines (a 64 × 64 aperture in 128 lines) but one-way focusing per dimension and a rectilinear field of view, a poor fit for a sector through a small window. Micro-beamforming: subarrays of 5 × 5 to 6 × 6 cut channels 25–36× with full sensitivity, at the cost of an ASIC in the tip (power ≈ sub-mW per element against a temperature limit), fine delays with limited range, and pre-steering limits on multiline. That is why matrix TEE uses micro-beamforming. Follow-up: "how do you transmit?" Pulsers in the ASIC with per-element delays, or a smaller directly wired transmit sub-aperture as in the 32 × 32 academic designs.

I5Derive the micro-beamformer pattern and say where its errors come from. · onsite

Element position = subarray centre \(X_k\) + offset \(\xi_i\). Fine delays steer to \(u_0\), coarse delays steer centres to \(u_s\): \(P(u) = \mathrm{CAF}(u - u_s) \cdot \mathrm{SF}(u - u_0)\). CAF has grating lobes at \(u_s \pm \lambda /(md)\); SF has nulls at \(u_0 \pm \lambda /(md)\). Equal steering cancels the lobes. Errors: (1) \(\delta = \theta_s - \theta_0 \ne 0\) (parallel beams, near-field focusing with shared pre-steer): gain \(\mathrm{SF}(\delta)\) and lobes \(\mathrm{SF}(\delta - \lambda /(md))\); 3 × 3 at \(\lambda /2\): −0.9 dB and −8 dB lobe at 10°. (2) Curvature inside the subarray: \(\pi (md/2)^{2}/(\lambda R)\), negligible for small m beyond a few mm. (3) Fine-delay range and quantization, and channel gain mismatch in the ASIC. Follow-up: "why do grating lobes matter less in pulse-echo?" The transmit beam multiplies them; they hurt most with broad transmits and many parallel receive beams.

I6How does a row-column array work, what is its PSF, and what are the ghosts? · onsite

Top electrodes in rows, bottom in columns: 2N connections for an N × N footprint. Transmit on rows focuses in y only, receive on columns focuses in x only, so \(\mathrm{PSF} = T(y)R(x)\): one-way per dimension, −13.3 dB axis sidelobes (uniform) and 1.21 \(\lambda\)F# width instead of −26.5 dB and 0.89 \(\lambda\)F#. Each strip acts as three sources (the closest point and two ends), so a scatterer gives up to nine echoes; the end contributions beamform into ghosts (the strongest edge echo is −26 dB in Rasmussen et al.'s single row/column simulation, their Fig. 4). Fix: weighting built into the strip ends (roll-off apodization), which keeps the centre at unit weight; Hann along the whole strip also works but kills off-centre sensitivity. Follow-up: "how do you get high volume rates?" Orthogonal plane waves: rows transmit tilted plane waves, columns receive, then swap, and compound.

I7A random sparse 2D array uses 256 of 1024 positions. Predict its sidelobes and say how you would improve it. · onsite

Mean sidelobe power ≈ \((N^2 - K)/(K(N^2 - 1)) = -25.3\) dB; the peak sidelobe of a draw is 10–15 dB higher, near −10 to −15 dB, comparable to a uniform aperture's −13 dB but everywhere. Improvements: independent transmit and receive layouts so the two-way floor multiplies (or vernier/coarray designs with no gaps), density tapering (spirals) to apodize the floor, optimization of the layout, broadband pulses which decorrelate the random sum across frequency. Follow-up: "what does thinning cost besides sidelobes?" Receive \(\mathrm{SNR} \propto K\) and transmitted energy ∝ K: −6 dB each for \(K = N^2 /4\).

I8Why does a CMUT need a DC bias, and what happens at pull-in? · onsite

Electrostatic force ∝ \(V^2\): without bias an AC drive produces force at 2f and the receive sensitivity \((\partial Q/\partial x \propto V)\) is zero. The bias linearizes transmit and sets the receive gain. The bias also softens the spring: \(k_{\mathrm{eff}} = k(1 - 2x/(g_0 - x))\), zero at \(x = g_0 /3\), the pull-in voltage \(\sqrt{8kg_0^3 /(27\varepsilon_0 A)}\). Approaching it, \(k_T^2 \to 1\) and the resonance drops (0.75 \(f_0\) at 0.9 \(V_{\mathrm{PI}}\)). Beyond it the membrane collapses onto the insulator; collapse-mode operation gives higher coupling and pressure with hysteresis. Follow-up: "reliability issues?" Dielectric charging shifts the effective bias, high DC voltage in the patient-applied part, and nonlinearity at large AC drive.

I9Estimate the receive loss of connecting one 2D element directly to a 2 m cable. · screen

0.2 × 0.2 mm, 0.3 mm thick, \(\varepsilon_r^S \approx 1500\): \(C_e = 8.85 \times 10^{-12} \cdot 1500\) \(\cdot\, 4 \times 10^{-8}/3 \times 10^{-4} \approx 1.8\) pF. Cable ≈ 75–100 pF/m × 2 m ≈ 150–200 pF. Divider \(C_e /(C_e + C_c) \approx 0.009\text{–}0.012\), −39 to −41 dB before the first amplifier. A 1D element of 60 pF loses ≈ 11–13 dB. Follow-up: "and with a preamp in the handle?" The element then sees only a few pF of trace and input capacitance, the loss drops to a few dB, and the cable is driven from a low impedance.

I10You want 3D ULM in a rat brain with one 256-channel system. Which array architecture, and what do you give up? · research

Options: (a) RCA 128 + 128 with orthogonal plane waves: full footprint and kHz volume rates, but one-way focusing per dimension, cross-shaped PSF and edge ghosts; ULM localizes microbubbles to a fraction of the PSF, so a separable but larger PSF mainly costs SNR and increases overlap between bubbles, and ghosts must be suppressed (roll-off apodization, ghost-aware tracking). (b) 256-element sparse (spiral) array: true 2D focusing, low channel count, but low sensitivity \((K = 256)\) and a high clutter floor, which raises false detections. (c) Multiplexed 1024-element matrix: four firings per volume, motion between them. A strong answer quantifies: localization precision scales as PSF width / \(\sqrt{\mathrm{SNR}}\) with SNR as a power ratio (Module 12), so compare width × \(\mathrm{SNR}^{-1/2}\) across options at the same volume rate. Follow-up: "how do you validate localization in 3D?" Microchannel or wire phantoms with known geometry, and Fourier ring/shell correlation on split datasets.

I11What is crosstalk in an array, how does it show up, and how is it measured? · onsite

Coupling of one element's drive into its neighbours: mechanical (kerf filler, backing, matching layers), guided waves in the lens and at the lens–tissue interface, and electrical (capacitive coupling in flex and cable). It widens the effective element, so the measured directivity is narrower than sinc predicts; guided modes add lobes or notches at specific angles and frequencies; it also adds a correlated component to neighbouring channels that biases coherence-based measures. Measurement: drive one element, record neighbours electrically in a water tank or measure surface velocity with a laser vibrometer; report nearest-neighbour levels in dB versus frequency. Follow-up: "design countermeasures?" Dice through the matching layers, soft kerf filler, sub-dicing, lens and matching-layer materials with high attenuation for the guided modes.

I12A square matrix at 0.6λ pitch must steer a 40° cone in every azimuth. Any grating lobes? Would a hexagonal grid help? · onsite

Replicas sit at \(\lambda/d = 1.667\) from the main lobe along the axes; the cone is lobe-free while \(1.667 \ge 1 + \sin\theta_{\max}\), i.e. up to 41.8°, so 40° passes with a thin margin (skirts of the nearest replica approach the edge at 40° along an axis). A triangular grid with the same element density has spacing \(0.6\sqrt{2/\sqrt{3}} = 0.645\lambda\) and six replicas at \(2\lambda/(\sqrt{3}\cdot 0.645\lambda) = 1.79\): lobe-free to 52.3°. Equivalently, for the 40° cone it allows 0.70\(\lambda\) spacing against 0.61\(\lambda\) for the square grid, 13 % fewer elements, or the same count over a larger aperture. Follow-up: "why are most matrix probes still rectangular?" Dicing, interconnect to a rectangular ASIC cell grid, and the fact that the micro-beamformer, not the element pitch, usually sets the usable steering per acquisition.

I13A 4D TEE probe with 3 × 3 micro-beamforming at λ/2 forms 16 parallel receive lines across ±20° around one pre-steer. What do you expect, and how do you fix it? · research

Lines at 20° from the pre-steer carry a subarray grating lobe at +0.4 dB relative to their own main lobe, which is itself 3.7 dB down; the −20 dB window is only ±3.0°, the −10 dB window ±8.5° (19.10). The outer lines show ghost copies of bright structures displaced by \(\lambda/D = 0.67\) in \(u\), partly suppressed by the transmit beam in two-way. Fixes: narrow the fan of lines per pre-steer (more transmit events, or one pre-steer zone per group of lines), reprogram fine delays per zone (14 zones cover ±45° per axis at −20 dB), use 2 × 2 subarrays if the cable budget allows (window ±3.6° at −20 dB), or broaden the transmit only as far as the window. Follow-up: "can retrospective beamforming of the subarray data fix it?" No: the lobe is baked into the analog sum; only the coarse lattice can be re-steered.

20Front-end electronics and system architecture

11 questions · open in the module

I1Walk me through the receive chain of one channel. Where is the noise floor set? · screen

Element → cable → T/R switch → LNA → VGA (TGC) → anti-aliasing filter → ADC → serializer → beamformer. The noise floor is set at the LNA input: the element's thermal noise \(4kT \operatorname{Re} Z_e\) (divided by the cable together with the signal) plus the LNA's \(e_n\) and \(i_n \lvert Z_s \rvert\). With 12–24 dB of LNA gain, later stages are divided down; the VGA's job is to put the signal above the ADC noise without clipping. Typical numbers: 0.8 nV/√Hz LNA, 3.5 MHz band, ≈ 1.5–2 µV rms input noise. Follow-up: "How do you check in a running system that the ADC isn't limiting?" Disconnect the probe or mute transmit, set maximum gain, and compare the measured noise in LSB with the ADC's own noise; the analog noise should exceed it by ≥ 6–10 dB.

I2Why do modern systems use 3- or 5-level pulsers instead of simple bipolar ones? · screen

A bipolar square burst has odd harmonics at −9.5 dB (3rd) and −14 dB (5th) and only one degree of freedom. A 3-level pulse with 120° conduction nulls the 3rd for 1.25 dB of fundamental; 5-level switching at 12° and 48° nulls both 3rd and 5th. Extra levels also allow amplitude shaping (tapered envelopes, transmit apodization, amplitude-modulation contrast sequences) without a power-hungry linear amplifier. Harmonic imaging needs the \(2f_0\) band clean, which is about duty symmetry (−36 dBc per ns at 5 MHz) and envelope shape. Follow-up: "Why not always use a linear amplifier?" Dissipation: a linear stage drops (rail − output) × current continuously, which is expensive across 128–256 channels.

I3A 30 pF element drives 150 pF of cable into a high-impedance LNA. How much signal do you lose, and how much SNR? · onsite

Capacitive divider estimate: \(30/(30 + 150 + 15)\) = −16.3 dB; with the motional branch at resonance (BVD, \(R_m\) ≈ 800 Ω) −12.0 dB. SNR drops much less, because the element's thermal noise is divided equally; only the LNA's noise is added after the divider. With 0.8 nV/√Hz and 1.5 pA/√Hz the NF goes from 0.88 to 3.72 dB: 2.8 dB of SNR. Follow-up: "And with a 1 pF matrix element?" Then the divider is ≈ −41 dB, the element's noise falls far below the LNA, and the NF is above 10 dB: amplify in the handle.

I4The input dynamic range is about 100 dB. Why is a 12-bit (74 dB) ADC enough? · onsite

Three reasons. (1) TGC: the 100 dB is mostly a depth-dependent mean level (\(2\alpha f_0 z\)), which the VGA removes; the instantaneous range at a given depth is smaller. (2) Processing gain: quantization noise spreads over \(f_s/2\), the echo occupies \(B\), so filtering gains \(10 \log_{10}(f_s/2B)\), ≈ 7 dB at 40 MS/s and 4 MHz. (3) Beamforming: uncorrelated per-channel noise averages, \(+10 \log_{10} N\) (21 dB for 128). The displayed image then needs only ≈ 60 dB of log-compressed dynamic range. Follow-up: "When would you want 14 bits?" Strong near-field clutter next to weak signals (cardiac apical views), Doppler with large wall signals, and systems with less TGC flexibility (plane-wave acquisitions with fixed gain).

I5What clock jitter do you need for a 14-bit ADC digitizing a 15 MHz harmonic band? · onsite

Set \(\mathrm{SNR}_{\mathrm{jitter}} \ge 6.02 \cdot 14 + 1.76\) = 86.0 dB: \(\sigma_t \le 10^{-86/20}/(2\pi \cdot 15\ \mathrm{MHz})\) = 0.53 ps rms. In practice the ADC's own SNR is ≈ 75–77 dBFS, so 1 ps (80.5 dB at 15 MHz) is acceptable; 5 ps (66.5 dB) is not. Follow-up: "Does beamforming average jitter?" Per-channel random jitter, yes; jitter of a shared clock, no (all channels err together for a coherent target), so the master clock needs the spec.

I6Size the data path for a 256-channel ultrafast research system, 62.5 MS/s, 14 bit, 10 cm, 5 kHz. · onsite

At the ADCs: 256 × 62.5 MHz × 14 bit = 224 Gb/s. Per event: \(2 \cdot 0.1/1540\) = 130 µs → 8117 samples × 256 × 2 B = 4.16 MB; at 5 kHz, 20.8 GB/s to the host, beyond PCIe Gen3 ×8 (7.9 GB/s) and within Gen4 ×16 (31.5 GB/s) with little margin. IQ at \(f_{\mathrm{IQ}}\) = 5 MHz complex cuts it by 62.5/10 = 6.25 to 3.3 GB/s. Follow-up: "Where would you beamform?" On the GPU for flexibility (ULM, adaptive beamforming), with IQ demodulation and decimation in the FPGA to fit the link.

I7Contrast images show tissue signal that should have been cancelled by pulse inversion. Diagnose. · research

Residual fundamental = \(\sqrt{\delta a^2 + (2\pi f_0 \Delta t)^2}\) plus motion. Check (1) pulser amplitude symmetry (1 % → −40 dB), (2) timing skew between the + and − pulses (1 ns at 5 MHz → −30 dB), including the slew-limited edges on the actual cable, (3) tissue motion between firings (axial displacement \(d\) gives \(\varphi = 4\pi f_0 d/c\)), (4) receive saturation that is polarity dependent. Measure with the probe in a static, linear phantom at low MI: whatever survives is the electronics. Follow-up: "How would amplitude modulation behave?" It cancels with half-amplitude pulses, so it depends on the pulser's amplitude linearity rather than polarity symmetry, and 5-level or linear drivers help.

I8Why can't you cable a 32 × 32 matrix probe element by element to the system? · research

Three budgets fail at once. Channel count: 1024 coax lines and 1024 system channels (492 Gb/s at 40 MS/s, 12 bit). Signal: a ≈ 0.2 mm element has ≈ 1 pF, so a 150 pF cable divider is ≈ −41 dB and the NF exceeds 10 dB. Cable bulk and cost for a TEE or handheld probe. The fix is in-probe electronics: preamplifiers and sub-array (micro-)beamformers that delay and sum groups of elements so only ≈ 100–200 partially beamformed channels travel down the cable (Savord and Solomon 2003; Module 19). Follow-up: "What do you lose?" Per-element channel data, so software beamforming, SoS estimation and aberration correction must work with sub-array sums.

I9What TGC slope do you program at 7.5 MHz, and how much VGA range do you need for 6 cm? · screen

\(2\alpha f_0 = 2 \times 0.5 \times 7.5\) = 7.5 dB/cm, i.e. \(\alpha f_0 c\) = 0.58 dB/µs. Over 6 cm: 45 dB, so a 40 dB attenuator plus some LNA/PGA gain margin is needed; deeper, the gain saturates and the image becomes noise-limited. Follow-up: "Why do users still need TGC sliders?" Attenuation varies (fluid, fat, liver), and the downshifted echo spectrum at depth reduces the effective slope.

I10You measure only 19 dB of beamforming SNR gain with 128 channels on a speckle phantom. Is the hardware broken? · research

Not necessarily. For diffuse scatterers the per-channel echoes are only partially coherent: with equal transmit and receive apertures the van Cittert–Zernike coherence is triangular and the gain is \((2N^2 + 1)/(3N)\) → 19.3 dB, not 21.1 dB. Check the noise side too: correlated interference (supply ripple, clock spurs) does not average. Test with a point target in water and with transmit off; a point should give ≈ 21 dB. Follow-up: "How does aberration show up?" Lower coherence, hence lower gain; that is the basis of coherence-based beamformers (SLSC) and sound-speed estimators.

I11What is active termination and why use it? · screen

A defined LNA input impedance made with feedback rather than a shunt resistor: \(R_{\mathrm{in}} = R_f/(1 + A)\). It damps the element and cable (shorter ring-down, flatter response), but the noise current of \(R_f\) is \((1 + A)\) times smaller than that of a physical \(R_{\mathrm{in}}\). In the model, a 200 Ω resistor gives NF 8.8 dB, a 200 Ω active termination with \(A = 10\) gives 5.9 dB. Follow-up: "When would you leave the input high-impedance?" When SNR matters most and the element is already well damped acoustically; high-Z gives the lowest NF (3.7 dB here).

21Simulation and characterization tools

10 questions · open in the module

I1What is a spatial impulse response, and why is it the core of Field II? · screen

It is the velocity potential at a point when the aperture moves with an impulsive velocity: \(h(\mathbf{r}, t) = \int \delta(t - R/c)/(2\pi R)\, dS\). In a linear homogeneous medium every field quantity follows from it by convolution with the excitation (\(p = \rho_0 v' * h\)) and pulse-echo is \(h_t * h_r\). Field II splits the aperture into small rectangles and sums their far-field responses, so it never solves a wave equation. Follow-up: "What does it cost you?" No heterogeneity, no nonlinearity, no multiple scattering.

I2Sketch h(t) for a rectangular element at a point on axis and at a point beyond its edge. · onsite

On axis: zero until \(z/c\), a flat plateau at \(c\) until the circle reaches the nearer edge (\(\sqrt{z^2 + a^2}/c\)), a decay through the farther edge, and zero after the corners at \(\sqrt{z^2 + a^2 + b^2}/c\). Beyond the edge: starts when the circle touches the nearest edge, never reaches \(c\) (only part of the circle is inside), shows knees at the corner arrivals, and lasts much longer. For 4 × 6 mm at z = 15 mm: on axis 0.28 µs at height \(c\); at x = 6 mm, 1.13 µs with peak 0.20c. Follow-up: "What does a long h do to the pulse?" It low-pass filters it, so off-axis pulses are longer and weaker.

I3When would you use k-Wave instead of Field II for a sound-speed estimation study? · research

Whenever the estimator's behaviour depends on the heterogeneity: layered fat and muscle, refraction at oblique interfaces, aberration-induced coherence loss, reverberation clutter. Field II with a global c only tests the homogeneous mismatch problem. I would validate the k-Wave setup in homogeneous water against Field II first, choose PPW from a convergence test, and smooth sharp interfaces. Follow-up: "Cost?" A 2D study at 5 MHz is cheap; 3D needs a GPU and tens of GB at 3 PPW.

I4Your FDTD simulation of a homogeneous medium shows the focus 1.5 mm shallower than expected. What do you check? · onsite

Numerical dispersion first: at 7–8 PPW a second-order scheme has phase-speed errors of 2–5 % across the band and they depend on direction, which distorts focusing (Demo 21.4). Then CFL, source discretization, absorbing boundary reflections, and whether the delays use the same c as the grid. Refine \(\Delta x\) by 2 and see if the focus moves; switch to fourth order or pseudospectral. Follow-up: "What is the convergence rate?" \(\Delta x^2\) for FD2, so halving the grid should cut the error by about 4.

I5Define centre frequency and fractional bandwidth from a pulse-echo test. What mistakes are common? · screen

Take the amplitude spectrum of the gated echo, find the −6 dB points \(f_l\), \(f_u\); \(f_c = (f_l + f_u)/2\), \(\mathrm{FBW} = (f_u - f_l)/f_c\). Mistakes: reporting the peak as the centre (wrong by 10–20 % for double-humped spectra), mixing −3 dB one-way with −6 dB two-way, gates that include reverberations (ripple) or cut the echo (broadening), power versus amplitude dB, and not stating arithmetic versus geometric centre. Follow-up: "How long is a 5 MHz, 65 % pulse?" About 0.27 µs at −6 dB for a Gaussian.

I6What is loop sensitivity, and how would you compare it between two probes measured in different labs? · onsite

\(20 \log_{10}(V_{\mathrm{echo,pp}}/V_{\mathrm{tx,pp}})\) for a flat reflector. To compare, normalize the reflector (steel −0.57 dB, quartz differs), the water path (\(2.2 \times 10^{-4}\) dB/(mm·MHz²) round trip), diffraction to the reflector distance, the pulser source impedance and the receiver load (50 Ω versus high impedance changes the voltage), and the cable. Follow-up: "Is loop sensitivity a transducer property?" No, it is a property of transducer + cable + electronics + target geometry.

I7How large may a hydrophone be for a focused 10 MHz beam at F# 2? · research

\(\lambda F\# = 0.31\) mm. For under 0.5 dB peak loss the radius should be below about \(0.3\, \lambda F\#\): \(b \lesssim 0.09\) mm, a diameter under about 0.2 mm, and harmonics need smaller still. The loss law is \((\pi b/\lambda F\#)^2/16\). Then check the hydrophone's directivity, its effective (not geometric) size at the frequency, and its bandwidth for the harmonic content. IEC 62127-1 gives the formal size criterion and corrections. Follow-up: "And for \(I_{\mathrm{SPTA}}\)?" The intensity error is twice the pressure error in dB.

I8Walk me through how MI is computed for a 510(k). · onsite

Hydrophone scans in water under IEC 62127-1 conditions; deconvolve or correct for sensitivity, bandwidth and spatial averaging; derate the peak rarefactional pressure by 0.3 dB/cm/MHz; find the depth defined by IEC 62359 (via the derated pulse-intensity integral); \(\mathrm{MI} = p_{\mathrm{r.3}}/\sqrt{f_{\mathrm{awf}}}\) with p in MPa and f in MHz; search operating modes for the global maximum; report against the FDA Track 3 limit MI ≤ 1.9 with uncertainty. Follow-up: "Why \(\sqrt{f}\)?" The cavitation threshold data scale roughly as \(p \propto \sqrt{f}\), so MI approximates the cavitation likelihood.

I9A probe returns from the field with two dead elements. Is it still usable? · screen

Quantify: for 64 active elements, uniform weights, the main lobe drops \(20 \log(62/64)\) = 0.28 dB and a far-sidelobe floor appears near \(10 \log(2/62^2)\) = −32.8 dB; with an apodized aperture the floor is relative to a much lower sidelobe level and is more visible. Two adjacent dead elements produce structured lobes rather than a floor. Whether that fails depends on the application's contrast requirement and the QA limit; many services set a maximum number of dead elements and forbid adjacent ones. Follow-up: "How do you detect them?" Per-element pulse-echo or impedance; a phantom image shows vertical dropout streaks in linear arrays.

I10How do you turn an impedance measurement into a KLM model? · research

Start with the bare plate in air: \(f_r\) and \(f_a\) give \(k_t\) through the IEEE 176 relation, the capacitance well below resonance times \(1 - k_t^2\) (or the capacitance well above resonance) gives \(\varepsilon^S\), the \(Q\) of the peaks gives the losses; \(f_a\) fixes the plate's sound speed for its thickness. Then add layers and backing and fit their impedances and thicknesses to the finished element's impedance in air and water, and finally fit the electrical network to the pulse-echo spectrum. Peaks the 1D model cannot produce are lateral modes: hand those to FE. Follow-up: "Example?" \(f_r\) = 3.00, \(f_a\) = 3.45 MHz gives \(k_t\) = 0.53.

22Acoustic output, bioeffects and regulatory measurement

11 questions · open in the module

I1Define MI and explain every factor in it. · screen

\(\mathrm{MI} = p_{r,\alpha}(z_{\mathrm{MI}})/(C_{\mathrm{MI}}\sqrt{f_{\mathrm{awf}}})\) with \(C_{\mathrm{MI}} = 1\ \mathrm{MPa\cdot MHz^{-1/2}}\) (22.13): the peak rarefactional pressure derated by 0.3 dB/cm/MHz, taken (IEC 62359 \(z_{\mathrm{MI}}\), FDA \(z_{\mathrm{sp}}\)) at the depth where the derated pulse-intensity integral peaks, divided by the root of the acoustic working frequency. Rarefaction because bubbles grow under tension; \(\sqrt{f}\) because the inertial-cavitation threshold of an optimally sized free nucleus rises roughly as \(\sqrt{f}\) (Apfel & Holland 1991). MI 1.9 at 3 MHz is 3.29 MPa derated.

Follow-up: does MI predict contrast-bubble destruction? Only loosely: shelled bubbles respond by resonance and shell properties, and the pressure at the bubble uses the real path attenuation.

I2Why is the derating 0.3 dB/cm/MHz when soft tissue is closer to 0.5–0.7? · screen

It is a deliberately low average so that exposure is not underestimated for paths with long fluid segments (full bladder, amniotic fluid) where attenuation is far below tissue values. The FDA states it as an assumed average attenuation along the beam axis. Pressure is derated by \(10^{-0.3fz/20}\) and intensity by \(10^{-0.3fz/10}\) (22.12): 4.5 dB at 3 MHz and 5 cm, × 0.596 and × 0.355.

Follow-up: when does derating still underestimate? Very low-loss paths and nonlinear water measurements at high drive, where energy already went into harmonics before derating.

I3State the Track 3 limits exactly, including the ophthalmic case. · regulatory

FDA guidance (February 2023): derated global maximum \(I_{\mathrm{SPTA.3}} \le\) 720 mW/cm², and either \(\mathrm{MI} \le 1.9\) or derated \(I_{\mathrm{SPPA.3}} \le\) 190 W/cm². Ophthalmic: \(\mathrm{TI} = \max(\mathrm{TIS}_{\mathrm{as}}, \mathrm{TIC}) \le 1\), \(I_{\mathrm{SPTA.3}} \le\) 50 mW/cm², \(\mathrm{MI} \le 0.23\). Any TI above 6.0 should be explained. Track 3 applies to devices conforming to the Output Display Standard in IEC 60601-2-37.

Follow-up: a short-pulse 3.5 MHz mode at MI 1.85 has \(I_{\mathrm{SPPA.3}}\) = 300 W/cm². Compliant? Yes, through the MI branch, provided \(I_{\mathrm{SPTA.3}} \le 720\).

I4Track 1 or Track 3: what is the difference and when would you choose Track 1? · regulatory

Track 1 is for devices that do not conform to the IEC 60601-2-37 output display; their derated output is held to application-specific preamendments levels (\(I_{\mathrm{SPTA.3}}\) 720 peripheral vessel, 430 cardiac, 94 fetal imaging and other, 17 mW/cm² ophthalmic; \(I_{\mathrm{SPPA.3}}\) 190, or 28 W/cm² ophthalmic, or MI 1.9 / 0.23). Track 3 replaces these with global limits plus an on-screen MI/TI display. Track 1 suits simple, fixed-output devices (some Doppler-only or single-purpose devices) where a display adds cost but the low application limit is easy to meet.

Follow-up: a fetal-imaging device on Track 1 at 200 mW/cm²? Not acceptable: the fetal level is 94.

I5When must MI and TI be displayed, and what is reported if they never exceed 1? · regulatory

The display requirement comes from IEC 60601-2-37 (the Output Display Standard), which, as public secondary sources summarize it (the IEC text is not public), ties the obligation to whether the system can exceed an index value of 1.0. The FDA guidance follows the same threshold: if no system/transducer combination can exceed a TI of 1.0 or an MI of 1.0 in any operating mode, the global maximum values are submitted instead of per-mode reporting tables; tables are provided for combinations whose global maximum displayed MI or TI exceeds 1.0. Display accuracy against measurement is part of the submission.

Follow-up: which TI is displayed? The one appropriate to the application (TIS, TIB or TIC), with the user able to select it.

I6You add an SWE mode to a cleared system. Plan the acoustic-output work. · onsite

Treat push plus tracking as one combined mode. Measure the push waveform with a membrane hydrophone at the push focus and along the axis to find \(z_{\mathrm{MI}}\); derate; compute PII per push (a 400 µs push at 3 MPa is about 0.09 J/cm²), \(I_{\mathrm{SPTA.3}}\) from push and tracking rates, \(I_{\mathrm{SPPA.3}}\), MI, and the TIs at the push and tracking powers. Search the global maximum over push voltage, duration, focal depth, push rate and region-of-interest size. Measure surface temperature under the worst sequence. Expect TI and \(I_{\mathrm{SPTA.3}}\), not MI, to set the push repetition limit; check AIUM guidance on reduced dwell for ARFI near bone for labelling.

Follow-up: why can a compliant \(I_{\mathrm{SPTA.3}}\) still be a heating concern? The average hides heating within a push and at the face; the TI and surface tests catch it.

I7Estimate \(I_{\mathrm{SPTA.3}}\) for a 5 MHz plane-wave mode at 5 kHz PRF with \(p_{r.3}\) = 0.8 MPa and 2-cycle pulses. · onsite

\(\sigma_t = 2/(2.3548\cdot 5\ \mathrm{MHz}) = 0.170\ \mathrm{\mu s}\), \(\int \mathrm{env}^{2} = 0.301\ \mathrm{\mu s}\). \(\mathrm{PII} = \frac{p^{2}}{2\rho c}\cdot 0.301\ \mathrm{\mu s}\) \(= \frac{(0.8\ \mathrm{MPa})^{2}}{3\ \mathrm{MRayl}}\times 0.301\ \mathrm{\mu s}\) \(= 0.0642\ \mathrm{J/m^{2}}\) = 6.4 µJ/cm². Every transmit reaches the point: \(I_{\mathrm{SPTA.3}}\) = 6.4 µJ/cm² × 5000/s = 32 mW/cm², \(\mathrm{MI} = 0.8/\sqrt{5} = 0.36\). A focused B-mode at the same pressure and 113 pulses/s would be 0.7 mW/cm².

Follow-up: a 60 s ULM acquisition at the same settings: \(I_{\mathrm{SPTA}}\) is the same, but TI and dwell time now govern; the AIUM tables apply to the displayed TI.

I8How do you choose a hydrophone for a 15 MHz linear array, and what corrections do you apply? · onsite

At 15 MHz and F/2 the focal \(W_{-6}\) is about 0.2 mm, so the effective diameter should be at most 50–100 µm (a spot-poled membrane, or a fibre-optic or small needle hydrophone with known effective radius), and the calibrated bandwidth should extend to several harmonics (≥ 60 MHz) because the focal waveform is nonlinear. Deconvolve with the complex sensitivity, apply the spatial-averaging correction of IEC 62127-1:2022, and include both in the uncertainty budget. A 200 µm hydrophone at \(W_{-6}\) = 0.2 mm would under-read p by 2.8 dB and intensity by about 48 %.

Follow-up: why do harmonics make it worse? Their beams are narrower, so averaging removes more of them and distorts \(p_c\) and \(p_r\) differently.

I9TIS, TIB or TIC: which applies when, and why does Doppler TIB exceed TIS? · screen

TIS for soft tissue (abdomen, vessels, early pregnancy), TIB when bone may lie at the beam's focus (obstetrics after 10 weeks LMP, Doppler near bone), TIC when bone is under the transducer (transcranial). Bone absorbs nearly all incident power in a thin layer, so its \(W_{\mathrm{deg}}\) scales with beam diameter (≈ 40 mW/cm × \(d\)), whereas soft tissue \(W_{\mathrm{deg}} = 210/f\) mW. A narrow Doppler focus therefore gives a small bone \(W_{\mathrm{deg}}\); in the PW preset TIB = 1.5 against TIS = 0.34.

Follow-up: why is TIB in scanned modes set equal to TIS? Sweeping the beam spreads the bone heating, and the surface soft-tissue case dominates.

I10What does the ±30 % / ±15 % statement in the FDA guidance mean for your uncertainty budget? · regulatory

The Track 1 preamendments levels are themselves estimated to be uncertain by ±30 % for intensities and ±15 % for MI. A manufacturer whose expanded measurement uncertainty is within those margins does not have to subtract it from the limits; if it exceeds them, the levels are reduced by the excess (an intensity uncertainty of 40 % reduces the level by 10 %). The protocol must describe Type A and Type B uncertainties: hydrophone calibration, spatial averaging, bandwidth, positioning, temperature and the power measurement.

Follow-up: which term usually dominates at high frequency? Hydrophone calibration and spatial averaging.

I11For CEUS and ULM, what pressure matters at the bubble, and why is the displayed MI a poor guide? · research

The bubble responds to the in situ pressure at its depth, with the real attenuation and the local beam, and to its resonance relative to \(f\); the displayed MI is the 0.3-derated pressure at one depth divided by \(\sqrt{f}\), a scaling derived for free gas nuclei. Through a fluid path (\(\alpha \approx\) 0.05) at 3.5 MHz and 8 cm the in situ pressure exceeds the derated value by 7.0 dB; through liver-like tissue (\(\alpha \approx\) 0.7) it is 11.2 dB lower. Researchers calibrate the in situ pressure (hydrophone plus a tissue model) and report it with the frequency and pulse length.

Follow-up: what MI threshold do safety statements give for contrast? The AIUM reports bioeffects above MI 0.4 with gas-body agents and recommends minimal MI and time; BMUS uses 0.7 as the cavitation-risk threshold.

23Beam patterns, k-space and compounding

14 questions · open in the module

I1Why can't you focus a 19.2 mm aperture at 400 mm at 5 MHz? · screen

The near-field length \(D^2/(4\lambda)\) is 299 mm, so 400 mm is beyond the last axial maximum: the Fresnel number \(N_F = D^2/(4\lambda F) = 0.75\) is below 1, there is less than one Fresnel zone of phase to cancel and the focusing law changes almost nothing. The intensity peak would sit well short of 400 mm (peak/F ≈ 0.6 already at \(N_F = 1\)). Fix: a larger aperture or a higher frequency (\(N_F \propto D^2 f\)). Follow-up: "What is the far field then?" \(D^2/\lambda\) = 1.2 m or \(2D^2/\lambda\) = 2.4 m, depending on the convention; say which.

I2Your paraxial simulation at F# 1.2 shows first sidelobes at −13.3 dB; a full field simulation shows −12.5 dB. Which is wrong? · onsite

Neither is buggy; the paraxial model is the approximation. The Fourier relation drops non-paraxial path terms and the 1/r amplitude variation across a wide aperture; at F# 1.56 the direct point-source sum already gives −12.7 dB and a 3.5 % wider main lobe, and the error grows as F# drops. For Hann the gap is larger (−28.8 vs −31.5 dB), because the error is a small additive field that matters more under low sidelobes. Follow-up: "Would element directivity change it?" Yes, it acts as an extra taper and usually lowers the far sidelobes; include it in the reference model.

I3You want lower sidelobes. Do you apodize on transmit or on receive, and what does each cost? · onsite

The two-way pattern is the same either way (product of factors), but the price differs. Receive apodization costs SNR against channel noise: \((\sum w)^2/(N\sum w^2)\), −1.76 dB for Hann. Transmit apodization at fixed peak drive costs radiated field: focal amplitude \(\sum w/N\), −6.0 dB for Hann, energy −4.3 dB. If SNR-limited, apodize on receive (or not at all); if MI-limited at shallow depth with SNR to spare, transmit apodization is acceptable. Follow-up: "What do you report for clutter?" ISLR or 2D integrated clutter and cyst gCNR, not just PSL.

I4A 128-element array at 7.5 MHz has pitch 1.46λ. The CW pattern has a 0 dB grating lobe at 43°. Why is the artifact in the image so faint? · onsite

Three factors. (1) Broadband: at the lobe the element pulses arrive one period apart and form a train; only \(\approx M\) of N overlap, so the peak is \(\approx 1.064\,M/N\): with 60 % band (M = 1.47) that is −38 dB one-way and about the same two-way. (2) Element factor: a 0.9d-wide element (w = 1.31λ) is about −19 dB at 43° one-way, −38 dB two-way. (3) The lobe must hit a strong reflector to show. Follow-up: "Energy pattern?" \(1.505\,M/N\), about −17.6 dB (power): hydrophone energy plots look much worse than the image.

I5Explain the coarray and use it to design a sparse array with fewer channels but no grating lobes. · research

Each Tx/Rx pair measures spatial frequency \((\xi_t + \xi_r)/(\lambda z)\); the two-way PSF is the Fourier transform of the coarray \(a * b\). Design goal: a filled, smooth coarray. Example: vernier, Tx every 3rd and Rx every 2nd element of 64: 22 + 32 elements, coarray without periodic gaps (only the lag next to each end is missing), two-way PSL −26.5 dB like the full aperture, whereas Tx = Rx every 2nd has a 0 dB lobe at \(\lambda z/(2d)\). Alternatives: random thinning (floor \(\approx 1/K\), peak ≈ 7 dB higher), or synthetic aperture over sub-apertures. Follow-up: "Cost?" SNR (fewer elements), and the vernier null alignment is exact only at \(f_0\).

I6What does a single plane wave give you laterally, and what exactly does compounding add? · screen

One plane wave is one virtual transmit element: the coarray is the receive aperture, so the lateral width is receive-only, \(1.21\,\lambda F_\#\) (0.56 mm at 5 MHz, F# 1.5), and the 2D PSF has X-shaped edge arms (PSL ≈ −17 dB). Compounding N angles builds a virtual transmit array: span → transmit \(F_\# = 1/(2\tan\theta_{\max})\) and narrower main lobe; step → replica position \(\lambda/\Delta(\sin\theta)\); count → clutter and SNR (\(10\log_{10} N\)). Follow-up: "Does more angles in the same span improve resolution?" No; it lowers clutter and moves replicas out.

I7Derive the plane-wave angle-step condition. Why is Δθ ≤ λ/L conservative? · onsite

The transmit factor \(T_N(x) = \frac{1}{N}\sum e^{j2\pi x\sin\theta_n/\lambda}\) is periodic in x with period \(\lambda/\Delta(\sin\theta)\), independent of depth. Requiring the replica of any target to fall outside an image of width W gives \(\Delta\theta \approx \Delta(\sin\theta) \le \lambda/W\); with W = L = 38.4 mm at 5 MHz, 0.46°, and with span \(1/F_\#\) the count \(N \approx L/(\lambda F_\#) = 83\) (Montaldo et al., 2009). Conservative because the ghost is also multiplied by the receive pattern at \(x_{\mathrm{rep}}\) and by the broadband factor \(\approx 1.06\,M/N\): 11 angles at 3° (replica 5.95 mm) leave a −53 dB ghost in simulation. Follow-up: "When is it not conservative?" Very bright specular targets and narrowband pulses.

I8With 11 angles at 3° you see a faint copy of a bright wire 6 mm to its side, at every depth. Diagnose and fix. · onsite

A transmit replica: \(\lambda/\Delta(\sin\theta) = 0.308/0.0523\) = 5.9 mm at 5 MHz, and its depth independence identifies it (a focused-array grating lobe would scale with z, a receive grating lobe would follow \(\lambda z/d\)). Fixes: a smaller step (more angles in the same span), receive apodization or a lower receive F# so that \(R(x_{\mathrm{rep}})\) is lower there (a higher F# makes it worse), or a nonuniform angle set that spreads the replica into a floor. Follow-up: "Cost of halving the step at fixed span?" Twice the angles: half the frame rate, +3 dB SNR, and a sequence twice as long, so twice the displacement for the same velocity.

I9Compounding 31 angles: what happens to SNR, cyst contrast and speckle, and why? · screen

SNR against electronic noise: \(+10\log_{10} 31\) = 14.9 dB (coherent signal, incoherent noise; simulated 14.86 dB). Cyst contrast: much better (−15.8 → −37.7 dB in the simulation, gCNR 0.87 → 0.995) because the synthesized transmit focus removes the edge arms and diffuse sidelobe energy. Speckle: unchanged statistics, \(\mathrm{SNR}_0 \approx\) 1.8–1.9, because coherent summation only changes the PSF; speckle reduction needs incoherent compounding of decorrelated looks. Follow-up: "Why does CNR improve only from 1.53 to 1.78?" CNR is bounded by background speckle variance (≈ 1.91 for a perfectly black cyst on the envelope); gCNR saturates near 1.

I10In cardiac plane-wave imaging the compounded image loses signal on the moving wall. Quantify and propose mitigations. · research

Axial motion \(\delta\) per firing gives a phase ramp \(\varphi \approx 4\pi\delta/\lambda\) per firing; at the true position the loss is \(\lvert\sin(N\varphi/2)/(N\sin(\varphi/2))\rvert\): λ/4 over the sequence costs 3.9 dB, λ/2 nulls it. At 5 MHz, 31 angles at 10 kHz, λ/4 is 25 mm/s. With sequential angle order part of the ramp turns into a lateral mislocation (0.12 mm, −2.2 dB peak loss in simulation). Mitigations: fewer angles or a higher PRF (shorter sequence), lower frequency, motion estimation and phase compensation per angle before summation, or diverging waves with few virtual sources for the heart. Follow-up: "Why does angle order matter?" A ramp linear in \(\sin\theta\) is equivalent to a lateral shift of the transmit factor.

I11Why use diverging waves on a phased array, and what changes in the k-space picture? · research

A phased array is narrow (≈ 20 mm), so a plane wave insonifies only a strip as wide as the probe, not a sector. A diverging wave from a virtual source behind the array covers the sector. Transmit delay to a pixel = distance to the virtual source (minus a constant); moving the virtual source tilts the wave. Each virtual source contributes the transmit direction from source to pixel, so the synthetic transmit aperture is the set of those directions as seen from the pixel, and its span shrinks with depth, unlike a plane-wave set. The same coarray reasoning applies: span sets \(F_{\#,\mathrm{tx}}\), spacing sets replicas, SNR is lower than plane waves because the energy spreads over the sector. Follow-up: Papadacci et al. (2014) for cardiac use.

I12Your 0.3 mm linear array shows a comet-tail streak under a wire only when you steer plane waves to 15° at 7.5 MHz. Explain and quantify. · onsite

A transmit grating lobe of the steered plane wave. At 7.5 MHz, \(d = 1.46\lambda\), above the threshold \(\lambda/(1+\sin 15^\circ) = 0.794\lambda\); the second wave leaves at \(\sin\theta_g = 0.259 - 0.685\), −25.2°. It reaches the wire late (7.8 mm of path along its ray at 30 mm) and as a pulse train, so DAS images it as a streak below the wire, about −23 dB for a single angle with 60 % band. Not \(M/N\): in the near field only about one Fresnel zone of elements builds the main arrival. Fixes: steer less, lower the transmit frequency, compound more angles (−35 dB with 11), finer pitch. Follow-up: "Why not at 5 MHz?" The lobe goes to −50° and its ray misses a 30 mm target (it comes from 36 mm off axis); it shows at 15 mm depth instead.

I13Derive the lateral resolution of coherent plane-wave compounding in terms of the angle span and the receive F-number. · research

Each angle contributes \(e^{j2\pi x\sin\theta/\lambda}\), so dense angles fill transmit k-space uniformly on \(\pm\sin\theta_{\max}/\lambda\); the receive aperture fills \(\pm 1/(2F_\#\lambda)\). The PSF is the product of the two sincs, support the trapezoid convolution: \(w_{-6} \approx \kappa\lambda/(\sin\theta_{\max} + 1/(2F_\#))\), κ from 0.603 (one support dominates, e.g. one angle: \(1.207\lambda F_\#\)) to 0.886 (equal supports, focused two-way). 5 MHz, F# 1.5, ±15°: 0.456 mm. Follow-up: "And the number of angles?" It sets the replica position \(\lambda/\Delta(\sin\theta)\), not the width; three angles over ±15° even give a narrower fringe (0.42 mm) with a replica shoulder.

I14Explain the plane-wave angle-step rule as a grating-lobe condition. · screen

Angle \(\theta_n\) acts as a virtual element at \(\xi = z\sin\theta_n\), so a uniform angle set is an array of pitch \(d_v = z\Delta(\sin\theta)\). Its grating lobe sits at \(\lambda z/d_v = \lambda/\Delta(\sin\theta)\), independent of depth. If \(d_v \le z/F_\#\) the coarray is gap-free and the replica falls beyond the first receive null, under receive sidelobes; \(\Delta\theta \le \lambda/W\) moves it out of the image (0.46° for 38.4 mm at 5 MHz). Follow-up: "Diagnose a ghost 6 mm to the side at every depth with 3° steps." \(0.308/0.0523 = 5.9\) mm: the angle-domain replica.

24Fourier-domain beamforming

11 questions · open in the module

I1Why does Stolt migration of pulse-echo data use \(c/2\)? · screen

Zero-offset delays are \(2r/c\), identical to one-way delays \(r/v\) with \(v = c/2\) from sources placed at the reflectors and fired at t = 0 (exploding reflector, (24.4)). The data are then a one-way field, and migration back-propagates it to t = 0 with speed \(c/2\). Using c doubles depths (10 → 19.94 mm in the lab) and defocuses by about 21 dB. Follow-up: "And for plane waves?" There is no exact equivalent speed; Garcia's fit gives \(\hat{v} = c/\sqrt{1 + \cos\theta + \sin^2\theta}\), \(c/\sqrt{2}\) at 0°.

I2What is the evanescent region of a \((k_x, f)\) spectrum, and why is it empty in channel data? · screen

\(\lvert k_x\rvert > f/c\), where \(k_z\) is imaginary and the field decays as \(e^{-2\pi z\sqrt{k_x^2 - (f/c)^2}}\): 81 dB per mm at \(1.1\,f/c\) at 5 MHz. Scatterers are millimetres away, so nothing evanescent reaches the array; what shows up there is truncation leakage (−22 dB for a 5 mm target, rectangular aperture). Follow-up: "What if you back-propagate it?" The decay becomes a gain on noise; mask it.

I3Derive Stolt's mapping and its Jacobian. What happens if you drop the Jacobian? · onsite

\(r(x, z) = \iint D(k_x, f)\, e^{\,j2\pi(k_x x + k_z z)}\, df\, dk_x\) with \(k_z = \sqrt{(f/v)^2 - k_x^2}\). Substitute \(f = v\sqrt{k_x^2 + k_z^2}\), \(df = v k_z/\sqrt{k_x^2 + k_z^2}\, dk_z\); the image is an inverse 2D FFT of D resampled at that f times the Jacobian (24.6). Dropping it changes the k-space weighting (a \(\cos\varphi\) obliquity), hence slightly the PSF shape and amplitude, not the focusing and not the \(\sqrt{z}\) depth gain of unnormalized f-k. Follow-up: "Where does Stolt need interpolation and how fine?" Along f; keep \(2\pi\,\Delta f\,\lvert t - t_c\rvert \lesssim 1\).

I4Why can't zero-offset Stolt be applied to plane-wave data directly? Derive Garcia's ERM velocity. · onsite

The plane-wave delay \([x_0\sin\theta + z_0\cos\theta + \sqrt{(x - x_0)^2 + z_0^2}]/c\) is not a hyperbola and its transmit part depends on \(x_0\). Steering compensation (shift by \(-x\sin\theta/c\)) makes it depend on \(x - x_0\) only; then fit \(\sqrt{(u - \hat{u})^2 + \hat{z}^2}/\hat{v}\) in value, slope and curvature at u = 0. That gives \(R^2 = z_0^2(1 + \cos\theta)^2/(1 + \cos\theta + \sin^2\theta)\), \(\hat{v} = c/\sqrt{1 + \cos\theta + \sin^2\theta}\), \(\beta = (1 + \cos\theta)^{3/2}/(1 + \cos\theta + \sin^2\theta)\), \(\gamma = \sin\theta/(2 - \cos\theta)\) (24.10)–(24.11). Follow-up: "How big is the error?" \(z\tan^4\alpha/16\) at 0°: F# ≳ 1.1 at 30 mm, 5 MHz.

I5How do you implement an F-number and apodization in f-k? · onsite

Receive angle α corresponds to data wavenumber \(k_x = (f/c)\sin\alpha\), so the F# is a fan filter \(\lvert k_x\rvert \le (f/c)\sin(\arctan(1/(2F\#)))\) applied before steering compensation and mapping, and apodization is a window across the fan. It equals a constant-F# dynamic aperture. Follow-up: "Is it identical to DAS?" Not at the array edges (DAS's aperture saturates) and not for aliased data, which can fold into the fan.

I6Your f-k images show a lateral haze at the depth of every bright reflector, absent in DAS. Diagnose. · onsite

Interpolation along f is too coarse: the error \(w(1 - w)\varphi^2/2\) with \(\varphi = 2\pi\,\Delta f\, t\) varies from column to column and spreads each echo across x at its depth (rms \(0.037\,\varphi^2\); −10 dB at φ = 3). Fix: zero-pad the time FFT (2–4× the record) and centre the time origin. In the lab, doubling the FFT length took beside-block clutter from −10.6 to −18.2 dB (DAS −18.7). Second suspect: lateral wrap or aliasing; check padding in x and the fan. Follow-up: "Why only at the depth of the echo?" The error multiplies that echo's own spectrum.

I7Compare the cost of f-k and DAS. When is f-k not worth it? · onsite

DAS: \(N_{\mathrm{pix}} \times N_{\mathrm{active}}\) interpolations per angle (3.4 M in the lab). f-k: FFTs of the padded data (N log N), one interpolation per output sample, one IFFT; compounding adds spectra and needs one IFFT per frame. Measured 2.1–2.5× in Octave; the asymptotic gain is about \(N_{\mathrm{active}}/\log_2 N\). f-k loses for small ROIs or a few lines, arbitrary pixel grids, heterogeneous media, and when a GPU DAS is memory-bound anyway. Follow-up: "And for 3D with a matrix array?" The element factor grows to thousands, which is where Fourier methods pay most.

I8The exact plane-wave map is closed form. Why would anyone use the ERM fit? · research

Both are one 1D interpolation per column after the same steering compensation, so cost is not the reason. The ERM keeps the Stolt square-root structure, shares code with zero-offset migration, and its error is negligible at moderate F# and steering. The exact map (24.15) removes the fit error (35 µm lateral offset at 9° in the lab, none for the exact map). A defensible choice today: the exact map, unless the code base is built around ERM. Follow-up: "What does the exact map still assume?" Born, homogeneous c, an infinite plane wave and point elements.

I9How would you extend f-k imaging to a subcutaneous fat layer, and then to lateral heterogeneity? · research

Horizontal layers: phase-shift migration, stepping Ψ by \(e^{\,j2\pi(k_z(z) + f/c(z))\Delta z}\) (24.17); \(k_x\) is conserved by Snell's law, so it is exact. An 8 mm layer at 1450 m/s shifts targets by 0.497 mm under single-speed Stolt; phase shift removes it. Lateral variation: split-step Fourier (a phase screen \(e^{\,j2\pi f\Delta s(x)\Delta z}\) per slab, then a homogeneous step) or phase shift plus interpolation between reference speeds; beyond that, time-domain delays from an eikonal solver. Follow-up: "Where does the speed map come from?" From a focus or coherence sweep (cheap in f-k), or tomography (Module 13).

I10How can Fourier-domain imaging help sound-speed estimation and aberration correction research? · research

At broadside the map is \(f = c\,g(\mathbf{K})\), so each trial c reuses the data FFTs: a dense global-c sweep costs one interpolation and one IFFT per value. For local aberrations, keep the full reflection matrix (all transmit angles and receive elements) instead of the confocal sum: propagating both sides to a depth in the angular-spectrum domain gives focused-basis matrices whose off-confocal structure reveals the aberration (distortion-matrix approach). Follow-up: "And for focused-beam data?" REFoCUS recovers the synthetic-aperture data set first.

I11A 128-element probe at 0.30 mm pitch runs at 7.5 MHz. Will f-k images alias, and what do you do? · screen

Nyquist is \(1/(2 \times 0.30) = 1.67\) /mm; the receive cone at 7.5 MHz reaches 4.87 /mm, so arrivals steeper than \(\arcsin(c/(2pf)) = 20.0°\) fold. With F# 1.5 the fan needs \(0.316 \times 4.87 = 1.54\) /mm, still inside Nyquist, but steep energy folded from beyond can land in the fan. Use the fan filter, let element directivity help, and limit the angles that matter, or accept DAS where the aperture window keeps steep elements out. Follow-up: "Zero-offset at the same pitch?" Cone \(2f/c\), hopeless without a small aperture.

25Adaptive and coherence-based beamforming

14 questions · open in the module

I1Derive the MV weights. When is MV identical to DAS? · screen

Minimize \(\mathbf{w}^H\mathbf{R}\mathbf{w}\) subject to \(\mathbf{w}^H\mathbf{a} = 1\). The Lagrangian gradient gives \(\mathbf{R}\mathbf{w} \propto \mathbf{a}\), and the constraint fixes the scale: \(\mathbf{w} = \mathbf{R}^{-1}\mathbf{a}/(\mathbf{a}^H\mathbf{R}^{-1}\mathbf{a})\), output power \(1/(\mathbf{a}^H\mathbf{R}^{-1}\mathbf{a})\) (25.2). After the delays \(\mathbf{a} = \mathbf{1}\), so with \(\mathbf{R} = \sigma^2\mathbf{I}\) (white noise) the weights are 1/N: DAS. The same holds for L = 1 (scalar covariance) and for Δ → ∞ (then the smoothed MV is DAS with a trapezoidal taper). Follow-up: "What does the steering vector assume?" Correct delays: a sound-speed error makes \(\mathbf{a}\) wrong and MV suppresses the signal itself.

I2Why does textbook MVDR fail on ultrasound echoes, and what are the three standard fixes? · onsite

(1) The interfering echoes are coherent copies of the same transmit pulse, so MV uses them to cancel the desired signal: with an equal-amplitude echo 1.5 mm away the gain is −66 dB at L = N. Forward subarray smoothing (\(L \le N/2\)) decorrelates them through the factor \(D_P(\psi)\); at L = 48 the gain is back to −0.3 dB. (2) One time sample is one snapshot: averaging ±K samples over about a pulse length stabilizes \(\hat{\mathbf{R}}\) (speckle SNR 1.02 → 1.53 in the lab). (3) Model errors and noise: diagonal loading \(\varepsilon = \Delta\cdot\mathrm{tr}(\hat{\mathbf{R}})/L\) with \(\Delta \approx 10^{-2}\) to \(10^{-1}\) (25.3). Follow-up: "What is the cost of smoothing?" The adaptive aperture shrinks to L elements, so resolution and null depth drop.

I3A paper shows an MV point 0.03 mm wide against 0.48 mm for DAS. Is the resolution 16 times better? · onsite

No. For a lone point in white noise the MV width is \(0.914/\sqrt{N\cdot\mathrm{SNR}}\) times the DAS width: it measures SNR (capped by loading near 1/Δ), not the ability to separate two scatterers. The honest test is two points closer than the DAS width: in the lab, a 0.5 mm pair shows a −12 dB dip with MV against −1.3 dB for DAS, which is a real gain, but nowhere near 16×. Also ask about lateral sampling: an MV peak narrower than the pixel pitch reads −23 dB half a pixel off the grid. Follow-up: "What would you ask the authors to show?" Pair dip versus separation, loading and subarray sweeps, and gCNR.

I4What CF do you expect for a point, for speckle and for noise? Why is our speckle CF 0.23 and not 2/3? · screen

Point at focus: 1. Noise: 1/N (0.0156 for 64 elements). Speckle: the ratio of means is \(\sum(N - \lvert m\rvert)C(m)/N^2\) with C the VCZ coherence. A fully focused transmit with the full receive aperture gives \(C(m) = 1 - \lvert m\rvert/N\) and CF ≈ 2/3. Nine plane waves at 1° synthesize only a 4.7 mm transmit aperture at 30 mm, so C falls to zero at \(m_{\mathrm{tx}} = 15.7\) lags and the expected CF is 0.225 (simulated 0.23). Follow-up: "So what does CF do to the speckle brightness?" It lowers speckle by about 13 dB and noise by 36 dB here, a 23 dB stretch.

I5How are CF and the Wiener postfilter related? · onsite

Write \((1/N)\sum\lvert s_i\rvert^2 = \lvert\bar{y}\rvert^2 + \hat{\sigma}^2\). Then \(\mathrm{CF} = \lvert\bar{y}\rvert^2/(\lvert\bar{y}\rvert^2 + \hat{\sigma}^2)\): signal over signal plus channel noise. The MMSE gain for the DAS output needs the output noise \(\hat{\sigma}^2/N\), so \(H = N\cdot\mathrm{CF}/((N - 1)\mathrm{CF} + 1)\) (25.6) (Nilsen and Holm, 2010). For noise CF = 1/N (−36 dB) but H ≈ 0.5 (−6 dB). Follow-up: "Is H optimal per pixel?" Only with good estimates; single-sample estimates are noisy, so average over an axial kernel or subarrays.

I6Explain GCF's \(M_0\) physically. How would you choose it? · onsite

\(S(k)\) is the DFT of the aperture vector; bin k is a wave arriving from lateral offset \(\Delta x = k\lambda z/(Np)\), 0.48 mm per bin at 30 mm. GCF keeps \(\lvert k\rvert \le M_0\), so \(M_0\) sets how far from the focus incoming energy still counts as signal. Speckle occupies \(\lvert k\rvert \lesssim N/m_{\mathrm{tx}} \approx 4\) here: \(M_0 = 0\) (CF) keeps 22 % of it, \(M_0 = 1\) keeps 60 %. Choose \(M_0\) small enough to exclude the clutter directions you care about and large enough to pass speckle and residual aberration; the lab finds gCNR flat at 0.975 for \(M_0\) = 1–4. Follow-up: "What does aberration do to \(S(k)\)?" It spreads every bin, so GCF degrades more gently than CF but still degrades.

I7DMAS costs \(N(N - 1)/2\) products per sample. Is that true, and where does its gain come from? · onsite

No: \(\sum_{i<j}\hat{s}_i\hat{s}_j = \frac12[(\sum\hat{s})^2 - \sum\hat{s}^2]\) (25.9), so it is O(N) (64 square roots instead of 2016 products). The \(2f_0\) part of each pair carries \(\varphi_i + \varphi_j\), a virtual element at the mid-point receiving at λ/2: the sum coarray, a triangle. In narrowband terms the beam is the DAS beam squared: −26.5 dB sidelobes and 0.74× width, the same as a p = 2 power law for an isolated point. The signed square root keeps amplitude linearity, which is what separates it from DAS². Follow-up: "Why the band-pass?" The pair sum also has a DC (difference-coarray) part that only reproduces the DAS intensity; it must be removed, and the products need RF sampled well above \(4f_0\).

I8Why is a point target wide in SLSC, and what sets the speckle brightness? · research

SLSC normalizes each channel pair, so it reports coherence, not amplitude. Wherever a point dominates the local noise, even far down its DAS sidelobes, the aperture signal is a coherent wavefront and \(R(m) \approx 1\): the point stays bright over about 1.9 mm here (M = 8). Larger M includes lags where the off-focus wavefront decorrelates, so points narrow (5.3 mm at M = 1 to 0.63 mm at M = 24). In speckle \(R(m)\) follows the VCZ curve, the autocorrelation of the transmit aperture, so \(R_{\mathrm{sl}} \approx \sum(1 - m/m_{\mathrm{tx}}) = 5.7\) for M = 8, and noise multiplies every lag ≥ 1 by SNR/(1 + SNR). Follow-up: "Why does SLSC beat DAS in clutter?" Reverberation and off-axis clutter in a dark region are incoherent there and contribute R ≈ 0 regardless of brightness.

I9A reviewer sees CF with CR 52 dB against DAS 25 dB. What do you say? · onsite

CR on the linear output is not a fair comparison for any [0, 1] weight. The weight adds \(20\log_{10}W\) per pixel and is smallest where the signal is incoherent, which stretches the dynamic range: DAS raised to the power p multiplies a dB-domain contrast by p (the power-ratio CR goes from 24.7 to 46.4 dB at p = 2) and leaves gCNR unchanged (25.15). On the same data CF's gCNR is 0.94 against DAS's 0.96, and its speckle SNR falls to 0.91. Report gCNR (Rindal et al., 2019), and CR or CNR only at a matched displayed dynamic range. Follow-up: "Is there any real CF benefit?" Suppression of incoherent noise and clutter where it is the dominant contrast limit, which gCNR will then show.

I10FXPF and ADMIRE both filter channel data before DAS. What can each remove? · research

FXPF fits a linear predictor across the aperture at each temporal frequency. A focused echo and any off-axis plane wave are predictable, so both survive; what goes is spatially incoherent energy (noise, diffuse reverberation): about −14 dB on white noise at p = 4. Its coefficients shrink at low SNR, so weak echoes lose too (−2 dB at 0 dB channel SNR). ADMIRE fits a dictionary of modelled scatterer signatures, inside and outside the region of interest, with elastic-net regularization and reconstructs only the in-region part, so it can remove coherent off-axis and multipath clutter, at a much higher cost. Follow-up: "Why put them before aberration correction?" They clean the data that the coherence or delay estimators use, without changing the delays.

I11You want to estimate sound speed or a phase screen from coherence. Which coherence measure, and where does MV go in the pipeline? · research

Use a region-averaged measure that falls smoothly and monotonically with the residual delay error: mean CF or GCF over a speckle region, or SLSC (here SLSC's gCNR fell only from 0.99 to 0.87 at 60 ns rms). For a point with phase errors the CF-weighted brightness falls three times as fast in dB as DAS (\(\lvert\bar{y}\rvert^3\)), which makes CF sensitive but also noisy as a per-pixel weight. Maximize over the assumed speed or the phase law, as in Module 14. Apply MV or ESBMV after the correction: they amplify residual delay errors into resolution loss (MV's 0.03 mm point widens to 1.1 mm at 50 ns). Follow-up: "Why not use MV's output power as the focus metric?" It depends on loading and SNR as much as on focus.

I12An ESBMV image shows a black horizontal band next to a bright point target that DAS and MV do not show. Explain and fix. · research

Near the bright scatterer its wavefront dominates the subarray covariance, so with a relative threshold \(\lambda_i \ge \delta\lambda_1\) the signal subspace holds only that eigenvector. It is nearly orthogonal to \(\mathbf{a}\), and the projection \(\mathbf{E}_s\mathbf{E}_s^H\mathbf{w}_{\mathrm{MV}}\) drops the unit-gain constraint (\(\lvert\mathbf{w}_{\mathrm{ESB}}^H\mathbf{a}\rvert \le 0.002\) in our band): the local tissue vanishes over the pulse length plus the temporal average, 52 dB below DAS in Demo 25.6. Fixes: a lower δ (−36 dB at δ = 0.1, −11.5 dB at 0.01, at the price of less contrast stretch), or a threshold referenced to the noise floor instead of \(\lambda_1\). Follow-up: "Does it affect the metrics?" Only where the band meets a ROI; here it touches the cyst rim and inflates CR slightly.

I13A colleague adds noise at 0 dB SNR to every channel of a 64-element, 9-angle plane-wave dataset and says the noise "did not work" because the DAS image still looks clean. What do you tell them? · onsite

Per-channel SNR is not image SNR. Independent noise adds incoherently, a power gain of \(64 \times 9 = 576\) (27.6 dB), while speckle adds partly coherently; in Demo 25.6 speckle gains about 41 dB, so 0 dB per channel becomes +12 to +14 dB speckle-to-noise in the image, depending on the speckle realization (+13.9 dB in the displayed run). In this plane-wave dataset what remains also looks like speckle: every pixel is a weighted sum of the same band-limited channel samples, so neighbouring pixels share most of their noise (\(R_n(\mathbf{r}_1,\mathbf{r}_2) \propto \sum_e h_e(\mathbf{r}_1)h_e^{*}(\mathbf{r}_2)\)) and the noise becomes a complex Gaussian field with grains of about the PSF size and a Rayleigh envelope. To check, beamform the noise-only data and compare ROI powers, look in an echo-free region, or compare two frames: noise changes from frame to frame, the speckle of stationary tissue does not. Follow-up: "Thermal noise never interacts with the sound field, so how can it look like speckle?" Speckle-like texture comes from any random complex field passing through a band-limited linear system, not from acoustics; the beamformer is that system. "Would a focused line-by-line scan look the same?" No: each line has its own transmit and recording, so noise is independent from line to line and correlated only over about a pulse length along depth, giving a fine, streaky texture that is nearly salt and pepper laterally. True per-pixel salt and pepper needs noise entering after the band-limited chain (post-detection electronics, quantization, image-domain processing) or a pixel grid that undersamples the PSF.

I14A paper reports that ADMIRE improves cyst contrast by 20 dB over DAS. What do you check before you believe it, and what does ADMIRE actually decide per pixel? · research

Per depth window and frequency bin it fits the delayed aperture vector with modelled wavefronts of scatterers inside a small ROI around the line and outside it, by an elastic net (α = 0.9, \(\lambda = 0.0189\,\mathrm{rms}(\mathbf{y})\) in the released code), and keeps only the ROI part (25.14). It is underdetermined by a factor of hundreds (about 31 000 complex predictors for 64 complex data with the released model space), so sparsity decides. Checks: (1) CR on the linear output is meaningless if the cyst is set to exactly zero; in our full-model run 67 % of the cyst pixels are exactly zero at −20 dB channel noise (CR 35.7 dB against 24.7 dB for DAS, set by the remaining third), so ask for gCNR and speckle SNR. (2) Is the comparison at matched dynamic range? (3) Was the model calibrated to the system (the released code carries per-probe wavenumber factors)? (4) What happens to speckle near bright scatterers (λ scales with the window's rms, so weak tissue next to a bright target is shrunk away; this motivated iADMIRE, Schlunk et al. 2021)? (5) Cost: about 0.05 s per fit (10 min per 81 × 161 image) in our Node.js run, real time only with GPU and ICA-compressed models (Khan et al. 2021). Follow-up: "Why can it remove off-axis clutter when FXPF cannot?" FXPF keeps anything predictable, and an off-axis plane wave is; ADMIRE has an explicit outer model that claims that wavefront, so it is dropped.

26Speckle statistics and image-quality metrics

11 questions · open in the module

I1Derive the speckle SNR of 1.91. What do you need to assume? · screen

Many scatterers with independent uniform phases per cell: by the central limit theorem \(z\) is circular complex Gaussian with \(\mathrm{E}\lvert z\rvert^2 = \langle I\rangle\). Then \(R = \lvert z\rvert\) is Rayleigh, \(\mathrm{E}R = (\sqrt{\pi}/2)\sqrt{\langle I\rangle}\), \(\mathrm{E}R^2 = \langle I\rangle\), \(\mathrm{var}\,R = (1 - \pi/4)\langle I\rangle\), so \(\mathrm{SNR}_0 = \sqrt{\pi/4}/\sqrt{1 - \pi/4} = \sqrt{\pi/(4-\pi)}\) = 1.913 (26.1). Intensity is exponential with contrast 1.

Follow-up: what is it on the intensity, and on the log image? Intensity mean/std = 1; log image std 5.57 dB, independent of gain.

I2Why is the standard deviation of log-compressed speckle 5.57 dB whatever the gain, and what is its mean? · screen

For exponential \(I\), \(u = \ln(I/\langle I\rangle)\) has a fixed (Gumbel) distribution with mean \(-\gamma\) and variance \(\pi^2/6\). Gain multiplies \(I\), which adds a constant to \(u\). Times \(10/\ln 10\): std = \(4.343\cdot\pi/\sqrt{6}\) = 5.57 dB, mean = −2.51 dB re \(10\log_{10}\langle I\rangle\) (26.3).

Follow-up: so how should you compute lesion contrast from a dB image? From mean intensities (linearize first), or accept the −2.51 dB offset cancels only if both regions are Rayleigh.

I3You measure \(\mathrm{SNR}_0\) = 2.4 in a liver ROI. List the explanations in the order you would check them. · onsite

(1) Processing: was it computed on the linear envelope, before log compression, spatial filtering, persistence or scan conversion? Smoothing raises \(\mathrm{SNR}_0\). (2) Compounding: incoherent compounding raises it by \(\sqrt{N_{\mathrm{eff}}}\). (3) Physics: a coherent component (specular structures, regular spacing), Rician-like; \(\mathrm{SNR}_0\) = 2.4 matches a Rician \(r \approx 1.75\) (2.4 dB), i.e. \(m \approx 1.7\) (invert a measured \(\hat m\) with \(r = m - 1 + \sqrt{m^2 - m}\)). (4) Check the same pipeline on a phantom: it must read \(1.91 \pm 1.36/\sqrt{n}\).

Follow-up: how would you tell coherent structure from smoothing? Smoothing widens the autocovariance beyond \(\lvert\mu\rvert^2\) of the PSF; a coherent component does not.

I4What is the speckle size at 5 MHz, 60% bandwidth, F#2, and how many independent cells are in a 10 mm × 10 mm ROI? · onsite

Axial FWHM of \(\lvert\mu\rvert^2\) = \(0.441\,c/B\) = 0.441 × 1540/3×10⁶ = 0.226 mm. Lateral (rect, focal zone) \(0.835\,\lambda F_\#\) = 0.835 × 0.308 × 2 = 0.514 mm. Correlation area \(\iint\lvert\mu\rvert^2\) = 0.241 × 0.554 mm², i.e. 7.5 cells/mm², so about 750 cells in 1 cm². \(\mathrm{SNR}_0\) from that ROI scatters by about \(1.36/\sqrt{750}\) = 0.05.

Follow-up: what changes off focus? The lateral size follows the local beam width, so the count falls with distance from focus.

I5Why does coherent plane-wave compounding not reduce speckle, while spatial compounding does? · screen

Coherent compounding sums IQ: the result is a linear combination of circular Gaussian fields, hence circular Gaussian, so \(C_I = 1\) and \(\mathrm{SNR}_0\) = 1.91. It widens the k-space support, so speckle gets finer and resolution and clutter improve. Spatial compounding averages detected images from looks with low k-space overlap, so contrast falls as \(1/\sqrt{N_{\mathrm{eff}}}\) (26.11).

Follow-up: does coherent compounding change gCNR of a cyst? Yes, through contrast: lower sidelobes and clutter mean less energy in the cyst, so a higher intensity ratio; at a given ratio, gCNR is the same function of \(r\).

I6Three incoherently compounded plane-wave angles at −10°, 0°, +10°, receive F# 1.5. How many effective looks? · onsite

Field correlation \(r = 1 - \lvert\Delta\sin\theta\rvert\cdot F_\#\) (26.13): adjacent pairs \(\Delta\sin\theta\) = 0.174, \(r\) = 0.74, \(\rho_I\) = 0.55; outer pair \(\Delta\sin\theta\) = 0.347, \(r\) = 0.48, \(\rho_I\) = 0.23. \(\sum\rho_I\) = 3 + 2(0.55 + 0.55 + 0.23) = 5.65, so \(N_{\mathrm{eff}}\) = 9/5.65 = 1.6 and \(C_I\) = 0.79, not 0.58.

Follow-up: what angle span gives independent looks? \(\Delta\sin\theta \ge 1/F_\#\) = 0.67 between neighbours, about 40°, which is why speckle reduction needs large angles or sub-aperture translation of half the aperture or more.

I7A reviewer notes that your CNR rose 40% between two figures and the only change was the display dynamic range. Respond. · onsite

The reviewer is right that a dB-image CNR depends on the clipping floor: for an anechoic cyst it is ≈ \((\mathrm{DR} - 2.51)/5.57\). State the CNR definition (we use the linear envelope, pooled variance), recompute on the envelope, which has a ceiling of 1.91 for an anechoic cyst in fully developed speckle, and add gCNR, which is invariant to strictly monotone display maps.

Follow-up: can gCNR also be gamed? Not by the display map; it can be changed by ROI choice, bin rule and sample size (positive bias around 0.1 with about 100 cells), and by beamformers that change the output distribution's shape.

I8Define gCNR, give its operational meaning and its value for a −6 dB lesion in fully developed speckle. · research

\(\mathrm{gCNR} = 1 - \int\min(p_{\mathrm{in}}, p_{\mathrm{out}})\), the total-variation distance between the pixel-value distributions inside and outside. An ideal pixel classifier with equal priors is correct with probability \((1 + \mathrm{gCNR})/2\). For two exponential intensities with ratio \(r = 10^{0.6}\): \(\mathrm{gCNR} = r^{-1/(r-1)}(1 - 1/r)\) = 0.471 (classifier 74% correct).

Follow-up: how do you estimate it? Histograms with a fixed rule (quantile bins of the pooled data make it exactly display invariant), enough independent cells, and a null measurement on two same-statistics regions.

I9You want a Nakagami-m image of liver with ±0.05 precision. What window size, and what limits it? · research

\(\mathrm{std}\,\hat m \approx 2/\sqrt{n}\), so \(n\) ≈ 1600 independent cells. At 7.5 cells/mm² (5 MHz, 60%, F#2) that is about 210 mm², roughly a 15 mm window: far coarser than the B-mode. Limits: depth normalization (TGC, attenuation, diffraction) inside the window, mixtures at boundaries, small-\(n\) bias (\(\hat m\) is biased upward), and the lateral cell growing off focus.

Follow-up: how to do better? Compound independent looks or frequency bands, use 3D data, or use estimators with better efficiency (maximum likelihood) and report precision maps.

I10How should a ULM pipeline set the bubble detection threshold after SVD filtering? · research

Model the background as complex Gaussian: \(P_{fa} = e^{-T/\langle I\rangle}\), so \(T = 13.8\langle I\rangle\) (11.4 dB) for \(10^{-6}\) (26.16). Because \(\langle I\rangle\) varies with depth and region, normalize by a local noise estimate (CFAR) rather than a global threshold. Check the background histogram: residual clutter is heavier-tailed (K with \(\alpha = 1\) gives 2×10⁻³ at that threshold). Detection probability for a bubble at SNR \(S\) is Marcum Q: 0.87 at 13 dB.

Follow-up: why does a \(10^{-6}\) rate still matter? 10⁹ pixel tests per acquisition give about 1000 false detections, which tracking (track length, velocity limits) must remove.

I11How many scatterers per resolution cell do you need for fully developed speckle, and why is "10" an incomplete answer? · onsite

\(C_I^2 = 1 + \kappa/N_c\) with \(N_c\) counted in the correlation cell \((\int\lvert h\rvert^2)^2/\int\lvert h\rvert^4\), which for a Gaussian PSF is 1.44 times the −6 dB cell (26.6). Ten per −6 dB cell gives \(m\) = 0.935 (equal amplitudes) or 0.878 (Rayleigh amplitudes), \(\mathrm{SNR}_0\) ≈ 1.88 or 1.84. Incomplete because the answer depends on the cell definition, the amplitude distribution, the dimensionality (elevation) and on how precisely you can measure \(m\) (std \(2/\sqrt{n}\)).

Follow-up: can density alone be estimated from \(m\)? Only if there is no coherent component; otherwise use homodyned K.

27Spectral estimation & QUS

11 questions · open in the module

I1Why doesn't a longer gate make the periodogram less noisy, and what does? · screen

Each DFT bin of Gaussian data is circular complex Gaussian, so \(\lvert X\rvert^2\) is exponential with std equal to its mean. A longer record adds bins at spacing \(1/T\); it does not average within a bin. Variance falls only by averaging independent looks: segments (Bartlett/Welch), tapers (multitaper), adjacent bins (smoothing), or in QUS independent A-lines. Each costs resolution or spatial extent. Follow-up: "how many independent looks does a Welch estimate with Hann 50 % overlap have?" \(K_{\mathrm{eff}}\) from the overlap formula (27.7), e.g. 14.3 for 15 segments.

I2Derive the factor in the log spectral ratio and convert a measured slope to dB/cm/MHz. · screen

Amplitude decays as \(e^{-\alpha_0 f z}\) one way (\(\alpha_0\) in Np/cm/MHz); the echo travels \(2z\), and power squares it: \(e^{-4\alpha_0 f z}\). Between gates \(\Delta z\) apart, \(\ln S(z_2) - \ln S(z_1) = -4\alpha_0 f\Delta z\). If that difference has slope −0.46 per MHz for \(\Delta z\) = 2 cm: \(\alpha_0\) = 0.46/8 = 0.0575 Np/cm/MHz = 0.50 dB/cm/MHz. Follow-up: "and with amplitude spectra?" The slope is −0.23 per MHz and you divide by \(2\Delta z\): same answer.

I3What does a reference phantom cancel, and what does it not? · onsite

It cancels the transducer response, focusing/diffraction, gain, TGC and gate/window factors, because they multiply both spectra identically at the same depth (27.17). It does not cancel a sound-speed mismatch (the focus and beam shape move), differences in overlying attenuation (critical for BSC, irrelevant to the depth slope inside a homogeneous ROI), errors in the phantom's own \(\alpha_r\) and \(\mathrm{BSC}_r\) (they pass straight into the result), nonlinear propagation differences at high pressure, or electronic noise. Follow-up: "how would you check?" Scan a second phantom of known properties as a sample, and check the ln ratio for curvature in depth.

I4You need σ(α) ≤ 0.05 dB/cm/MHz with a 3–7 MHz band. Size the ROI. · onsite

Use \(\sigma \approx \sqrt{4.5\gamma\psi_1(K)c/(B^3L^3)}\) in Np (27.26), Hann \(\gamma = 1.75\), \(c\) = 0.154 cm/µs, \(B\) = 4 MHz. Target 0.05/8.686 = 0.00576 Np. With \(K = 32\) (\(\psi_1\) = 0.0317): \(L^3\) = 4.5·1.75·0.0317·0.154/(64·0.00576²) ≈ 18.2 → \(L\) ≈ 2.6 cm. With \(K = 16\), \(L\) ≈ 3.3 cm. Lines must be independent, i.e. spaced by about a beam width (\(\lambda F_\#\) ≈ 0.6 mm at 5 MHz, F/2): 32 independent lines span about 2 cm laterally; denser lines are correlated and count for less than their number. Follow-up: "what if the band shrinks to 2 MHz at depth?" \(\sigma\) grows by \(2^{1.5}\) = 2.8×, so \(L\) must grow by 2.

I5A 2 mm gate: Welch, multitaper or AR? · onsite

A 2 mm gate is about 104 samples at 40 MHz, so Fourier resolution is ≈ 0.4 MHz. For calibrated power (BSC, SLD) use multitaper (\(NW\) ≈ 3–4, \(K\) = 5–7): full use of the gate, \(\nu \approx 2K\), controlled leakage; Welch would split an already short gate. For locating features below \(1/T\) or for moments, Burg AR with AIC-chosen order; never use AR peak heights as power. In all cases the dominant averaging still comes from lateral lines. Follow-up: "what does \(NW\) cost?" Spectral smoothing over \(2W = 2NW f_s/N\).

I6Centroid downshift or spectral log difference? · onsite

\(\partial f_c/\partial z = -4\alpha_0\sigma^2(z)\) holds exactly (27.18), so the centroid needs the spectral variance at every depth; using the transmit \(\sigma_p^2\) assumes a Gaussian, unchanged by BSC, diffraction and noise. The log-ratio needs a reference but no shape model. On the same data both reached 0.030 dB/cm/MHz std; the centroid kept a 6 % bias. Choose the centroid when no reference exists and the shape is stable (or estimate \(\sigma^2\) per depth), the log ratio otherwise. Follow-up: "how does the BSC frequency dependence affect each?" It cancels in the SLD depth slope, but changes \(\sigma^2\) and biases the centroid with a fixed \(\sigma_p^2\).

I7Why does electronic noise bias attenuation estimates low, and how do you guard against it? · research

Noise adds a roughly flat power floor. Deep gates and high frequencies, where the signal is weakest, are lifted the most, so the distal spectrum looks less attenuated and the ln-ratio slope shrinks. Guards: measure the noise spectrum (no transmit) and subtract it in power before taking logs; restrict the band per depth to SNR above ~10 dB; weight the regression by the estimated variance; report the usable depth. Follow-up: "does noise subtraction bias anything?" It can produce negative powers at low SNR; clip or drop those bins, which reintroduces a small bias.

I8When can you estimate an effective scatterer diameter, and what correlates with it? · research

When the band covers \(ka\) of roughly 0.5 to 1.5, where the Gaussian form factor bends the \(f^4\) trend (local exponent \(4 - 1.654(ka)^2\), (27.25)). Below \(ka\) ≈ 0.3 the fit is noise (36 % failures for 20 µm at 3–7 MHz). The estimate correlates with the attenuation compensation (an error of 0.1 dB/cm/MHz at 3 cm takes 100 µm to 78 µm) and with the assumed form factor. Follow-up: "how would you pick the transducer?" Put the band so \(ka\) spans the size range of interest, then check SNR at the needed depth.

I9How would you turn SLD into an attenuation image with better resolution? · research

Write the log-spectral data from many small gates as a linear model \(y = A\alpha + b + \varepsilon\), with \(\alpha\) a spatial map (and optionally BSC-ratio terms as nuisance parameters), and solve it with regularization: Tikhonov or total variation in space, the same machinery as Module 13. Regularized SLD (Coila & Lavarello, IEEE TUFFC, 2018) is one published example. Report bias and resolution of the regularized map (model-resolution matrix), not just its smoothness. Follow-up: "what is the price?" Bias toward the prior: TV shrinks contrast of small inclusions, Tikhonov blurs edges.

I10Why might a log-averaged spectrum differ systematically between two depths even with no attenuation difference? · onsite

The expected log of an average of \(K\) exponentials, \(\psi(K) - \ln K\), depends on the effective number of independent looks. If the distal lines are more correlated (beam widens past focus) or contain noise, \(K_{\mathrm{eff}}\) differs between depths and the offsets no longer cancel: −2.51 dB at \(K = 1\) vs −0.07 dB at \(K = 32\). Fix by averaging power before the log, matching \(K_{\mathrm{eff}}\), or correcting with \(\psi(K_{\mathrm{eff}})\). Follow-up: "average log or log average?" Log of the average: lower variance and a bias that vanishes as \(1/(2K)\).

I11How would you validate a new attenuation estimator before a 510(k) or a paper? · research

Phantoms with \(\alpha\) measured independently by insertion loss (narrowband substitution) over the same band; several \(\alpha\) values and depths; bias and standard deviation versus ROI size, not only means; a layered phantom to test overlying-layer immunity; a sound-speed mismatch case; and in vivo repeatability (same subject, repeated scans, different operators). Report units, band, ROI dimensions and the number of independent lines. Public benchmarks that pair attenuation with sound-speed ground truth help comparison across methods. Follow-up: "what accuracy is enough?" Set it from the clinical decision threshold and the population spread, then size the ROI with the precision formula.

28Microbubble physics, contrast imaging and the ULM signal source

11 questions · open in the module

I1Derive the resonance frequency of a gas bubble in water. What is it for \(R_0\) = 2 µm? · screen

Radial flow \(u = R^{2}\dot R/r^{2}\) has kinetic energy \(\tfrac12(4\pi R^{3}\rho)\dot R^{2}\): effective mass \(4\pi R_0^{3}\rho\). Polytropic gas: stiffness \(12\pi\kappa P_0R_0\). \(\omega_0^{2} = 3\kappa P_0/(\rho R_0^{2})\), \(f_0R_0 \approx\) 3.3 MHz·µm for air, so 1.64 MHz at 2 µm. Then I would add surface tension (2.04 MHz for a free 2 µm air bubble) and say which \(\kappa\) I assumed. Follow-up: "And for a lipid-coated agent?" Add \(4\chi/R_0\): with \(\chi\) = 0.5 N/m and a perfluorocarbon-like \(\kappa\) about 3 MHz.

I2Why do microbubbles scatter so much more than tissue or red cells of similar size? · screen

Rayleigh scattering from a small sphere has a compressibility (monopole) term \(((\kappa_s - \kappa)/\kappa)^{2}\) and a density (dipole) term. Gas is about 15 000 times more compressible than water, so the monopole term is about \(2.4\times 10^{8}\), against order \(10^{-3}\) for a tissue inhomogeneity of a few percent: some \(10^{11}\) per particle of equal size. Near resonance there is another factor of order \(1/\delta^{2}\) (about 20–200 for real bubbles). Follow-up: "Then why not use bigger bubbles?" They do not pass the lung capillaries, and \(\sigma_s\) above resonance saturates at \(4\pi R_0^{2}\).

I3Write the Rayleigh–Plesset equation with a Marmottant shell and explain each term. How would you integrate it? · onsite

\(\rho(R\ddot R + 1.5\dot R^{2}) = p_g(R)(1 - 3\kappa\dot R/c) - P_0\) \(- 2\sigma(R)/R - 4\mu\dot R/R - 4\kappa_s\dot R/R^{2} - P_{\mathrm{ac}}(t)\) (28.6): liquid inertia; polytropic gas with first-order radiation damping; ambient pressure; radius-dependent tension (0 buckled, \(\chi(R^{2}/R_b^{2} - 1)\) elastic, \(\sigma_w\) ruptured); liquid viscosity; shell viscosity; drive. Integrate \((R, \dot R)\) with RK4 or an adaptive RK45, 100–200 steps per period with refinement during collapse, and check convergence by halving the step (\(R_{\mathrm{min}}\) should not move). Follow-up: "Why not Euler?" Collapse is stiff; explicit Euler needs orders of magnitude more steps and loses energy accuracy.

I4Compare pulse inversion and amplitude modulation. What does each cancel and keep, and where does the signal live? · onsite

With \(e(a) = \sum a_n a^{n}p^{n}\): PI sums \(e(1) + e(-1) = 2a_2p^{2}\), the even orders, at \(2f_0\). AM forms \(e(1) - 2e(\tfrac12) = \tfrac12 a_2p^{2} + \tfrac34 a_3p^{3}\), which includes an odd-order nonlinear signal in the fundamental band: better penetration. AM needs an exact half amplitude (0.1 dB error leaves −38.7 dB of tissue); PI needs exact inversion. Both are hurt by motion: \(2\lvert\sin(\pi f\tau)\rvert\) with \(\tau = 2v/(c\cdot\mathrm{PRF})\). Follow-up: "Which removes tissue harmonics better?" AM, because the half pulse carries a quarter of the tissue harmonic; PI keeps it entirely.

I5In a contrast mode you see bright tissue around the focus before the agent arrives. Explain and fix. · onsite

Nonlinear propagation: near the focus the tissue-generated second harmonic is largest (\(B_2/B_1 \approx \beta\omega P x/2\rho c^{3}\), −28 dB at 100 kPa, 2.5 MHz, 40 mm), and it is even in amplitude, so PI keeps it. Fixes: lower MI (the residue falls 6 dB per halving of \(P\), the bubble signal much less), move the focus deeper or use unfocused transmits, switch to AM or AM + PI, and check for motion residue and transmit asymmetry. Follow-up: "How would you measure it?" Image a tissue-mimicking phantom without agent in contrast mode and plot the residue against MI.

I6What is compression-only behaviour, and why does it matter for low-MI imaging? · research

A phospholipid bubble near its buckling radius has no surface tension in compression and a rising one in expansion (Marmottant). It compresses several times further than it expands (de Jong et al. 2007 observed it optically; the model gives a ratio of 3 at 10 kPa). That asymmetry is a strong even-order response at very low pressure, which is why PI and AM work at MI below 0.1, and it varies from bubble to bubble with the initial surface tension. Follow-up: "Consequence for quantification?" The echo is not proportional to concentration times incident pressure, so calibrations at one MI do not transfer to another.

I7How would you measure perfusion with flash-replenishment, and what limits its accuracy? · onsite

Constant infusion, low-MI imaging, a few high-MI frames to clear the slice, then fit \(I(t) = A(1 - e^{-\beta t})\) on linearized intensity: \(A \propto\) blood volume, \(\beta \propto\) velocity/slice thickness, \(A\beta \propto\) flow. Limits: incomplete destruction, bubble destruction by the imaging frames (biases \(\beta\) upward), attenuation and shadowing, receiver saturation of the plateau, beam elevation profile, speckle noise (\(\beta\) is the noisiest parameter, 12 % in the MATLAB example). Follow-up: "Why not fit the log-compressed curve?" The model is linear in concentration; the log map changes the shape (MTT doubles in the bolus example).

I8Why is the subharmonic attractive for contrast imaging, and where is its threshold lowest? · research

Tissue propagation does not generate \(f/2\), so a subharmonic image has a very high CTR. Period doubling has a threshold that is lowest for a drive near \(2f_0\), where \(f/2\) hits the resonance; for a free 2 µm bubble the model gives \(f/2\) at −30 dB at 100 kPa and −4 dB at 200 kPa. Costs: long bursts (poor axial resolution), the threshold behaviour, and the need for a transducer with bandwidth at \(f\) and \(f/2\). Follow-up: "What else is it used for?" Ambient-pressure estimation, because the subharmonic amplitude of some agents depends on hydrostatic pressure.

I9How do you choose the microbubble concentration for ULM? · research

Model bubbles in the image as a Poisson process with areal density \(n\); a bubble is usable if no other lies within \(k\) FWHM: \(P_{\mathrm{iso}} = e^{-nA}\), \(A = \pi k^{2}F_xF_z\) (28.12). The isolated count \(n\,e^{-nA}\) peaks at \(n^{*} = 1/A\) (\(P_{\mathrm{iso}} = 1/e\)). At 5 MHz with \(F_x = \lambda\), \(F_z = 0.75\lambda\): 4.5 per mm². Convert \(n = C\) × slice × vascular fraction, then remember that large vessels are locally much denser. Follow-up: "And the price of a lower concentration?" Capillaries are sampled more slowly in proportion (next question).

I10Why does ULM take minutes to map capillaries, and how would you shorten it? · research

A capillary of diameter \(d\) and speed \(v\) receives \(C\cdot\pi d^{2}v/4\) bubbles per second: about one per 140 s for 8 µm, 1 mm/s and 140 per µL. Frame rate does not help; the concentration is capped by overlap at \(n^{*} = 1/A\). Ways out: localize at higher densities (overlap-tolerant or sparsity-based estimators, learned localizers), shrink \(A\) (higher frequency, 3D removes elevation overlap), accept a coarser map for capillaries (density rather than tracks), or separate subpopulations (for example by acoustic signature). Follow-up: "What did Hingot et al. show?" That microvascular flow sets the compromise between resolution and acquisition time.

I11Your SVD-filtered ULM data miss the slow capillary flow seen in histology. Why, and what would you change? · research

The SVD cut separates by slow-time spectrum and spatial coherence. A bubble at 1 mm/s has a Doppler of 6.5 Hz at 5 MHz, inside the tissue-motion band, so it is projected out with tissue. Options: an adaptive cut chosen from the spatial-similarity of singular vectors, lower cut with better motion correction, longer ensembles for finer frequency resolution, or a nonlinear sequence (AM) that removes tissue regardless of velocity, accepting the motion residue and the extra transmits. Follow-up: "How would you validate?" A flow phantom with known slow flow, or a microchannel, reporting detection rate versus velocity.

29Doppler and flow imaging

10 questions · open in the module

I1Where does the factor 2 in the Doppler equation come from, and how large is the shift for carotid flow? · screen

The scatterer is a moving receiver on the way in and a moving source on the way back, so the path changes at \(2v\cos\theta\) and the echo phase rotates at \(f_D = 2vf_0\cos\theta/c\). For 1 m/s peak systolic flow at 60° and 5 MHz: 3.25 kHz. Follow-up: "is it a frequency shift of the echo?" In PW it is a phase change from pulse to pulse, sampled at the PRF; the shift of the RF spectrum within one pulse (\(v/c \approx 10^{-3}\)) is far below the bandwidth and is not what is measured.

I2Why does PW Doppler alias and CW not? Give the maximum unaliased velocity for a carotid at 3 cm, 5 MHz, 60°. · screen

PW samples the slow-time signal at the PRF; CW demodulates continuously. \(\mathrm{PRF}_{\max} = c/(2z)\) = 25.7 kHz at 3 cm, so \(v_{\max} = c^{2}/(8f_0\cos\theta\,z)\) = 3.95 m/s (29.4): no problem for a normal carotid, but a severe stenosis jet can exceed it. Follow-up: "what if the gate were at 15 cm?" 0.79 m/s; use baseline shift, lower \(f_0\), high-PRF or CW.

I3Sketch the Doppler spectrum of laminar flow in a vessel and explain why the maximum-velocity envelope overestimates. · onsite

Uniform insonation of a parabolic profile gives a flat spectrum from 0 to \(f_{D,\max}\) (equal area per velocity bin); a thin beam or a central gate tilts it toward the maximum. The aperture spreads each velocity over \(f_D[\cos(\theta + \alpha), \cos(\theta - \alpha)]/\cos\theta\): at F# 2 and 60° the edge reaches 1.39 \(f_{D,\max}\), so an envelope tracer reads 20–40 % high. The mean is unbiased to first order. Follow-up: "how would you correct it?" Calibrate the envelope for the aperture, use a smaller receive aperture, or use the mean.

I4Derive the Kasai estimator. Why is it the mean frequency, and when is it biased? · onsite

\(R(T) = E[z_{n+1}z_n^{*}] = \int S(f)e^{j2\pi fT}df\); for a spectrum symmetric about \(\bar f\), \(R(T) = e^{j2\pi\bar fT}\int S(f')\cos(2\pi f'T)\,df'\), so \(\arg R(T) = 2\pi\bar fT\) (29.8). Biases: asymmetric spectra (small), clutter residue (pulls toward tissue), aliasing, and the RF downshift: using \(f_{\mathrm{dem}}\) instead of the mean RF frequency scales \(\hat v\) by \(\bar f_{\mathrm{RF}}/f_{\mathrm{dem}}\), −10 % for 4.5 vs 5 MHz. Follow-up: Loupas 2D autocorrelation (29.9), which also lowers the variance by normalizing each estimate by the local mean RF frequency of the same block (depth averaging of \(\hat R(1)\) helps both estimators).

I5The variance map lights up in a vein with laminar flow. What is going on? · onsite

Low SNR: noise is white, adds to \(R(0)\) only and reduces \(\lvert R(1)\rvert/R(0)\) by \(\mathrm{SNR}/(1+\mathrm{SNR})\) (29.10). At PRF 4 kHz and 10 dB SNR a 150 Hz spectrum reads 306 Hz. Slow venous flow is also close to the wall filter, which shapes the residual spectrum. Fixes: subtract a noise estimate from \(R(0)\), threshold the variance by power, average over more samples. Follow-up: geometric broadening and shear also widen the spectrum without turbulence.

I6Design a wall filter: N = 10, PRF 3 kHz, 5 MHz, tissue moving at 5 mm/s, clutter 40 dB above blood. · onsite

Tissue Doppler is 32.5 Hz. Regression filters attenuate it by 14 dB (\(p\) = 0), 36 dB (\(p\) = 1) and 61 dB (\(p\) = 2), so \(p\) = 2 is needed to bring the residue below blood. Its cutoff is 380 Hz (rule: 378 Hz), i.e. flow slower than 0.117 m/s along a vessel at 60° is lost. To recover slow flow: longer \(N\) (cutoff ∝ PRF/N), or a lower PRF if the Nyquist limit allows, or an adaptive (SVD, eigen) filter. Follow-up: "why not an IIR filter?" Its transient eats most of a 10-sample ensemble unless it is initialized (projection initialization).

I7Why does an SVD filter beat a high-pass filter for microvascular imaging, and how do you pick K? · research

Slow microvascular flow and tissue motion overlap in Doppler frequency (1 mm/s at 15 MHz is 19.5 Hz, breathing and pulsatility give tens of Hz), so no frequency cutoff separates them. They differ in space: tissue moves coherently and is low rank in the Casorati matrix, blood is not (29.12). A short-ensemble high-pass also leaks motion through the window sidelobes (m29_ultrafast_power_doppler.m: background 11.5 dB above noise). K: from the knee of the singular spectrum, from the spatial similarity of singular vectors (Baranger et al.), or by a power threshold; too small leaves flash, too large removes slow blood that has not decorrelated. Follow-up: large fields with non-uniform motion need block-wise SVD; the noise subspace can also be cut at the high end.

I8Derive how power Doppler sensitivity scales with ensemble length. What stops you from making N arbitrarily large? · research

The background value is a mean of \(N - K\) exponential variables: mean \(P_n\), relative spread \(1/\sqrt{N-K}\); hence \(\mathrm{CNR} \approx (P_b/P_n)\sqrt{N-K}\) (29.13), ×2 per ×4 in \(N\). Limits: the stationarity of the scene (motion moves vessels between pixels and raises the clutter rank), the temporal resolution needed (fUS wants a power value every 0.2–1 s), data rate and memory (\(P\times N\) complex samples per block) and SVD cost (\(\propto PN^{2}\)). Follow-up: compounding adds \(10\log_{10}n_a\) dB to \(P_b/P_n\) at the cost of frame rate.

I9How does cross-beam vector Doppler work, and why is the lateral component noisier? Compare with transverse oscillation. · onsite

Each beam pair measures a projection \(f_k\) (29.14); least squares gives \(\mathbf{v}\) with covariance \(\sigma_f^{2}(\mathbf{A}^{T}\mathbf{A})^{-1}\). For \(\pm\gamma\) the lateral row coefficient is \(\propto\sin\gamma\), the axial \(\propto\cos\gamma\): ratio \(\cot\gamma\), 5.7 at 10°. Transverse oscillation creates a lateral wavelength \(\lambda_x = 2\lambda z/d\) (1.85 mm at 30 mm with \(d\) = 10 mm, six times \(\lambda\)), so the lateral phase per unit motion is six times smaller, the same geometric penalty in another form. Follow-up: both degrade with depth because the usable angle shrinks as \(d/z\).

I10How do sound-speed errors and aberration enter Doppler, and how is Doppler used in ULM? · research

\(v = cf_D/(2f_0\cos\theta)\) scales with the assumed \(c\) (a 3 % error is a 3 % velocity error), refraction changes the true \(\theta\), and defocusing (wrong \(c\), aberrating layers) widens the beam, so more angles contribute (more broadening) and the compounded SNR drops (Modules 32, 33). Phase aberration that is constant over the ensemble cancels in \(\arg R(1)\), which is why Doppler is robust to it while B-mode is not. In ULM the SVD filter of 29.6 isolates bubbles; the sign of the bubble's Doppler (or the direction of its track) separates up- and down-going flow, and track velocities give the velocity maps. Follow-up: motion between compounded angles creates a velocity-dependent phase error that biases both Doppler and ULM localization.

30Elastography: strain, ARFI and shear-wave imaging

11 questions · open in the module

I1Why can a scanner whose pulses travel at 1540 m/s measure tissue stiffness at all? · screen

Because it does not measure stiffness with its own pulses: it uses them as a motion sensor. Compressional speed depends on \(K + 4\mu/3\) with \(K/\mu \sim 10^{6}\), so it is insensitive to \(\mu\). Elastography deforms the tissue (compression, radiation force, vibration) and tracks the displacement with speckle tracking at kHz rates; the shear wave moves at 1–10 m/s, about 1000 times slower, so each frame freezes it. Follow-up: what frame rate? Several samples per shear period: 5–10 kHz for 200–800 Hz content.

I2Convert 1.8 m/s to kPa and list every assumption you just made. · screen

\(E = 3\rho c^{2}\) = 3 × 1000 × 3.24 = 9.7 kPa (μ = 3.24 kPa). Assumptions: linear elastic (no dispersion, so phase = group), isotropic, incompressible (ν = ½), unbounded (no guided modes), ρ = 1000 kg/m³, and a plane shear wave measured along its propagation direction. Follow-up: which fails first in liver? Dispersion (viscosity), then near-capsule effects; in muscle, anisotropy; in arteries, guided waves.

I3Derive the radiation force and estimate the displacement of a 200 µs push. · onsite

Power absorbed per volume \(2\alpha I\), divided by \(c\) for momentum flux: \(F = 2\alpha I/c\). At 4 MHz, 0.5 dB/cm/MHz (23 Np/m), 500 W/cm²: 1.5×10⁵ N/m³. During a short push the column accelerates at \(F/\rho\) = 150 m/s², so \(u(T) = FT^{2}/2\rho\) = 3.0 µm; it coasts a little after (3.5 µm peak in the axisymmetric model at 1.5 m/s). Follow-up: heating, 2αIT/ρC = 12 mK per push; and why the in-situ force is unknown (α, derating, focal gain).

I4Your strain images look great at 1 % compression and terrible at 5 %. Why, and what do you change? · onsite

The strain filter: at 5 % the kernel is compressed by εT relative to its partner (0.1 mm over a 2 mm kernel, a third of a wavelength at 5 MHz), the echoes decorrelate and peak hopping starts. Fixes: global or adaptive companding (stretch the post frame by the expected strain), shorter kernels, multi-step (accumulate small inter-frame strains), guided search and outlier rejection. Follow-up: and at 0.05 %? Then jitter dominates; use longer kernels, larger LSQ span or more compression.

I5Explain supersonic shear imaging. Why does the Mach number matter for estimation? · onsite

Pushes are fired at successive depths faster than the shear wave can leave, so the source moves at \(V > c_s\) and the wavelets add on a Mach cone, giving two quasi-plane waves of larger amplitude and depth extent than one push. The fronts are tilted by \(\varphi = \arcsin(1/M)\); a lateral-only TOF reads \(c_s/\cos\varphi\): 0.5 % high at M = 10, 15.5 % at M = 2. Follow-up: how to fix? Higher M, 2D direction estimation, or directional filters.

I6TTP regression, cross-correlation or k–f: which would you put in a product, and why? · onsite

Depends on the claim. For a robust single number in liver (pSWE), TTP or cross-correlation regression with outlier rejection (RANSAC, Radon sum): cheap, short kernels, stable at low SNR; XC is more precise than TTP because it uses the whole trace. For maps, local XC between x ± Δ or local phase gradient with directional filtering. For viscosity or thin structures, k–f phase velocity, because it gives \(c(f)\) and lets you fit a model. And whichever it is, specify the frequency band, because in dispersive tissue they disagree (1.85 / 1.55 / 1.53 m/s in m30_shear_wave_tof_kf.m). Follow-up: how do you validate? Viscoelastic phantoms plus digital phantoms through your exact pipeline.

I7Derive the Kelvin–Voigt phase velocity and explain why group velocity exceeds it. · onsite

\(\mu^{*} = \mu + j\omega\eta\), \(k = \omega\sqrt{\rho/\mu^{*}}\); write \(\mu^{*} = \lvert\mu^{*}\rvert e^{j\theta}\), Re k = \(\omega\sqrt{\rho/\lvert\mu^{*}\rvert}\cos(\theta/2)\), and \(\cos(\theta/2) = \sqrt{(\lvert\mu^{*}\rvert+\mu)/2\lvert\mu^{*}\rvert}\) gives (30.16). Since \(c_{\mathrm{ph}}\) rises with ω, \(d\omega/dk > \omega/k\): \(c_g = c_{\mathrm{ph}}/(1 - (\omega/c)\,dc/d\omega)\). Numbers: 1.5 m/s, 1 Pa·s, 200 Hz: 1.66 and 2.01 m/s. Follow-up: why do people prefer power laws? KV disperses as √ω at high frequency, much faster than tissue.

I8Your 2D SWE map shows a bright halo and stripes upstream of a stiff lesion. Diagnose. · research

Reflections from the lesion boundary: the reflected wave travels back through the background, superposes on the incident one and biases arrival times or phases (stripes at half-wavelength spacing upstream), while the halo is the kernel straddling the boundary (partial volume over 2Δ). Fix with directional filtering in k–f (or angular sectors in 2D), smaller kernels near edges, and compounding of left and right pushes; check the residual with a simulated inclusion run through the exact pipeline. Follow-up: cost of the filter? Edge smearing over about a wavelength and distortion of near-field components.

I9How would you measure the stiffness of the carotid wall with SWE? · research

Not with \(3\rho c^{2}\): the 0.6–1 mm wall carries guided (Lamb, tube) modes. Push in the wall, track along it at high frame rate, compute the k–f spectrum, extract \(c_{\mathrm{ph}}(f)\), and fit a plate or tube model: the immersed-plate fit \(c^{2} = \omega h c_t/(2\sqrt3)\) (Couade et al. 2010) in the high-frequency part, or the full leaky-Lamb or tube dispersion. Measure thickness from B-mode, synchronize with the ECG because pressure changes \(c_t\), and report the band. Follow-up: error from thickness? \(c_t \propto c^{2}/h\), so a 10 % thickness error is a 10 % \(c_t\) and 21 % \(E\) error.

I10Two vendors measure the same patient at 1.45 and 1.62 m/s. Is one of them wrong? · research

Not necessarily. 12 % is the size of the dispersion effects in liver: different push durations and tracking bands sample different frequencies, TTP on displacement versus TOF on particle velocity weights different spectra, ROI depth and focal configuration differ, and kPa conversion doubles any gap. The QIBA phantom work found exactly such system dependence growing in viscoelastic phantoms. The right comparison is c(f) over a stated band, ideally from a k–f or spectroscopy method, plus a phantom cross-calibration. Follow-up: what would you put in the 510(k) bench testing? Elastic and viscoelastic phantoms over depth, with the estimator's frequency response stated.

I11Where do elastography, sound-speed estimation and aberration correction meet? · research

All three are delay estimation problems on the same channel data. Sound-speed errors defocus the tracking beams and bias the displacement estimates (and the depth scale, so the measured shear speed scales with the assumed c). Aberration lowers tracking SNR and correlation, raising jitter. Conversely, a measured SoS map (Module 32) and an aberration correction (Module 33) improve tracking, and steatosis raises attenuation and lowers SoS while it changes shear dispersion, so multiparametric QUS + SWE models share one forward model. Follow-up: how much does a 2 % c error change the reported shear speed? Lateral distances are set by the array geometry and frame times by the clock, so to first order little; depth registration of the ROI shifts by 2 %.

31ULM end to end (capstone)

11 questions · open in the module

I1Walk me through a ULM pipeline and name the number that limits each stage. · screen

Acquire compounded plane waves (ENR per frame and displacement per frame, \(\delta = v/\mathrm{FR}\)); SVD clutter filter (slowest bubble kept, about \(\lambda/(2T_b)\) axially); motion correction (residual adds in quadrature); detection (false-alarm rate and merging below one FWHM); localization (\(w/\sqrt{\mathrm{ENR}}\) plus estimator bias); tracking (\(e^{-\pi n d^{2}}\), gate, minimum length); rendering (saturation \(T = \max(d,l)/(\eta CQl)\), FRC). Follow-up: which stage would you improve first for capillary maps? Acquisition time, through η: more isolated bubbles per frame (overlap-tolerant localization) and fewer slow bubbles lost in the filter.

I2How do you choose the SVD cut-off, and what goes wrong on each side? · screen

Look for the drop in the singular spectrum, the jump in the mean Doppler of the temporal vectors (tissue at a few hertz, bubbles at tens), and spatial vectors that stop looking like the B-mode (Baranger et al. 2018 automate this). Too low: residual clutter dominates the noise estimate and the detector finds nothing or fires on tissue. Too high: slow bubbles are removed, and their removed mean leaves negative imprints that become false alarms and bias neighbours (m31_ulm_pipeline.m: r = 3 gives 0.26 false alarms per frame, r = 4 gives 2.45). Follow-up: what if tissue moves a lot? The tissue rank grows; use shorter or block-wise SVD, or register before filtering.

I3Derive the localization CRLB for a Gaussian PSF and give numbers at 5 MHz. · onsite

For \(s = a\,g(x - x_0)\) in complex white noise of density \(N_0\), the Fisher information on \(x_0\) is \((2/N_0)\lvert a\rvert^{2}\int\!\!\int (\partial_x g)^{2}\); with \(g = e^{-x^{2}/2w_x^{2} - z^{2}/2w_z^{2}}\) that is \(\mathrm{ENR}/w_x^{2}\) with \(\mathrm{ENR} = \lvert a\rvert^{2}\pi w_x w_z/N_0\), and the unknown phase costs nothing because \(\partial_{x_0}s\) is orthogonal to \(s\). So \(\sigma_x \ge w_x/\sqrt{\mathrm{ENR}}\): 13 µm at 20 dB, 4 µm at 30 dB (\(w_x\) = 131 µm). Follow-up: why do real localizers miss it? Few-sample estimators, pixel locking, overlap, and the threshold region below about 15 dB.

I4Your Gaussian-fit localizer runs at three times the CRLB. Is the CRLB wrong? · onsite

Probably not: a 3-point fit on λ/4 pixels uses three samples of a PSF four pixels wide, with independent noise in each. m31_localization_methods.m gives 3.4× at 25 dB for white pixel noise, 1.3× at λ/2, while the ML estimate (normalized matched-filter maximum) reaches 0.98× and radial symmetry 1.8×. Check that it is not overlap (split errors by nearest-neighbour distance) and not bias (sub-pixel histogram). Follow-up: is ML affordable? Separable templates and a two-stage grid make it a few thousand multiply-adds per bubble.

I5Nearest neighbour or Hungarian for ULM tracking? · onsite

Depends on \(\rho = n d^{2}\). NN is right with probability \(e^{-\pi\rho}\); at 1 kHz, 5 localizations per mm² and 10 mm/s, ρ = 5×10⁻⁴ and every method is above 99.8 %. Hungarian matters at low frame rates, fast flow or high density: at ρ = 0.2 NN gets 52 %, Hungarian 72 %, Hungarian with a constant-velocity prediction 96 % (m31_tracking_hungarian.m); greedy global assignment is worse than NN (42 %). Follow-up: what breaks links in practice then? Missed detections, merged bubbles and false alarms: gap closing and a Kalman gate help more than the assignment rule.

I6How long must I acquire to map 10 µm vessels with 5 µm pixels? · onsite

\(T = \max(d,l)/(\eta C Q l)\) (Hingot et al. 2019). With \(C\) = 200 /µL, \(v\) = 2 mm/s: \(Q\) = 1.57×10⁻⁴ µL/s, flux 0.031 bubbles/s, two columns, so \(T\) = 64 s at η = 1 and 90 % after 2.3 T = 2.4 min; with η = 0.1 about 25 minutes. Follow-up: does a faster frame rate help? No; flux sets new bubbles per second. Higher concentration helps until overlap caps η (Section 31.3).

I7A paper claims "λ/10 resolution." What do you ask? · research

Is it precision (from repeated localizations or track smoothness) or resolution (separation of two vessels, or FRC)? If FRC: split by tracks or by localizations, which threshold (half-bit, 2σ, 1/7), how many tracks, and was the image saturated? m31_saturation_frc.m shows the localization split reporting 28–31 µm when the track split gives 64 µm with 300 tracks. Then: motion correction residual, aberration (bias, not blur), and whether vessels below the saturation diameter are missing rather than resolved. Follow-up: one number you would add? The saturation curve.

I8Why does uncorrected motion matter more than localization precision in vivo? · research

Errors add in quadrature: \(\sigma_{\mathrm{eff}}^{2} = \sigma_{\mathrm{loc}}^{2} + \sigma_{\mathrm{m}}^{2} + \sigma_{\mathrm{reg}}^{2}\). A breathing excursion of 100 µm is 71 µm rms against 5–15 µm of localization error, so a vessel appears 170 µm wide. Correction by speckle registration of the tissue signal per block reduces it to the registration error (a few micrometres at good tissue SNR), but out-of-plane motion and deformation remain. Motion along a vessel is harmless; across it is fatal. Follow-up: how would you validate? Measure the spread across a vessel of known size: \(s^{2} = R^{2}/5 + \sigma_{\mathrm{eff}}^{2}\) for a 2D parabolic profile counted once per bubble (\(R^{2}/6\) to \(R^{2}/4\) for a cylinder, depending on how localizations are counted).

I9What changes in transcranial ULM, and why is aberration worse than blur? · research

Lower frequency (larger PSF, so worse \(w/\sqrt{\mathrm{ENR}}\)), skull attenuation (lower ENR), longer acquisitions, and a phase screen. The quadratic part of the screen defocuses (precision); the linear part steers the beam and displaces every bubble behind it by \(z c \alpha\): 77 µm for 10 ns across 10 mm at 50 mm. A bias does not average out, so the map is warped with full confidence. Demené et al. (2021) corrected aberration and brain motion with adaptive speckle tracking. Follow-up: how would you estimate the screen? From the bubbles themselves as guide stars, or from the distortion matrix (Module 33).

I10Make the case for, and against, deep-learning localization at high concentration. · research

For: the density cap \(1/(eA)\) limits isolated bubbles per frame; a localizer that separates bubbles at half an FWHM raises the cap fourfold and cuts acquisition time fourfold. Deep-ULM trained on densities up to 260 /cm² and reports 20–30 µm in vivo where centroid methods break down. Against: it inherits its simulator (PSF shape, clutter residue, noise), errors become domain gap, and it is hard to give a per-bubble uncertainty. I would ask for PALA-style benchmarks on shared data, recall and precision separately, and a test on a PSF it was not trained on. Follow-up: a physics-anchored alternative? Sparse recovery or unrolled networks with the beamformer as forward model.

I11Design a 3D ULM system: matrix or row–column array? · research

Matrix (e.g. 32 × 32 = 1024 elements, Heiles et al. 2019): good focusing in both directions, but 1024 channels or multiplexing, small aperture per channel budget, lower ENR per element. Row–column (Jensen et al. 2020: 62 + 62 channels): \(2N\) channels for an \(N \times N\) aperture, but one-way focusing in each direction, edge waves and lower contrast, so stronger clutter filtering and careful apodization. Either way saturation is slower by \(\max(d,l)/l\), volume rates are lower (larger ρ, so assignment and prediction matter), and the data rate is the system bottleneck. Follow-up: what does 3D fix? Out-of-plane bubbles and vessels, so tracks are longer and η rises.

32Sound-speed estimation (capstone)

11 questions · open in the module

I1Why does a 1540 m/s assumption hurt a high-frequency, low-F-number image more than a low-frequency, high-F-number one? · screen

The speed error appears as a curvature error, \(\delta\tau(X) = (\Delta c/c)X^{2}/(zc)\); in phase that is \(2\pi f_0\delta\tau\). Requiring less than π/4 at the edge gives \(\lvert\Delta c\rvert < c^{2}F^{2}/(2f_0z)\): inversely proportional to frequency and proportional to F². At 5 MHz and 30 mm: 7.9 m/s at F/1, 71 m/s at F/3. Follow-up: and the depth error? \((\Delta c/c)z\), independent of aperture: 0.3 mm per percent at 30 mm.

I2You sweep the assumed speed and maximize coherence in a liver ROI at 50 mm under 10 mm of subcutaneous fat. What number did you measure? · screen

An average of the path, specifically the RMS speed \(V_{\mathrm{rms}} = \sqrt{\sum w_i c_i^{2}}\) (time weights), because small-offset moveout under flat layers is hyperbolic with \(V_{\mathrm{rms}}\) (Dix). With 1450 over 1580 m/s it reads about 1553, 27 m/s below the liver, and only about 1 m/s above \(z/T\). Follow-up: how do you get the liver itself? Measure at several depths and invert (Dix or regularized), or use a local method (tomography, layer fit with the B-mode interfaces).

I3Derive the noise of a local speed obtained by differencing two average speeds. · onsite

\(z\bar s(z) = \int_0^{z}s\), so \(s_k = (z_k\bar s_k - z_{k-1}\bar s_{k-1})/\Delta z\). With independent errors \(\sigma_{\bar s} = \sigma_{\bar c}/c^{2}\) on each average, \(\sigma_{s} = \sigma_{\bar s}\sqrt{z_k^{2} + z_{k-1}^{2}}/\Delta z\), i.e. \(\sigma_{c,\mathrm{loc}} \approx \sqrt2(z/\Delta z)\sigma_{\bar c}\). 2 m/s averages 1 mm apart at 30 mm give 85 m/s. Follow-up: what fixes it? Larger Δz (resolution), Tikhonov with a discrepancy-chosen λ, or a layer prior from the image, each a different bias.

I4Explain CUTE to a beamforming engineer in two minutes, including what limits it. · onsite

Beamform the same region with \(c_0\) from several steered plane waves (and receive angles); for each pixel the echo phase changes between steering angles in proportion to the difference of the travel-time errors along the two paths, \(\Delta\tau = L_\theta - L_{\theta'}\), \(L_\theta = \sec\theta\int\Delta s\). Stack all pixels and angles into \(\mathbf{d} = \mathbf{A}\Delta\mathbf{s}\) and invert with regularization. Limits: phase wraps at \(1/(2f_0)\) (small steps, accumulate), speckle decorrelation between angles (common mid-angle pairing), limited angle (axial blur, layers and the mean nearly invisible), and the forward model (straight rays, echo-position shift). Follow-up: what did Stähli et al. change? Common mid-angle pairs and the echo-position term.

I5Why is a laterally uniform fat layer hard for echo-mode tomography? Quantify. · onsite

All steered paths cross it, with lengths \(d/\cos\theta\); the difference to 0° is \((\sec\theta - 1)S\), where \(S\) is the vertical delay. For 4 mm at 1450 m/s, \(S\) = 161 ns, but the 15° datum moves by 5.7 ns (11.4 ns for a symmetric common-mid-angle pair), comparable to a few times the 1 ns jitter of one datum. A constant offset of the mean behaves the same way. Follow-up: so how do you anchor the mean? A global coherence estimate, a calibrated a priori value, or physics beyond straight rays.

I6What is the phase jitter of a CUTE datum, and what is the cheapest way to halve it? · onsite

\(\sigma_\phi \approx \sqrt{1-\rho^{2}}/(\rho\sqrt{2N})\), the interferometric-phase CRLB; at ρ = 0.95, N = 64: 0.029 rad, 0.93 ns at 5 MHz. Halving it by kernel size needs 4N (resolution loss); raising ρ from 0.95 to 0.987 does the same. Hence pairing schemes that keep speckle correlated (common mid-angle, common midpoint), small angle steps and good SNR. Follow-up: when does the formula fail? Low ρ, small N: the distribution grows tails and wraps.

I7Your SoS tomography gives 25 m/s RMSE on a phantom. More angles or a better forward model? · research

Check which term dominates first. Rerun the inversion on data synthesized by your own forward model with your measured noise level: if that RMSE is small (a few m/s, as in m32_sos_tomography.m: 3.8 and 1.9 m/s), the 25 m/s is model error (straight rays, refraction, echo-position shift, discretization, lens) and more angles will not help. Then look at the resolution matrix to see whether the phantom features are resolvable at all. Follow-up: how would you validate the forward model alone? Full-wave simulation of the phantom geometry, compared datum by datum.

I8What is the relationship between differentiable-beamforming autofocus and CUTE? · research

With straight rays, each beamforming delay is \(\sum_j \ell_{ij}s_j\); the chain rule makes the gradient of any delay-based loss \(\mathbf{A}^{T}\nabla_\tau\mathcal{L}\), a back-projection along the same rays CUTE uses. Gradient descent is a nonlinear iterative tomography that recomputes the delays (and could recompute bent rays) at each step; the iteration count and TV play the role of λ. It shares CUTE's missing cone and benefits from the same correlated pairings (common midpoint). Follow-up: what does it buy? Losses evaluated on the actual beamformed data, any differentiable forward model, and easy priors.

I9Where do phase-based autofocus losses get stuck, and how would you initialize? · research

Wherever a residual delay difference exceeds half a period, the wrapped phase reverses and the loss has a secondary minimum: cycle skipping (Module 14). Long subaperture lags and high frequency make it worse. Initialize with a coherence- or CF-based global speed (smooth, wide basin), use lag-1 pairs with a correlation threshold (as in CMPE), and go from low to high frequency. Follow-up: can coherence alone be the loss? Yes, it is smooth but flat near the optimum and biased toward the path average.

I10Write the error budget for a liver sound-speed number reported to a hepatologist. · research

Random: curvature or phase CRLB divided by √M (sub-m/s after averaging). Biases: layer bias if an average is reported (17–45 m/s under 10 mm of fat at 30–80 mm), forward-model error (refraction, straight rays, echo position: tens of m/s in tomography), calibration (≈ 0.8 m/s per ns of edge delay error for 64 elements at 30 mm; lens speed and thickness), regularization and the a priori mean. Compare with 14 m/s between the Imbault cut-offs. Follow-up: how do you validate? Phantoms with through-transmission references (±5 m/s), then patients against MRI-PDFF.

I11Would you trust a deep network that outputs a sound-speed map from channel data? · research

Only inside its training distribution. Feigin et al. (2020) trained on 6,026 k-Wave simulations of random ellipses and reported 20.5 m/s RMSE on simulated test data, with qualitative results in vivo: the forward model is now the simulator, and anatomy, lens, attenuation and noise mismatches become an unseen domain gap. I would want physics anchoring (unrolled or model-based inversion), uncertainty estimates and benchmark results on independent data such as QUASOS. Follow-up: what is a good sanity test? Feed a homogeneous phantom with known speed and a layered phantom; check the mean and the layer bias.

33Aberration correction (capstone)

11 questions · open in the module

I1Model an abdominal-wall aberrator in two numbers and predict the focal loss at 5 MHz for 40 ns rms. · screen

A thin delay screen at the array: rms \(\sigma_\tau\) and correlation length \(\ell\) (Gaussian). \(\sigma_\phi = 2\pi f_0\sigma_\tau\) = 1.26 rad. Infinite-aperture Strehl \(e^{-1.58}\) = 0.21; with 128 elements and \(\ell\) = 4 mm the exact mean is 0.315 because nearby elements stay in phase; two way about 0.1. The −6 dB width hardly changes until about 30 ns, then the halo takes over. Follow-up: what if the frequency halves? \(\sigma_\phi\) halves and the exponent quarters: about 0.67.

I2Why does nearest-neighbour cross-correlation drift, and how do you fix it? · screen

Integrating \(N-1\) independent differences is a random walk: error \(\propto\sqrt{k}\), \(\sigma_\varepsilon\sqrt{N/15}\) after removing the mean and ramp. Fixes: weighted least squares over many lags (graph Laplacian plus a constant constraint), better SNR, longer kernels. Element noise telescopes and does not drift. Follow-up: in bright speckle the jitter is sub-ns but the residual is 5.7 ns; why? A bias from the blurred transmit spot, removed by transmitting with the estimate and re-measuring.

I3Show that speckle-brightness maximization and DORT give the same aberration law. · onsite

Brightness of the beamformed pixel is \(\mathbf{w}^{H}\mathbf{C}\mathbf{w}\) with \(\mathbf{C}\) the aperture covariance, maximized over \(\lVert\mathbf{w}\rVert^{2} = N\) by the dominant eigenvector. Iterative time reversal is power iteration on \(\mathbf{K}^{H}\mathbf{K}\), and DORT diagonalizes it; for speckle behind a focused transmit, \(\mathbf{C} = \mathbf{\Phi}\mathbf{C}_{\mathrm{vr}}\mathbf{\Phi}^{H}\), so the eigenvector is the screen times the eigenvector of the virtual reflector. Follow-up: when is that second factor flat? When the insonified spot is compact and symmetric, which is why you iterate.

I4Derive the isoplanatic patch for a thin screen at depth d. When is a correction worse than none? · onsite

The straight ray from \(x_u\) to \((x_p, z)\) crosses the screen at \(x_u(1-d/z) + x_p d/z\); moving the point by \(\Delta x\) shifts the footprint by \(\delta = \Delta x\,d/z\). The borrowed law leaves a residual of variance \(2\sigma_\phi^{2}(1-\rho(\delta))\): Strehl \(e^{-2\sigma_\phi^{2}(1-\rho)}\), worse than uncorrected when \(\rho < 1/2\), at \(\Delta x_\times = \ell z\sqrt{\ln 2}/d\) = 13.3 mm for \(\ell\) 4, z 40, d 10 mm. Follow-up: finite aperture? The crossover moves in (11.0 mm) because the uncorrected Strehl is higher than \(e^{-\sigma_\phi^{2}}\).

I5You have a CUTE sound-speed map from Module 32. Turn it into an aberration correction and write its error budget. · onsite

Integrate slowness from every element to every pixel (straight rays or eikonal) and beamform with those times. Errors: map resolution (a low-pass truth: 3 mm blur cost 8.2 against 5.2 ns rms in m33_sos_map_correction.m), map bias (a 10 m/s error in the fat speed: 7.3 ns, mostly defocus), path model (straight rays always overestimate times: 3–11 ns behind a wavy fat layer), and the mean speed CUTE cannot see. Follow-up: how would you check it in vivo? Measure the residual law with a matrix method; a large smooth residual points to the mean or the path model.

I6What is the focused reflection matrix, and how does it measure focusing without a point target? · onsite

\(\mathbf{R}_{xx} = \mathbf{G}_0^{H}\mathbf{K}\mathbf{G}_0^{*}\): responses between virtual transducers at \(x_{\mathrm{in}}\) and \(x_{\mathrm{out}}\) after focusing in software. The diagonal is the confocal image; in speckle the mean intensity versus offset is the cross-correlation of transmit and receive PSF intensities, so its width (Lambert et al.'s \(F = w_0/w\)) or its confocal energy fraction measures the local focus anywhere. Follow-up: what spoils it? Multiple scattering adds a broad background (useful in itself) and noise adds a floor.

I7Walk through distortion-matrix correction. Why re-centre, why SVD, why iterate, and how big should the window be? · research

Focus on transmit, keep element receive, multiply each column by the conjugate geometric wavefront of its focal point: every point of a patch now looks like one virtual source at the origin through the same screen. \(\mathbf{D}\mathbf{D}^{H}\) averages the speckle over the columns; its first singular vector is screen × virtual-reflector eigenvector. Correcting the input shrinks the reflector, so iterate (0.83 → 0.018 rad in three steps in the script). Window: many resolution cells for averaging, smaller than a patch; Lambert et al. shrank it from 12 × 16 to 6.2 × 8.1 mm. Follow-up: transmit vs receive laws with plane-wave transmits? Estimate both, alternating sides.

I8Your speckle-based estimator converges, but the image is shifted relative to CT. What happened? · research

A residual linear phase (tilt) is unobservable from speckle: it moves both foci together, so focus metrics (Γ, F, CF) look perfect while every point is displaced, a bias field as in Module 31's transcranial ULM. Remedies: anchor the tilt with known targets, a sound-speed map, or bubbles as guide stars with known geometry; in matrix methods watch for linear ramps removed during the estimation. Follow-up: how big? 10 ns across 10 mm at 50 mm depth moves a point by 77 µm (Module 31).

I9What breaks when you move from abdominal wall to transcranial imaging, and what do you do instead? · research

Bone is about twice as fast, attenuating and variable in thickness, so amplitude matters as much as phase, the aberrator is thick (finite patches in depth) and \(d/z\) is large for shallow targets (small patches). Options: CT-based priors (map route), guide stars (Demené et al. 2021 derived one law per patch from microbubbles before ULM), 3D matrix imaging with 2D arrays (Bureau et al. 2023). Follow-up: what is the cheapest first step? Lower frequency: the same delays cost a quarter of the phase variance at half the frequency.

I10Which metrics would you report for a new aberration-correction method, and which would you avoid? · research

A Strehl-like or energy-fraction focus measure (Γ or the confocal energy), point-target FWHM where available, gCNR with region sizes (Module 26), and target positions for geometric fidelity; all before and after, at several depths and lateral positions to show the patch. Avoid CF or SLSC images as the main evidence (they reward noise suppression too) and contrast in dB on a display range chosen afterwards. Follow-up: why not FWHM alone? It saturates: 0.38 → 0.40 mm at 20 ns while the Strehl is already 0.73.

I11Propose a method that combines a sound-speed map and a matrix estimate. What does each contribute? · research

Use the map (Module 32) to compute space-variant starting delays, so the residual aberration is small and the patch large; then estimate the residual laws with local distortion matrices, which converge fast because the virtual reflector is already compact. Feed the residual delays back as tomographic data with the correct (bent-ray) forward model to update the map, and iterate. The map supplies locality and the tilt/mean anchor; the matrix supplies the measured truth where the map's resolution, bias or path model fail. Follow-up: what would you validate first? On a phantom with a known aberrator, compare map-only, matrix-only and combined Γ maps and target positions.