Piezoelectric transduction and equivalent circuits: Mason, KLM
Every echo you have ever beamformed passed twice through a piezoelectric plate. This module builds that plate from its constitutive equations up to a transfer-matrix model that predicts the electrical impedance, the pulse-echo spectrum and the impulse response of a real stack, so you can read a probe datasheet, an impedance-analyzer trace or a KLM paper and know where every number comes from.
You will be able to
- Write the constitutive equations in the form that suits a bar, a plate or an array element, and say what \(k_{33}\), \(k_t\), \(k_{31}\) and \(k^2\) mean as energy ratios
- Choose between PZT, single crystal, 1-3 composite and PVDF with numbers for Z, \(k_t\) and \(\varepsilon\)
- Extract \(k_t\) from \(f_r\) and \(f_a\) with the IEEE Std 176 relation and know when it fails
- Draw the Mason and KLM circuits, state every element value and prove they agree
- Chain ABCD matrices into impedance, transmit, receive and pulse-echo transfer functions, and predict how backing, matching layers, tuning and cables move bandwidth and insertion loss
Assumes
- Phasors (Module 01, Complex numbers & phasors) and transfer functions (Module 03, Convolution, LTI & filters)
- Pulse spectra, bandwidth and the RF line as pulse ∗ reflectivity (Module 04, Sampling, Fourier, LTI)
- Plane-wave reflection and transmission at an impedance step (Module 15, Wave physics & harmonics)
16.1The piezoelectric effect and the constitutive equations
A poled ferroelectric ceramic (PZT) or crystal (PMN-PT) carries a remanent polarization along its poling axis, labelled 3. Stress changes that polarization, and the change appears as charge on the electrodes: the direct effect, used on receive. A field changes the strain: the converse effect, used on transmit. For small signals both are linear and, by thermodynamics, governed by the same coefficient, which is why one device is both the speaker and the microphone and why its transmit and receive responses are linked by reciprocity.
Four field variables are in play: stress T (Pa), strain S, electric field E (V/m) and electric displacement D (C/m²). Pick any two as independent and you get one of four equivalent forms. Superscripts say what is held constant: E (short circuit), D (open circuit), T (free), S (clamped).
- \(\mathbf{S}\)
- strain, dimensionless
- \(\mathbf{T}\)
- stress, Pa
- \(\mathbf{E}\)
- electric field, V/m
- \(\mathbf{D}\)
- electric displacement, C/m²
- \(\mathbf{s}^{E}\)
- compliance at constant field (short circuit), m²/N
- \(\mathbf{d}\)
- piezoelectric strain constants, C/N = m/V; superscript \(t\) = transpose
- \(\varepsilon^{T}\)
- free (constant-stress) permittivity, F/m
- \(\mathbf{c}^{E}\)
- stiffness at constant field, Pa
- \(\mathbf{e}\)
- piezoelectric stress constants, C/m²
- \(\varepsilon^{S}\)
- clamped (constant-strain) permittivity, F/m
- \(\mathbf{S},\ \mathbf{D}\)
- strain and electric displacement, as in (16.1)
- \(\mathbf{e}^{t}\)
- transpose of \(\mathbf{e}\)
- \(c_{33}^{D}\)
- open-circuit (stiffened) thickness stiffness, Pa
- \(h_{33}\)
- piezoelectric constant, V/m
- \(\varepsilon_{33}^{S}\)
- clamped permittivity, F/m
- \(T_3,\ S_3,\ D_3,\ E_3\)
- thickness-direction (axis 3, poling axis) components of \(\mathbf{T},\mathbf{S},\mathbf{D},\mathbf{E}\) in (16.1), (16.2)
Why D is the right variable inside a plate. The electrodes are equipotential and there is no free charge inside the dielectric, so Gauss's law gives \(\partial D_3 /\partial z = 0\): \(D_3\) is uniform through the thickness while the strain varies with z. Two consequences carry the whole module. First, in the wave equation \(\partial T_3 /\partial z = \rho \partial^2 u/\partial t^2\), the uniform \(h_{33} D_{3}\) term drops out, so waves inside the plate travel at the stiffened speed \(v^D\) and the plate is simply an acoustic transmission line. The piezoelectric coupling acts only at the two faces, where the term \(h_{33} D_{3}\) jumps to zero outside the plate. Second, integrating \(E_3\) across the thickness t gives the terminal voltage:
- \(V,\ I\)
- terminal voltage (V) and current (A)
- \(t\)
- plate thickness, m
- \(A\)
- electrode area, m²
- \(u(z)\)
- particle displacement, m
- \(v_t,\ v_0\)
- velocities of the two faces, m/s
- \(\omega\)
- angular frequency, rad/s
- \(C_0\)
- clamped capacitance, F
- \(z\)
- coordinate through the thickness, m
- \(D_3,\ E_3\)
- thickness components of displacement (C/m²) and field (V/m)
- \(\varepsilon^{S},\ h_{33}\)
- clamped permittivity and \(h_{33}\), as in (16.3)
- \(j\)
- imaginary unit
Open versus short circuit: two stiffnesses, two capacitances
Exact formulaAn open-circuited piezoelectric is stiffer because straining it builds charge whose field pushes back. Shorting the electrodes lets the charge flow away and removes that extra stiffness. The difference between the two stiffnesses is the coupling: \(c^E = c^D (1 - k^2)\). Dually, a clamped plate cannot move, so it stores less charge per volt: \(\varepsilon^S = \varepsilon^T (1 - k^2)\) for that mode.
16.2Coupling coefficients and what k² means
A coupling coefficient is a ratio of energies. Drive the material quasi-statically through one cycle on one port, and \(k^2\) is the fraction of the stored energy that can be taken out on the other port. Equivalently it is the normalized cross term of the energy density, so it depends on the mode, that is, on which stresses and strains the geometry lets be zero.
- \(k_{33},\ k_{31},\ k_t\)
- bar, lateral-bar and thickness (plate) coupling factors, dimensionless
- \(d_{33},\ d_{31}\)
- piezoelectric strain constants, C/N
- \(s_{33}^{E},\ s_{11}^{E}\)
- short-circuit compliances, m²/N
- \(\varepsilon_{33}^{T},\ \varepsilon_{33}^{S}\)
- free and clamped permittivity, F/m
- \(e_{33}\)
- piezoelectric stress constant, C/m²
- \(c_{33}^{D}\)
- open-circuit stiffness, Pa
- \(h_{33}\)
- piezoelectric constant, V/m
| Mode | Geometry and boundary condition | Coupling | PZT-5H | Where it appears |
|---|---|---|---|---|
| \(k_{31}\) | Long thin bar poled across its thickness, vibrating along its length; only \(T_1 \ne 0\) | \(d_{31} /\sqrt{s_{11}^E \varepsilon_{33}^T }\) | 0.39 | Lateral (spurious) modes, bimorphs, PVDF films used in 31 mode |
| \(k_{33}\) | Long bar poled and vibrating along its length; \(T_1 = T_2 = 0\) | \(d_{33} /\sqrt{s_{33}^E \varepsilon_{33}^T }\) | 0.75 | Pillars of a 1-3 composite; upper bound for any 33 device |
| \(k_{33}^{\prime}\) | Array sliver: thin in azimuth \((T_1 = 0)\), long in elevation \((S_2 = 0)\) | from d, s, \(\varepsilon\) with \(S_2 = 0\) | 0.70 | Sub-diced or narrow array elements (width/thickness ≲ 0.6) |
| \(k_t\) | Thin plate, lateral size ≫ thickness; \(S_1 = S_2 = 0\) | \(e_{33} /\sqrt{c_{33}^D \varepsilon_{33}^S }\) | 0.51 | Single-element and wide-element transducers, KLM model |
| \(k_p\) | Thin disc, radial vibration | \(\lvert d_{31} \rvert \sqrt{2/((s_{11}^E +s_{12}^E)\varepsilon_{33}^T)}\) | 0.65 | Low-frequency radial resonance of a disc (\(s_{12}^E = -4.78\) pm²/N) |
The \(k_{33}^{\prime}\) value, 0.70, is computed from the same datasheet constants with \(T_1 = 0\) and \(S_2 = 0\). The ordering \(k_t\) < \(k_{33}^{\prime}\) < \(k_{33}\) is the geometric story of transducer design: the more the geometry clamps the lateral directions, the more the lateral (\(d_{31}\)) contraction fights the thickness motion and the lower the coupling. Two useful identities for a single-mode, one-dimensional description:
- \(C_{\mathrm{LF}}\)
- low-frequency (free) capacitance of the 1D plate, F
- \(k_{\mathrm{eff}}\)
- effective (lumped) coupling of one resonance, dimensionless
- \(f_r,\ f_a\)
- resonance and antiresonance frequencies, Hz
- \(c^{E},\ c^{D}\)
- stiffness at constant field and at constant displacement, Pa
- \(k\)
- coupling factor of the mode considered, dimensionless
- \(C_0,\ k_t\)
- as in (16.4) and (16.5)
Quoting \(k_{33} = 0.75\) (or 0.92 for PMN-PT) for an array element is the most common overstatement in transducer talk. A wide element works at \(k_t\), a well-diced sliver approaches \(k_{33}^{\prime}\), and only a pillar reaches \(k_{33}\). Also keep \(k_{\mathrm{eff}}\) (from \(f_r /f_a\) of any resonance) apart from the material constant \(k_t\).
16.3Materials: PZT, single crystals, 1-3 composites, PVDF
Three numbers decide most choices: the acoustic impedance \(Z = \rho v^D\) (how hard it is to get energy into 1.5 MRayl tissue), \(k_t\) (how much of the energy converts, hence bandwidth and sensitivity), and the clamped permittivity \(\varepsilon^S\) (which sets \(C_0\) and therefore the electrical impedance of a small element). The table gives representative ranges; check the datasheet of the actual lot before a design.
| Material | \(\rho\) (kg/m³) | \(v^D\) (m/s) | Z (MRayl) | \(k_t\) | \(k_{33}\) | \(\varepsilon^S /\varepsilon_0\) | \(\varepsilon^T /\varepsilon_0\) | \(d_{33}\) (pC/N) | Notes |
|---|---|---|---|---|---|---|---|---|---|
| PZT-5A (Navy II) | 7750 | ≈ 4350 | 33–34 | 0.47–0.49 | 0.70–0.71 | 800–920 | 1700–1900 | 370–400 | High Curie point (≈ 350 °C), stable; lower \(\varepsilon\) |
| PZT-5H (Navy VI) | 7450–7500 | ≈ 4550–4600 | 33–35 | 0.49–0.52 | 0.75 | 1200–1500 | 3200–3400 | 590–650 | Workhorse for imaging arrays; Curie ≈ 190–230 °C |
| PMN-PT [001] (28–33 % PT) | 8000–8100 | ≈ 4000–4600 | 32–37 | 0.58–0.63 | 0.88–0.94 | 700–900 | 5000–8000 | 1200–2800 | \(T_{\mathrm{rt}} \approx 80\text{–}100\) °C, \(E_c \approx 2.5\) kV/cm: limits on heat and drive |
| PIN-PMN-PT [001] | ≈ 8100 | – | ≈ 32–37 | 0.55–0.58 | 0.89–0.92 | 850–950 | 4500–4800 | > 1000 | \(T_{\mathrm{rt}} \approx 100\text{–}125\) °C, \(E_c \approx 4.5\text{–}6\) kV/cm: more robust |
| 1-3 PZT/epoxy, \(\nu = 0.3\text{–}0.7\) | 3000–5600 | 3500–4000 | 10–22 | 0.60–0.70 | n/a | ≈ \(\nu \cdot \varepsilon_{\mathrm{cer}}^S\) | – | close to ceramic | Lower Z, higher \(k_t\), suppressed lateral modes; lossier |
| 1-3 crystal/epoxy | – | – | ≈ 15–22 | 0.75–0.85 | n/a | ≈ \(\nu \cdot \varepsilon_{\mathrm{cr}}^S\) | – | ≈ 1000 | Widest-band, highest-sensitivity option; thermal limits of the crystal remain |
| PVDF film | 1780 | 2200–2250 | ≈ 4 | 0.10–0.20 | n/a | ≈ 8–12 | 8–12 | \(\lvert d_{33} \rvert \approx 20\text{–}35\) | Matches water, very low C and k; \(\tan \delta \approx 0.1\text{–}0.25\) at MHz; hydrophones |
Single-crystal values are for [001]-poled rhombohedral crystals (manufacturer data for PMN-28PT and PMN-32PT list \(k_t = 0.60\text{–}0.62\), \(\varepsilon^S /\varepsilon_0 = 700\text{–}895\), \(\varepsilon^T /\varepsilon_0 = 5500\text{–}7000\)); \(d_{33}\) varies strongly with composition and supplier, so read the table as ranges. Composite rows come from the Smith–Auld model below and published 1-3 devices.
Why 1-3 composites win. Thin ceramic pillars in a soft polymer matrix are laterally unclamped, so they work closer to the bar mode \((k_{33} \approx 0.75)\) than to the plate mode (0.51), while the polymer lowers the density and stiffness. Smith and Auld (1991) model a fine-scale composite as an effective homogeneous plate: uniform strain along the thickness, uniform lateral stress, and polymer and ceramic sharing the lateral strain. The result is closed-form expressions for \(\bar{c}_{33}^D\), \(\bar{e}_{33}\), \(\bar{\varepsilon}_{33}^S\) and \(\bar{\rho}\) versus volume fraction \(\nu\). The demo implements them for PZT-5H or PZT-5A in an epoxy.
- \(\nu\)
- ceramic volume fraction, 0 to 1
- \(\mathcal{D}\)
- lateral-stiffness denominator, Pa (written D in the original paper; not the electric displacement)
- \(c_{ij}^{p},\ \varepsilon_{11}^{p},\ \rho^{p}\)
- polymer stiffness (Pa), permittivity (F/m), density (kg/m³)
- \(c_{ij}^{E},\ e_{3j},\ \rho^{c}\)
- ceramic short-circuit stiffness (Pa), stress constants (C/m²), density (kg/m³)
- \(\bar{c}_{33}^{D} = \bar{c}_{33}^{E} + \bar{e}_{33}^2/\bar{\varepsilon}_{33}^{S}\)
- effective stiffened stiffness, Pa (overbar: effective composite property)
- \(\bar{c}_{33}^{E},\ \bar{e}_{33},\ \bar{\varepsilon}_{33}^{S},\ \bar{\rho},\ \bar{k}_t\)
- effective composite stiffness (Pa), stress constant (C/m²), clamped permittivity (F/m), density (kg/m³) and thickness coupling
- \(e_{31}\)
- lateral piezoelectric stress constant of the ceramic, C/m²
1-3 composite designer (Smith–Auld effective medium)
Approx. modelSmith–Auld effective medium: pillar period ≪ thickness and polymer shear wavelength; uniform strain along z, lateral stress shared.Imaging arrays commonly use either PZT-5H-class ceramic with two or three matching layers, or single crystal (often as a composite), the crystal mainly for wider bandwidth: harmonic imaging and higher penetration at the same resolution. CMUTs (Module 18) reach low impedance by a different route: a thin membrane radiating directly.
16.4Thickness-mode resonance and the impedance of a plate
Solve the 1D wave equation in a plate of thickness t with both faces free (zero stress) and a sinusoidal current, then integrate \(E_3\) for the voltage. The algebra (Kino 1987, ch. 1) gives the classic result:
- \(Z(f)\)
- electrical input impedance, Ω
- \(C_0\)
- clamped capacitance, F
- \(k_t\)
- thickness coupling factor
- \(x\)
- half the phase thickness of the plate, rad
- \(t\)
- plate thickness, m
- \(v^{D}\)
- stiffened longitudinal speed, m/s
- \(f_a\)
- antiresonance (half-wave) frequency, Hz
- \(\omega,\ f\)
- angular frequency (rad/s) and frequency (Hz)
- \(j\)
- imaginary unit
- \(\tan\)
- tangent (argument in rad)
- \(f_r\)
- resonance (series, \(f_s\)): minimum \(\lvert Z\rvert\), Hz
- \(f_a\)
- antiresonance (parallel, \(f_p\)): maximum \(\lvert Z\rvert\), Hz
- \(x_r\)
- normalized resonance frequency, rad
- \(k_t\)
- thickness coupling factor, as in (16.5)
- \(\tan,\ \cot\)
- tangent and cotangent (argument in rad)
Worked example. PZT-5H, \(v^D = 4600\) m/s, target \(f_a = 5\) MHz: \(t = 4600/(2 \times 5\times 10^{6}) = 0.460\) mm. With \(k_t = 0.51\) the root of \(x \cot x = 0.2601\) is \(x_r = 1.385\), so \(f_r /f_a = 0.882\) and \(f_r = 4.41\) MHz; the frequency constant \(f_r t = 2028\) Hz·m. A 0.3 × 5 mm element area (1.5 mm²) has \(C_0 = 1470 \varepsilon_0 \times 1.5\) mm²/0.46 mm = 42.4 pF, i.e. \(\lvert Z\rvert = 1/(\omega C_0) = 750\) Ω at 5 MHz.
Impedance explorer: find fr and fa, extract kt
Approx. model1D KLM/ABCD: thickness mode only, plane waves at normal incidence, no diffraction, lateral modes or electrode mass; losses as complex constants (\(Q_m\), \(\tan \delta\)). Loads are semi-infinite, so no back-echo.IEEE 176 assumes an unloaded, low-loss, laterally infinite plate. Measure \(k_t\) on the bare plate before bonding backing or matching layers, use \(f_s /f_p\) from conductance and resistance maxima when loss is significant (with \(Q_m = 10\) they still give \(k_t = 0.507\) where the \(\lvert Z\rvert\) extrema give 0.563; on a 5 MRayl backing 0.508 versus 0.571, m16_impedance_kt.m), and check that the plate is wide enough (lateral dimension ≳ 10–20 × thickness) that radial or width modes do not sit on top of the thickness resonance.
16.5The Mason and KLM equivalent circuits
Both circuits are three-port networks (one electrical port, two acoustic ports for the back and front faces) that reproduce the plate equations exactly in 1D. In the acoustic ports "voltage" is force F = pA and "current" is particle velocity u, so an acoustic impedance in MRayl becomes a mechanical impedance \(Z = \rho vA\) in kg/s. Mason (1942/1948) puts the coupling at the electrical side with a transformer and a negative capacitance; Krimholtz, Leedom and Matthaei (Electronics Letters, 1970) move it to the middle of the acoustic line.
- \(N\)
- transformer ratio, N/V
- \(Z_0\)
- mechanical impedance of the plate \(\rho v^{D} A\), kg/s
- \(k\)
- wavenumber \(\omega/v^{D}\), rad/m
- \(Z_T,\ Z_S\)
- T-network arms, kg/s
- \(\varepsilon^{S},\ A,\ t\)
- as in (16.4)
- \(h_{33}\)
- as in (16.3)
- \(C_0\)
- clamped capacitance, F
- \(j\)
- imaginary unit
- \(\tan,\ \csc\)
- tangent and cosecant (argument in rad)
- \(X_1\)
- series reactance, Ω
- \(\varphi\)
- transformer ratio, V/N (\(V = \varphi F\), \(I = u/\varphi\))
- \(\omega_0\)
- angular antiresonance frequency, rad/s
- \(Z_0\)
- plate mechanical impedance \(\rho v^{D} A\), kg/s
- \(\operatorname{sinc}(x)\)
- \(\sin(\pi x)/(\pi x)\)
- \(C_0,\ k_t,\ f_a\)
- as in (16.4), (16.5) and (16.8)
- \(f\)
- frequency, Hz
- \(\omega\)
- angular frequency, rad/s
- \(j\)
- imaginary unit
The schematic shows the form the ABCD code implements: the transformer secondary connects the centre node of the line to the common return, with \(\varphi\) in V/N as above. Published drawings differ in how they draw this connection and in whether the ratio is written 1 : \(\varphi\) or \(\varphi\) : 1, so read element values together with the drawing they come from; the test is always that the free plate reproduces the closed form of 16.4.
Proof of agreement in one line. For a free plate each half line is shorted at its far end, so the centre node sees two impedances \(jZ_0 \tan (kt/2)\) in parallel. The KLM electrical branch then gives \(X_1 + \varphi^2 (Z_0/2)\tan (kt/2)\) \(= (2h^2/\omega^2 Z_0)\,[\sin (kt/2)\cos (kt/2)\) \(+ \sin^{2}(kt/2)\tan (kt/2)]\) \(= (2h^2/\omega^2 Z_0)\tan (kt/2)\). Using \(h^2 /Z_0 = k_t^2 v^D /(C_0 t)\) this is \((k_t^2 /\omega C_0)\tan (x)/x\), which is the free-plate formula of 16.4. The code on this page goes further: it solves the Mason circuit by nodal analysis and the KLM circuit by ABCD matrices, independently, for loaded and lossy plates; the two agree to 10⁻¹² (cards below, and m16_klm_abcd.m).
In KLM the acoustic path is an unbroken transmission line, so backing, bond lines, matching layers, a lens and the load just hang off its ends as more line sections. The electromechanical physics sits in two frequency-dependent elements (\(X_1\), \(\varphi\)) at one node. The sinc in \(\varphi\) also tells you at a glance which harmonics are excited: \(\varphi = 0\) at \(f = 2f_a\), \(4f_a\), … because a uniform field across the thickness cannot drive a mode whose strain integrates to zero.
16.6Transfer matrices: impedance, transmit, receive and pulse-echo
Every element of the stack is a two-port with an ABCD matrix that maps (voltage, current) or (force, velocity) at its output to its input. Cascading is matrix multiplication, which is why the transducer, a matching layer, a cable and a tuning inductor can be modelled with one routine (the same algebra as Module 03, with 2 × 2 matrices instead of scalars).
- \(Z,\ Y\)
- impedance and admittance of the element (Ω or kg/s, and inverse)
- \(k\)
- wavenumber in the layer, rad/m
- \(\ell\)
- layer thickness or line length, m
- \(\varphi\)
- transformer ratio, as in (16.11)
- \(j\)
- imaginary unit
- \(\cos,\ \sin\)
- argument \(k\ell\) in rad
- \(\mathrm{S},\ \mathrm{Sh},\ \mathrm{T},\ \mathrm{L}\)
- series, shunt, transformer and line matrices of (16.12)
- \(Z_{\mathrm{back}}\)
- input impedance of the rear half line plus backing, kg/s
- \(Z_{m1},\ Z_{m2}\)
- matching-layer impedances, kg/s
- \(k_i\ell_i\)
- phase thickness of layer \(i\), rad
- \(\mathbf{M}\)
- overall chain (ABCD) matrix from the electrical port to the load port
- \(\omega,\ C_0,\ X_1,\ \varphi\)
- as in (16.4), (16.11)
- \(Z_0,\ k,\ t\)
- as in (16.10); \(kt/2\) is the phase of half the plate
- \(j\)
- imaginary unit
- \(\cdot\)
- matrix product, left to right from the electrical port
- \(A,\ B,\ C,\ D\)
- elements of the chain matrix \(\mathbf{M}\)
- \(Z_L\)
- load (tissue) mechanical impedance, kg/s
- \(R_s,\ R_r\)
- source and receiver resistance, Ω
- \(F_L,\ F_{\mathrm{inc}}\)
- force on the load and incident echo force, N
- \(V_s,\ V_{\mathrm{out}}\)
- source and received voltage, V
- \(H\)
- pulse-echo transfer function, dimensionless
- \(Z_{\mathrm{in}}\)
- impedance seen at the electrical port, Ω
- \(\Delta(R)\)
- common denominator of the transfer functions for terminating resistance \(R\), Ω
- \(R\)
- terminating resistance (\(R_s\) or \(R_r\)), Ω
KLM / ABCD transducer calculator
Approx. model1D KLM/ABCD: thickness mode only, plane waves at normal incidence, no diffraction, lateral modes or electrode mass; losses as complex constants (\(Q_m\), \(\tan \delta\)). Layers lossless \(\lambda /4\) at \(f_m\), unit planar reflector, ideal inductor; the stack strip is drawn to scale from the computed thicknesses.Four independent checks, run in the page and in m16_klm_abcd.m: (1) free plate, KLM-ABCD versus the closed form of 16.4: maximum relative difference below 10⁻¹² over 0.2–20 MHz; (2) KLM-ABCD versus the Mason circuit solved by nodal analysis, loaded (\(Z_B = 5\), \(Z_F = 1.5\) MRayl) and lossy (\(Q_m = 30\), \(\tan \delta = 0.02\)): 10⁻¹³; (3) energy: for a lossless chain (layers and inductor included) the real power entering the electrical port equals the power radiated into \(Z_L\) to 10⁻¹³ over 0.05–1.85 \(f_a\) (close to \(2f_a\), where \(\varphi \to 0\), the matrices become ill-conditioned); (4) analytic limits: the IEEE 176 inversion of the simulated \(\lvert Z\rvert\) returns the input \(k_t\) (0.5100 for 0.51), and the open-circuit transmit Q equals \((\pi /2)Z_p /(Z_B + Z_F)\) within 0.2–2 % for \(Z_B \le 5\) MRayl (16.15).
Reference design, numbers to remember. Pulse-echo centre 4.30 MHz, −6 dB fractional bandwidth 67.5 %, insertion loss −10.1 dB at 50 Ω with series tuning (−21.6 dB untuned), envelope widths 0.38 µs at −6 dB, 0.87 µs at −20 dB and 2.2 µs at −40 dB. The centre lands below \(f_a = 5\) MHz for three reasons you can isolate in the calculator: a low-impedance drive pulls the response toward \(f_r\) (4.41 MHz for the bare plate), the matching layers were designed at 4.5 MHz, and the series inductor forms an electrical resonance with \(C_0\).
16.7Backing and front load: Q, bandwidth and sensitivity
A half-wave plate is an acoustic resonator. Each face reflects with amplitude coefficient \(r = (Z_p - Z)/(Z_p + Z) \approx 1 - 2Z/Z_p\) for \(Z \ll Z_p\). One round trip lasts \(2t/v = 1/f_a\), one period, and loses the energy fraction \(1 - r_B^2 r_F^2 \approx 4(Z_B + Z_F)/Z_p\). Since \(Q = 2\pi \times\) stored/lost per cycle:
- \(Q_m\)
- mechanical quality factor of the loaded plate
- \(Z_p\)
- acoustic impedance of the plate, MRayl
- \(Z_B,\ Z_F\)
- backing and front-load impedance, MRayl
Energy that goes into the backing is absorbed there (a good backing is a lossy tungsten- or alumina-loaded epoxy thick enough that its echo never returns), so backing buys bandwidth by throwing away sensitivity. Matching layers instead transform the 1.5 MRayl load up toward \(Z_p\), so the front face sees a heavier load and the energy leaves forward. The table is the reference design without tuning (lossy PZT-5H, 50 Ω):
| \(Z_B\) (MRayl) | No matching: FBW | IL (dB) | −40 dB ring-down (µs) | Two layers: FBW | IL (dB) | −40 dB ring-down (µs) |
|---|---|---|---|---|---|---|
| 0 (air) | 7 % | −5.6 | > 8 | 65 % | −17.2 | 1.9 |
| 5 | 19 % | −22.7 | 3.0 | 68 % | −21.6 | 1.4 |
| 10 | 30 % | −31.2 | 1.8 | 68 % | −24.5 | 1.0 |
| 20 | 58 % | −41.0 | 0.9 | 70 % | −29.0 | 1.0 |
| 34.5 (= \(Z_p\)) | 118 % | −48.1 | 0.6 | 80 % | −33.9 | 1.0 |
Backing trade-off: bandwidth versus insertion loss
Approx. model1D KLM/ABCD: thickness mode only, plane waves at normal incidence, no diffraction, lateral modes or electrode mass; losses as complex constants (\(Q_m\), \(\tan \delta\)). Semi-infinite absorbing backing, no tuning, 50 Ω.16.8Electrical tuning, cables and source impedance
A small array element is a high-impedance capacitor (750 Ω at 5 MHz for the reference element) facing a 50 Ω pulser and preamplifier through a cable. The electrical port wastes power by mismatch exactly as an unmatched acoustic face does, and three standard fixes attack different parts of the mismatch:
- Series inductor \(L = 1/(\omega_m^2 C_0)\) cancels the series reactance and leaves the series resistance (346 Ω at 4.5 MHz for the reference element). Right when the source is low impedance.
- Shunt inductor, same formula with \(C_0\) (plus the cable capacitance if it sits at the far end), resonates the parallel capacitance. Right when the receiver is high impedance; across a 50 Ω source it changes the reference design by only 0.03 dB because a 50 Ω source already behaves like a voltage source to a 750 Ω element.
- Transformer (or an L-network) scales the remaining real part toward 50 Ω. A 1 : 2.5 turns ratio (\(n = 0.4\) system : element, impedance ratio 6.25) on top of the series inductor takes the reference design from −10.1 dB to −2.6 dB.
Cables: coaxial cable of characteristic impedance \(Z_c\) and velocity \(v_c\) has capacitance \(1/(Z_c v_c)\) per metre: 100 pF/m for 50 Ω at 2×10⁸ m/s, so a 2 m probe cable carries 200 pF, about five times \(C_0\). If the system impedance equals \(Z_c\), a lossless cable only adds delay: the source sees the same available power and the mismatch at the element is unchanged, so \(\lvert H\rvert\) does not move. If the receiver is high impedance, the cable capacitance divides the voltage: \(C_0 /(C_0 + C_{\mathrm{cable}}) = -15\) dB for 2 m in the lumped open-circuit limit, −9.6 dB in the full line model with a 1 kΩ receiver. That is the reason for preamplifiers in the probe handle or micro-beamforming ASICs at the element (Module 19).
- \(\omega_m\)
- angular tuning frequency, rad/s
- \(C\)
- capacitance to be tuned out (\(C_0\), plus cable if at the far end), F
- \(\ell\)
- cable length, m
- \(Z_c,\ v_c\)
- cable characteristic impedance (Ω) and wave speed (m/s)
- \(n\)
- turns ratio, system : element
- \(L_{\mathrm{tune}}\)
- tuning inductance, H
- \(C_{\mathrm{cable}}\)
- cable capacitance, F
- \(Z_{\mathrm{seen}},\ Z_{\mathrm{element}}\)
- impedance seen by the system and impedance of the element, Ω
Electrical tuning, transformer and cable
Approx. model1D KLM/ABCD: thickness mode only, plane waves at normal incidence, no diffraction, lateral modes or electrode mass; losses as complex constants (\(Q_m\), \(\tan \delta\)). Ideal transformer and inductor, lossless 50 Ω coax (100 pF/m, 2×10⁸ m/s).With \(R_s \ne R_r\) the quantity \(20 \log_{10} \lvert 2V_{\mathrm{out}} /V_s \rvert\) is a two-way voltage gain, not an insertion loss, and it can exceed 0 dB into a high-impedance receiver without violating energy conservation. Compare probes by insertion loss at a stated impedance (usually 50 Ω) or by loop sensitivity with a stated pulser and receiver.
16.9Visual summary
Fourteen pictures that carry the module, each computed from the formulas and the model on this page (PZT-5H, \(f_a = 5\) MHz, 1.5 mm², water load, 50 Ω unless stated).
16.10How experts think about it
A transducer engineer sees the plate as a half-wave acoustic line whose two faces are driven by the uniform charge, coupled to a capacitor \(C_0\) through a frequency-dependent transformer. Bandwidth comes from how hard the faces are loaded (backing, and the load transformed by matching layers) and from \(k_t\), sensitivity from how much of that load is useful (the front) and from how well \(C_0\) is matched to the electronics. Every number on a datasheet is a consequence of \(Z_p\), \(k_t\), \(\varepsilon^S\), thickness, area and those loads.
Hold a strain S open-circuited: stored energy \(\tfrac{1}{2}c^D S^2\). Short the electrodes at fixed S: the stress drops to \(c^E S\) and the energy \(\tfrac{1}{2}(c^D - c^E)S^2 = k^2 \cdot \tfrac{1}{2}c^D S^2\) leaves as charge (Demo 16.1). In a resonant, lossless, matched device the conversion can be complete at one frequency, so \(k^2\) is not an efficiency; it sets the \(f_r \text{–}f_a\) gap, \(f_a - f_r \approx (4/\pi^2)k_t^2 f_a\) for small \(k_t\) (exact: 11.8 % of \(f_a\) at \(k_t = 0.51\)), and with it the band over which the plate can be loaded and matched.
In a wide plate \(S_1 = S_2 = 0\), so the \(d_{31}\) lateral contraction that accompanies thickness expansion is blocked and stores energy that is not converted. A sliver (width/thickness ≲ 0.6) frees one lateral direction \((T_1 = 0)\) and keeps \(S_2 = 0\), giving \(k_{33}^{\prime} = d^{\prime} /\sqrt{s^{\prime} \varepsilon^{\prime} } = 0.70\) from datasheet constants; a pillar frees both \((k_{33} = 0.75)\). 1-3 composites exploit exactly this.
Inside the plate D is uniform, so waves run at \(v^D = \sqrt{c^D /\rho }\), and with free faces the displacement has antinodes at both faces: the fundamental is \(t = \lambda /2\). Under short-circuit conditions the effective stiffness is softer, so the electrical resonance \(f_r\) sits below \(f_a\) (4.41 MHz for \(k_t = 0.51\)). The frequency constant \(N_t = f_r t\), about 2000 Hz·m for PZT-5H, is what datasheets list.
Set the free-plate impedance to zero: \(k_t^2 \tan (x)/x = 1\) at \(x = (\pi /2)f_r /f_a\), i.e. \(k_t^2 = x \cot x\), which is the IEEE Std 176 form. The small-gap version \(k_t^2 \approx (\pi^2 /4)(f_a - f_r)/f_a\) is 6 % high at \(k_t = 0.5\) and 12 % high at 0.7. The lumped \(k_{\mathrm{eff}}^2 = 1 - (f_r /f_a)^{2}\) gives 0.47 for the same plate: a different quantity.
\(C_0\) uses the clamped permittivity because above the thickness resonance the plate cannot follow the field. Impedance scales as \(t/(\varepsilon^S A \omega)\): halving the element width doubles it, and raising frequency at fixed area lowers it as \(1/f^2\) (\(t \propto 1/f\) and \(\omega \propto f\)). That is why small, high-frequency 2D-array elements (kΩ range) need electronics at the element, while large single elements need a step-up.
Both reproduce \(V = I/(j\omega C_0) - (h/j\omega)(v_t - v_0)\) coupled to an acoustic line. Mason puts a T-network for the line and a transformer plus \(-C_0\) on the electrical side; KLM splits the line at its centre and couples there through \(\varphi\) with a series reactance \(X_1\). Solved independently here (nodal analysis versus ABCD) they agree to round-off for loaded, lossy plates.
The uniform field drives the plate through the face terms only; a mode whose two faces move in the same direction (even harmonic) gets no net drive, so \(\varphi = (2h/\omega Z_0)\sin (kt/2)\) is zero there. The odd harmonics remain with transformer power factor \(\varphi^2 \propto \operatorname{sinc} ^{2}(f/2f_a)\): sinc²(1.5)/sinc²(0.5) = 1/9 of the fundamental value at \(3f_a\), because |sin(kt/2)| = 1 at both and \(\varphi \propto 1/\omega\). This third-harmonic passband is what some dual-frequency designs and superharmonic contrast imaging exploit, and what a receive filter removes.
Each round trip (one period at \(f_a\)) loses \(1 - r_B^2 r_F^2 \approx 4(Z_B + Z_F)/Z_p\) of the energy, so \(Q = 2\pi /[4(Z_B + Z_F)/Z_p ]\). The KLM model with an open-circuit electrical port gives \(Q = 36.06\) (formula 36.13) on air and 8.19 (8.34) with 5 MRayl, the dots in the figure. Under voltage drive the peak moves to \(f_r\) and Q falls by ≈ 20 %.
The fraction of acoustic power leaving forward is ≈ \(Z_F /(Z_B + Z_F)\) per face, and the receive path pays it again, so the two-way efficiency of an unmatched plate falls roughly as \([Z_F /(Z_B + Z_F)]^{2}\) while Q falls as \(1/(Z_B + Z_F)\). Reference plate into water: bandwidth 7 %, 19 %, 30 %, 58 % at \(Z_B = 0\), 5, 10, 20 MRayl, insertion loss −5.6, −22.7, −31.2, −41.0 dB (Demo 16.5).
A \(\lambda /4\) layer of impedance \(Z_m\) presents \(Z_m^2 /Z_L\) to the plate at its design frequency, so the front face sees 12–35 MRayl (22 for the two-layer set) instead of 1.5 and the forward load dominates the damping. With two Desilets-type layers (9.0 and 2.35 MRayl) the untuned reference plate reaches 65 % bandwidth at −17 dB on an air backing, whereas the unmatched plate needs \(Z_B \approx 20\) MRayl and −41 dB for similar bandwidth. Layer design is Module 17.
With the electrical port shorted the plate resonates where its input impedance is zero (\(f_r\)); with it open, where the stiffened line is a half wave (\(f_a\)). A 50 Ω system on a 750 Ω element is close to the short-circuit case on both transmit and receive, so the pulse-echo response centres near \(f_r\); a high-impedance preamplifier pushes the receive response toward \(f_a\). The figure shows the transmitted force per volt of a water-loaded, air-backed plate for \(R_s = 0.01\) Ω and 10 MΩ.
The element's series model is \(R + 1/(j\omega C_0)\) with \(R = 346\) Ω at 4.5 MHz; a series inductor cancels the reactance and a transformer with impedance ratio 1/6.25 brings 346 Ω near 50 Ω. A shunt inductor across a 50 Ω source changes almost nothing (0.03 dB here) because the source already pins the element voltage; across a high-impedance preamp it removes the capacitive divider.
\(C^{\prime} = 1/(Z_c v_c) = 1/(50 \times 2\times 10^{8}) = 100\) pF/m. If \(R_s = R_r = Z_c\), the lossless line rotates the impedance seen at the system but leaves the available power and the reflection magnitude unchanged, so \(\lvert H\rvert\) is identical (checked: 0 to 3 m, same IL). Into a high-impedance input the element's charge shares with the cable capacitance: V scales by \(C_0 /(C_0 + C_{\mathrm{cable}})\).
Smith–Auld effective-medium model, PZT-5H pillars in a soft epoxy (\(v_L\) 2500, \(v_S\) 1100 m/s, \(\rho\) 1150). The pillars are laterally free, so their coupling approaches the bar value, while the polymer lowers \(\rho\) and c. \(k_t\) has a broad plateau of 0.67–0.69 for \(\nu = 0.3\text{–}0.7\); Z falls almost linearly with \(\nu\). Lower Z means easier matching and wider bandwidth; \(\varepsilon\) falls ∝ \(\nu\), raising the electrical impedance.
[001]-poled PMN-PT reaches \(k_{33} \approx 0.9\) and \(k_t \approx 0.6\), so a crystal stack gives more bandwidth and sensitivity at equal loading. Its rhombohedral-to-tetragonal transition at 80–100 °C caps processing and bonding temperatures, and its low coercive field limits drive under bipolar excitation. PIN-PMN-PT raises \(T_{\mathrm{rt}}\) to ≈ 100–125 °C and \(E_c\) to 4.5–6 kV/cm at slightly lower coupling.
The −6 dB width is set by the bandwidth (≈ 0.38 µs for 67 % at 4.3 MHz, about 1.6 cycles); the tail is set by the sharpness of the band edges and by residual resonances (series tuning, light backing). A long −40 dB tail becomes clutter behind bright reflectors and limits contrast in harmonic and CEUS modes where 40 dB matters. The figure is the reference design's impulse response after a 2–2.5 \(f_a\) low-pass.
16.11Cheat sheet
| Concept | Formula | Ultrasound use |
|---|---|---|
| Thickness-mode equations | \(T = c^D S - hD\), \(E = -hS + D/\varepsilon^S\), \(h = e/\varepsilon^S\) | Basis of Mason and KLM |
| Thickness coupling | \(k_t^2 = e_{33}^2 /(c_{33}^D \varepsilon_{33}^S)\) | Bandwidth and sensitivity potential of a plate |
| Bar coupling | \(k_{33}^2 = d_{33}^2 /(s_{33}^E \varepsilon_{33}^T)\) | Composite pillars, upper bound |
| Stiffening | \(c^E = c^D (1 - k^2)\), \(v^D /v^E = 1/\sqrt{1 - k_t^2 }\) | Waves inside the plate run at \(v^D\) |
| Clamped capacitance | \(C_0 = \varepsilon^S A/t\) | Electrical impedance, tuning, cable loading |
| Antiresonance | \(f_a = v^D /(2t)\) | Plate thickness for a target frequency |
| Free-plate impedance | \(Z = (1/j\omega C_0)[1 - k_t^2 \tan x/x]\), \(x = (\pi /2)f/f_a\) | Impedance-analyzer check of a bare plate |
| IEEE Std 176 | \(k_t^2 = (\pi /2)(f_r /f_a)\tan [(\pi /2)(f_a - f_r)/f_a ]\) | Measuring \(k_t\) |
| KLM elements | \(X_1 = (k_t^2 /\omega C_0)\operatorname{sinc} (f/f_a)\), \(\varphi = k_t \sqrt{\pi /\omega_0 C_{0}Z_{0}} \operatorname{sinc} (f/2f_a)\) | Stack modelling with layers as lines |
| Mason elements | \(N = h_{33} C_{0}\), \(Z_T = jZ_0 \tan (kt/2)\), \(Z_S = -jZ_0 \csc (kt)\) | Same physics, lumped T-network |
| Line ABCD | \(\begin{bmatrix} \cos k\ell & jZ\sin k\ell \\ j\sin k\ell/Z & \cos k\ell \end{bmatrix}\) | Layers, backing, cables |
| Pulse-echo | \(H = 2RZ_L /[AZ_L + B + R(CZ_L + D)]^{2}\), \(\mathrm{IL} = 20 \log \lvert 2H\rvert\) | Bandwidth, insertion loss, impulse response |
| Mechanical Q | \(Q \approx (\pi /2)Z_p /(Z_B + Z_F)\) | Backing choice, ring-down |
| Tuning | \(L = 1/(\omega^2 C_0)\); \(Z_{\mathrm{seen}} = n^2 Z\) | Matching an element to 50 Ω |
| Cable | \(C^{\prime} = 1/(Z_c v_c) \approx 100\) pF/m | Receive loss into high-Z preamps |
| Composite (Smith–Auld) | \(\bar{k}_t^2 = \bar{e}_{33}^2 /(\bar{c}_{33}^D \bar{\varepsilon}_{33}^S)\), \(\bar{\rho} = \nu \rho_c + (1 - \nu)\rho_p\) | Choosing volume fraction |
16.12Interview and qualifying-exam questions
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.
16.13MATLAB lab
Base MATLAB only (no toolboxes); each script runs as-is in MATLAB R2019b+ and GNU Octave 8. Paste into the editor and press Run. All four share the same KLM/ABCD conventions as the page, written with the matrices stored as \(N_f \times 4\) arrays [A B C D] so the frequency loop is vectorized.
m16_klm_abcd.mKLM plate + backing + two matching layers + tuning: impedance, pulse-echo spectrum, impulse response, with the four validation checks · change materials, \(Z_B\), layers, R · runtime ≈ 2 s
%% m16_klm_abcd.m KLM transducer model with ABCD matrices: impedance, pulse-echo, impulse response
% Purpose : build the KLM model of a thickness-mode plate (Krimholtz, Leedom & Matthaei 1970) as a chain
% of 2x2 ABCD matrices, add backing, two quarter-wave matching layers, a series tuning inductor
% and a 50-ohm source/receiver, and compute the electrical input impedance, the pulse-echo
% transfer function H = 2 R ZL / Delta^2 and the impulse response.
% Four validation checks run first: free plate vs closed form, KLM vs Mason (independent nodal
% solution), energy balance of a lossless chain, IEEE Std 176 kt recovery and the Q formula.
% Change : material constants, fa, area A, backing ZB, layer set, tuning, R.
% Expected: validation errors ~1e-12 or smaller; reference design fc ~ 4.30 MHz, FBW ~ 67.5 %,
% IL ~ -10.1 dB, envelope widths ~ 0.38 / 0.87 / 2.2 us at -6 / -20 / -40 dB.
% Conventions: acoustic "voltage" = force F (N), "current" = particle velocity u (m/s);
% matrices are stored as Nf x 4 arrays [A B C D], one row per frequency.
clear; close all; rng(1);
%% ABCD building blocks (anonymous functions only: runs in MATLAB and Octave)
e0 = 8.854187817e-12; % vacuum permittivity (F/m)
mm = @(M,N) [M(:,1).*N(:,1)+M(:,2).*N(:,3), M(:,1).*N(:,2)+M(:,2).*N(:,4), ...
M(:,3).*N(:,1)+M(:,4).*N(:,3), M(:,3).*N(:,2)+M(:,4).*N(:,4)];
ser = @(Z) [ones(size(Z)), Z, zeros(size(Z)), ones(size(Z))]; % series impedance
shu = @(Y) [ones(size(Y)), zeros(size(Y)), Y, ones(size(Y))]; % shunt admittance
xfm = @(p) [p, zeros(size(p)), zeros(size(p)), 1./p]; % ideal transformer V = p F, I = u/p
tl = @(Z,th) [cos(th), 1i*Z.*sin(th), 1i*sin(th)./Z, cos(th)]; % transmission line
zlin = @(Z,th,ZT) Z.*(ZT.*cos(th)+1i*Z.*sin(th))./(Z.*cos(th)+1i*ZT.*sin(th)); % line input impedance
zin = @(M,ZL) (M(:,1).*ZL+M(:,2))./(M(:,3).*ZL+M(:,4));
%% Piezo plate: KLM elements from (rho, vD, kt, epsS_r, Qm, tan delta, thickness d, area A)
kof = @(f,P) 2*pi*f ./ (P.v*sqrt(1+1i/P.Qm)); % complex wavenumber (1/m)
Z0of = @(P) P.rho*P.v*sqrt(1+1i/P.Qm)*P.A; % plate mechanical impedance (kg/s)
C0of = @(P) P.er*e0*(1-1i*P.tand)*P.A/P.d; % clamped capacitance (F)
hof = @(P) P.kt*sqrt(P.rho*P.v^2*P.er*e0)/(P.er*e0*(1-1i*P.tand)); % h33 = e33/epsS (V/m)
X1of = @(f,P) hof(P)^2*sin(kof(f,P)*P.d)./((2*pi*f).^2*Z0of(P)); % KLM series reactance (ohm)
phiof = @(f,P) 2*hof(P)*sin(kof(f,P)*P.d/2)./(2*pi*f*Z0of(P)); % KLM transformer (V/N)
klm = @(f,P,ZB) mm(mm(mm(mm(ser(1./(1i*2*pi*f*C0of(P))), ser(1i*X1of(f,P))), xfm(phiof(f,P))), ...
shu(1./zlin(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2, ZB*ones(size(f))))), tl(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2));
fa = 5e6; % plate antiresonance vD/(2t) (Hz)
P = struct('rho',7500,'v',4600,'kt',0.51,'er',1470,'Qm',65,'tand',0.02,'d',4600/(2*fa),'A',1.5e-6); % PZT-5H, 0.3 x 5 mm
PL = P; PL.Qm = Inf; PL.tand = 0; % lossless copy for the analytic checks
Zp = P.rho*P.v/1e6; % plate impedance (MRayl)
fprintf('PZT-5H plate: t = %.3f mm, Zp = %.1f MRayl, C0 = %.1f pF\n', P.d*1e3, Zp, real(C0of(P))*1e12);
%% Validation 1: free lossless plate vs closed form Z = (1/(j w C0)) (1 - kt^2 tan(x)/x)
f = (0.2e6:13.7e3:20e6)';
Zk = zin(klm(f,PL,0), 0);
x = 2*pi*f*PL.d/(2*PL.v);
Za = (1 - PL.kt^2*tan(x)./x) ./ (1i*2*pi*f*real(C0of(PL)));
fprintf('V1 free plate, KLM vs closed form : max rel diff = %.1e\n', max(abs(Zk-Za)./abs(Za)));
%% Validation 2: KLM vs Mason (nodal analysis of the T-network + transformer N = h C0 and -C0)
Pm = P; Pm.Qm = 30; % lossy, loaded plate
ZB = 5e6*P.A; ZF = 1.5e6*P.A; % backing and front loads (kg/s)
Zk = zin(klm(f,Pm,ZB), ZF);
k = kof(f,Pm); Z0 = Z0of(Pm); C0 = C0of(Pm); N = hof(Pm)*C0;
ZT = 1i*Z0*tan(k*Pm.d/2); ZS = Z0./(1i*sin(k*Pm.d)); % Mason arms
Y = 1./(ZT+ZB) + 1./(ZT+ZF); G = Y./(1+ZS.*Y); % velocity into the T-network per transformer force
jwC = 1i*2*pi*f*C0;
Ip = N^2*G./(1 - N^2*G./jwC); % primary current per volt
Zm = 1./(jwC + Ip);
fprintf('V2 loaded lossy plate, KLM vs Mason : max rel diff = %.1e\n', max(abs(Zk-Zm)./abs(Zm)));
%% Reference stack: backing, two Desilets-type layers at fm = 0.9 fa, series L, R = 50 ohm, water load
ZBm = 5; ZLm = 1.5; R = 50; fm = 0.9*fa;
Zl = [Zp^(4/7)*ZLm^(3/7), Zp^(1/7)*ZLm^(6/7)]; % layer impedances (MRayl): 9.00, 2.35
vl = [2800, 2400]; tlay = vl/(4*fm); % layer speeds (m/s), quarter-wave thicknesses (m)
Lser = 1/((2*pi*fm)^2*real(C0of(P))); % series tuning inductor (H)
stack = @(f,Pp,ZBk,Lt) mm(ser(1i*2*pi*f*Lt), mm(mm(klm(f,Pp,ZBk*1e6*Pp.A), tl(Zl(1)*1e6*Pp.A*ones(size(f)), 2*pi*f/vl(1)*tlay(1))), ...
tl(Zl(2)*1e6*Pp.A*ones(size(f)), 2*pi*f/vl(2)*tlay(2))));
fprintf('Layers: Z = %.2f, %.2f MRayl, t = %.3f, %.3f mm; series L = %.1f uH\n', Zl, tlay*1e3, Lser*1e6);
%% Validation 3: energy balance of the lossless chain: Re{V I*} = Re{ZL} |u|^2 with u = 1
fe = fa*(0.05 + 1.8*(0:159)'/160) + 917;
M = stack(fe, PL, 0, Lser); ZL = ZLm*1e6*P.A;
V = M(:,1)*ZL + M(:,2); I = M(:,3)*ZL + M(:,4);
fprintf('V3 energy balance, lossless chain : max rel diff = %.1e\n', max(abs(real(V.*conj(I)) - real(ZL))/real(ZL)));
%% Validation 4: IEEE Std 176 on the simulated free plate, and the open-circuit Q formula
imZ = @(ff) imag(zin(klm(ff,PL,0),0)); imY = @(ff) imag(1./zin(klm(ff,PL,0),0));
a = 0.3*fa; b = 0.999*fa; for it = 1:60, c = (a+b)/2; if imZ(a)*imZ(c) <= 0, b = c; else a = c; end, end, fr = (a+b)/2;
a = 1.001*fr; b = 1.5*fa; for it = 1:60, c = (a+b)/2; if imY(a)*imY(c) <= 0, b = c; else a = c; end, end, fan = (a+b)/2;
kt176 = sqrt(pi/2*fr/fan*tan(pi/2*(fan-fr)/fan));
fprintf('V4 fr = %.4f MHz, fa = %.4f MHz, IEEE 176 kt = %.4f (input %.2f)\n', fr/1e6, fan/1e6, kt176, PL.kt);
fq = (3e6:2.5e3:7e6)';
for ZBq = [0 5]
Mq = klm(fq, PL, ZBq*1e6*P.A); Dq = Mq(:,1)*ZL + Mq(:,2) + 1e7*(Mq(:,3)*ZL + Mq(:,4));
Ft = abs(ZL./Dq); [pk, ip] = max(Ft); lev = pk/sqrt(2);
i1 = find(Ft(1:ip) < lev, 1, 'last'); i2 = ip - 1 + find(Ft(ip:end) < lev, 1, 'first');
fl = fq(i1) + (lev - Ft(i1))*(fq(i1+1) - fq(i1))/(Ft(i1+1) - Ft(i1)); % interpolated -3 dB crossings
fu = fq(i2-1) + (lev - Ft(i2-1))*(fq(i2) - fq(i2-1))/(Ft(i2) - Ft(i2-1));
Qs = fq(ip)/(fu - fl);
fprintf(' open-circuit Q, ZB = %g MRayl: simulated %.2f, (pi/2) Zp/(ZB+ZF) = %.2f\n', ZBq, Qs, pi/2*Zp/(ZBq+ZLm));
end
%% Reference design: spectrum on a grid scaled to fa, -6 dB band, impulse response
Nfft = 1024; fs = 25.6*fa; df = fs/Nfft; fg = (0:Nfft/2)'*df; fg(1) = 1e-3*df;
M = stack(fg, P, ZBm, Lser);
Dn = M(:,1)*ZL + M(:,2) + R*(M(:,3)*ZL + M(:,4));
H = 2*R*ZL./Dn.^2; % pulse-echo V_out/V_source, unit planar reflector
Zi = zin(M, ZL);
lim = round(2*fa/df); mag = abs(H(2:lim)); fb = fg(2:lim);
[pk, ip] = max(mag); lev = pk/2;
i1 = find(mag(1:ip) <= lev, 1, 'last'); i2 = ip - 1 + find(mag(ip:end) <= lev, 1, 'first');
fl = fb(i1) + (lev-mag(i1))/(mag(i1+1)-mag(i1))*(fb(i1+1)-fb(i1));
fu = fb(i2-1) + (lev-mag(i2-1))/(mag(i2)-mag(i2-1))*(fb(i2)-fb(i2-1));
fc = (fl+fu)/2; fbw = (fu-fl)/fc; IL = 20*log10(2*pk);
wlp = ones(size(fg)); band2 = fg > 2*fa & fg < 2.5*fa; wlp(band2) = 0.5*(1+cos(pi*(fg(band2)-2*fa)/(0.5*fa))); wlp(fg >= 2.5*fa) = 0;
Hs = zeros(Nfft,1); Hs(2:Nfft/2) = 2*H(2:Nfft/2).*wlp(2:Nfft/2); % analytic spectrum, low-passed 2-2.5 fa
asig = ifft(Hs); env = abs(asig); ir = real(asig); [epk, ie] = max(env);
dur = zeros(1,3); lv = [0.5 0.1 0.01];
for q = 1:3
a = ie; while a > 1 && env(a) > epk*lv(q), a = a-1; end
b = ie; while b < Nfft && env(b) > epk*lv(q), b = b+1; end
dur(q) = (b-a)/fs;
end
fprintf('Reference design: fc = %.2f MHz, -6 dB FBW = %.1f %%, IL = %.1f dB at %.2f MHz\n', fc/1e6, 100*fbw, IL, fb(ip)/1e6);
fprintf(' envelope widths -6/-20/-40 dB = %.2f / %.2f / %.2f us\n', dur*1e6);
ifc = round(fc/df)+1;
fprintf(' Zin(fc) = %.0f %+.0fj ohm (untuned element alone at fm: ', real(Zi(ifc)), imag(Zi(ifc)));
Mu = stack(fm, P, ZBm, 0); zu = zin(Mu, ZL); fprintf('%.0f %+.0fj ohm)\n', real(zu), imag(zu));
%% Figures
figure('Name','KLM: impedance');
sel = fg <= 2.2*fa & fg > 0.05*fa;
subplot(2,1,1); semilogy(fg(sel)/1e6, abs(Zi(sel)), 'b', 'LineWidth', 1.5); grid on;
xlabel('frequency (MHz)'); ylabel('|Z_{in}| (\Omega)'); title('Electrical input impedance, reference stack (series L)');
subplot(2,1,2); plot(fg(sel)/1e6, angle(Zi(sel))*180/pi, 'r', 'LineWidth', 1.2); grid on;
xlabel('frequency (MHz)'); ylabel('phase (deg)');
figure('Name','KLM: pulse-echo');
sel = fg <= 3.5*fa & fg > 0;
subplot(2,1,1); plot(fg(sel)/1e6, 20*log10(2*abs(H(sel))), 'b', 'LineWidth', 1.5); grid on; hold on;
plot([fl fl]/1e6, [-80 0], 'k--', [fu fu]/1e6, [-80 0], 'k--'); ylim([IL-60, IL+5]);
xlabel('frequency (MHz)'); ylabel('20 log_{10}|2H| (dB)'); title(sprintf('Pulse-echo: f_c = %.2f MHz, FBW = %.1f %%, IL = %.1f dB', fc/1e6, 100*fbw, IL));
idx = mod((ie-60:ie+300)-1, Nfft) + 1; tt = ((-60:300)/fs)*1e6;
subplot(2,1,2); plot(tt, ir(idx)/epk, 'b', tt, env(idx)/epk, 'r', 'LineWidth', 1.2); grid on;
xlabel('time re envelope peak (\mus)'); ylabel('normalized'); title('Pulse-echo impulse response (low-passed at 2-2.5 f_a)');
What to look for: validation errors of order 10⁻¹² to 10⁻¹³; reference design \(f_c \approx 4.30\) MHz, FBW ≈ 67.5 %, IL ≈ −10.1 dB, −6/−20/−40 dB envelope widths ≈ 0.38/0.87/2.2 µs.
m16_impedance_kt.mFree-plate impedance, \(f_r /f_a\) extraction and IEEE 176 versus approximations; bias from loss and loading · change \(k_t\), \(Q_m\), loads · runtime ≈ 2 s
%% m16_impedance_kt.m Plate impedance, fr/fa, and the IEEE Std 176 kt extraction
% Purpose : (1) tabulate fr/fa versus kt from the free-plate resonance condition kt^2 = x cot x,
% x = (pi/2) fr/fa, and compare the IEEE 176 formula, the small-gap approximation and the
% lumped keff; (2) "measure" fr and fa on simulated |Z| curves (KLM model) and show how loss
% and acoustic loading bias the extracted kt.
% Change : kt list, Qm, tan delta, front/back loads.
% Expected: IEEE 176 returns kt exactly on a lossless free plate (fr/fa = 0.9959, 0.9621, 0.8870,
% 0.7487 for kt = 0.1, 0.3, 0.5, 0.7); the small-gap form is ~6 % high at 0.5; loading biases it.
clear; close all; rng(1);
e0 = 8.854187817e-12;
mm = @(M,N) [M(:,1).*N(:,1)+M(:,2).*N(:,3), M(:,1).*N(:,2)+M(:,2).*N(:,4), ...
M(:,3).*N(:,1)+M(:,4).*N(:,3), M(:,3).*N(:,2)+M(:,4).*N(:,4)];
ser = @(Z) [ones(size(Z)), Z, zeros(size(Z)), ones(size(Z))];
shu = @(Y) [ones(size(Y)), zeros(size(Y)), Y, ones(size(Y))];
xfm = @(p) [p, zeros(size(p)), zeros(size(p)), 1./p];
tl = @(Z,th) [cos(th), 1i*Z.*sin(th), 1i*sin(th)./Z, cos(th)];
zlin = @(Z,th,ZT) Z.*(ZT.*cos(th)+1i*Z.*sin(th))./(Z.*cos(th)+1i*ZT.*sin(th));
zin = @(M,ZL) (M(:,1).*ZL+M(:,2))./(M(:,3).*ZL+M(:,4));
kof = @(f,P) 2*pi*f ./ (P.v*sqrt(1+1i/P.Qm));
Z0of = @(P) P.rho*P.v*sqrt(1+1i/P.Qm)*P.A;
C0of = @(P) P.er*e0*(1-1i*P.tand)*P.A/P.d;
hof = @(P) P.kt*sqrt(P.rho*P.v^2*P.er*e0)/(P.er*e0*(1-1i*P.tand));
X1of = @(f,P) hof(P)^2*sin(kof(f,P)*P.d)./((2*pi*f).^2*Z0of(P));
phiof = @(f,P) 2*hof(P)*sin(kof(f,P)*P.d/2)./(2*pi*f*Z0of(P));
klm = @(f,P,ZB) mm(mm(mm(mm(ser(1./(1i*2*pi*f*C0of(P))), ser(1i*X1of(f,P))), xfm(phiof(f,P))), ...
shu(1./zlin(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2, ZB*ones(size(f))))), tl(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2));
kt176 = @(fr,fa) sqrt(pi/2*fr./fa.*tan(pi/2*(fa-fr)./fa));
%% (1) fr/fa from the resonance condition, and three ways back to k
fprintf(' kt fr/fa IEEE176 small-gap keff\n');
ktl = [0.1 0.3 0.5 0.51 0.6 0.7];
rr = zeros(size(ktl));
for q = 1:numel(ktl)
a = 1e-6; b = pi/2 - 1e-12; % bisection on x cot x = kt^2
for it = 1:80, c = (a+b)/2; if c/tan(c) > ktl(q)^2, a = c; else b = c; end, end
rr(q) = 2*((a+b)/2)/pi;
fprintf(' %.2f %.4f %.4f %.4f %.4f\n', ktl(q), rr(q), kt176(rr(q),1), sqrt(pi^2/4*(1-rr(q))), sqrt(1-rr(q)^2));
end
%% (2) simulated impedance curves and extraction
fa = 5e6; base = struct('rho',7500,'v',4600,'kt',0.51,'er',1470,'Qm',Inf,'tand',0,'d',4600/(2*fa),'A',1.5e-6);
f = (2.5e6:1e3:6.5e6)';
cases = {'lossless, free', Inf, 0, 0, 0; 'Qm 65, tan d 0.02', 65, 0.02, 0, 0; 'Qm 10', 10, 0.02, 0, 0; ...
'Qm 65, front 1.5 MRayl', 65, 0.02, 0, 1.5; 'Qm 65, back 5 MRayl', 65, 0.02, 5, 0};
fprintf('\n case f_m (MHz) f_n (MHz) kt(|Z|) f_s f_p kt(G,R)\n');
Zc = zeros(numel(f), size(cases,1));
for q = 1:size(cases,1)
P = base; P.Qm = cases{q,2}; P.tand = cases{q,3};
Z = zin(klm(f, P, cases{q,4}*1e6*P.A), cases{q,5}*1e6*P.A); Zc(:,q) = Z;
[~, im] = min(abs(Z)); [~, iM] = max(abs(Z)); [~, iG] = max(real(1./Z)); [~, iR] = max(real(Z));
if isinf(P.Qm) && P.tand == 0 % lossless: Re Z = 0, G and R maxima undefined
fprintf(' %-26s %8.4f %8.4f %6.4f (no loss: G, R undefined)\n', cases{q,1}, f(im)/1e6, f(iM)/1e6, kt176(f(im),f(iM)));
else
fprintf(' %-26s %8.4f %8.4f %6.4f %7.4f %7.4f %6.4f\n', cases{q,1}, f(im)/1e6, f(iM)/1e6, kt176(f(im),f(iM)), f(iG)/1e6, f(iR)/1e6, kt176(f(iG),f(iR)));
end
end
%% Figures
figure('Name','impedance curves');
semilogy(f/1e6, abs(Zc(:,1)), 'k', f/1e6, abs(Zc(:,2)), 'b', f/1e6, abs(Zc(:,3)), 'g', f/1e6, abs(Zc(:,4)), 'r', 'LineWidth', 1.2); grid on;
xlabel('frequency (MHz)'); ylabel('|Z| (\Omega)'); title('PZT-5H plate, f_a = 5 MHz, 1.5 mm^2');
legend('lossless free', 'Q_m 65', 'Q_m 10', 'Q_m 65 + 1.5 MRayl front', 'Location', 'southwest');
figure('Name','k from fr/fa');
r = linspace(0.6, 0.9999, 400);
plot(r, kt176(r,1), 'b', r, sqrt(pi^2/4*(1-r)), 'r--', r, sqrt(1-r.^2), 'k:', 'LineWidth', 1.5); grid on;
xlabel('f_r / f_a'); ylabel('coupling'); legend('IEEE 176 k_t', 'small-gap (\pi^2/4)(1 - f_r/f_a)', 'k_{eff} = (1 - (f_r/f_a)^2)^{1/2}');
title('Reading k from the resonance gap');
What to look for: the \(k_t\) table reproduces 0.10, 0.30, 0.50, 0.70 from \(f_r /f_a = 0.9959\), 0.9621, 0.8870, 0.7487; the small-gap formula overestimates; with a 1.5 MRayl front load the extracted \(k_t\) is visibly off.
m16_backing_tradeoff.mSweep of backing impedance with 0, 1 or 2 matching layers: bandwidth, insertion loss, ring-down, and the Q formula check · change the layer set and material · runtime ≈ 3 s
%% m16_backing_tradeoff.m Backing impedance vs bandwidth, insertion loss and ring-down (KLM/ABCD)
% Purpose : sweep the backing impedance ZB of a 5 MHz PZT-5H plate radiating into water, with no
% matching layer, one layer (Zp^1/3 ZL^2/3) or two layers (Zp^4/7 ZL^3/7, Zp^1/7 ZL^6/7),
% untuned, 50 ohm source and receiver. Report the -6 dB fractional bandwidth, the peak
% insertion loss IL = 20 log10|2H| and the -40 dB envelope length, and check the mechanical
% Q formula Q = (pi/2) Zp/(ZB + ZF) against the open-circuit KLM response.
% Change : the layer set, material, R, the ZB grid.
% Expected: no matching: FBW 7 -> 19 -> 30 -> 58 -> 118 % and IL -5.6 -> -22.7 -> -31.2 -> -41.0 -> -48.1 dB
% at ZB = 0, 5, 10, 20, 34.5 MRayl; two layers: 65-80 % at -17 to -34 dB.
clear; close all; rng(1);
e0 = 8.854187817e-12;
mm = @(M,N) [M(:,1).*N(:,1)+M(:,2).*N(:,3), M(:,1).*N(:,2)+M(:,2).*N(:,4), ...
M(:,3).*N(:,1)+M(:,4).*N(:,3), M(:,3).*N(:,2)+M(:,4).*N(:,4)];
ser = @(Z) [ones(size(Z)), Z, zeros(size(Z)), ones(size(Z))];
shu = @(Y) [ones(size(Y)), zeros(size(Y)), Y, ones(size(Y))];
xfm = @(p) [p, zeros(size(p)), zeros(size(p)), 1./p];
tl = @(Z,th) [cos(th), 1i*Z.*sin(th), 1i*sin(th)./Z, cos(th)];
zlin = @(Z,th,ZT) Z.*(ZT.*cos(th)+1i*Z.*sin(th))./(Z.*cos(th)+1i*ZT.*sin(th));
kof = @(f,P) 2*pi*f ./ (P.v*sqrt(1+1i/P.Qm));
Z0of = @(P) P.rho*P.v*sqrt(1+1i/P.Qm)*P.A;
C0of = @(P) P.er*e0*(1-1i*P.tand)*P.A/P.d;
hof = @(P) P.kt*sqrt(P.rho*P.v^2*P.er*e0)/(P.er*e0*(1-1i*P.tand));
X1of = @(f,P) hof(P)^2*sin(kof(f,P)*P.d)./((2*pi*f).^2*Z0of(P));
phiof = @(f,P) 2*hof(P)*sin(kof(f,P)*P.d/2)./(2*pi*f*Z0of(P));
klm = @(f,P,ZB) mm(mm(mm(mm(ser(1./(1i*2*pi*f*C0of(P))), ser(1i*X1of(f,P))), xfm(phiof(f,P))), ...
shu(1./zlin(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2, ZB*ones(size(f))))), tl(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2));
fa = 5e6; P = struct('rho',7500,'v',4600,'kt',0.51,'er',1470,'Qm',65,'tand',0.02,'d',4600/(2*fa),'A',1.5e-6);
Zp = P.rho*P.v/1e6; ZLm = 1.5; ZL = ZLm*1e6*P.A; R = 50; fm = 0.9*fa;
sets = {'none', [], []; 'one layer', Zp^(1/3)*ZLm^(2/3), 2500; 'two layers', [Zp^(4/7)*ZLm^(3/7), Zp^(1/7)*ZLm^(6/7)], [2800 2400]};
Nfft = 1024; fs = 25.6*fa; df = fs/Nfft; fg = (0:Nfft/2)'*df; fg(1) = 1e-3*df; lim = round(2*fa/df);
wlp = ones(size(fg)); bnd = fg > 2*fa & fg < 2.5*fa; wlp(bnd) = 0.5*(1+cos(pi*(fg(bnd)-2*fa)/(0.5*fa))); wlp(fg >= 2.5*fa) = 0;
ZBs = 0:0.5:34.5;
res = zeros(numel(ZBs), 3, size(sets,1)); % [FBW %, IL dB, -40 dB length us]
%% Sweep
for s = 1:size(sets,1)
for q = 1:numel(ZBs)
M = klm(fg, P, ZBs(q)*1e6*P.A);
for L = 1:numel(sets{s,2})
Zl = sets{s,2}(L); vl = sets{s,3}(L);
M = mm(M, tl(Zl*1e6*P.A*ones(size(fg)), 2*pi*fg/vl*vl/(4*fm)));
end
Dn = M(:,1)*ZL + M(:,2) + R*(M(:,3)*ZL + M(:,4)); H = 2*R*ZL./Dn.^2;
mag = abs(H(2:lim)); fb = fg(2:lim); [pk, ip] = max(mag); lev = pk/2;
i1 = find(mag(1:ip) <= lev, 1, 'last'); i2 = ip - 1 + find(mag(ip:end) <= lev, 1, 'first');
if isempty(i1), fl = fb(1); else fl = fb(i1) + (lev-mag(i1))/(mag(i1+1)-mag(i1))*(fb(i1+1)-fb(i1)); end
fu = fb(i2-1) + (lev-mag(i2-1))/(mag(i2)-mag(i2-1))*(fb(i2)-fb(i2-1));
Hs = zeros(Nfft,1); Hs(2:Nfft/2) = 2*H(2:Nfft/2).*wlp(2:Nfft/2); env = abs(ifft(Hs)); [epk, ie] = max(env);
a = ie; while a > 1 && env(a) > 0.01*epk, a = a-1; end
b = ie; while b < Nfft && env(b) > 0.01*epk, b = b+1; end
res(q,:,s) = [100*(fu-fl)/((fu+fl)/2), 20*log10(2*pk), (b-a)/fs*1e6];
end
end
fprintf(' ZB (MRayl) | none: FBW IL t40 | one layer: FBW IL t40 | two layers: FBW IL t40\n');
for zb = [0 5 10 20 34.5]
q = find(abs(ZBs-zb) < 1e-9);
fprintf(' %5.1f | %6.1f %6.1f %5.2f | %8.1f %6.1f %5.2f | %8.1f %6.1f %5.2f\n', zb, res(q,:,1), res(q,:,2), res(q,:,3));
end
fprintf(' (t40 = 8.00 us means the ring-down exceeds the %0.f us analysis window)\n', Nfft/fs*1e6);
%% Q check: open-circuit (R = 10 Mohm) transmit force of the bare, lossless plate
PL = P; PL.Qm = Inf; PL.tand = 0; fq = (3e6:2.5e3:7e6)';
ZBq = [0 1.5 3 5 8 12]; Qs = zeros(size(ZBq));
for q = 1:numel(ZBq)
M = klm(fq, PL, ZBq(q)*1e6*P.A); D = M(:,1)*ZL + M(:,2) + 1e7*(M(:,3)*ZL + M(:,4)); F = abs(ZL./D);
[pk, ip] = max(F); i1 = find(F(1:ip) < pk/sqrt(2), 1, 'last'); i2 = ip - 1 + find(F(ip:end) < pk/sqrt(2), 1, 'first');
lv = pk/sqrt(2); % interpolate the -3 dB crossings (grid samples alone bias Q low by ~1 %)
fl = fq(i1) + (lv - F(i1))*(fq(i1+1) - fq(i1))/(F(i1+1) - F(i1));
fu = fq(i2-1) + (lv - F(i2-1))*(fq(i2) - fq(i2-1))/(F(i2) - F(i2-1));
Qs(q) = fq(ip)/(fu - fl);
end
Qth = pi/2*Zp./(ZBq + ZLm);
fprintf('\n ZB Q sim Q = (pi/2) Zp/(ZB+ZF) ratio\n');
fprintf(' %4.1f %6.1f %6.1f %5.3f\n', [ZBq; Qs; Qth; Qs./Qth]);
%% Figures
figure('Name','backing trade-off');
subplot(2,1,1); plot(ZBs, res(:,1,1), 'k', ZBs, res(:,1,2), 'm', ZBs, res(:,1,3), 'b', 'LineWidth', 1.5); grid on;
xlabel('backing Z_B (MRayl)'); ylabel('-6 dB FBW (%)'); legend('no layer', 'one layer', 'two layers', 'Location', 'northwest'); title('Pulse-echo bandwidth');
subplot(2,1,2); plot(ZBs, res(:,2,1), 'k', ZBs, res(:,2,2), 'm', ZBs, res(:,2,3), 'b', 'LineWidth', 1.5); grid on;
xlabel('backing Z_B (MRayl)'); ylabel('IL (dB)'); title('Peak insertion loss, 50 \Omega, untuned');
figure('Name','Q formula');
zz = linspace(0, 12.5, 200);
plot(zz, pi/2*Zp./(zz+ZLm), 'b', 'LineWidth', 1.5); hold on; plot(ZBq, Qs, 'ro', 'MarkerFaceColor', 'r'); grid on;
xlabel('backing Z_B (MRayl)'); ylabel('mechanical Q'); legend('(\pi/2) Z_p/(Z_B + Z_F)', 'KLM, open circuit'); title('Q of a loaded half-wave plate (water front)');
What to look for: the table in 16.7 (no matching: 7 % → 118 % FBW, −5.6 → −48 dB) and simulated open-circuit Q within 2 % of \((\pi /2)Z_p /(Z_B + Z_F)\) at light backing.
m16_electrical_tuning.mSeries and shunt inductors, a transformer and a coaxial cable as ABCD factors; 50 Ω versus high-impedance receiver · change n, cable length, receiver · runtime ≈ 2 s
%% m16_electrical_tuning.m Tuning inductors, a transformer and a coaxial cable as ABCD factors
% Purpose : take the reference stack (PZT-5H plate, fa = 5 MHz, 1.5 mm^2, ZB = 5 MRayl, two layers at
% 4.5 MHz, water load) and compare electrical networks between it and the system:
% none, series L, shunt L, series L + ideal transformer, and a 50-ohm coaxial cable
% (100 pF/m) into a 50-ohm or a 1-kohm receiver. Prints peak two-way gain 20 log10|2 Vout/Vs|
% (equal to the insertion loss when the source and receiver impedances are equal).
% Change : transformer ratio n, cable length, receiver impedance, which capacitance L resonates.
% Expected: untuned -21.6 dB, series L -10.1 dB, shunt L within 0.03 dB of untuned (50 ohm source),
% series L + n = 0.4 about -2.6 dB; 50-ohm system: cable length has no effect on |H|;
% 1-kohm receiver: 2 m of cable costs about 9.6 dB.
clear; close all; rng(1);
e0 = 8.854187817e-12;
mm = @(M,N) [M(:,1).*N(:,1)+M(:,2).*N(:,3), M(:,1).*N(:,2)+M(:,2).*N(:,4), ...
M(:,3).*N(:,1)+M(:,4).*N(:,3), M(:,3).*N(:,2)+M(:,4).*N(:,4)];
ser = @(Z) [ones(size(Z)), Z, zeros(size(Z)), ones(size(Z))];
shu = @(Y) [ones(size(Y)), zeros(size(Y)), Y, ones(size(Y))];
xfm = @(p) [p, zeros(size(p)), zeros(size(p)), 1./p];
tl = @(Z,th) [cos(th), 1i*Z.*sin(th), 1i*sin(th)./Z, cos(th)];
zlin = @(Z,th,ZT) Z.*(ZT.*cos(th)+1i*Z.*sin(th))./(Z.*cos(th)+1i*ZT.*sin(th));
zin = @(M,ZL) (M(:,1).*ZL+M(:,2))./(M(:,3).*ZL+M(:,4));
kof = @(f,P) 2*pi*f ./ (P.v*sqrt(1+1i/P.Qm));
Z0of = @(P) P.rho*P.v*sqrt(1+1i/P.Qm)*P.A;
C0of = @(P) P.er*e0*(1-1i*P.tand)*P.A/P.d;
hof = @(P) P.kt*sqrt(P.rho*P.v^2*P.er*e0)/(P.er*e0*(1-1i*P.tand));
X1of = @(f,P) hof(P)^2*sin(kof(f,P)*P.d)./((2*pi*f).^2*Z0of(P));
phiof = @(f,P) 2*hof(P)*sin(kof(f,P)*P.d/2)./(2*pi*f*Z0of(P));
klm = @(f,P,ZB) mm(mm(mm(mm(ser(1./(1i*2*pi*f*C0of(P))), ser(1i*X1of(f,P))), xfm(phiof(f,P))), ...
shu(1./zlin(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2, ZB*ones(size(f))))), tl(Z0of(P)*ones(size(f)), kof(f,P)*P.d/2));
fa = 5e6; P = struct('rho',7500,'v',4600,'kt',0.51,'er',1470,'Qm',65,'tand',0.02,'d',4600/(2*fa),'A',1.5e-6);
Zp = P.rho*P.v/1e6; ZLm = 1.5; ZL = ZLm*1e6*P.A; Rs = 50; fm = 0.9*fa;
Zl = [Zp^(4/7)*ZLm^(3/7), Zp^(1/7)*ZLm^(6/7)]; vl = [2800 2400];
Nfft = 1024; fs = 25.6*fa; df = fs/Nfft; f = (0:Nfft/2)'*df; f(1) = 1e-3*df; lim = round(2*fa/df);
Mt = mm(mm(klm(f, P, 5e6*P.A), tl(Zl(1)*1e6*P.A*ones(size(f)), 2*pi*f/vl(1)*vl(1)/(4*fm))), tl(Zl(2)*1e6*P.A*ones(size(f)), 2*pi*f/vl(2)*vl(2)/(4*fm)));
C0 = real(C0of(P)); Lopt = 1/((2*pi*fm)^2*C0); w = 2*pi*f; I4 = [ones(size(f)) zeros(size(f)) zeros(size(f)) ones(size(f))];
cable = @(len) tl(50*ones(size(f)), w*len/2e8); % 50-ohm coax, 2e8 m/s -> 100 pF/m
fprintf('C0 = %.1f pF, Lopt(4.5 MHz) = %.1f uH, cable C = 100 pF/m\n', C0*1e12, Lopt*1e6);
% gain(E, Rr): two-way voltage gain spectrum with electrical network E in front of the transducer
Dfun = @(M,R) M(:,1)*ZL + M(:,2) + R*(M(:,3)*ZL + M(:,4));
gain = @(E,Rr) (ZL./Dfun(mm(E,Mt),Rs)) .* (2*Rr./Dfun(mm(E,Mt),Rr));
peak = @(H) 20*log10(2*max(abs(H(2:lim))));
%% Cases
Hn = gain(I4, Rs);
Hse = gain(ser(1i*w*Lopt), Rs);
Hsh = gain(shu(1./(1i*w*Lopt)), Rs);
Hx = gain(mm(xfm(0.4*ones(size(f))), ser(1i*w*Lopt)), Rs);
fprintf('\n 50-ohm system peak gain (dB)\n');
fprintf(' untuned %7.2f\n', peak(Hn));
fprintf(' series L %7.2f\n', peak(Hse));
fprintf(' shunt L %7.2f (difference from untuned %.4f dB)\n', peak(Hsh), peak(Hsh)-peak(Hn));
for n = [0.5 0.4 0.35]
fprintf(' series L + transformer n = %.2f %7.2f\n', n, peak(gain(mm(xfm(n*ones(size(f))), ser(1i*w*Lopt)), Rs)));
end
fprintf('\n cable length (m) 50-ohm rx, series L 1-kohm rx, untuned 1-kohm rx, shunt L(C0+Ccab)\n');
for len = [0 1 2 3]
Lsh = 1/((2*pi*fm)^2*(C0 + len*100e-12));
fprintf(' %3.1f %7.2f %7.2f %7.2f\n', len, peak(gain(mm(cable(len), ser(1i*w*Lopt)), Rs)), ...
peak(gain(cable(len), 1000)), peak(gain(mm(cable(len), shu(1./(1i*w*Lsh))), 1000)));
end
fprintf(' lumped open-circuit estimate for 2 m: 20 log10(C0/(C0+200 pF)) = %.1f dB\n', 20*log10(C0/(C0+200e-12)));
Zu = zin(Mt, ZL); Zs = zin(mm(ser(1i*w*Lopt), Mt), ZL); Zx = zin(mm(mm(xfm(0.4*ones(size(f))), ser(1i*w*Lopt)), Mt), ZL);
i45 = round(fm/df) + 1;
fprintf('\n Z seen at 4.5 MHz: untuned %.0f%+.0fj, series L %.0f%+.0fj, + n = 0.4 %.0f%+.0fj ohm\n', real(Zu(i45)), imag(Zu(i45)), real(Zs(i45)), imag(Zs(i45)), real(Zx(i45)), imag(Zx(i45)));
%% Figures
sel = f > 0 & f <= 10e6;
figure('Name','tuning: gain');
plot(f(sel)/1e6, 20*log10(2*abs(Hn(sel))), 'k', f(sel)/1e6, 20*log10(2*abs(Hse(sel))), 'b', f(sel)/1e6, 20*log10(2*abs(Hx(sel))), 'g', f(sel)/1e6, 20*log10(2*abs(Hsh(sel))), 'r--', 'LineWidth', 1.4);
grid on; ylim([-60 0]); xlabel('frequency (MHz)'); ylabel('20 log_{10}|2V_{out}/V_s| (dB)');
legend('untuned', 'series L', 'series L + n = 0.4', 'shunt L', 'Location', 'south'); title('Electrical tuning of the reference element, 50 \Omega');
figure('Name','tuning: impedance seen');
subplot(2,1,1); plot(f(sel)/1e6, real(Zu(sel)), 'k', f(sel)/1e6, real(Zs(sel)), 'b', f(sel)/1e6, real(Zx(sel)), 'g', 'LineWidth', 1.3); grid on; ylim([0 1500]);
ylabel('Re Z (\Omega)'); title('Impedance seen by the system'); legend('untuned', 'series L', '+ n = 0.4');
subplot(2,1,2); plot(f(sel)/1e6, imag(Zu(sel)), 'k', f(sel)/1e6, imag(Zs(sel)), 'b', f(sel)/1e6, imag(Zx(sel)), 'g', 'LineWidth', 1.3); grid on; ylim([-2000 1500]);
xlabel('frequency (MHz)'); ylabel('Im Z (\Omega)');
What to look for: untuned −21.6 dB, series −10.1 dB, series + \(n = 0.4\) −2.6 dB, shunt within 0.03 dB of untuned at 50 Ω; with a 1 kΩ receiver, 2 m of cable costs ≈ 9.6 dB, and a shunt L sized for \(C_0 + C_{\mathrm{cable}}\) recovers the peak.
16.14Self-check
Q1What PZT-5H thickness gives \(f_a = 7.5\) MHz, and where will the bare-plate \(f_r\) be?
\(t = 4600/(2 \times 7.5\times 10^{6}) = 0.307\) mm. \(f_r = 0.882 \times 7.5 = 6.61\) MHz for \(k_t = 0.51\).
Q2\(C_0\) and \(\lvert Z\rvert\) at 5 MHz for a 10 mm diameter, 0.46 mm thick PZT-5H disc?
\(A = 78.5\) mm², \(C_0 = 1470 \times 8.854\times 10^{-12} \times 78.5\times 10^{-6}\) \(/\, 0.46\times 10^{-3} = 2.22\) nF, \(\lvert Z\rvert = 1/(2\pi \times 5\times 10^{6} \times 2.22\times 10^{-9}) = 14.3\) Ω: too low for 50 Ω, so a large element wants a step-up transformer (n > 1).
Q3A crystal plate has \(f_r /f_a = 0.749\). What is \(k_t\), and what does the small-gap formula say?
\(x = (\pi /2)(0.749) = 1.176\), \(k_t^2 = x \cot x = 0.490\), \(k_t = 0.70\). Small gap: \((\pi^2 /4)(0.251) = 0.620\), \(k_t = 0.79\), 12 % high.
Q4Estimate Q for a PZT-5H plate (34.5 MRayl) into water with air backing, and with a 5 MRayl backing.
\((\pi /2)(34.5/1.5) = 36\); \((\pi /2)(34.5/6.5) = 8.3\). Fractional −3 dB bandwidths ≈ 3 % and 12 % (one-way, open circuit).
Q5Why does a 2 m, 50 Ω cable not change \(\lvert H\rvert\) in a 50 Ω system, yet costs ≈ 10 dB into a 1 kΩ preamp?
A lossless line terminated at the system end by its own impedance only rotates phase: available power and reflection magnitude at the element are unchanged. Into a high-impedance input the line looks like 200 pF in parallel with the element, dividing its voltage by about \(C_0 /(C_0 + C_{\mathrm{cable}})\) (−15 dB lumped, −9.6 dB with 1 kΩ in the full model).
Q6A PZT-5H/soft-epoxy composite at \(\nu = 0.3\): \(k_t\) and Z from Smith–Auld? Why is \(\varepsilon\) lower?
\(k_t \approx 0.67\), \(Z \approx 11.0\) MRayl (\(v^D \approx 3610\) m/s). \(\bar{\varepsilon}^S \approx 457 \varepsilon_0\) because only 30 % of the area is ceramic, so \(C_0\) and the electrical impedance change by ≈ 3× for the same element.
Q7Which tuning inductor do you use in front of a 50 Ω pulser, and which at the input of a high-impedance preamp at the end of a cable?
Series \(L = 1/(\omega^2 C_0)\) for the 50 Ω source (cancel the series reactance), then a transformer for the real part. Shunt \(L = 1/(\omega^2 (C_0 + C_{\mathrm{cable}}))\) at the high-impedance input, to resonate the total capacitance.
16.15Figures from the literature
Measured data to hold the models of this page against. Links only (no images reproduced); each figure number was checked against the published text.
A measured thickness-mode impedance spectrum of a PZT plate, with a fitted model
S. Sherrit, B. K. Mukherjee, "Characterization of piezoelectric materials for transducers," arXiv:0711.2657, 2007. Fig. 3(a). doi:10.48550/arXiv.0711.2657
Thickness-extensional resonance of CTS 3203HD PZT (\(\rho = 7800\) kg/m³), measured, with a complex-constant fit to the fundamental (solid) and to the next resonance (broken). Look for the \(f_r /f_a\) pair, the peak heights set by loss, and the weaker higher resonance: the same features as Demo 16.3 with finite \(Q_m\) and \(\tan \delta\). Figs. 3(b–e) show the other modes (shear, length, radial) that a 1D model leaves out.
Mason and KLM drawn side by side, and shown to coincide
Same source. Figs. 5, 6 and 9. doi:10.48550/arXiv.0711.2657
Fig. 5 is Mason's network for the thickness mode, Fig. 6 the KLM circuit, and Fig. 9 the impedance of a plate on stainless-steel and epoxy backings from the analytic solution, Mason and KLM: the three curves overlap. This is the published counterpart of validation checks (1) and (2) in 16.6. Compare the drawing conventions (transformer sense, where \(-C_0\) sits) with Figure 16.D.
Measured impedance of a single-crystal (PIN-PMN-PT) transducer
P. Sun, Q. Zhou, B. Zhu, D. Wu, C. Hu, J. M. Cannata, J. Tian, P. Han, G. Wang, K. K. Shung, "Design and fabrication of PIN-PMN-PT single-crystal high-frequency ultrasound transducers," IEEE Trans. Ultrason. Ferroelectr. Freq. Control 56(12), 2760–2763, 2009. Fig. 3 (also Figs. 2 and 5). doi:10.1109/TUFFC.2009.1367 (open copy: PMC2849290)
Measured \(\lvert Z\rvert\) and phase in air of the 35 MHz transducer; the authors report \(k_t = 0.56\) and ≈ 80 Ω near resonance: an element sized so its impedance sits near 50 Ω, the opposite end of the 750 Ω array element of 16.8. Fig. 2 is the KLM-modelled pulse-echo response and Fig. 5 the measured one: the measured bandwidth comes out slightly below the model, the usual direction (unmodelled loss, bond lines, lateral modes).
Equivalent-circuit impedance against measurement, with and without the cable
Z. Shen, J. Xu, Z. Li, Y. Chen, Y. Cui, X. Jian, "An improved equivalent circuit simulation of high frequency ultrasound transducer," Frontiers in Materials 8, 663109, 2021 (CC BY). Figs. 7, 8 and 9. doi:10.3389/fmats.2021.663109
Three PZT-5H transducers (11.3–19.6 MHz, 100–185 µm plates, one matching layer): simulated versus measured impedance magnitude and phase without cable (Fig. 7), with cable (Fig. 8) and for 0.5, 1.5 and 2.5 m of cable (Fig. 9). Watch how the cable moves and reshapes the resonance seen at the connector, the line-model effect of 16.8 and Demo 16.6; the paper also reports where a KLM-based package missed the centre frequency by more than 10 %.
1-3 composite transducer: KLM prediction versus the built device
J.-M. Zhang, G.-C. Bao, W. Gao, R.-Q. Lin, F. Yang, K.-H. Lam, "Miniature ultrasound transducer incorporating Sm-PMN-PT 1-3 composite," J. Compos. Sci. 8(3), 80, 2024 (CC BY). Fig. 8 versus Fig. 10. doi:10.3390/jcs8030080
Fig. 8 is the simulated pulse-echo response and electrical spectrum, Fig. 10 the measured ones: centre frequency 14.3 MHz modelled versus 12.4 MHz measured, −6 dB bandwidth 79 % versus ≈ 70 %, \(k_{\mathrm{eff}} = 0.66\) from \(f_r\), \(f_a\). The authors suggest, as a possible cause, a pillar aspect ratio near the limit where lateral modes appear, the fine-scale condition behind the Smith–Auld model of Demo 16.2.
16.16Go deeper
- W. P. Mason, Electromechanical Transducers and Wave Filters, Van Nostrand, 1942 (2nd ed. 1948). The original equivalent circuit.
- R. Krimholtz, D. A. Leedom, G. L. Matthaei, "New equivalent circuits for elementary piezoelectric transducers," Electronics Letters 6(13), 398–399, 1970. The KLM model.
- D. A. Leedom, R. Krimholtz, G. L. Matthaei, "Equivalent circuits for transducers having arbitrary even- or odd-symmetry piezoelectric excitation," IEEE Trans. Sonics Ultrason. SU-18(3), 128–141, 1971.
- IEEE Standard on Piezoelectricity, ANSI/IEEE Std 176-1987. Constitutive equations, coupling factors, resonance measurement of kt.
- G. S. Kino, Acoustic Waves: Devices, Imaging, and Analog Signal Processing, Prentice-Hall, 1987. Chapter 1: plate impedance, Mason model, Q of loaded transducers.
- C. S. Desilets, J. D. Fraser, G. S. Kino, "The design of efficient broad-band piezoelectric transducers," IEEE Trans. Sonics Ultrason. SU-25(3), 115–125, 1978.
- W. A. Smith, B. A. Auld, "Modeling 1-3 composite piezoelectrics: thickness-mode oscillations," IEEE Trans. Ultrason. Ferroelectr. Freq. Control 38(1), 40–47, 1991.
- S. Sherrit, S. P. Leary, B. P. Dolgin, Y. Bar-Cohen, "Comparison of the Mason and KLM equivalent circuits for piezoelectric resonators in the thickness mode," Proc. IEEE Ultrasonics Symposium, 1999.
- S. Zhang, F. Li, "High performance ferroelectric relaxor-PbTiO₃ single crystals: status and perspective," J. Appl. Phys. 111, 031301, 2012.
- T. L. Szabo, Diagnostic Ultrasound Imaging: Inside Out, 2nd ed., Academic Press, 2014. Transducer chapter with KLM and ABCD modelling.