*2.2. Solution*

As shown in Figure 1, the boundary and continuity conditions of the cylindrical transducer consist of one innermost displacement boundary condition, one outermost stress boundary condition and eight continuous conditions, which are given as follows:

$$\begin{cases} \left. u\_{rE(1)} \right|\_{r=0} = 0\\ \left. \sigma\_{rE(3)} \right|\_{r=R\_{\mathbb{S}}} = 0 \end{cases} \tag{16}$$

$$\begin{cases} \left. u\_{rP(i)} \right|\_{r=R\_{2i-1}} = \left. u\_{rE(i)} \right|\_{r=R\_{2i-1}} \\ \left. \sigma\_{rP(i)} \right|\_{r=R\_{2i-1}} = \left. \sigma\_{rE(i)} \right|\_{r=R\_{2i-1}} \end{cases} \text{ (\$i=1, 2)}, \tag{17}$$

$$\left. \begin{array}{l} \left. u\_{rP(i)} \right|\_{r=R\_{2i}} = \left. u\_{rE(i+1)} \right|\_{r=R\_{2i}} \right|\_{r=R\_{2i}} \quad (i=1,2). \tag{18} \\ \left. \sigma\_{rP(i)} \right|\_{r=R\_{2i}} = \left. \sigma\_{rE(i+1)} \right|\_{r=R\_{2i}} \end{array} \text{ ( $i=1,2$ )}. \tag{18}$$

Combining Equations (2), (3), (10), (11), and (16)–(18), 10 coefficients can be derived, which are listed in Equations (A1)–(A5) (Appendix A). Further, the total electrical charge *Qtotal*(*t*) and the total current *Itotal*(*t*) can be expressed by the following expressions:

$$Q\_{\rm total}(t) = Q\_{(1)}(t) + Q\_{(2)}(t) = \int\_0^{2\pi} \int\_{R\_1}^{R\_2} D\_{z(1)}r d\theta \, dr + \int\_0^{2\pi} \int\_{R\_3}^{R\_4} D\_{z(2)}r d\theta \, dr = (\widetilde{\mathbf{C}}\_1 + \widetilde{\mathbf{C}}\_2)V\_0 \mathbf{e}^{j\omega\_1 t}, \text{ (19)}$$

$$I\_{\text{total}}(t) = -dQ\_{\text{total}}(t)/dt = -j\omega(\tilde{\mathcal{C}}\_1 + \tilde{\mathcal{C}}\_2)V\_0e^{j\omega\cdot t},\tag{20}$$

where

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$$\dot{C}\_{1} = 2\pi \{ (a\_{1}\upsilon\_{9} + \upsilon\_{1})[f\_{9}(R\_{2}, 1) - f\_{9}(R\_{1}, 1)] + (a\_{2}\upsilon\_{9} + \upsilon\_{2})[f\_{10}(R\_{2}, 1) - f\_{10}(R\_{1}, 1)] \} - C\_{1},\tag{21}$$

$$\mathcal{L}\_2 = 2\pi \{ (a\mathfrak{z}\mathfrak{z}\mathfrak{y} + \mathfrak{v}\mathfrak{z}) [f\mathfrak{s}(R\_4, 2) - f\mathfrak{s}(R\_3, 2)] + (a\_6\mathfrak{z}\mathfrak{y} + \mathfrak{v}\mathfrak{e}) [f\_{10}(R\_4, 2) - f\_{10}(R\_3, 2)] \} - \mathcal{L}\_2,\tag{22}$$

$$f\_{\theta}(r, i) = \varepsilon\_{31(i)} r l\_1(k\_{P(i)} r) (i = 1, 2), \tag{23}$$

$$f\_{10}(r,i) = \varepsilon\_{31(i)} r Y\_1(k\_{P(i)}r)(i=1,2). \tag{24}$$

In Equations (23) and (24), *C*1 = *<sup>π</sup>*(*κ<sup>ε</sup>*33(1)/*h*)(*R*22 − *R*21) and *C*2 = *<sup>π</sup>*(*κ<sup>ε</sup>*33(2)/*h*)(*R*24 − *R*23) are the clamped electric capacitances of piezoceramic rings #1 and #2 in radial vibration, respectively. *C* % 1 and *C* % 2 are the effective electric capacitances of piezoceramic rings #1 and #2 in radial vibration, respectively. Then, the electrical impedance of the transducer *Z* can be given as:

$$Z = V(\mathbf{t}) / l\_{\text{total}}(\mathbf{t}) = -1 / [j\omega (\tilde{\mathbf{C}}\_1 + \tilde{\mathbf{C}}\_2)],\tag{25}$$

Subsequently, by letting |*Z*| = 0 and |*Z*| = <sup>∞</sup>, we can obtain the resonance frequency *fr* and anti-resonance frequency *fa*, respectively. Based on these frequencies, the electromechanical coupling factor of the transducer is obtained as [55]:

$$k\_d^2 = (f\_a^2 - f\_r^2) / f\_{a\prime}^2 \tag{26}$$
