Nonlinearity reduction in MEMS resonators based on design of H-shaped beams

Abstract

Structural nonlinearity for conventional double-end fixed beam MEMS resonators considerably restricts their performance because of large displacement instabilities and excessive frequency noise. To solve this problem, a new resonator design based on H-shaped structural beams is proposed to alleviate the axial force caused by a large deflection of the beam structure. An analytical expression for the nonlinear stiffness of the proposed beam is derived using the Rayleigh–Ritz energy method. Nonlinear vibration equations of the resonant system are then formulated and solved analytically via the method of multiple scales. The results show that the H-shaped beam design can effectively reduce the nonlinearity without significantly decreasing the linear stiffness. The solution to the nonlinear stiffness of a given electrically excited resonant system is quantitatively verified by finite element analysis. The analytical solution to the nonlinear vibration of the system is verified by numerical simulation. The obtained dynamic results indicate that the H-shaped beam design can effectively improve the stability of the resonant frequency and it can also depress the cross-sensitivity of the resonator.

Appendix

1.1 Perturbation analysis of nonlinear response of the main resonance

For the main resonance, ${\omega }_{0}\approx \omega$. In order to analyze the response of system near natural frequency, it is necessary to specify the order of damping, nonlinearity and excitation so that they can appear simultaneously in the perturbation method [28]. If ${y}_{e}=\varepsilon \mathrm{cos}\left({\omega }_{0}t\right)$ is assumed, the nonlinearity will produce a term that is proportional to the ${\rm O}\left({\varepsilon }^{3}\right)$ order of $\mathrm{cos}{\omega }_{0}t$. Therefore, if ${y}_{e}=\varepsilon u$ is assumed, then the order of $\zeta {\omega }_{0}\dot{{y}_{e}}$ needs to be ${\varepsilon }^{2}\mu \dot{{y}_{e}}$, and the order of $\left({f}_{0}/m\right)\mathrm{cos}\left(\omega t\right)$ is ${\varepsilon }^{3}f\mathrm{cos}\left(\omega t\right)$, the governing equation (26) in Sect. 2.2 becomes

$$\ddot{u}+2{\varepsilon }^{2}\mu \dot{u}+{\omega }_{0}^{2}u+\varepsilon {k}_{2}{u}^{2}+{\varepsilon }^{2}{k}_{3}{u}^{3}={\varepsilon }^{2}f\mathrm{cos}\left(\omega t\right),$$

(A.1)

where

$$k_{2} = \frac{{3K_{3} y_{0} }}{m},{\text{~}}k_{3} = \frac{{K_{3} }}{m},{\text{~}}\mu = \zeta \omega _{0} ,{\text{~}}f = \frac{{f_{0} }}{m}.$$

The multiscale method is used to find the approximate solution to eq. (A.1).

Assume

$${T}_{n}={\varepsilon }^{n}t \left(n=\mathrm{0,1},2,\dots \right),$$

$$\frac{d}{{dt}} = \frac{\partial }{{\partial T_{0} }}\frac{{dT_{0} }}{{dt}} + \frac{\partial }{{\partial T_{1} }}\frac{{dT_{1} }}{{dt}} + \ldots = D_{0} + \varepsilon D_{1} + \ldots ,$$

then

$$\frac{{d^{2} }}{{dt^{2} }} = D_{0}^{2} + 2\varepsilon D_{0} D_{1} + \varepsilon ^{2} \left( {D_{1}^{2} + 2D_{0} D_{2} } \right) + \ldots .$$

Because the excitation is the order of ${\rm O}({\varepsilon }^{2})$, for the sake of consistency, we assume

$$\omega ={\omega }_{0}+{\upvarepsilon }^{2}\upsigma .$$

(A.2)

In this paper, the problem is solved by quadratic approximation:

$$u={u}_{0}\left({T}_{0},{T}_{1},{T}_{2}\right)+\varepsilon {u}_{1}\left({T}_{0},{T}_{1},{T}_{2}\right)+{\varepsilon }^{2}{u}_{2}\left({T}_{0},{T}_{1},{T}_{2}\right)\dots$$

(A.3)

Substituting Eqs. (A.2) and (A.3) into Eq. (A.1), and making the coefficients of $\varepsilon$, ${\varepsilon }^{2}$, ${\varepsilon }^{3}$ on both sides of the equation equal,

$$D_{0}^{2} u_{0} + \omega_{0}^{2} u_{0} = 0,$$

(A.4)

$$D_{0}^{2} u_{1} + \omega_{0}^{2} u_{1} = - 2D_{0} D_{1} u_{0} - k_{2} u_{0}^{2},$$

(A.5)

$$D_{0}^{2} u_{2} + \omega_{0}^{2} u_{2} = - 2D_{0} D_{1} u_{1} - 2D_{0} D_{1} u_{0} - D_{1}^{2} u_{0} - 2\mu D_{0} u_{0}$$

$$-2{k}_{2}{u}_{0}{u}_{1}-{k}_{3}{u}_{0}^{3}+f\mathrm{cos}\left({\omega }_{0}{T}_{0}+\sigma {T}_{2}\right)$$

$$-2{k}_{2}{u}_{0}{u}_{1}-{k}_{3}{u}_{0}^{3}+f\mathrm{cos}\left({\omega }_{0}{T}_{0}+\sigma {T}_{2}\right).$$

(A.6)

The general solution of Eq. (A.4) can be written as

$${u}_{0}=Q\left({T}_{1},{T}_{2}\right){e}^{i{\omega }_{0}{T}_{0}}+\overline{Q }\left({T}_{1},{T}_{2}\right){e}^{-i{\omega }_{0}{T}_{0}}.$$

(A.7)

In the formula, $Q\left({T}_{1},{T}_{2}\right)$ is still an undetermined function, which will be determined by eliminating the long-term terms in ${u}_{1}$ and ${u}_{2}$. Substituting Eq. (A.7) into Eq. (A.5), we obtain.

$$D_{0}^{2} u_{1} + \omega_{0}^{2} u_{1} = - 2i\omega_{0} D_{1} Qe^{{i\omega_{0} T_{0} }} - k_{2} \left( {Q^{2} e^{{2i\omega_{0} T_{0} }} + Q\overline{Q}} \right) + cc.$$

(A.8)

where cc represents the conjugate complex number of the preceding items. By eliminating those items that cause ${u}_{1}$ to produce long-term terms, ${D}_{1}Q=0$ or $Q=Q({T}_{2})$ will be obtained. Therefore, the solution of Eq. (A.8) is

$$u_{1} = \frac{{k_{2} }}{{\omega _{0}^{2} }}\left( { - 2Q\bar{Q} + \frac{1}{3}Q^{2} e^{{2i\omega _{0} T_{0} }} + \frac{1}{3}\bar{Q}^{2} e^{{ - 2i\omega _{0} T_{0} }} } \right).$$

(A.9)

Substituting Eqs. (A.7) and (A.9) into Eq. (A.6), we obtain.

$$D_{0}^{2} u_{2} + \omega _{0}^{2} u_{2} = - \left[ {2i\omega _{0} \left( {Q^{\prime} + \mu Q} \right) + \left( {3k_{3} - \frac{{10k_{2}^{2} }}{{3\omega _{0}^{2} }}} \right)Q^{2} \bar{Q} - \frac{1}{2}fe^{{i\sigma T_{2} }} } \right]e^{{i\omega _{0} T_{0} }} + cc + NST,$$

(A.10)

where $Q\mathrm{^{\prime}}$ represents the derivative of ${T}_{2}$, and NST represents those terms containing ${e}^{3i{\omega }_{0}{T}_{0}}$. If

$$2i\omega _{0} \left( {Q^{\prime} + \mu Q} \right) + \left( {3k_{3} - \frac{{10k_{2}^{2} }}{{3\omega _{0}^{2} }}} \right)Q^{2} \bar{Q} - \frac{1}{2}fe^{{i\sigma T_{2} }} = 0,$$

(A.11)

the long-term terms in ${u}_{2}$ will be eliminated. To solve the Eq. (A.11), $Q$ is expressed as

$$Q = \frac{1}{2}ae^{{i\beta }} ,$$

(A.12)

where $\alpha$ and $\beta$ are both real numbers, so the result is divided into real and imaginary parts.

$$a^{\prime} = - \mu a + \frac{1}{2}\frac{f}{{\omega _{0} }}\sin \left( \lambda \right),$$

(A.13)

$$a\beta ^{\prime} = \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a^{3} - \frac{1}{2}\frac{f}{{\omega _{0} }}\cos \left( \lambda \right),$$

(A.14)

where

$$\lambda =\sigma {T}_{2}-\beta .$$

(A.15)

The simultaneous equations (A.13) and (A.14) can eliminate $\beta$, then

$$a\lambda ^{\prime} = a\sigma - \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a^{3} + \frac{1}{2}\frac{f}{{\omega _{0} }}\cos \left( \lambda \right).$$

(A.16)

Substituting Eq. (A.12) into Eq. (A.7) and (A.9), and substituting the result into Eq. (A.3), the quadratic approximation is obtained:

$$u = a\cos \left( {\omega t - \lambda } \right) + \frac{1}{2}\varepsilon k_{2} \omega _{0}^{{ - 2}} a^{2} \left[ { - 1 + \frac{1}{3}\cos \left( {2\omega t - 2\lambda } \right)} \right] + O\left( {\varepsilon ^{2} } \right).$$

(A.17)

When ${a}^{\mathrm{^{\prime}}}={\lambda }^{\mathrm{^{\prime}}}=0$, steady-state motion exists, corresponding to the singularities of Eqs. (A.13) and (A.16), where the amplitude and phase are unchanged, which is

$$\mu a = \frac{1}{2}\frac{f}{{\omega _{0} }}\sin \left( \lambda \right),$$

(A.18)

$$a\sigma - \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a^{3} = - \frac{1}{2}\frac{f}{{\omega _{0} }}\cos \left( \lambda \right).$$

(A.19)

The frequency response function can be obtained by adding the squares of the Eqs. (A.18) and (A.19), as follows.

$$\left[ {\mu ^{2} + \left( {\sigma - \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a^{2} } \right)^{2} } \right]a^{2} = \frac{{f^{2} }}{{4\omega _{0}^{2} }},$$

(A.20)

which is

$$\sigma = \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a^{2} \pm \left[ {\frac{{f^{2} }}{{4\omega _{0}^{2} a^{2} }} - \mu ^{2} } \right]^{{\frac{1}{2}}} .$$

(A.21)

The phase response function can be obtained by comparing Eqs. (A.18) to (A.19),

$$- \lambda = \tan ^{{ - 1}} \frac{\mu }{{\sigma - \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a^{2} }}.$$

(A.22)

The stability of different parts of the frequency response curve can be obtained by analyzing the properties of their motion in the neighborhood of each singular point.

Assuming $\left({a}_{0},{\theta }_{0}\right)$ is the singularity of Eqs. (A.18) and (A.19), then

$$\mu a_{0} = \frac{1}{2}\frac{f}{{\omega _{0} }}\sin \left( {\theta _{0} } \right),$$

(A.23)

$$a_{0} \sigma - \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a_{0}^{3} = - \frac{1}{2}\frac{f}{{\omega _{0} }}\cos \left( {\theta _{0} } \right).$$

(A.24)

Combined with the Eqs. (A.23) and (A.24), the Taylor series expansion of the Eqs. (A.13) and (A.14) in the neighborhood of $\left({a}_{0},{\theta }_{0}\right)$ is carried out, and the equation is retained to the linear term. If $\left({a}_{0}+{a}_{1},{\theta }_{0}+{\theta }_{1}\right)$ is any point in this neighborhood, then there is

$$a^{\prime} = - \mu a_{1} + \left[ {\frac{f}{{2\omega _{0} }}\cos \left( {\theta _{0} } \right)} \right]\theta _{1},$$

(A.25)

$$\theta ^{\prime} = - \left[ {\frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{12\omega _{0}^{3} }}a_{0} + \frac{f}{{2\omega _{0} a_{0}^{2} }}\cos \left( {\theta _{0} } \right)} \right]a_{1} - \left[ {\frac{f}{{2\omega _{0} a_{0} }}\sin \left( {\theta _{0} } \right)} \right]\theta _{1}.$$

(A.26)

Therefore, the stability problem of steady-state motion is transformed into the eigenvalue problem of the coefficient matrix at the right end of the above equation. Using Eqs. (A.25) and (A.26), the following eigenvalue equation can be obtained:

$$\left| {\begin{array}{*{20}c} { - \mu - \lambda } & { - a_{0} \sigma + \frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a_{0}^{3} } \\ { - \left( {\frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{8\omega _{0}^{3} }}a_{0} - \frac{\sigma }{{a_{0} }}} \right)} & { - \mu - \lambda } \\ \end{array} } \right| = 0,$$

(A.27)

which is

$$\lambda ^{2} + 2\mu \lambda + \mu ^{2} + \left( {\frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{24\omega _{0}^{3} }}a_{0}^{2} - \sigma } \right)\left( {\frac{{9k_{3} \omega _{0}^{2} - 10k_{2}^{2} }}{{8\omega _{0}^{3} }}a_{0}^{2} - \sigma } \right) = 0.$$

(A.28)

Thus,

$${\lambda }_{\mathrm{1,2}}=-\mu \pm {\frac{1}{2}}\sqrt{\Delta },$$

(A.29)

where

$$\Delta =-4\left({\frac{9{k}_{3}{\omega }_{0}^{2}-10{k}_{2}^{2}}{24{\omega }_{0}^{3}}{a}_{0}^{2}-\sigma} \right)\left({\frac{9{k}_{3}{\omega }_{0}^{2}-10{k}_{2}^{2}}{8{\omega }_{0}^{3}}{a}_{0}^{2}-\sigma} \right).$$

(A.30)

If $\Delta >0$, the eigenvalues are two unequal real roots. When ${\lambda }_{1}$ and ${\lambda }_{2}$ are both positive, the singularity is an unstable point. When ${\lambda }_{1}$ and ${\lambda }_{2}$ are both negative, the singularity is a stable point. When the signs of ${\lambda }_{1}$ and ${\lambda }_{2}$ are opposite, the singularity is an unstable saddle point. If $\Delta =0$, the eigenvalues are two equal real roots. At this time, if the eigenvalue is less than 0, the node is stable, otherwise, it is unstable. If $\Delta <0$, the eigenvalue is a conjugate complex. When the real part of the eigenvalue is less than 0, the singularity is a stable focus. When it is greater than 0, the singularity is an unstable point. When it is equal to 0, the singularity is called the center.

In summary, regardless of the value of $\Delta$, the singularity is a stable point as long as the eigenvalue (or the real part of the eigenvalue) is less than zero. For this problem, the damping $\mu$ is always positive, so for the case of $\Delta \le 0$, the steady-state motion is stable. For $\Delta >0$, only the relationship $-\mu +\sqrt{\Delta }/2<0$ is satisfied, and the steady-state motion is stable. There is only one case in which steady-state motion is unstable, which is

$$\left\{\begin{array}{l}\Delta >0\\ -\mu +{\frac{1}{2}}\sqrt{\Delta }>0\end{array}\right..$$

(A.31)

Therefore, we can obtain

$$\Gamma ={\mu }^{2}+\left({\frac{9{k}_{3}{\omega }_{0}^{2}-10{k}_{2}^{2}}{24{\omega }_{0}^{3}}{a}_{0}^{2}-\sigma} \right)\left({\frac{9{k}_{3}{\omega }_{0}^{2}-10{k}_{2}^{2}}{8{\omega }_{0}^{3}}{a}_{0}^{2}-\sigma} \right)<0.$$

(A.32)

Therefore, when $\Gamma <0$, the steady-state motion is unstable, otherwise it is stable. In a nonlinear system, if there are multiple steady-state solutions, the initial conditions determine which steady-state solution the system physically realizes.