Mechanical Impedance Analysis

Impedance analysis is commonly used to analyze electrical circuits [110]. By means of equivalent circuits, we can use the same analysis methods for mechanical systems.

For example, referring to Fig.7.9, the Laplace transform of the force on the spring $k$ is given by the so-called voltage divider relation:^8.2

$$F_k(s) = F_{\mbox{ext}}(s) \frac{R_k(s)}{R_m(s)+R_k(s)} = F_{\mbox{ext}}(s) \frac{\frac{k}{s}}{ms+\frac{k}{s}}$$

Similarly, the Laplace transform of the force on the mass $m$ is given by

$$F_m(s) = F_{\mbox{ext}}(s) \frac{R_m(s)}{R_m(s)+R_k(s)} = F_{\mbox{ext}}(s) \frac{ms}{ms+\frac{k}{s}}. \protect$$ (8.1)

As a simple application, let's find the motion of the mass $m$ , after time zero, given that the input force is an impulse at time 0:

$$f_{\mbox{ext}}(t)=\delta(t) \;\leftrightarrow\; F_{\mbox{ext}}(s)=1$$

Then, by the ``voltage divider'' relation Eq.(7.1), the Laplace transform of the mass force $f_m(t)$ after time 0 is given by

$$F_m(s) = \frac{ms}{ms+\frac{k}{s}} = \frac{s^2}{s^2+\frac{k}{m}} \isdef \frac{s^2}{s^2+\omega_0^2},$$

where we have defined $\omega_0^2\isdef k/m$ . The mass velocity Laplace transform is then

$$\begin{eqnarray*} V_m(s) &=& \frac{F_m(s)}{ms} \;=\; \frac{1}{m} \cdot \frac{s}{s^2+\omega_0^2}\\ [5pt] &=& \frac{1}{m} \left[\frac{1/2}{s+j\omega_0} + \frac{1/2}{s-j\omega_0}\right]\\ [5pt] &\leftrightarrow& \frac{1}{m} \cos(\omega_0 t). \end{eqnarray*}$$

Thus, the impulse response of the mass oscillates sinusoidally with radian frequency $\omega_0=\sqrt{k/m}$ , and amplitude $1/m$ . The velocity starts out maximum at time $t=0$ , which makes physical sense. Also, the momentum transferred to the mass at time 0 is $m\,v(0+) = 1$ ; this is also expected physically because the time-integral of the applied force is 1 (the area under any impulse $\delta(t)$ is 1).