Frequency-Domain Response of Second-Order Systems

Second order systems have follow the form:

$$\begin{array}{rl}m\ddot{x}+c\dot{x}+kx& =f(t)\\ (m{s}^2+cs+k)X(s)& =F(s)\end{array}$$

Thus, the transfer function is:

$$H(s)=\frac{X(s)}{F(s)}=\frac{1}{m{s}^2+cs+k}$$

Overdamped Case

In the overdamped case, where $\zeta >1$, there are two real poles $s_1$ and $s_2$. We have:

$$H(s)=\frac{1/k}{\frac{m}{s}{s}^2+\frac{c}{k}s+1}=\frac{1/k}{(s-s_1)(s-s_2)}$$

For frequency-domain system analysis, we have $s=j\omega$, such that:

$$H(j\omega )=\frac{1/k}{(j\omega -s_1)(j\omega -s_2)}$$

Underdamped Case

In the underdamped case, we have $0<\zeta <1$. This time, we divide by $m$ instead of $k$:

$$\begin{array}{rl}H(s)& =\frac{1/m}{s^2+\frac{c}{m}s+\frac{k}{m}}\\ & =\frac{1/m}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}\end{array}$$

where ${\omega }_n=\sqrt{\frac{k}{m}}$ and $\zeta =\frac{c}{2\sqrt{km}}$.

Then, if we divide by ${\omega }_n^2$, we have:

$$H(s)=\frac{1/m{\omega }_n^2}{\frac{s^2}{{\omega }_n^2}+\frac{2\zeta s}{{\omega }_n}+1}$$

Using $s=j\omega$ for frequency domain analysis (steady response to a sinusoid):

$$H(j\omega )=\frac{\frac{1}{m{\omega }_n^2}}{\frac{(j\omega {)}^2}{{\omega }_n^2}+2\zeta j\frac{\omega }{{\omega }_n}+1}$$

where $r=\frac{\omega }{{\omega }_n}$ is called the frequency ratio.

This gives:

$$\begin{array}{rl}H(j\omega )& =\frac{\frac{1}{m\left(\frac{k}{m}\right)}}{(-1)r^2+2\zeta rj+1}\\ & =\frac{1/k}{-r^2+2\zeta rj+1}\\ & =\frac{1/k}{1-r^2+j_2\zeta r}\\ & =\frac{\frac{1}{k}}{\sqrt{(1-r^2{)}^2+(2\zeta r{)}^2}}\end{array}$$

and

$$\begin{array}{rl}\angle H(j\omega )& =\angle \frac{1}{k}-\angle (1-r^2+j2\zeta r)\\ & =0-\arctan \left(\frac{2\zeta r}{1-r^2}\right)\\ & =-\arctan \left(\frac{2\zeta r}{1-r^2}\right)\end{array}$$

If we have $\omega =0$, we have $r=\frac{\omega }{{\omega }_n}=0$, which would give:

$$\begin{array}{rl}|H|& =\frac{1/k}{\sqrt{1-0}}=\frac{1}{k}\\ \angle H& =0\end{array}$$

If we have $\omega ={\omega }_n$ and $r=\frac{\omega }{{\omega }_n}=1$, we have:

$$\begin{array}{rl}|H|& =\frac{1/k}{\sqrt{0+(2\zeta {)}^2}}=\frac{1}{2\zeta k}\\ \angle H& =-\tan \left(\frac{2\zeta }{1-1}\right)=-\frac{\pi }{2}\end{array}$$

When $r\to \infty$, we have:

$$\begin{array}{rl}|H|& =0\\ \angle H& =-\pi \end{array}$$