Second-Order System

A prototype 2nd-order system has the form:

$$H(s)=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$$

By the quadratic formula, the poles are:

$$\begin{array}{rl}s& =-\zeta {\omega }_n\pm {\omega }_n\sqrt{{\zeta }^2-1}\\ & =-{\omega }_n(\zeta \pm \sqrt{{\zeta }^2-1})\end{array}$$

The nature of the poles depends on $\zeta$:

• $\zeta >1$ means both poles are real and negative
• $\zeta =1$ means one negative pole
• $\zeta <1$ means two complex poles with negative real parts

In general, the poles take the form:

$$s=-\sigma \pm j{\omega }_d$$

where $\sigma =\zeta {\omega }_n$ and ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$ (damped frequency).

Note that

$$\begin{array}{rl}{\sigma }^2+{\omega }_d^2& ={\zeta }^2{\omega }_n^2+{\omega }_n^2-{\zeta }^2{\omega }_n^2\\ & ={\omega }_n^2\end{array}$$

and

$$\cos \varphi =\frac{\zeta {\omega }_n}{{\omega }_n}=\zeta$$

2nd-Order Response

Let’s compute the system’s impulse and step response:

$$H(s)=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}=\frac{{\omega }_n^2}{(s+\sigma {)}^2+{\omega }_d^2}$$

Impulse response

$$\begin{array}{rl}h(t)& ={\mathcal{L}}^{-1}\{H(s)\}\\ & ={\mathcal{L}}^{-1}\left\{\frac{({\omega }_n^2/{\omega }_d){\omega }_d}{(s+\sigma {)}^2+{\omega }_d^2}\right\}\\ & =\frac{{\omega }_n^2}{{\omega }_d}{e}^{-\sigma t}\sin ({\omega }_{d}t)\end{array}$$

Step response

$$\begin{array}{rl}{\mathcal{L}}^{-1}\left\{\frac{H(s)}{s}\right\}& ={\mathcal{L}}^{-1}\left\{\frac{{\sigma }^2+{\omega }_d^2}{s[(s+\sigma {)}^2+{\omega }_d^2]}\right\}\\ & =1-e^{-\sigma t}\left(\cos ({\omega }_{d}t)+\frac{\sigma }{{\omega }_d}\sin ({\omega }_{d}t)\right)\end{array}$$

We can see the variation in behavior with respect to the $\zeta$ here: