Stability Analysis Examples

Example: Determine the stability of $G(s)=\frac{1}{s+\gamma }$. We let the input $u(t)$ be the unit step function $\sigma (t)$ to analyze the system.

Let $u(t)=\sigma (t)$. Then, we have have:

$$Y(s)=G(s)U(s)=\frac{1}{s+\gamma }\cdot \frac{1}{s}=\frac{1}{s(s+\gamma )}$$

For $\gamma <0$, we have:

$$\begin{array}{rl}y(t)& ={\mathcal{L}}^{-1}\left[\frac{1}{s(s+\gamma )}\right]\\ & =\frac{1}{\gamma }(1-e^{-\gamma t})u(t)\end{array}$$

When $t\to \infty$, $y(t)\to \infty$, so the system is unstable.

For $\gamma =0$. we have

$$y(t)={\mathcal{L}}^{-1}\left[\frac{1}{s^2}\right]=t\sigma (t)$$

When $t\to \infty ,y(t)\to \infty$, so the system is unstable.

For $\gamma >0$, the impulse response of $G(s)$ is $g(t)=e^{-\gamma t}\sigma (t)$. Then,

$$|y(t)|=\left|{\int }_0^{t}g(\tau )u(t-\tau )d\tau \right|\le {\int }_0^t|g(\tau )||u(t-\tau )|d\tau$$

If $|u(t)\le M|$ for all $t\in [0,\infty )$, then

$$|y(t)|\le M{\int }_0^{\infty }|g(\tau )|d\tau =M{\int }_0^{\infty }{e}^{-\gamma \tau }d\tau =\frac{M}{\gamma }$$

As long as $u(t)$ is bounded by $M$, then $y(t)$ is bounded by $\frac{M}{\gamma }$. This is stable.