3 Engineering Example 3

3.1 System response

An engineering system is modelled by the block diagram in Figure 29:

Figure 29

Determine the system response $v_0\left(t\right)$ when the input function is a unit step function when $K=2.5$ and $a=0$ .

Solution

If the system has an overall transfer function $H\left(s\right)$ then $V_0\left(s\right)=H\left(s\right)V_1\left(s\right)$ . But this particular system is the negative feedback loop described earlier and so

$H\left(s\right)=\frac{G\left(s\right)}{1+G\left(s\right)}=\frac{\frac{K}{1+as}}{1+\frac{K}{1+as}}=\frac{K}{K+1+as}$

In this particular case

$H\left(s\right)=\frac{2.5}{3.5+0.5s}=\frac{5}{7+s}$

Thus the impulse response $h\left(t\right)$ is

$h\left(t\right)={\mathcal{L}}^{-1}\left\{H\left(s\right)\right\}={\mathcal{L}}^{-1}\left\{\frac{5}{\left(7+s\right)}\right\}=5{\text{e}}^{-7t}u\left(t\right)$

and so the response to a step input $u\left(t\right)$ is given by the convolution of $h\left(t\right)$ with $u\left(t\right)$