Work out transfer function of dynamic process.

Question

Hi, I am having problem in solving exercise visible on picture. I don't really know how to begin. I know that I have to use Laplace transform and:

$$\mathcal{L}\{1(t)\} = 1/s $$

Did anyone face same problem before and can help me? :)

Answer 1

First, there are some properties from Signals and Systems you should keep in mind:

Considering $1(t)=1$ and that this is a causal system, which means $\neq 0 \iff t\geq0$, we have:

$$1'(t)=\delta(t)$$

and

$$\mathcal{L}\{\delta(t)\}=1$$

So, if you have the equation: $y''(t)+3y'(t)+2y(t)=u'(t)+u(t)$

Laplace-transforming the equation gives you:

$$ \begin{align} s^2Y(s)+3sY(s)+2Y(s)&=sU(s)+U(s)\\ Y(s)\left[s^2+3s+2\right]&=U(s)\left[s+1\right] \end{align} $$

Then, the transfer function for this system is:

$$\frac{Y(s)}{U(s)}=\frac{s+1}{s^2+3s+2}$$

When your input $u(t)=1(t)$, $U(s)=1/s$, so:

$$Y(s)=\frac{s+1}{s(s^2+3s+2)}$$

To find $y(t)$ apply the inverse Laplace transform. I suggest partial fractions expansion:

$$Y(s)=\frac{s+1}{s(s+1)(s+2)}=\frac{K_{1}}{s}+\frac{K_{2}}{s+1}+\frac{K_{3}}{s+2}$$

You should be able to go on from here!

Hope it was helpful.