1
$\begingroup$

I have an exercise that gives me the transfer function of a system

$$H(s) = \frac{3s^2+27}{s^4+8s^3 + 16s^2} $$

and an input

$$x(t) = \frac13 cos(3t) $$

An ask's what is the output

I don't whant the answer but the steps to calculate it.

Where to start? I need to find the inverse or there's a shorcut?

$\endgroup$
  • $\begingroup$ I assume that the definition of $x(t)$ holds only for $t>0$, and that $x(t)=0$ for $t<0$, right? $\endgroup$ – Matt L. Jun 4 '14 at 10:23
  • $\begingroup$ Are you sure that the numerator isn't $3s^2+27$ (instead of $s^3$)? This would simplify the problem quite a bit. $\endgroup$ – Matt L. Jun 4 '14 at 10:41
  • $\begingroup$ Yes you are right its $$s^2$$ and it can be written as $$ 3 (s^2 + 3^2) $$ $\endgroup$ – Giannis Foulidis Jun 4 '14 at 12:30
  • 1
    $\begingroup$ OK, then things cancel nicely and the result can be obtained quite easily. $\endgroup$ – Matt L. Jun 4 '14 at 12:47
4
$\begingroup$

For an LTI system, output $y(t)$ is given by $$y(t) = h(t)\otimes x(t)$$ Where $x(t)$ is input and $h(t)$ is impulse response of the system. The operator $\otimes$ represents convolution.

Convolution operation is mapped into multiplication in Laplace domain. ie, $$Y(s) = H(s)\times X(s)$$ Where, $Y(s)$, $H(s)$ and $X(s)$ are the Laplace transform of $y(t)$, $h(t)$ and $x(t)$ respectively.

You can use any one of the above equation to solve your problem.

Two ways:

  1. Find $h(t)$ from $H(s)$ and use first equation. OR
  2. Find $X(s)$ from $x(t)$, use second equation to find $Y(s)$ and then find $y(t)$ from it.

I prefer second method.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.