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?

  • $\begingroup$ 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.
    Commented Jun 4, 2014 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.
    Commented Jun 4, 2014 at 10:41
  • $\begingroup$ Yes you are right its $$s^2$$ and it can be written as $$ 3 (s^2 + 3^2) $$ $\endgroup$ Commented Jun 4, 2014 at 12:30
  • 1
    $\begingroup$ OK, then things cancel nicely and the result can be obtained quite easily. $\endgroup$
    – Matt L.
    Commented Jun 4, 2014 at 12:47

1 Answer 1


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.


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