If we give a sinusoidal $\cos(\omega t)$ to a 1/s, it means that you just integrate the given input.

What if we have the same input but Transfer Funtion with a pole, let's say 1/(s+1), what will be the output of the system ?

Edit #1

According to my hand calculations, I guess there no way rather than dealing with Laplace Transform.

Hint: As you may know,

$$\begin{gather} \cos(\omega t) \rightarrow s/(s^2+w^2) \\ h(s) \rightarrow 1/(s+1) \end{gather}$$

output --> multiply them, and then apply partial fraction expansion, and then Laplace^-1

P.S.(Took approx. 30 min)


For a sinusoidal input signal at frequency $\omega$ in rad/s you can use that the steady state response of the system would be another sinusoidal signal whose amplitude is scaled by $|H(j\,\omega)|$ and its phase is shifted by $\angle H(j\,\omega)$. The total output of the system would be the sum of the steady state response and the transient response. The transient response only depends on the initial conditions and $H(s)$.

For example when the input is $\cos (\omega\,t)$ and $H(s)=1/(1+s)$ you get that the steady state response is equal to $|H(j\,\omega)|\cos(\omega\,t+\angle H(j\,\omega))$. The values for $|H(j\,\omega)|$ and $\angle H(j\,\omega)$ can be found by multiplying and dividing $H(j\,\omega)$ by the complex conjugate of its denominator

$$ H(j\,\omega) = \frac{1}{1+j\,\omega}\frac{1-j\,\omega}{1-j\,\omega} = \frac{1-j\,\omega}{1+\omega^2} $$

such that

$$ |H(j\,\omega)| = \frac{\sqrt{1 + \omega^2}}{1+\omega^2} $$ $$ \angle H(j\,\omega) = atan2(-\omega, 1) $$

The transient response of $H(s)$ are a scaled summation of $e^{\lambda\,t}$, where $\lambda$ are the poles of $H(s)$. So for this example the transient would be $C\,e^{-t}$ with $C$ depending on the initial conditions.

  • $\begingroup$ Thanks for the explanation. That's what I am looking for $\endgroup$ – cooleng Oct 22 '18 at 18:05

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