# Input-Output Relation [closed]

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)

## 1 Answer

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.

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