2
$\begingroup$

The question I'm trying to understand is as follows: A linear time-invariant continuous-time system has the frequency response function $$H(\omega)=\frac{1}{j\omega+1} $$

Compute the output response $y(t)$ for $-\infty<t<\infty$ when the input $$x(t)=\cos(t),\quad -\infty<t<\infty$$

I know the answer, but I don't understand how to get there.

I know $H(\omega)=\frac{Y(\omega)}{X(\omega)}$, and from that $y(t)=h(t)*x(t)$. Taking the inverse Fourier transform of $H(\omega)$ I get $e^{-t}$. If I try to compute the convolution I just get a divergent integral. Is this wrong? The answer is supposedly $$y(t)=|H(1)|\cos(t+\angle H(1))$$ but this seems to come from nowhere. Can someone please explain?

$\endgroup$
2
$\begingroup$

The easiest way to solve the problem is using the knowledge of eigenfunctions of LTI system and the consequence that an LTI system's response to a sinusoidal input $x(t)=A\cos(\omega_0t+\phi)$ is given by

$$y(t)=A\big|H(\omega_0)\big|\cos\big(\omega_0t+\phi+\angle H(\omega_0)\big)\tag{1}$$

where $H(\omega)$ is the system's frequency response. This is explained in detail in Fat32's answer.

However, I would like to add that there must be some mistake in your derivation if you end up with a divergent integral. Note that the system's impulse response is $h(t)=e^{-t}u(t)$, where $u(t)$ is the unit step function. So the convolution integral becomes

$$\begin{align}(x\star h)(t)&=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\\&=\int_{-\infty}^{\infty}\cos(\tau)e^{-(t-\tau)}u(t-\tau)d\tau\\&=e^{-t}\int_{-\infty}^{t}\cos(\tau)e^{\tau}d\tau\tag{2}\end{align}$$

which is clearly not divergent. If you solve $(2)$ you get the correct result. I think that you forgot about the step function $u(t)$ in your derivation of the system's impulse response.

$\endgroup$
  • $\begingroup$ I did forget the $u(t)$! Thanks for pointing that out. $\endgroup$ – griffin175 Nov 25 '19 at 20:19
2
$\begingroup$

Apply the eigenfunction property of the LTI system after decomposing the sinusoidal input by Euler identity.

The eigenfunction property of the LTI system states that

$$ x(t) = e^{j \omega_0 t} \implies y(t) = H(\omega_0) e^{j \omega_0 t} $$

where $H(w)$ is the frequency response of the LTI system.

Euler identity states that :

$$ x(t) = \cos(t) = 0.5 \{ e^{jt} + e^{-jt} \} $$

Then due to system being LTI, you can define the output as:

$$ y(t) = 0.5 \{ H(1) e^{jt} + H(-1) e^{-jt} \} .$$

Since the system impulse response was a real function, its frequency response $H(w)$ will be conjugate symmetric; i.e., $H(w) = H^*(-w)$.

Let's call $H(1) = A e^{j\phi} $ where $A = |H(1)|$ is the magnitude and $\phi = \angle{H(1)}$ is the phase of the frequency response at the frequency $\omega = 1$.

Hence the output will be : $$ y(t) = 0.5 \{ H(1) e^{jt} + H^*(1) e^{-jt} \} $$

consider the term $z = H(1)e^{jt}$ as a complex variable, then the second term will be $z^*$, hence the output is:

$$y(t) = 0.5 \{ z + z^* \} = \mathcal{Re} \{ z \} $$

$$y(t) = \mathcal{Re} \{ A e^{j\phi} e^{jt} \} = A \cos(t + \phi) = |H(1)| \cos(t + \angle{H(1)})$$

$\endgroup$

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.