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?


2 Answers 2


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.

  • $\begingroup$ I did forget the $u(t)$! Thanks for pointing that out. $\endgroup$
    – griffin175
    Commented Nov 25, 2019 at 20:19

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:

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

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


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