Cheers, I am given the input signal of $x(t) = A \cos(2πft + \frac{\pi}{4})$ which is linear and time invariant, and I am asked to find the output, if I know that, $f = 500hz$ and $|H(f)| = 1, \angle H(f) = \frac{\pi}{4}$

What I know, is that when given an exponential input, I can compute the output as $x(t) = A e^{j\omega_0t}$, then $y(t) = H(\omega_0)x(t)$. $H$ is a function that has an imaginary part, so we can say that: $H(f) = |H(f)|e^{j \arg H(f)}$.Thus I would say that the output is: $1 \times A \cos(2πft + \frac{\pi}{4}) \times e^{j\frac{π}{4}}$

My professor however states that the answer is: $1 \times A \cos(2πft + \frac{\pi}{4} + \frac{\pi}{4}) $, but I can't understand how he simply puts the $\frac{\pi}{4}$ inside the output of cosine. I tried to work it out using the euler's formula, but that doesn't seem to do the trick. Can anyone explain the correct answer? Thanks =)

  • 2
    $\begingroup$ Signals are not linear or time-invariant; systems are. $\endgroup$ Commented Feb 5, 2022 at 19:56

1 Answer 1


which is linear and time invariant,

linearity and time invariance is NOT a property of a signal, it only applies systems. In order for a system to have a transfer function in the first place, it needs to be LTI.

If you are not familiar with phasor notation, than can use indeed Euler.

$$x(t) = A\cos(\omega_0t + pi/4) = \frac{A}{2}\left( e^{j\omega_0t+\pi/4} + e^{-j\omega_0t+\pi/4}\right)$$

Assuming that the system has is real valued (i.e. has a real valued impulse response), we can use the symmetry property of the transfer function

$$H(-\omega) = H^*(\omega) = |H(\omega)|e^{-j\phi_H}$$

At this point we have an input that consists of two complex exponentials and we know the transfer function at both frequencies, so we get

$$y(t) = \frac{A}{2}\left( e^{j\omega_0t+\pi/4}H(\omega) + e^{-j(\omega_0t+\pi/4)} H(-\omega)\right) = \frac{A}{2}\left( e^{j\omega_0t+\pi/4}e^{j\pi/4} + e^{-j(\omega_0t+\pi/4)} e^{-j\pi/4}\right) $$ $$ = \frac{A}{2}\left( e^{j\omega_0t+\pi/4+\pi/4}+ e^{-j(\omega_0t+\pi/4+\pi/4)} \right) = A\cos(\omega_0 t + \pi/4 + \pi/4) = A\cos(\omega_0 t +\pi/2) $$

  • $\begingroup$ Thanks for your answer. I think in your last to equations you forgot the over 2, as it will get eliminated when you go from $e^{jθ} + e^{-jθ}$ to the cosine form. Other than that, I think that clears a lot. Thanks! $\endgroup$ Commented Feb 6, 2022 at 19:41

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