Consider an LTI system with impulse response $$h[n] = \frac{1}{5^n} u[n].$$

(a) Determine and sketch the magnitude and phase response $|H(\omega)|$ and $\arg(H(\omega))$ respectively.

(b) Determine and sketch the magnitude and phase spectra for the input and outPUt signals for the following inputs:

  1. $x[n] = \cos(6\pi n/10)$ for $-\infty < n < \infty$
  2. $x[n] = \{ ... ,1,0,0,1,1,1,0,1,1,1,0,1, ... \}$ |
  • 1
    $\begingroup$ Can I please ask what have you tried so far? $\endgroup$ – A_A Dec 28 '17 at 9:37

For part a,

given the impulse response as $$h[n] = \left( \frac{1}{5} \right)^n u[n] $$ its frequency response is the DTFT of $h[n]$ which is:

$$ h[n] = \left( \frac{1}{5} \right)^n u[n] \implies H(\omega) = \frac{1}{1 - \frac{1}{5} e^{-j\omega} } $$

For the part b,

you must use the convolution property which says that: $$ y[n] = x[n] \star h[n] \implies Y(\omega) = X(\omega) H(\omega) $$

| improve this answer | |
  • $\begingroup$ thanx , but what about the magnitude and phase of H(w) $\endgroup$ – eng.fawzy Dec 27 '17 at 21:55
  • $\begingroup$ $H(\omega)$ is a complex valued function So you can plot its magnitude and phase just as if you were computing the magnitude and phase of an ordinary complex number, but, for each $\omega$ instead of a single one. $\endgroup$ – Fat32 Dec 27 '17 at 22:22
  • $\begingroup$ Fat32: There's another question with a similar result and they call it the transfer function but use z instead. dsp.stackexchange.com/questions/33858/…. is it correct to say that when $z$ is used, it's referred to as the transfer function and when $e^{jw}$ is used, it's referred to as the frequency response ? As you can tell, I find the terminology confusing. thanks. $\endgroup$ – mark leeds Dec 28 '17 at 7:33
  • $\begingroup$ FAt32: I ask this question because I $\endgroup$ – mark leeds Dec 28 '17 at 7:37
  • $\begingroup$ have one book where they use poliynomials in z to obtain the output. but there's also this approach that you talk about here. clearly, the 2 approaches must be equivalent but maybe there are cases where one is easier to use than the other or one is inappropriate. thanks. $\endgroup$ – mark leeds Dec 28 '17 at 7:46

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.