# The Spectrum of the Impulse Response of Linear Time Invariant (LTI) System

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. $x[n] = \{ ... ,1,0,0,1,1,1,0,1,1,1,0,1, ... \}$ |
• Can I please ask what have you tried so far? – 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: $$y[n] = x[n] \star h[n] \implies Y(\omega) = X(\omega) H(\omega)$$

• thanx , but what about the magnitude and phase of H(w) – eng.fawzy Dec 27 '17 at 21:55
• $H(\omega)$ 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. – Fat32 Dec 27 '17 at 22:22
• 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. – mark leeds Dec 28 '17 at 7:33
• FAt32: I ask this question because I – mark leeds Dec 28 '17 at 7:37
• 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. – mark leeds Dec 28 '17 at 7:46