I am studying for my DSP final and I came across this question from the Oppenhiem and Schafer book 3rd edition. The question says

3.18 A casual LTI system has the system function

$$H(z)=\dfrac{1+2z^{-1}+z^{-2}}{\left(z+\frac{1}{2}z^{-1} > \right)\left(1-z^{-1} \right)} $$

(a) Find the impulse response of the system, $h[n]$

(b) Find the output of this system, $y[n]$, for the input

$$x[n]=e^{j(\pi/2)n} $$

For the first part, I got $$h[n] = -2 \delta[n]+\frac{1}{3} \left(\frac{-1}{2}\right)^2 u[n]+\frac{8}{3} u[n]$$

which matches the answer at the back of the book, but the second part is difficult. We are asked to find the function $y[n]$ if $x[n] = 2^n$

I know that $2^n$ is an eigenfunction of $y[n]$, but I am not sure how to substitute it in $H(e^{j\omega})$. The answer at the back of the book is $y[n] = \frac{18}{5} 2^n$

Can anyone help me? Thanks in advance


Hint: $y[n] = H(e^{j (\pi / 2)}) x[n]$.

Since $x[n]$ is a complex exponential, it is an eigenfunction of a LTI system with eigenvalue given by the frequency response at the frequency of $x[n]$ (which in this case is $H(e^{j (\pi/2)})$. In less fancy words, if you put a complex exponential into a LTI system, you get the same complex exponential out, except scaled by the transfer function evaluated at the frequency of the complex exponential.

  • $\begingroup$ Yes, thanks that makes sense for x[n] = e^j(pi/2)n but in the newer edition version of the the same question. They put x[n] = 2^n. I know that 2^n is an eigenfunction of the system, but how do I find y[n] ? Is it y[n] = H(e^jw) * x[n] ? If yes what do I substitute for w ? $\endgroup$ – Mona Dec 6 '14 at 17:15

Y[z] = H[z]X[z], and y[n] is found by taking the inverse Z-transform of Y[z].

You have H[z] already, so the next step is finding the Z-transform of x[n] = 2^n. Assuming you need to find the unilateral Z-transform of x[n], you can use a table of Z-transform pairs (such as this one), you can see that X[z] = z/(z-2)... again, this assumes that you need to take the unilateral transform or that x[n] = (2^n)u[n].

Then, multiply H[z] by X[z] to get Y[z]. Once you have Y[z], you take the inverse Z-transform and you get y[n].

Hope that helps,

  • $\begingroup$ Thank you very much for your reply. I appreciate it, but my problem is that x[n] = 2^n and not 2^n u[n] . I think it has to do with eigenfunction stuff. How can I find X(z) in this case, or how can I substitute 2^n into H(e^jw) $\endgroup$ – Mona Dec 7 '14 at 5:26

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