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I have a few questions about this assignment I am solving. It is q3.33(a) from oppenheim's DSP book 2nd edition.

basically, the question gives you a pole zero plot, from which I was able to determine the correct x[n]. (Couple classmates confirmed my answer on that).

and it also gives you the y[n] of the form $(0.5)^n \cdot x[n]$.

And wants you to work out a pole-zero of y[n] in the z plane.

my question is this: given that I already know X[z], can i just do z transform of $0.5^n$ as if it was $a^n \cdot u[n]$?

Then I will have $[\frac{1}{1-0.5z^{-1}}] \cdot X[z]$

Thanks everyone.

D.Zou
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  • are you sure that "$$" ain't a convolution symbol? $$ y[n] = \left(\frac12\right)^n \circledast x[n] $$ might very well be LTI with a Z Transform and poles and zeros (it might need a unit step function in there). but $$ y[n] = \left(\frac12\right)^n \cdot x[n] $$ is not LTI. i was* gonna edit your question for clarity, but i better not if i don't know what you're saying. – robert bristow-johnson Oct 05 '14 at 19:05
  • @robertbristow-johnson thanks, I am pretty new at the formatting of equation SE uses and had no idea how to do the dot or the convolution sign. And my DSP book uses * for convolution so I just wanted to avoid confusion. – D.Zou Oct 05 '14 at 19:08
  • @robertbristow-johnson It's fixed now. – D.Zou Oct 05 '14 at 19:11
  • well, there is no Z transform transfer function if it's multiplication. if it's multiplication in one domain, it has to be convolution in the other domain. "poles and zeros" mean multiplication in normally the "frequency domain" or the "Z domain", but that would mean it's convolution in the $x[n],y[n]$ domain. – robert bristow-johnson Oct 05 '14 at 19:23
  • so I can't do Z[ab] = Z[a]Z[b] – D.Zou Oct 05 '14 at 19:49
  • @D.Zou: No, you can't. You can use a basic property of the $\mathcal{Z}$-transform to solve this problem. See the hint in my answer below. – Matt L. Oct 06 '14 at 10:39

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HINT:

Use the property

$$z_0^nx[n]\Longleftrightarrow X(z/z_0)$$

with $z_0=1/2$. You can find this property in Table 3.2 on page 126. It's quite straightforward to show this property by using the definition of the $\mathcal{Z}$-transform.

Matt L.
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