Difference between revisions of "ECE438 Week10 Quiz Q4sol" - Rhea

Revision as of 18:23, 27 October 2010

ECE438 Lab Week10 Quiz Question 4 Solution

a. By computing X(z) and Y(z), we can obtain H(z)=Y(z)/X(z)

$ \begin{align} X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}&=\sum_{n=-\infty}^{\infty}(\frac{1}{2})^nu[n]z^{-n}+\sum_{n=-\infty}^{\infty}2^nu[-n-1]z^{-n} \\ &=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=-\infty}^{-1}2^nz^{-n} \\ &=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=1}^{\infty}2^{-n}z^{n} \\ &=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=0}^{\infty}2^{-n}z^{n}-1 \\ &=\frac{1}{1-\frac{1}{2}z^{-1}}+\frac{1}{1-\frac{z}{2}}-1\text{ ,if }|z|>\frac{1}{2}\text{ and }|z|<2 \\ &=\frac{\frac{3}{4}z^{-1}}{(1-\frac{1}{2}z^{-1})(z^{-1}-\frac{1}{2})}\text{ ,ROC: }\frac{1}{2}<|z|<2 \end{align} $

$ \begin{align} Y(z)=\sum_{n=-\infty}^{\infty}y[n]z^{-n}&=\sum_{n=-\infty}^{\infty}6(\frac{1}{2})^nu[n]z^{-n}-\sum_{n=-\infty}^{\infty}6(\frac{3}{4})^nu[n]z^{-n} \\ &=\sum_{n=0}^{\infty}6(\frac{1}{2})^nz^{-n}-\sum_{n=0}^{\infty}6(\frac{3}{4})^nz^{-n} \\ &=\sum_{n=0}^{\infty}6(\frac{1}{2})^nz^{-n}-\sum_{n=0}^{\infty}6(\frac{3}{4})^nz^{-n} \\ &=\frac{6}{1-\frac{1}{2}z^{-1}}-\frac{6}{1-\frac{3}{4}z^{-1}}\text{ ,if }|z|>\frac{1}{2}\text{ and }|z|>\frac{3}{4} \\ &=\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}\text{ ,ROC: }|z|>\frac{3}{4} \end{align} $

Thus

$ H(z)=\frac{Y(z)}{X(z)}=\frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}\text{ ,ROC: }\frac{3}{4}<|Z|<2 $

b. By computing the inverse Z transform of H(z), we can obtain the impulse response h[n]

$ H(z)=\frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}=\frac{1}{1-\frac{3}{4}z^{-1}}-2*\frac{z^{-1}}{1-\frac{3}{4}z^{-1}} $

Given the ROC of $ \frac{3}{4}<|Z|<2 $, the z inverse transform of $ \frac{1}{1-\frac{3}{4}z^{-1}} $ can be obtained as $ (\frac{3}{4})^nu[n] $

using time shifting property of Z transform we can get

$ h[n]=(\frac{3}{4})^nu[n]-2(\frac{3}{4})^{n-1}u[n-1] $

c.

Back to Quiz Pool

Back to Lab wiki