m
Line 63: Line 63:
 
X(z) = \left\{
 
X(z) = \left\{
 
   \begin{array}{l l}
 
   \begin{array}{l l}
     \frac{z}{z-\frac{1}{5}} + \frac{z}{5-z} &, if \quad \frac{1}{5} < |z| < 5 </math> \\
+
     \frac{z}{z - \frac{1}{5}} + \frac{z}{5-z} &, if \quad \frac{1}{5} < |z| < 5 </math> \\
 
     \text{diverges} &, \quad \text{otherwise}
 
     \text{diverges} &, \quad \text{otherwise}
 
   \end{array} \right.
 
   \end{array} \right.

Revision as of 21:27, 2 November 2014


Homework 7 Solution, ECE438 Fall 2014, Prof. Boutin

Questions 1

Compute the z-transform of the signal

$ x[n]= \left( \frac{1}{2} \right)^n u[-n] $

$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} = \sum_{n=-\infty}^{\infty} (\frac{1}{2})^n u[-n] z^{-n} = \sum_{n=-\infty}^{\infty} (2z)^{-n} u[-n] $

Let k=-n, then

$ X(z) = \sum_{k=-\infty}^{\infty} (2z)^k u[k] = \sum_{k=0}^{\infty} (2z)^k $

$ X(z) = \left\{ \begin{array}{l l} \frac{1}{1-2z} &, if \quad |z| < \frac{1}{2}\\ \text{diverges} &, \quad \text{otherwise} \end{array} \right. $


Questions 2

Compute the z-transform of the signal

$ x[n]= 5^n u[n-3] \ $

$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} = \sum_{n=-\infty}^{\infty} 5^n u[n-3] z^{-n} = \sum_{n=-\infty}^{\infty} (\frac{5}{z})^{n} u[n-3] $

$ X(z) = \sum_{n=3}^{\infty} (\frac{5}{z})^{n} = \frac{(\frac{5}{z})^3}{1-\frac{5}{z}} = (\frac{5}{z})^3 \frac{z}{z-5} , if \quad |z| > 5 $

$ X(z) = \left\{ \begin{array}{l l} (\frac{5}{z})^3 \frac{z}{z-5} &, if \quad |z| > 5\\ \text{diverges} &, \quad \text{otherwise} \end{array} \right. $


Questions 3

Compute the z-transform of the signal

$ x[n]= 5^{-|n|} \ $

$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} = \sum_{n=-\infty}^{\infty} 5^{-|n|} z^{-n} = \sum_{n=0}^{\infty} (\frac{1}{5z})^n + \sum_{m=-\infty}^{-1} (\frac{5}{z})^m $

Let k=-m, then

$ X(z) = \sum_{n=0}^{\infty} (\frac{1}{5z})^n + \sum_{k=1}^{\infty} (\frac{5}{z})^k $

$ X(z) = \frac{1}{1-\frac{1}{5z}} + \frac{ \frac{z}{5}}{1-\frac{z}{5}} = \frac{z}{z-\frac{1}{5}} + \frac{z}{5-z} , if \quad \frac{1}{5} < |z| < 5 $

$ X(z) = \left\{ \begin{array}{l l} \frac{z}{z - \frac{1}{5}} + \frac{z}{5-z} &, if \quad \frac{1}{5} < |z| < 5 $ \\

   \text{diverges} &, \quad \text{otherwise}
 \end{array} \right.

</math>

$ X(z) = \left\{ \begin{array}{l l} \frac{z}{z-\frac{1}{5}} + \frac{z}{5-z} &, if \quad \frac{1}{5} < |z| < 5 $ \\

   \text{diverges} &, \quad \text{otherwise}
 \end{array} \right.

</math>


Question 4

Compute the z-transform of the signal

$ x[n]= 2^{n}u[n]+ 3^{n}u[-n+1] \ $

Question 4

Compute the inverse z-transform of

$ X(z)=\frac{1}{1+z}, \text{ ROC } |z|<1 $


Question 5

Compute the inverse z-transform of

$ X(z)=\frac{1}{1+2 z}, \text{ ROC } |z|> \frac{1}{2} $

Question 6

Compute the inverse z-transform of

$ X(z)=\frac{1}{1+2 z}, \text{ ROC } |z|< \frac{1}{2} $

Question 7

Compute the inverse z-transform of

$ X(z)=\frac{1}{(1+ z)(3-z)}, \text{ ROC } |z|<1 $


Question 8

Compute the inverse z-transform of

$ X(z)=\frac{1}{(1+ z)(3-z)}, \text{ ROC } |z|>3 $

Question 9

Compute the inverse z-transform of

$ X(z)=\frac{1}{(1+ z)(3-z)}, \text{ ROC } 1< |z|<3 $



Back to ECE438, Fall 2014, Prof. Boutin

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang