# 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

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

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

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

## Question 4

Compute the z-transform of the signal

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

$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} = \sum_{n=-\infty}^{\infty} (2^{n}u[n]+ 3^{n}u[-n+1]) z^{-n} = \sum_{n=-\infty}^{\infty} 2^{n}u[n] z^{-n} + \sum_{m=-\infty}^{\infty} 3^{m}u[-m+1] z^{-m}$

$X(z) = \sum_{n=-\infty}^{\infty} (\frac{2}{z})^n u[n] + \sum_{m=-\infty}^{\infty} (\frac{3}{z})^{m}u[-m+1]$

Let k = -m+1, then

$X(z) = \sum_{n=-\infty}^{\infty} (\frac{2}{z})^n u[n] + \sum_{k=-\infty}^{\infty} (\frac{z}{3})^{k-1}u[k] = \sum_{n=0}^{\infty} (\frac{2}{z})^n + \frac{3}{z} \sum_{k=0}^{\infty} (\frac{z}{3})^{k}$

$X(z) = \frac{1}{1-\frac{2}{z}} + \frac{3}{z} \frac{1}{1-\frac{z}{3}} = \frac{z}{z-2} + \frac{3}{z} \frac{3}{3-z} , if \quad 2 < |z| < 3$

$X(z) = \left\{ \begin{array}{l l} \frac{z}{z-2} + \frac{3}{z} \frac{3}{3-z} &, if \quad 2 < |z| < 3 \\ \text{diverges} &, \quad \text{otherwise} \end{array} \right.$

## Question 4

Compute the inverse z-transform of

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

$X(z)=\frac{1}{1+z}=\frac{1}{1-(-z)}$

So for $|z|<1$, we have

$X(z)=\sum_{k=0}^{\infty} (-z)^k = \sum_{k=-\infty}^{\infty} (-z)^k u[k] = \sum_{k=-\infty}^{\infty} (-1)^k u[k] (z)^k$

Let n = -k, then

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

$x[n]=(-1)^n u[-n] \$

## Question 5

Compute the inverse z-transform of

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

$X(z)=\frac{1}{1+2z}=\frac{1}{2z} \frac{1}{1+\frac{1}{2z}} =\frac{1}{2z} \frac{1}{1-(-\frac{1}{2z})}$

So for $|z| > \frac{1}{2}$, we have

$X(z)=\frac{1}{2z} \sum_{k=0}^{\infty} (-\frac{1}{2z})^k =\sum_{k=0}^{\infty} \frac{1}{2z} (-2z)^{-k} = \sum_{k=-\infty}^{\infty} \frac{1}{2} (-2)^{-k} z^{-k-1} u[k]$

Let n = k+1, then

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

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

## Question 6

Compute the inverse z-transform of

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

$X(z)=\frac{1}{1+2 z} =\frac{1}{1-(-2z)}$

So for $|z| < \frac{1}{2}$, we have

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

Let n = -k, then

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

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

## Question 7

Compute the inverse z-transform of

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

$X(z)=\frac{1}{(1+ z)(3-z)} =\frac{1}{4} (\frac{1}{1+z} + \frac{1}{3-z}) =\frac{1}{4} \frac{1}{1-(-z)} + \frac{1}{12} \frac{1}{1-\frac{z}{3}}$

So for $|z| < 1$, we have

$X(z)=\frac{1}{4} \sum_{k=0}^{\infty} (-z)^k + \frac{1}{12} \sum_{k=0}^{\infty} (\frac{z}{3})^k$

$X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^k z^k u[k] + \frac{1}{12} \sum_{k=-\infty}^{\infty} (\frac{1}{3})^k z^k u[k]$

Let n = -k, then

$X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^n u[-n] z^{-n} + \frac{1}{12} \sum_{k=-\infty}^{\infty} 3^n u[-n] z^{-n}$

$x[n]=[\frac{1}{4} (-1)^n + \frac{1}{12} 3^n] u[-n]$

## Question 8

Compute the inverse z-transform of

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

$X(z)=\frac{1}{(1+ z)(3-z)} =\frac{1}{4} (\frac{1}{1+z} + \frac{1}{3-z}) =\frac{1}{4} (\frac{1}{z} \frac{1}{1+\frac{1}{z}} - \frac{1}{z} \frac{1}{1-\frac{3}{z}}) =\frac{1}{4z} (\frac{1}{1-(-\frac{1}{z})} - \frac{1}{1-\frac{3}{z}})$

So for $|z| > 3$, we have

$X(z)=\frac{1}{4z} [\sum_{k=0}^{\infty} (-\frac{1}{z})^k - \sum_{k=0}^{\infty} (\frac{3}{z})^k]$

$X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^k z^{-k-1} u[k] - \frac{1}{4} \sum_{k=-\infty}^{\infty} 3^k z^{-k-1} u[k]$

Let n = k+1, then

$X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^{n-1} u[n-1] z^{-n} + \frac{1}{4} \sum_{k=-\infty}^{\infty} 3^{n-1} u[n-1] z^{-n}$

$x[n]=[-\frac{1}{4} (-1)^n - \frac{1}{12} 3^n] u[n-1]$

## Question 9

Compute the inverse z-transform of

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

$X(z)=\frac{1}{(1+ z)(3-z)} =\frac{1}{4} (\frac{1}{1+z} + \frac{1}{3-z}) =\frac{1}{4} \frac{1}{z} \frac{1}{1+\frac{1}{z}} + \frac{1}{12} \frac{1}{1-\frac{z}{3}}$

Similarly we can have

$X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^{n-1} u[n-1] z^{-n} + \frac{1}{12} \sum_{k=-\infty}^{\infty} 3^n u[-n] z^{-n}$

when $1<|z|<3$,

$X(z)=\sum_{k=-\infty}^{\infty} [-\frac{1}{4} (-1)^n u[n-1] + \frac{1}{12} 3^n u[-n]] z^{-n}$

So, $x[n]=-\frac{1}{4} (-1)^n u[n-1] + \frac{1}{12} 3^n u[-n]$

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett