Practice Question, ECE438 Fall 2013, Prof. Boutin

On computing the inverse z-transform of a discrete-time signal.

Compute the inverse z-transform of

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

(Write enough intermediate steps to fully justify your answer.)

Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

Answer 1

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

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

$  = \sum_{n=-\infty}^{+\infty} -3^{k-1} u[k-1] z^{-k} $ (compare with $ \sum_{n=-\infty}^{+\infty} x[n] z^{-k} $) 

Therefore, $  x[n]= -3^{n-1} u[n-1]  $

Answer 2

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

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

Let k = n+1, so n = k-1

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

By comparison with the z-transform equation

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

Answer 3

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

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

Let k = n+1 then n = k-1

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

By comparison,

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

Answer 4

$ X(Z) = \frac{1}{3-Z} $

$ X(Z) = \frac{-1}{Z} \frac{1}{1-\frac{3}{Z}} $

Since,$ |3|<Z $

$ \frac{1}{1-\frac{3}{Z}} = \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $

Thus,

$ X(Z) = \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $

$ X(Z) = -Z^{-1} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $

$ X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] 3^{n}Z^{-n-1} $

Let k=n+1, then -k=-n-1,n=k-1

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

Therefore, $ x(n) = -u[n-1] 3^{n-1} $

Answer 5

By Yixiang Liu

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

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

$ X(z) = \frac{1}{-z}* \frac{1} {1-\frac{3}{Z}} $

$ X(z) = \frac{1}{-Z} \sum_{n=0}^{+\infty} (\frac{3}{3})^n $

$ X(z) = \sum_{n=0}^{+\infty} 3^{n} Z^{-n-1} $

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

Let -k = -n-1, so n = k-1

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

By comparison with the x-transform formula

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

Answer 6

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

we have |z| > 3,

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

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

Answer 7

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

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

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

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

$ X(z) = \frac{-1}{z} \frac{1-\left( \frac{3}{z} \right) ^{n}}{1-\frac{3}{z}}, \quad \text{As n goes to} \quad \infty \text{ since ROC} \quad |z|>3 $

$ X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} \left( \frac{3}{z} \right)^{n} $

$ X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} 3^n z^{-n} $

$ X(z) = -\sum_{n=0}^{\infty} 3^n z^{-\left(n+1 \right)}, \quad \text{Replace n+1 with k} $

$ X(z) = -\sum_{k=1}^{\infty} 3^{k-1} z^{-k} $

$ X(z) = -\sum_{-\infty}^{\infty} 3^{k-1} u[k-1] z^{-k}, \quad \text{By comparing with DTFT equation we get} $

$ x[n] = -3^{\left( n-1 \right) } u[n-1] $

Answer 5

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

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

Using the geometric series formula,

$ = -(\frac{1}{z})\sum_{n=0}^{\infty}(\frac{3}{z})^n $

$ = -(\frac{1}{z})\sum_{n=0}^{\infty}3^nz^{-n} $

$ = \sum_{n=-\infty}^{\infty}-u[n]3^nz^{-n-1} $

Let k = n+1,

$ = \sum_{k=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k} $

By comparison with the Z-transform formula,

x[n] = -u[n-1]3^n-1

Answer 9

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

$ = \frac{-1}{z}*\frac{1}{1-\frac{3}{z}} $

$ = \frac{-1}{z}\sum_{n=0}^{\infty}(\frac{3}{z})^n $

$ = \sum_{n=0}^{\infty}(-z^{-1})(3z^{-1})^n $

$ = \sum_{n=0}^{\infty}-3^n z^{-n-1} $

$ = \sum_{n=-\infty}^{\infty}-3^n u[n] z^{-n-1} $

substituting n+1 with k:

$ = \sum_{n=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k} $

Comparing to common Z-transform tables:

x[n] = (−3^(n−1))u[n−1]

Back to ECE438 Fall 2013 Prof. Boutin