Revision as of 00:54, 20 September 2013

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.)

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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

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