Revision as of 14:04, 19 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}<math> <math>X(Z) = \frac{-1}{Z} \frac{1}{1-\frac{3}{z}}<math> <math>X(Z) = \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{3}{z})^{n}<math> <math><math> ---- ---- [[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]] $