Practice Question inverse z transform 4 ECE438F13 - Rhea

Question

Compute the inverse z-transform of

$ X(z) =\frac{1}{(3-z)(2-z)}, \quad \text{ROC} \quad 2<|z|<3 $.

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

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Answer 1

Ruofei

$ X(Z) = \frac{1}{(3-Z) (2-Z)} $

$ X(Z) = -\frac{1}{3-Z} + \frac{1}{2-Z} $

$ X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} + \frac{1}{Z} \frac{1}{\frac{2}{Z}-1} $

$ X(Z) = -\frac{\frac{1}{3}}{1-\frac{Z}{3}} - \frac{1}{Z} \frac{1}{1-\frac{2}{Z}} $

Since $ |2|<Z<|3| $

$ \frac{1}{1-\frac{2}{Z}} = \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n} $

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

Thus,

$ X(Z) = -\frac{1}{3} \sum_{n=0}^{+\infty} (\frac{Z}{3})^{n} + \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{2}{Z})^{n} $

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

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

In $ -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} $, Let k=-n, then -k=n

In $ \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n} $, Let i=n+1, then n=i-1

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

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

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

Grader's comment: Made a mistake in the last step . It should be 2 instead of 3

Answer 2

Li-Pang Mo

$ X(z) =\frac{1}{(3-z)(2-z)} $

$ X(z) =\frac{-1}{3-z} + \frac{1}{2-z} $

$ X(z) =(\frac{-1}{3})(\frac{1}{1-\frac{z}{3}}) + (\frac{-1}{z})(\frac{1}{1-\frac{2}{z}}) $

$ |2|<Z<|3| $, which makes $ \frac{z}{3}<1, \frac{2}{z}<1 $

Use geometric series:

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

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

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

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

$ let p = -n , n = -p, q = n+1 , n = q-1 $

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

By observation:

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

Grader's comment: Answer is Correct

Answer 3

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Answer 4

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