Practice Question on "Digital Signal Processing"

Topic: Computing an inverse z-transform


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.)


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

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

Write it here.

Answer 4

Write it here.



Back to ECE438 Fall 2013 Prof. Boutin

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett