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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)(2-z)}, \quad \text{ROC} \quad |z|<2 $.
(Write enough intermediate steps to fully justify your answer.)
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{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)
$ = -\frac{1}{3-z}-\frac{1}{2-z} $
$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $
$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $
$ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $
Let k=-n
$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $
by comparison with z-transform formula
x[n] = u[ − n]( − 3n − 1 − 2n − 1)
- Grader's comment: Partial Fractions splitting is wrong
Answer 2
Using a partial fraction expansion, we can change the original equation to
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ Where A = 1, B = -1, so we get (Instructor's comment: You can skip this explanation and write the expansion directly)
$ = -\frac{1}{3-z}-\frac{1}{2-z} $
By factoring out 1/3 for the first term, and 1/2 for the second term, we can have both terms in form of (Instructor's comment: No need to explain this.)
$ \frac{1}{1-r} $, which is equal to $ \sum_{n=0}^{+\infty} (\frac{1}{r})^n $ (Instructor's comment: This is not true in general: only when |r|<1)
$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $
$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $
$ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $
Then let k=-n
$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $
Comparing it with z-transform formula, we can get
x[n] = u[ − n]( − 3n − 1 − 2n − 1)
- Grader's comment: Partial Fractions splitting is wrong
Answer 3
First, using partial fraction we get..
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)
A(2-z) + B(3-z) = 1
let z=2, then B=1
let z=3, then A=-1 (Instructor's comment: You do not need to explain how you got the A and the B. )
$ = -\frac{1}{3-z}+\frac{1}{2-z} $
$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})+\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $
$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $
$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{1}{3})^n(z)^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{1}{2})^n(z)^n $
now let n = -k
$ = -\frac{1}{3}\sum_{n=0}^{+\infty} 3^{k} z^{-k} +\frac{1}{2}\sum_{n=0}^{+\infty} 2^{k}z^{-k} $ (Instructor's comment: Your sum is over n, but the expression depends on k.)
by comparison with z-transfrom formula
x[n] = − 3n − 1u[ − n] + 2n − 1u[ − n]
x[n] = ( − 3n − 1 + 2n − 1)u[ − n]
Answer 4
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)
$ = -\frac{1}{3-z} - \frac{1}{2-z} $
$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $
$ = -\frac{1}{3}\sum_{k=0}^{+\infty} (\frac{z}{3})^k -\frac{1}{2}\sum_{k=0}^{+\infty} (\frac{z}{2})^k $
$ = \sum_{k=0}^{+\infty}[(-\frac{1}{3})(\frac{1}{3})^k + (-\frac{1}{2})(\frac{1}{2})^k]u[k] * z^k $
Substitute k with -n
$ = \sum_{n=-\infty}^{+\infty}[(-\frac{1}{3})3^{-n} + (-\frac{1}{2})2^{-n}]u[-n] * z^{-n} $
Look up Z transform equation on RHEA table and see that X(z) becomes...
x[n] = ( − 3n − 1 − 2n − 1)u[ − n]
- Grader's comment: Partial Fractions splitting is wrong
Answer 5
by partical fraction, we get,
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)
$ = -\frac{1}{3-z}+\frac{1}{2-z} $
For $ \quad \text{ROC} \quad |z|<2 $
$ X(z)= -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})+\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $
$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n +\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $
assume n=-k.
$ X(z)= -\frac{1}{3}\sum_{k=-\infty}^{0} 3^{k} z^{-k} +\frac{1}{2}\sum_{k=-\infty}^{0} 2^{k}z^{-k} $
$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + \frac{1}{2}2^k]z^{-k} $
So, x[n] = (−3n-1+2n-1)u[-n]
Answer 6
$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.)
$ = \frac{1}{2-z}-\frac{1}{3-z} $
$ = \frac{1}{2}\frac{1}{1-\frac{z}{2}}-\frac{1}{3}\frac{1}{1-\frac{z}{3}} $
By the geometric series formula,
$ X(z) = \frac{1}{2}\sum_{n=0}^{+\infty}(\frac{z}{2})^n - \frac{1}{3}\sum_{n=0}^{+\infty}(\frac{z}{3})^n $
$ = \sum_{n=0}^{+\infty}(\frac{1}{2}(\frac{1}{2})^n - \frac{1}{3}(\frac{1}{3})^n)z^n $
$ = \sum_{n=-\infty}^{+\infty}u[n]((\frac{1}{2})^{n+1} - (\frac{1}{3})^{n+1})z^n $
Substituting k = -n for n gives,
$ X(z) = \sum_{k=-\infty}^{+\infty}u[-k](\frac{1}{2}^{-k+1} - \frac{1}{3}^{k+1})z^{-k} $
$ = \sum_{k=-\infty}^{+\infty}u[-k](2^{k-1} - 3^{k-1})z^{-k} $
By comparison with the Z-transform formula,
x[n] = u[-n](2n-1-3n-1)
Answer 7
$ X(z) =\frac{1}{(3-z)(2-z)} $
$ = \frac{A}{3-z}+\frac{B}{2-z} $ (Instructor's comment: You can skip this step.) $ = -\frac{1}{3-z}-\frac{1}{2-z} $
$ = -\frac{1}{3}*(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}*(\frac{1}{1-\frac{z}{2}}) $ (Instructor's comment: Be careful! You do not mean convolution here, do you? Then you should use $ \times $ instead of $ * $.)
$ = -\frac{1}{3}\sum_{n=0}^{+\infty}(\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty}(\frac{z}{2})^n $ $ = -\frac{1}{3}\sum_{n=0}^{+\infty}(\frac{1}{3})^n(z)^n -\frac{1}{2}\sum_{n=0}^{+\infty}(\frac{1}{2})^n(z)^n $
substituting k for -n:
$ = -\frac{1}{3}\sum_{n=0}^{+\infty}3^{k} z^{-k}-\frac{1}{2}\sum_{n=0}^{+\infty}2^{k}z^{-k} $
Using the Z transform tables to find the common transformation: (Instructor's comment: I don't understand what you mean. You should rephrase this.)
x[n] = (−3^(n−1))u[−n] - (2^(n−1))u[− n]
- Grader's comment: Partial Fractions splitting is wrong
