Practice Question inverse z transform example S15 - Rhea

Question

Compute the inverse z-transform of

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

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

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Answer 1

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

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

$ X(z) = \frac{1}{2} \frac{1-\left( \frac{z}{2} \right) ^n}{1- \frac{z}{2}} $ $ , \quad \text{As n goes to} \quad \infty \quad \text{since} \quad |z|<2 $ (*)

$ X(z) = \frac{1}{2} \sum_{n=0}^{\infty} \left( \frac{z}{2} \right)^n $

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

$ X(z) = \frac{1}{2} \sum_{n=-\infty}^{\infty} \left(\frac{1}{2} \right)^{-n} u[-n] z^{-n} $ $ , \quad \text{By comparing with DTFT equation, we get} \quad $

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

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

Grader's comment: Correct Answer

* This line is incorrect. The left-hand-side does not depend on 'n', so you cannot have your right-hand-side depend on 'n'. Note that, in the following lines, 'n' is the summation index in the right-hand-side (a "dummy" index"), so that is fine. -pm

Answer 2

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

$ =\frac{1}{2}\frac{1}{1-\frac{z}{2}} $

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

let n=-k

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

by comparison to inverse z-transform formula,

x[n] = 2 ^{− 1 + k}u[ − k]

Grader's comment: Right-hand-side is a function of 'k' , but left-hand-side is a function of 'n'. This can't be correct...

Answer 3

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

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

Let n = -k

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

By comparison with the z-transform formula,

x[n] = 2^{n − 1}u[ − n]

Looks good! -pm

Answer 4

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

Domain confusion: you started out computing a function of 'k', but along the way it became a function of 'n'. While the final answer is correct, your explanation is not 100% correct. -pm