Topic: Computing an inverse z-transform

## Question

Compute the inverse z-transform of

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

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

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

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]

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

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]

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

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

$= \frac{A}{3-z}+\frac{B}{2-z}$ (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