Topic: Computing an inverse z-transform

## Question

Compute the inverse z-transform of

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

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

$=\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n = \sum_{n=0}^{+\infty} (-z^{-1}) (3z^{-1})^n$
NOTE: $(3z^{-1})^n = (3^n) (z^{-n})$
$= \sum_{n=0}^{+\infty} -3^n z^{-n-1} = \sum_{n=-\infty}^{+\infty} -3^n u[n] z^{-n-1}$
NOTE: Let k=n+1

$= \sum_{n=-\infty}^{+\infty} -3^{k-1} u[k-1] z^{-k}$ (compare with $\sum_{n=-\infty}^{+\infty} x[n] z^{-k}$)

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


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

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

Let k = n+1, so n = k-1

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

By comparison with the z-transform equation

$x[n] = -u[n-1] 3^{n-1}$

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

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

Let k = n+1 then n = k-1

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

By comparison,

$x[n] = -u[n-1] 3^{n-1}$

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

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

Since,$|3|<Z$

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

Thus,

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

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

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

Let k=n+1, then -k=-n-1,n=k-1

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

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

By Yixiang Liu

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

$X(z) = \frac{1}{-z}* \frac{1} {\frac{3}{-Z}+1}$  : Grader's comment: Use x instead of convolution symbol

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

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

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

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

Let -k = -n-1, so n = k-1

$X(z) = \sum_{-\infty}^{+\infty} u[k-1] 3^{k-1} Z^{-k}$

By comparison with the x-transform formula

$x[n] = 3^{n-1} u[n-1]$

Grader's comment: Forgot the negative sign

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

we have |z| > 3,

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

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

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

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

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

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

$X(z) = \frac{-1}{z} \frac{1-\left( \frac{3}{z} \right) ^{n}}{1-\frac{3}{z}}, \quad \text{As n goes to} \quad \infty \text{ since ROC} \quad |z|>3$

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

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

$X(z) = -\sum_{n=0}^{\infty} 3^n z^{-\left(n+1 \right)}, \quad \text{Replace n+1 with k}$

$X(z) = -\sum_{k=1}^{\infty} 3^{k-1} z^{-k}$

$X(z) = -\sum_{-\infty}^{\infty} 3^{k-1} u[k-1] z^{-k}, \quad \text{By comparing with DTFT equation we get}$

$x[n] = -3^{\left( n-1 \right) } u[n-1]$

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

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

Using the geometric series formula,

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

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

$= \sum_{n=-\infty}^{\infty}-u[n]3^nz^{-n-1}$

Let k = n+1,

$= \sum_{k=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k}$

By comparison with the Z-transform formula,

x[n] = -u[n-1]3n-1

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

$= \frac{-1}{z}*\frac{1}{1-\frac{3}{z}}$

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

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

$= \sum_{n=0}^{\infty}-3^n z^{-n-1}$

$= \sum_{n=-\infty}^{\infty}-3^n u[n] z^{-n-1}$

substituting n+1 with k:

$= \sum_{n=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k}$

Comparing to common Z-transform tables:

x[n] = (−3^(n−1))u[n−1]

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.