Inverse $z$ transform - contour integration

Question

Here is my task:

Find inverse $z$ transform of $$X(z)=\frac{1}{2-3z}$$if $$|z|>\frac{2}{3}$$ using definition formula.

I found that $$x(n)=\dfrac{1}{3}\left (\dfrac{2}{3}\right )^{n-1}*u(n-1)=\dfrac{1}{3}\left (\dfrac{2}{3}\right )^{n-1}, n=1,2,3,...$$ (using other method). But how can I find it using definition formula, $$x(n)=\frac{1}{2\pi j}\oint_{C} X(z)z^{n-1}dz?$$

Thanks in advance.

Answer 1

\begin{equation} X(z)=\frac{1}{2 - 3z} \end{equation}

The result could be found by using

(i) Geometrical Series:
(a) Left Series, \begin{equation} X(z) = \frac{1}{2} \frac{ 1}{1 - 3 z/2} =\frac{1}{2} \sum_{n=0}^{\infty} \left ( \frac{3}{2} \right )^n (z^{-1})^{-n} =\frac{1}{2} \sum_{n=0}^{-\infty} \left ( \frac{3}{2} \right )^n (z^{-1})^{n}, \end{equation}

from which the coefficients are

\begin{equation} x[n] = \frac{1}{2} \left ( \frac{3}{2} \right )^n \quad n < 1 \end{equation}

This infinite series converges for $|z|<2/3 \, , \, |1/z| > 2/3$.

  (b) Right Series 

\begin{eqnarray} X(z) &=& \frac{1}{z(2 \, z^{-1} - 3)} \\ &=& \frac{z^{-1}}{(2 z^{-1} -3)} \\ &=& -\frac{1}{3} \frac{z^{-1}}{( 1 - (2/3) z^{-1})} \\ &=& - \left ( \frac{1}{3} \right ) z^{-1} \sum_{i=0}^{\infty} \left (\frac{2}{3} \right )^i z^{-i} \end{eqnarray} from which \begin{eqnarray} X(z) = - \left ( \frac{1}{3} \right ) \sum_{n=0}^{\infty} \left (\frac{2}{3} \right )^n z^{-n-1} = - \left ( \frac{1}{3} \right ) \sum_{n=1}^{\infty} \left (\frac{2}{3} \right )^{n-1} z^{-n} \end{eqnarray} and then

\begin{eqnarray} x[n] = -\frac{1}{3} \left ( \frac{2}{3} \right )^{n-1} = -\frac{1}{2} \left ( \frac{2}{3} \right )^n \end{eqnarray} with $n \ge 1$, and converges for $|z^{-1}| < \frac{3}{2}$ or $|z| > \frac{3}{2}$.

(ii) Contour Integration: We know that \begin{equation} x[n] = \frac{1}{2 \pi \mathrm{i}} \oint_C X(z) z^{n-1} dz. \end{equation} We can use the Cauchy residue theorem. That is, \begin{equation} \frac{1}{2 \pi \mathrm{i}} \oint_C X(z) z^{n-1} dz = \sum_{k=1}^n R_k, \end{equation}

where $R_k$ is a residue.

We further consider several cases:

(a) $n \ge 1$

For a simple pole (and this is the case here) \begin{equation} R_k = \lim_{z \to p_k} (z - p_k) X(z) z^{n-1}. \end{equation} since $X(z)=-1/3(z - 2/3)$, then for $p_0=2/3$, we consider the contour as the unit circle, for which the only pole $p_0=2/3$ is inside. \begin{eqnarray*} R_0[2/3,n] &=& \lim_{z \to 2/3} (z - 2/3) X(z) z^{n-1} \\ &=& \lim_{z \to 2/3} \left [ (z - 2/3) \frac{-1}{3 (z - 2/3)} z^{n-1} \right ] \\ &=& -\frac{1}{3} \left ( \frac{2}{3} \right )^{n-1} \\ &=& -\frac{1}{2} \left ( \frac{2}{3} \right )^n \end{eqnarray*} from which the coefficients are \begin{equation} R_0[2/3,n] = -\frac{1}{2} \left ( \frac{3}{2} \right )^n \end{equation}

(b) $n < 1$ Since $n$ is integer we are talking about poles at 0 of multiplicity. If n=0, we get

\begin{equation} R_0[0] = \lim_{z \to 0} z \frac{1}{(2 - 3 z) z} = \lim_{z \to 0} \frac{1}{2 - 3 z} = \frac{1}{2}. \end{equation}

If $n=-1$ we have a pole of multiplicity 2, from which we apply the formula: \begin{equation} R_1[0] = \lim_{z \to 0} \frac{d}{dz} \left ( \frac{z^2 X(z)}{z^2} \right ) = \lim_{z \to 0} \frac{d}{d z} X(z) = \frac{3}{4}. \end{equation}

In general:

\begin{equation} R_n[0] = \left . \frac{1}{(n-1)!} \frac{d^{n-1}}{d^{n-1}} X(z) \right |_{z=0} \end{equation}

That is, $R_n[0]$ is the $(n-1)^{th}$ coefficient of the Taylor series expansion of $f(z)=1/(2 - 3 z)$.

This is:

\begin{equation} R_n[0] = \left ( \frac{1}{2} \right ) \left ( \frac{3}{2} \right )^{-n} \end{equation}

Now we apply the residue equation, and get \begin{eqnarray*} x[n]= \left \{ \begin{array}{lll} -\frac{1}{2} \left ( \frac{3}{2} \right )^n & n \ge 1 & |z^{-1}| < 3/2 \\ \\ \quad \frac{1}{2} \left ( \frac{3}{2} \right )^{-n} & n > 1 & |z^{-1}| > 2/3 \end{array} \right . \end{eqnarray*}

The bilateral Z transform converges in the annulus $2/3 < |z| < 3/2 $ Unilateral Z transforms converge for |z|<3/2 and |z|>2/3