z transform
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 $$x(n) = \frac{1}{3}(\frac{2}{3})^{n-1}u(n-1)$$ (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\$ ?
In a hybrid system, the sampler and ZOH perform the operations: a continuous signal is sampled (analogue to digital conversion, ADC) to produce a discrete signal, and the discrete signal is returned to continuous form (digital-to-analogue conversion, DAC). Usually there is some digital processing between these two operations, but the combined DAC/ADC operation is normally represented as a single Laplace TF in the block diagram.
For simplicity, consider the ZOH in isolation, i.e. no data processing between the sample and hold. Let the input signal = x(t), the output signal = y(t), and the sampling increment = T sec. At the k'th sampling instant (ie at t=kT), the input to the ADC may be denoted x(k) and the output from the DAC is held constant at x(k) until the next sample arrives.
Thus, the DAC output between x(k) and x(k+1) is a rectangular pulse of height, x(k), and duration, T. This can be modelled, for Laplace transform representation, as a step of height x(k) at t=kT and a step of height -x(k) at time t=(k+1)T, and may be realised via the LT delay operator, e^-skT, which delays any signal by kT sec
Hence the LT of the isolated pulse is {e^(-skT)} x(k)/s - {e^[-s(k+1)T} x(k)/s, or:
e^(-skT) {1-e^-sT} x(k)/s
For the entire signal, from t=0 to t=kT this function reduces to Y(s) = X(s) {1-e^-sT}/s where X(s) and Y(s) are the Laplace transformed ZOH input and output. Hence Y(s)/X(s) = (1-e^-sT)/s
If, now, we wish to proceed by connecting a G(s) block to the ZOH output, the overall TF becomes (1 - e^-sT) G(s)/s. But there's a problem: if we are required to close the loop around this TF, the resultant CLTF will not be analytic due to e^-sT appearing in the CLTF denominator. So we must revert to z-transforms to render it analytic.
The z-transform of the exponential bit is easy; it's (1 - z^-1), because z^-1 is the delay operator in the z-domain (and is therefore equivalent to e^-sT) and we must then find the z-transform of G(s)/s to complete the picture.
You can write your system as \$H(z)=\dfrac{B(z)}{A(z)}\$ the use a recursive formula for IIR filter where you have A and B coefficients. Before you have to change the form to have denominator like \$ A(z) = 1+a1\cdot z^{-1}+ a2\cdot z^{-2}+..\$ by multiplying your form
Best Answer
From definition of inverse z-transforn,
$$\begin{align} x(n)&= -\frac{1}{3}\times \frac{1}{2\pi j}\oint_{C}^{ } \frac{z^{n-1}}{z-\frac{2}{3}}dz\\ &= -\frac{1}{3}\times \frac{1}{2\pi j}\oint_{C}^{ } \frac{f(z)}{z-\frac{2}{3}}dz\end{align}$$
Where, \$f(z) = z^{n-1}\$. Using Cauchy's Integral formula:
$$x(n) = -\frac{1}{3}\times \left(\dfrac{2}{3}\right)^{n-1}$$
If \$n<1\$, then the function \$f(z) = z^{n-1}\$ will have singularity at \$z=0\$ and hence Cauchy's Integral formula can not be applied. So \$n\$ must be greater than or equal to \$1\$. Or we can write:
$$x(n) = -\frac{1}{3}\times \left(\dfrac{2}{3}\right)^{n-1}u(n-1)$$
PS: Read this page. They have given equations to find inverse z-transform directly (without using integration) using residue method.