Editing Z-transform (section)

==Inverse Z-transform==
The ''inverse'' Z-transform is:

{{Equation box 1 |title=
|indent =: |cellpadding= 6 |border |border colour = #0073CF |background colour=#F5FFFA
|equation = <math> x[n] = \mathcal{Z}^{-1} \{X(z) \}= \frac{1}{2 \pi j} \oint_{C} X(z) z^{n-1} dz</math>
}}

where <math>C</math> is a counterclockwise closed path encircling the origin and entirely in the [[Radius of convergence|region of convergence]] (ROC). In the case where the ROC is causal (see [[#Example 2 (causal ROC)|Example 2]]), this means the path <math>C</math> must encircle all of the poles of <math>X(z)</math>.

A special case of this [[contour integral]] occurs when <math>C</math> is the unit circle. This contour can be used when the ROC includes the unit circle, which is always guaranteed when <math>X(z)</math> is stable, that is, when all the poles are inside the unit circle. With this contour, the inverse Z-transform simplifies to the [[Discrete-time Fourier transform#Inverse transform|inverse discrete-time Fourier transform]], or [[Fourier series]], of the periodic values of the Z-transform around the unit circle:

{{Equation box 1 |title=
|indent =: |cellpadding= 6 |border |border colour = #0073CF |background colour=#F5FFFA
|equation = <math> x[n] = \frac{1}{2 \pi} \int_{-\pi}^{+\pi}  X(e^{j \omega}) e^{j \omega n} d \omega.</math>
}}

The Z-transform with a finite range of <math>n</math> and a finite number of uniformly spaced <math>z</math> values can be computed efficiently via [[Bluestein's FFT algorithm]]. The [[discrete-time Fourier transform]] (DTFT)—not to be confused with the [[discrete Fourier transform]] (DFT)—is a special case of such a Z-transform obtained by restricting <math>z</math> to lie on the unit circle.

The following three methods are often used for the evaluation of the inverse -transform,

=== Direct Evaluation by Contour Integration ===
This method involves applying the [[Residue theorem|Cauchy Residue Theorem]] to evaluate the inverse Z-transform. By integrating around a closed contour in the complex plane, the residues at the poles of the Z-transform function inside the ROC are summed. This technique is particularly useful when working with functions expressed in terms of complex variables.

=== Expansion into a Series of Terms in the Variables ''z'' and ''z''{{sup|-1}} ===
In this method, the Z-transform is expanded into a power series. This approach is useful when the Z-transform function is rational, allowing for the approximation of the inverse by expanding into a series and determining the signal coefficients term by term.

=== Partial-Fraction Expansion and Table Lookup ===
This technique decomposes the Z-transform into a sum of simpler fractions, each corresponding to known Z-transform pairs. The inverse Z-transform is then determined by looking up each term in a standard table of Z-transform pairs. This method is widely used for its efficiency and simplicity, especially when the original function can be easily broken down into recognizable components.

==== Example:<ref>{{Cite book |last1=Proakis |first1=John |title=Digital Signal Processing Principles, Algorithms and Applications |last2=Manolakis |first2=Dimitris |publisher=PRENTICE-HALL INTERNATIONAL, INC. |edition=3rd}}</ref> ====
A) Determine the inverse Z-transform of the following by series expansion method,

<math display=block>X(z) = \frac{1}{1 - 1.5 z^{-1} + 0.5 z^{-2}}</math>

Solution:

Case 1:

ROC: <math>\left\vert Z \right\vert > 1</math>

Since the ROC is the exterior of a circle, <math>x(n)</math> is causal (signal existing for n≥0).

<math display=block>X(z) = {1\over 1 - {3\over 2}z^{-1} + {1\over 2}z^{-2}} = 1 + {{3\over 2}z^{-1}} + {{7\over 4}z^{-2}} + {{15\over 8}z^{-3}} + {{31\over 16}z^{-4}} +....</math>

thus,

<math display=block>\begin{align} 
  x(n) &= \left\{1 , \frac{3}{2} , \frac{7}{4} , \frac{15}{8} , \frac{31}{16} \ldots \right\} \\ 
  & \qquad\! \uparrow \\ 
\end{align}</math> (arrow indicates term at x(0)=1)

Note that in each step of long division process we eliminate lowest power term of <math>z^{-1}</math>.

Case 2:

ROC: <math>\left\vert Z \right\vert < 0.5</math>

Since the ROC is the interior of a circle, <math>x(n)</math> is anticausal (signal existing for n<0).

By performing long division we get,

<math display=block>X(z) = \frac{1}{1 - \frac{3}{2}z^{-1} + \frac{1}{2}z^{-2} } = 2z^2 + 6z^3 +14z^4 + 30z^5 + \ldots</math>

<math>\begin{align} 
  x(n) & = \{30, 14, 6, 2, 0, 0\} \\ 
  & \qquad \qquad \qquad \quad\ \ \, \uparrow\\ 
\end{align}</math> (arrow indicates term at x(0)=0)

Note that in each step of long division process we eliminate lowest power term of <math>z</math>.

''Note:''

# ''When the signal is causal, we get positive powers of <math>z</math> and when the signal is anticausal, we get negative powers of <math>z</math>.''
# ''<math>z^k</math> indicates term at <math>x(-k)</math> and <math>z^{-k}</math> indicates term at <math>x(k)</math>.''

B) Determine the inverse Z-transform of the following by series expansion method,

Eliminating negative powers if <math>z</math> and dividing by <math>z</math>,

<math display=block>\frac{X(z)}{z} = \frac{z^2}{z(z^2 - 1.5z + 0.5)} = \frac{z}{z^2 - 1.5z + 0.5}  </math>

By Partial Fraction Expansion,

<math display=block>\begin{align}
  \frac{X(z)}{z} &= \frac{z}{(z-1)(z-0.5)} = \frac{A_1}{z-0.5} + \frac{A_2}{z-1} \\[4pt]
  & A_1 = \left. \frac{(z-0.5) X(z)}{z} \right\vert_{z=0.5} = \frac{0.5}{(0.5-1)} = -1 \\[4pt]
  & A_2 = \left. \frac{(z-1) X(z)}{z} \right\vert_{z=1} = \frac{1}{1-0.5} = {2} \\[4pt]
  \frac{X(z)}{z} &= \frac{2}{z-1} - \frac{1}{z-0.5} 
\end{align}</math>

Case 1:

ROC:<math>\left\vert Z \right\vert > 1  </math>

Both the terms are causal, hence <math>x(n)</math> is causal.

<math display=block>\begin{align} 
  x(n) &= 2{(1)^n}u(n) - 1{(0.5)^n}u(n) \\ 
  &= (2-0.5^n) u(n) \\ 
\end{align}</math>

Case 2:

ROC:<math>\left\vert Z \right\vert < 0.5  </math>

Both the terms are anticausal, hence <math>x(n)</math> is anticausal.

<math>\begin{align} 
  x(n) &= -2{(1)^n}u(-n-1) - (-1{(0.5)^n}u(-n-1) ) \\ 
  &= (0.5^n-2) u(-n-1) \\
\end{align}</math>

Case 3:

ROC:<math>0.5 < \left\vert Z \right\vert < 1</math>

One of the terms is causal (p=0.5 provides the causal part) and other is anticausal (p=1 provides the anticausal part), hence <math>x(n)</math> is both sided.

<math>\begin{align} 
  x(n) &= -2{(1)^n}u(-n-1) - 1{(0.5)^n}u(n) \\ 
  &= -2u(-n-1) - 0.5^n u(n) \\ 
\end{align}</math>