Editing Z-transform (section)

==== 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>