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INVERSE z-TRANSFORM BY PARTIAL FRACTION EXPANSION

Module by: Nguyen Huu Phuong

In the method of partial fraction expansion, after expanding the given z–transform expression into partial fractions we use the listed transform pairs table and transform properties (section 4.2) to find the corresrponding time expression . For those who are unfamliar with the concept the following is a good introduction .
Example 1 
Find the inverse z–transform of the system function H ( z ) = 1,6 z ( z − 0 . 8 ) ( 2 z − 1 ) H ( z ) = 1,6 z ( z − 0 . 8 ) ( 2 z − 1 ) size 12{H \( z \) = { {1,6} over { ital "z " \( ital "z " - 0 "." 8 \) \( 2 ital "z " - 1 \) } } } {}
Solution
H ( z ) = 0 . 8 z ( z − 0 . 8 ) ( z − 0 . 5 ) H ( z ) = 0 . 8 z ( z − 0 . 8 ) ( z − 0 . 5 ) size 12{H \( z \) = { {0 "." 8} over { ital "z " \( ital "z " - 0 "." 8 \) \( ital "z " - 0 "." 5 \) } } } {}
The system has 3 simple poles at z=0z=0 size 12{z=0} {} , 0.8 and 0.5 . The system is stable . Now express H(z)H(z) size 12{H \( z \) } {} as
H ( z ) = 0 . 8 z ( z − 0 . 8 ) ( z − 0 . 5 ) = A 1 z + A 2 z − 0 . 8 + A 3 z − 0 . 5 H ( z ) = 0 . 8 z ( z − 0 . 8 ) ( z − 0 . 5 ) = A 1 z + A 2 z − 0 . 8 + A 3 z − 0 . 5 size 12{H \( z \) = { {0 "." 8} over { ital "z " \( ital "z " - 0 "." 8 \) \( ital "z " - 0 "." 5 \) } } = { {A rSub { size 8{1} } } over {z} } + { {A rSub { size 8{2} } } over {z - 0 "." 8} } + { {A rSub { size 8{3} } } over {z - 0 "." 5} } } {}
The problem is to determine the constants A1,A2,A3A1,A2,A3 size 12{A rSub { size 8{1} } ,A rSub { size 8{2} } ,A rSub { size 8{3} } } {} . For this we use a common denominator for the LHS :
H ( z ) = A 1 ( z − 0 . 8 ) ( z − 0 . 5 ) + A 2 z ( z − 0 . 5 ) + A 3 z ( z − 0 . 8 ) z ( z − 1 ) ( z − 0 . 5 ) = ( A 1 + A 2 + A 3 ) z 2 + ( − 1 . 3A 1 − 0 . 5A 2 − 0 . 8A 3 ) z + 0 . 4A 1 z ( z − 0 . 8 ) ( z − 0 . 5 ) H ( z ) = A 1 ( z − 0 . 8 ) ( z − 0 . 5 ) + A 2 z ( z − 0 . 5 ) + A 3 z ( z − 0 . 8 ) z ( z − 1 ) ( z − 0 . 5 ) = ( A 1 + A 2 + A 3 ) z 2 + ( − 1 . 3A 1 − 0 . 5A 2 − 0 . 8A 3 ) z + 0 . 4A 1 z ( z − 0 . 8 ) ( z − 0 . 5 ) alignl { stack { size 12{H \( z \) = { {A rSub { size 8{1} } \( ital "z " - 0 "." 8 \) \( ital "z " - 0 "." 5 \) + ital " A" rSub { size 8{2} } ital " z" \( ital "z " - 0 "." 5 \) + ital " A" rSub { size 8{3} } ital " z" \( ital "z " - 0 "." 8 \) } over { ital "z " \( ital "z " - 1 \) \( ital "z " - 0 "." 5 \) } } } {} # matrix { {} # {} # {} } = { { \( A rSub { size 8{1} } + ital " A" rSub { size 8{2} } + ital " A" rSub { size 8{3} } \) z rSup { size 8{2} } + \( -1 "." 3A rSub { size 8{1} } - 0 "." 5A rSub { size 8{2} } - 0 "." 8A rSub { size 8{3} } \) ital "z "+ 0 "." 4A rSub { size 8{1} } } over { ital " z " \( ital "z " - 0 "." 8 \) \( ital "z " - 0 "." 5 \) } } {} } } {}
Equating the corresponding coefficients of both sides , we get 3 simultaneours equations
A 1 + A 2 + A 3 = 0 − 1 . 3A 1 − 0 . 5A 2 − 0 . 8A 3 = 0 0 . 4A 1 = 0 . 8 A 1 + A 2 + A 3 = 0 − 1 . 3A 1 − 0 . 5A 2 − 0 . 8A 3 = 0 0 . 4A 1 = 0 . 8 alignl { stack { size 12{ matrix { {} # {} # {} } A rSub { size 8{1} } +A rSub { size 8{2} } +A rSub { size 8{3} } =0} {} # - 1 "." 3A rSub { size 8{1} } - 0 "." 5A rSub { size 8{2} } - 0 "." 8A rSub { size 8{3} } =0 {} # matrix { matrix { matrix { {} # {} # {} } {} # {} # {} } {} # {} # {} } 0 "." 4A rSub { size 8{1} } =0 "." 8 {} } } {}
having the roots
A 1 = 2, A 2 = − 1, A 3 = − 1 A 1 = 2, A 2 = − 1, A 3 = − 1 size 12{A rSub { size 8{1} } =2, matrix { {} # {} } A rSub { size 8{2} } = - 1, matrix { {} # {} } A rSub { size 8{3} } = - 1} {}
Thus
H ( z ) = 2 z − 1 z − 0 . 8 − 1 z − 0 . 5 H ( z ) = 2 z − 1 z − 0 . 8 − 1 z − 0 . 5 size 12{H \( z \) = { {2} over {z} } - { {1} over {z - 0 "." 8} } - { {1} over {z - 0 "." 5} } } {}
Looking at the list of z transform pairs table we see that the terms on the RHS are not ready for the inverse transform , we then proceed to rearrange H(z)H(z) size 12{H \( z \) } {} :
H ( z ) = [ 2 + z z − 0 . 8 − z z − 0 . 5 ] z − 1 = [ 2 − 1 1 − 0 . 8z − 1 − 1 1 − 0 . 5z − 1 ] z − 1 H ( z ) = [ 2 + z z − 0 . 8 − z z − 0 . 5 ] z − 1 = [ 2 − 1 1 − 0 . 8z − 1 − 1 1 − 0 . 5z − 1 ] z − 1 size 12{H \( z \) = \[ 2+ { {z} over {z - 0 "." 8} } - { {z} over {z - 0 "." 5} } \] z rSup { size 8{ - 1} } = \[ 2 - { {1} over {1 - 0 "." 8z rSup { size 8{ - 1} } } } - { {1} over {1 - 0 "." 5z rSup { size 8{ - 1} } } } \] z rSup { size 8{ - 1} } } {}
The inverse transform of the expression within the square brackets is
2δ ( n ) − 0 . 8 n u ( n ) − 0 . 5 n u ( n ) 2δ ( n ) − 0 . 8 n u ( n ) − 0 . 5 n u ( n ) size 12{2δ \( n \) - 0 "." 8 rSup { size 8{n} } u \( n \) - 0 "." 5 rSup { size 8{n} } u \( n \) } {}
Finally , delay above function one sample due to the factor z−1z−1 size 12{z rSup { size 8{ - 1} } } {} to get
h ( n ) = 2δ ( n − 1 ) + ( 0 . 5 n − 1 + 0 . 8 n − 1 ) u ( n − 1 ) h ( n ) = 2δ ( n − 1 ) + ( 0 . 5 n − 1 + 0 . 8 n − 1 ) u ( n − 1 ) size 12{h \( n \) =2δ \( n - 1 \) + \( 0 "." 5 rSup { size 8{n - 1} } +0 "." 8 rSup { size 8{n - 1} } \) u \( n - 1 \) } {}
This example shows the basic principle of the method . Full treatment would cover different cases , depending on the order of the numerator compared to that of the denominator, types of poles , etc … One important advantage of the method of partial fraction expansion is that it usually leads to results in closed forms .

Proper rational polynomial and simple poles

The zeros of X(z)X(z) size 12{X \( z \) } {} or H(z)H(z) size 12{H \( z \) } {} (here we take X(z)X(z) size 12{X \( z \) } {} as the representative) are inrelevant in the method , only the poles need consideration . Thus we write the rational polynomial as
X(z)=N(z)D(z)=N(z)(z−p1)(z−p2)(z−p3)X(z)=N(z)D(z)=N(z)(z−p1)(z−p2)(z−p3) size 12{X \( z \) = { {N \( z \) } over {D \( z \) } } = { {N \( z \) } over { \( z - p rSub { size 8{1} } \) \( z - p rSub { size 8{2} } \) \( z - p rSub { size 8{3} } \) } } } {} (1)
where the poles are assumed simple. When the order of the numerator is less than that of the denominator , the rational polynomial is proper . The case of equal orders is also treated here .
The normal partial fraction expansion is
X ( z ) = A 1 z − p 1 + A 2 z − p 2 + A 3 z − p 3 + . . . X ( z ) = A 1 z − p 1 + A 2 z − p 2 + A 3 z − p 3 + . . . size 12{X \( z \) = { {A rSub { size 8{1} } } over {z - p rSub { size 8{1} } } } + { {A rSub { size 8{2} } } over {z - p rSub { size 8{2} } } } + { {A rSub { size 8{3} } } over {z - p rSub { size 8{3} } } } + "." "." "." } {}
However, on the knowledge of the transform pairs given in Equation (4.31) and Equation (4.32), we expand size 12{ {X \( z \) } wideslash {z} } {} rather than X(z)X(z) size 12{X \( z \) } {} :
X ( z ) z = N ( z ) zD ( z ) = N ( z ) z ( z − p 1 ) ( z − p 2 ) ( z − p 3 ) = A 0 z + A 1 z − p 1 + A 2 z − p 2 + A 3 z − p 3 + . . . X ( z ) z = N ( z ) zD ( z ) = N ( z ) z ( z − p 1 ) ( z − p 2 ) ( z − p 3 ) = A 0 z + A 1 z − p 1 + A 2 z − p 2 + A 3 z − p 3 + . . . alignl { stack { size 12{ { {X \( z \) } over {z} } = { {N \( z \) } over { ital "zD" \( z \) } } = { {N \( z \) } over {z \( z - p rSub { size 8{1} } \) \( z - p rSub { size 8{2} } \) \( z - p rSub { size 8{3} } \) } } } {} # matrix { {} # {} # {} } = { {A rSub { size 8{0} } } over {z} } + { {A rSub { size 8{1} } } over {z - p rSub { size 8{1} } } } + { {A rSub { size 8{2} } } over {z - p rSub { size 8{2} } } } + { {A rSub { size 8{3} } } over {z - p rSub { size 8{3} } } } + "." "." "." {} } } {}
Now let’s multiply both sides of above equation with each term (z−pi)(z−pi) size 12{ \( z - p rSub { size 8{i} } \) } {}, i = 1,2,…, and then evaluate the obtained expressions at the poles p1,p2...p1,p2... size 12{p rSub { size 8{1} } ,p rSub { size 8{2} } "." "." "." } {}This will lead to the following formula for the constants :
Ai=(z−pi)X(z)z∣z=pi,i=0,1,2...Ai=(z−pi)X(z)z∣z=pi,i=0,1,2... size 12{A rSub { size 8{i} } = \( z - p rSub { size 8{i} } \) { {X \( z \) } over {z} } \lline rSub { size 8{z=p rSub { size 6{i} } } } , matrix { {} # {} # {} } i=0,1,2 "." "." "." } {} (2)
Example 2 
Find the signal whose z – transform is given by X ( z ) = 2z 2 − 2 . 05 z z 2 − 2 . 05 z + 1 X ( z ) = 2z 2 − 2 . 05 z z 2 − 2 . 05 z + 1 size 12{X \( z \) = { {2z rSup { size 8{2} } - 2 "." "05"z} over {z rSup { size 8{2} } - 2 "." "05"z+1} } } {}
Solution
The poles are p1=0.8p1=0.8 size 12{p rSub { size 8{1} } =0 "." 8} {} and p1=1.25p1=1.25 size 12{p rSub { size 8{1} } =1 "." "25"} {}. Notice that the order of the numerator is equal to that of denominator so that the rational polynomial is not proper .
We take X(z)/zX(z)/z size 12{ {X \( z \) } slash {z} } {} , which is proper, for the expansion and write
X ( z ) z = 2z 2 − 2 . 05 z z ( z 2 − 2 . 05 z + 1 ) = 2z 2 − 2 . 05 z z ( z − 0 . 8 ) ( z − 1 . 25 ) = A 0 z + A 1 z − p 1 + A 2 z − p 2 X ( z ) z = 2z 2 − 2 . 05 z z ( z 2 − 2 . 05 z + 1 ) = 2z 2 − 2 . 05 z z ( z − 0 . 8 ) ( z − 1 . 25 ) = A 0 z + A 1 z − p 1 + A 2 z − p 2 alignl { stack { size 12{ { {X \( z \) } over {z} } = { {2z rSup { size 8{2} } - 2 "." "05"z} over {z \( z rSup { size 8{2} } - 2 "." "05"z+1 \) } } = { {2z rSup { size 8{2} } - 2 "." "05"z} over {z \( z - 0 "." 8 \) \( z - 1 "." "25" \) } } } {} # matrix { matrix { matrix { {} # {} # {} } {} # {} # {} # {} } {} # {} # {} # {} } = { {A rSub { size 8{0} } } over {z} } + { {A rSub { size 8{1} } } over {z - p rSub { size 8{1} } } } + { {A rSub { size 8{2} } } over {z - p rSub { size 8{2} } } } {} } } {}
The constrants are computed as follows.
A 0 = z X ( z ) z ∣ z = 0 = X ( z ) ∣ z = v = 0 A 1 = ( z − 0 . 8 ) X ( z ) z ∣ z = 0 . 8 = 2z 2 − 2 . 05 z z ( z − 1 . 25 ) ∣ z = 0 . 8 = 1 A 2 = ( z − 1 . 25 ) X ( z ) z ∣ z = 1 . 25 = 2z 2 − 2 . 05 z z ( z − 0 . 8 ) ∣ z = 1 . 25 = 1 A 0 = z X ( z ) z ∣ z = 0 = X ( z ) ∣ z = v = 0 A 1 = ( z − 0 . 8 ) X ( z ) z ∣ z = 0 . 8 = 2z 2 − 2 . 05 z z ( z − 1 . 25 ) ∣ z = 0 . 8 = 1 A 2 = ( z − 1 . 25 ) X ( z ) z ∣ z = 1 . 25 = 2z 2 − 2 . 05 z z ( z − 0 . 8 ) ∣ z = 1 . 25 = 1 alignl { stack { size 12{A rSub { size 8{0} } =z { {X \( z \) } over {z} } \lline rSub { size 8{z=0} } =X \( z \) \lline rSub { size 8{z=v=0} } } {} # A rSub { size 8{1} } = \( z - 0 "." 8 \) { {X \( z \) } over {z} } \lline rSub { size 8{z=0 "." 8} } = { {2z rSup { size 8{2} } - 2 "." "05"z} over {z \( z - 1 "." "25" \) } } \lline rSub { size 8{z=0 "." 8} } =1 {} # A rSub { size 8{2} } = \( z - 1 "." "25" \) { {X \( z \) } over {z} } \lline rSub { size 8{z=1 "." "25"} } = { {2z rSup { size 8{2} } - 2 "." "05"z} over {z \( z - 0 "." 8 \) } } \lline rSub { size 8{z=1 "." "25"} } =1 {} } } {}
Hence
X ( z ) = z z − 0 . 8 + z z − 1 . 25 X ( z ) = z z − 0 . 8 + z z − 1 . 25 size 12{X \( z \) = { {z} over {z - 0 "." 8} } + { {z} over {z - 1 "." "25"} } } {}
Depending on the ROC , we have three different time functions:
x ( n ) = 0 . 8 n u ( n ) + 1 . 25 n u ( n ) , ∣ z ∣ > 1 . 25 x ( n ) = 0 . 8 n u ( n ) + 1 . 25 n u ( − n − 1 ) , 0 . 8 < ∣ z ∣ < 1 . 25 x ( n ) = − 0 . 8 n u ( − n − 1 ) − 1 . 25 n u ( − n − 1 ) , ∣ z ∣ < 0 . 8 x ( n ) = 0 . 8 n u ( n ) + 1 . 25 n u ( n ) , ∣ z ∣ > 1 . 25 x ( n ) = 0 . 8 n u ( n ) + 1 . 25 n u ( − n − 1 ) , 0 . 8 < ∣ z ∣ < 1 . 25 x ( n ) = − 0 . 8 n u ( − n − 1 ) − 1 . 25 n u ( − n − 1 ) , ∣ z ∣ < 0 . 8 alignl { stack { size 12{x \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) +1 "." "25" rSup { size 8{n} } u \( n \) , matrix { {} # {} } lline z rline >1 "." "25"} {} # x \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) +1 "." "25" rSup { size 8{n} } u \( - n - 1 \) , matrix { {} # {} } 0 "." 8< lline z rline <1 "." "25" {} # x \( n \) = - 0 "." 8 rSup { size 8{n} } u \( - n - 1 \) - 1 "." "25" rSup { size 8{n} } u \( - n - 1 \) , matrix { {} # {} } lline z rline <0 "." 8 {} } } {}
Another way is to first express X(z)X(z) size 12{X \( z \) } {} in terms of :
X ( z ) = N ( z − 1 ) D ( z − 1 ) = N ( z − 1 ) ( 1 − p 1 z − 1 ) ( 1 − p 2 z − 1 ) ( 1 − p 3 z − 1 ) . . . = A 1 1 − p 1 z − 1 + A 2 1 − p 2 z − 1 + A 3 1 − p 3 z − 1 . . . X ( z ) = N ( z − 1 ) D ( z − 1 ) = N ( z − 1 ) ( 1 − p 1 z − 1 ) ( 1 − p 2 z − 1 ) ( 1 − p 3 z − 1 ) . . . = A 1 1 − p 1 z − 1 + A 2 1 − p 2 z − 1 + A 3 1 − p 3 z − 1 . . . alignl { stack { size 12{X \( z \) = { {N \( z rSup { size 8{ - 1} } \) } over {D \( z rSup { size 8{ - 1} } \) } } = { {N \( z rSup { size 8{ - 1} } \) } over { \( 1 - p rSub { size 8{1} } z rSup { size 8{ - 1} } \) \( 1 - p rSub { size 8{2} } z rSup { size 8{ - 1} } \) \( 1 - p rSub { size 8{3} } z rSup { size 8{ - 1} } \) "." "." "." } } } {} # matrix { matrix { {} # {} # {} } {} # {} # {} } = { {A rSub { size 8{1} } } over {1 - p rSub { size 8{1} } z rSup { size 8{ - 1} } } } + { {A rSub { size 8{2} } } over {1 - p rSub { size 8{2} } z rSup { size 8{ - 1} } } } + { {A rSub { size 8{3} } } over {1 - p rSub { size 8{3} } z rSup { size 8{ - 1} } } } _ "." "." "." {} } } {}
and then evaluate the constants as
Ai=(1−piz−1)X(z)∣z=piAi=(1−piz−1)X(z)∣z=pi size 12{A rSub { size 8{i} } = \( 1 - p rSub { size 8{i} } z rSup { size 8{ - 1} } \) X \( z \) \lline rSub { size 8{z=p rSub { size 6{i} } } } } {} (3)
Example 3 
Repeat previous example but expressing X(z)X(z) size 12{X \( z \) } {} in terms of z−1z−1 size 12{z rSup { size 8{ - 1} } } {}.
Solution
First
X ( z ) = 2z 2 − 2 . 05 z z 2 − 2 . 05 z − 1 = 2 − 2 . 05 z − 1 1 − 2 . 05 z − 1 + z − 2 X ( z ) = 2z 2 − 2 . 05 z z 2 − 2 . 05 z − 1 = 2 − 2 . 05 z − 1 1 − 2 . 05 z − 1 + z − 2 size 12{X \( z \) = { {2z rSup { size 8{2} } - 2 "." "05"z} over {z rSup { size 8{2} } - 2 "." "05"z - 1} } = { {2 - 2 "." "05"z rSup { size 8{ - 1} } } over {1 - 2 "." "05"z rSup { size 8{ - 1} } +z rSup { size 8{ - 2} } } } } {}
Notice in this way the order numerator is less than that of the denominator , i.e. the rational polynomial is proper . Next
X ( z ) = 2 − 2 . 05 z − 1 1 − 2 . 05 z − 1 + z − 2 = 2 − 2 . 05 z − 1 ( 1 − 0 . 8z − 1 ) ( 1 − 1 . 25 z − 1 ) = A 1 1 − 0 . 8z − 1 + A 2 1 − 1 . 25 z − 1 X ( z ) = 2 − 2 . 05 z − 1 1 − 2 . 05 z − 1 + z − 2 = 2 − 2 . 05 z − 1 ( 1 − 0 . 8z − 1 ) ( 1 − 1 . 25 z − 1 ) = A 1 1 − 0 . 8z − 1 + A 2 1 − 1 . 25 z − 1 alignl { stack { size 12{X \( z \) = { {2 - 2 "." "05"z rSup { size 8{ - 1} } } over {1 - 2 "." "05"z rSup { size 8{ - 1} } +z rSup { size 8{ - 2} } } } = { {2 - 2 "." "05"z rSup { size 8{ - 1} } } over { \( 1 - 0 "." 8z rSup { size 8{ - 1} } \) \( 1 - 1 "." "25"z rSup { size 8{ - 1} } \) } } } {} # matrix { {} # {} # {} } = { {A rSub { size 8{1} } } over {1 - 0 "." 8z rSup { size 8{ - 1} } } } + { {A rSub { size 8{2} } } over {1 - 1 "." "25"z rSup { size 8{ - 1} } } } {} } } {}
The constants are given by
A 1 = ( 1 − 0 . 8z − 1 ) X ( z ) ∣ z = 0 . 8 = 2 − 2 . 05 z − 1 1 − 1 . 25 z − 1 ∣ z = 0 . 8 = 1 A 2 = ( 1 − 1 . 25 z − 1 ) X ( z ) ∣ z = 1 . 25 = 2 − 2 . 05 z − 1 1 − 0 . 8z − 1 ∣ z = 0 . 8 = 1 A 1 = ( 1 − 0 . 8z − 1 ) X ( z ) ∣ z = 0 . 8 = 2 − 2 . 05 z − 1 1 − 1 . 25 z − 1 ∣ z = 0 . 8 = 1 A 2 = ( 1 − 1 . 25 z − 1 ) X ( z ) ∣ z = 1 . 25 = 2 − 2 . 05 z − 1 1 − 0 . 8z − 1 ∣ z = 0 . 8 = 1 alignl { stack { size 12{A rSub { size 8{1} } = \( 1 - 0 "." 8z rSup { size 8{ - 1} } \) X \( z \) \lline rSub { size 8{z=0 "." 8} } = { {2 - 2 "." "05"z rSup { size 8{ - 1} } } over {1 - 1 "." "25"z rSup { size 8{ - 1} } } } \lline rSub { size 8{z=0 "." 8} } =1} {} # A rSub { size 8{2} } = \( 1 - 1 "." "25"z rSup { size 8{ - 1} } \) X \( z \) \lline rSub { size 8{z=1 "." "25"} } = { {2 - 2 "." "05"z rSup { size 8{ - 1} } } over {1 - 0 "." 8z rSup { size 8{ - 1} } } } \lline rSub { size 8{z=0 "." 8} } =1 {} } } {}
The result is the same as before .
Example 4 
Find the inverse z – transform of X ( z ) = z 2 + z + 1 z 2 + 3z + 2 X ( z ) = z 2 + z + 1 z 2 + 3z + 2 size 12{X \( z \) = { {z rSup { size 8{2} } +z+1} over {z rSup { size 8{2} } +3z+2} } } {}
Solution
The poles of X(z)X(z) size 12{X \( z \) } {} are z=−1z=−1 size 12{z= - 1} {} and z=−2z=−2 size 12{z= - 2} {}. Now let’s write
X ( z ) z = 2z 2 + z + 1 z ( z + 1 ) ( z + 2 ) = A 0 Z + A 1 z + 1 + A 2 z + 2 X ( z ) z = 2z 2 + z + 1 z ( z + 1 ) ( z + 2 ) = A 0 Z + A 1 z + 1 + A 2 z + 2 size 12{ { {X \( z \) } over {z} } = { {2z rSup { size 8{2} } +z+1} over {z \( z+1 \) \( z+2 \) } } = { {A rSub { size 8{0} } } over {Z} } + { {A rSub { size 8{1} } } over {z+1} } + { {A rSub { size 8{2} } } over {z+2} } } {}
The constants are given by
A 0 = ( z ) X ( z ) z ∣ z = 0 = z 2 + z + 1 ( z + 1 ) ( z + 2 ) ∣ z = 0 = 0 . 5 A 1 = ( z + 1 ) X ( z ) z ∣ z =< 1 = z 2 + z + 1 z ( z + 2 ) ∣ z =+ 1 = − 1 A 2 = ( z + 2 ) X ( z ) z ∣ z = − 2 = z 2 + z + 1 z ( z + 1 ) ∣ z = − 2 = 1,5 A 0 = ( z ) X ( z ) z ∣ z = 0 = z 2 + z + 1 ( z + 1 ) ( z + 2 ) ∣ z = 0 = 0 . 5 A 1 = ( z + 1 ) X ( z ) z ∣ z =< 1 = z 2 + z + 1 z ( z + 2 ) ∣ z =+ 1 = − 1 A 2 = ( z + 2 ) X ( z ) z ∣ z = − 2 = z 2 + z + 1 z ( z + 1 ) ∣ z = − 2 = 1,5 alignl { stack { size 12{A rSub { size 8{0} } = \( z \) { {X \( z \) } over {z} } \lline rSub { size 8{z=0} } = { {z rSup { size 8{2} } +z+1} over { \( z+1 \) \( z+2 \) } } \lline rSub { size 8{z=0} } =0 "." 5} {} # A rSub { size 8{1} } = \( z+1 \) { {X \( z \) } over {z} } \lline rSub { size 8{z"=<"1} } = { {z rSup { size 8{2} } +z+1} over {z \( z+2 \) } } \lline rSub { size 8{z"=+"1} } = - 1 {} # A rSub { size 8{2} } = \( z+2 \) { {X \( z \) } over {z} } \lline rSub { size 8{z= - 2} } = { {z rSup { size 8{2} } +z+1} over {z \( z+1 \) } } \lline rSub { size 8{z= - 2} } =1,5 {} } } {}
Hence
H ( z ) = 0 . 5 z z + 1 + 1,5 z z + 2 H ( z ) = 0 . 5 z z + 1 + 1,5 z z + 2 size 12{H \( z \) =0 "." 5 { {z} over {z+1} } + { {1,5z} over {z+2} } } {}
The same as previous example , depending on the specified ROC we have different time functions or , in other word , depending on the type of time function we would like , we choose the appropriate ROC . For example in this case if we want a causal signal we would choose ∣z∣>1∣z∣>1 size 12{ lline z rline >1} {}, then
x ( n ) = 0 . 5δ ( n ) − ( − 1 ) n u ( n ) x ( n ) = 0 . 5δ ( n ) − ( − 1 ) n u ( n ) size 12{x \( n \) =0 "." 5δ \( n \) - \( - 1 \) rSup { size 8{n} } u \( n \) } {}

Proper rational polynomial and simple complex poles

The difference here is that the poles are complex . Remember complex poles always appear in complex conjugate pairs . General procedure is the same as above but the evaluation is more time – consuming due to complex numbers . The inverse z-transform are sinusoidal functions.
Example 5 
Find the inverse z–transform of X ( z ) = z 2 z 2 + 2z + 1 X ( z ) = z 2 z 2 + 2z + 1 size 12{X \( z \) = { {z rSup { size 8{2} } } over {z rSup { size 8{2} } +2z+1} } } {}
solution
The poles are given by
z 2 + 2z + 1 = 0 z 2 + 2z + 1 = 0 size 12{z rSup { size 8{2} } +2z+1=0} {}
whose roots are
p 1 = − 1 2 + j 3 2 = e j 2π 3 p 2 = − 1 2 − j 3 2 = e − j 2π 3 = p 1 p 1 = − 1 2 + j 3 2 = e j 2π 3 p 2 = − 1 2 − j 3 2 = e − j 2π 3 = p 1 alignl { stack { size 12{p rSub { size 8{1} } = - { {1} over {2} } +j { { sqrt {3} } over {2} } =e rSup { size 8{j { {2π} over {3} } } } } {} # p rSub { size 8{2} } = - { {1} over {2} } - j { { sqrt {3} } over {2} } =e rSup { size 8{ - j { {2π} over {3} } } } =p rSub { size 8{1} } rSup { size 8{*} } {} } } {}
Notice that ∣p1∣=∣p2∣=1∣p1∣=∣p2∣=1 size 12{ lline p rSub { size 8{1} } rline = lline p rSub { size 8{2} } rline =1} {}, i.e. the poles are right on the unit circle then the system is marginally stable, and we expect the time signal will be a stable sine wave. Let’s proceed with the expansion:
X ( z ) z = z ( z + 1 2 − j 3 2 ) ( z + 1 2 + j 3 2 ) = A 1 z + 1 2 − j 3 2 + A 2 z + 1 2 + j 3 2 X ( z ) z = z ( z + 1 2 − j 3 2 ) ( z + 1 2 + j 3 2 ) = A 1 z + 1 2 − j 3 2 + A 2 z + 1 2 + j 3 2 size 12{ { {X \( z \) } over {z} } = { {z} over { \( z+ { {1} over {2} } - j { { sqrt {3} } over {2} } \) \( z+ { {1} over {2} } +j { { sqrt {3} } over {2} } \) } } = { {A rSub { size 8{1} } } over {z+ { {1} over {2} } - j { { sqrt {3} } over {2} } } } + { {A rSub { size 8{2} } } over {z+ { {1} over {2} } +j { { sqrt {3} } over {2} } } } } {}
The constants are
A 1 z z + 1 2 + j 3 2 ∣ z = − 1 2 + j 3 2 = − 1 2 − j 1 2 3 = 1 3 e j A 2 = 0,5 z z + 1 − j 3 ∣ z = − 1 2 − j 3 2 = − 1 2 + j 1 2 3 = 1 3 e − j A 1 z z + 1 2 + j 3 2 ∣ z = − 1 2 + j 3 2 = − 1 2 − j 1 2 3 = 1 3 e j A 2 = 0,5 z z + 1 − j 3 ∣ z = − 1 2 − j 3 2 = − 1 2 + j 1 2 3 = 1 3 e − j alignl { stack { size 12{A rSub { size 8{1} } { {z} over {z+ { {1} over {2} } +j { { sqrt {3} } over {2} } } } \lline rSub { size 8{z= - { {1} over {2} } +j { { sqrt {3} } over {2} } } } = - { {1} over {2} } - j { {1} over {2 sqrt {3} } } = { {1} over { sqrt {3} } } e rSup { size 8{j {5π} wideslash {6} } } } {} # A rSub { size 8{2} } = { {0,5z} over {z+1 - j sqrt {3} } } \lline rSub { size 8{z= - { {1} over {2} } - j { { sqrt {3} } over {2} } } } = - { {1} over {2} } +j { {1} over {2 sqrt {3} } } = { {1} over { sqrt {3} } } e rSup { size 8{ - j {5π} wideslash {6} } } {} } } {}
The time signal is
X ( n ) = ( A 1 p 1 n + A 2 p 2 n ) u ( n ) = [ 1 3 e j ( e j ) n + 1 3 e − j ( e − j ) n ] u ( n ) = 2 3 cos ( 2π 3 n + 5π 6 ) u ( n ) X ( n ) = ( A 1 p 1 n + A 2 p 2 n ) u ( n ) = [ 1 3 e j ( e j ) n + 1 3 e − j ( e − j ) n ] u ( n ) = 2 3 cos ( 2π 3 n + 5π 6 ) u ( n ) alignl { stack { size 12{X \( n \) = \( A rSub { size 8{1} } p rSub { size 8{1} rSup { size 8{n} } } +A rSub { size 8{2} } p rSub { size 8{2} rSup { size 8{n} } } \) u \( n \) } {} # matrix { {} # {} # {} } = \[ { {1} over { sqrt {3} } } e rSup { size 8{j {5π} wideslash {6} } } \( e rSup { size 8{j {2π} wideslash {3} } } \) rSup { size 8{n} } + { {1} over { sqrt {3} } } e rSup { size 8{ - j {5π} wideslash {6} } } \( e rSup { size 8{ - j {2π} wideslash {3} } } \) rSup { size 8{n} } \] u \( n \) {} # matrix { {} # {} # {} } = { {2} over {3} } "cos" \( { {2π} over {3} } n+ { {5π} over {6} } \) u \( n \) {} } } {}
This is a causal stable sine wave .

Order of numerator is greater than order of denominator

In this situation one way is to proceed the long division (the power series method) to get the quotient and the remainder:
X ( z ) = N ( z ) D ( z ) ⇒ N ( z ) = Q ( z ) D ( z ) + R ( z ) X ( z ) = N ( z ) D ( z ) ⇒ N ( z ) = Q ( z ) D ( z ) + R ( z ) size 12{X \( z \) = { {N \( z \) } over {D \( z \) } } matrix { {} # {} } drarrow matrix { {} # {} } N \( z \) =Q \( z \) D \( z \) +R \( z \) } {}
The order of R(z)R(z) size 12{R \( z \) } {} must be smaller than that of D(z)D(z) size 12{D \( z \) } {} . Next
X(z)=N(z)D(z)=Q(z)D(z)+R(z)D(z)=Q(z)+R(z)D(z)X(z)=N(z)D(z)=Q(z)D(z)+R(z)D(z)=Q(z)+R(z)D(z) size 12{X \( z \) = { {N \( z \) } over {D \( z \) } } = { {Q \( z \) D \( z \) +R \( z \) } over {D \( z \) } } =Q \( z \) + { {R \( z \) } over {D \( z \) } } } {} (4)
We then expand the rational polynomial term as usual.
In another way , we first temporarily forget N(z)N(z) size 12{N \( z \) } {}and proceed to expand 1/D(z)1/D(z) size 12{ {1} slash {D \( z \) } } {}, afterwards multiply N(z−1)N(z−1) size 12{N \( z rSup { size 8{ - 1} } \) } {} in for the full expansion of X(z)X(z) size 12{X \( z \) } {}.

Multiple poles

For rational polynomials having multiple poles , the expansion is rather different . To be specific let’s consider
X(z)z=N(z)(z−p)mX(z)z=N(z)(z−p)m size 12{ { {X \( z \) } over {z} } = { {N \( z \) } over { \( z - p \) rSup { size 8{m} } } } } {} (5)
The multiple pole p may be real or complex . The expansion is
X(z)z=A1z−p+A2(z−p)2+...+Am−1(z−p)m−1+Am(z−p)mX(z)z=A1z−p+A2(z−p)2+...+Am−1(z−p)m−1+Am(z−p)m size 12{ { {X \( z \) } over {z} } = { {A rSub { size 8{1} } } over {z - p} } + { {A rSub { size 8{2} } } over { \( z - p \) rSup { size 8{2} } } } + "." "." "." + { {A rSub { size 8{m - 1} } } over { \( z - p \) rSup { size 8{m - 1} } } } + { {A rSub { size 8{m} } } over { \( z - p \) rSup { size 8{m} } } } } {} (6)
We should add expansion terms to take into accurnt the simple poles if any . The procedure is illustrated by the case of double pole (pole of order 2) as follows .
Example 6 
Given X ( z ) = z 2 ( z + 1 ) ( z − 1 ) 2 X ( z ) = z 2 ( z + 1 ) ( z − 1 ) 2 size 12{X \( z \) = { {z rSup { size 8{2} } } over { \( z+1 \) \( z - 1 \) rSup { size 8{2} } } } } {} Find x(n)x(n) size 12{x \( n \) } {}.
Solution
X(z)X(z) size 12{X \( z \) } {} has a simple pole at p0=−1p0=−1 size 12{p rSub { size 8{0} } = - 1} {} , and double pole p1=p2=1p1=p2=1 size 12{p rSub { size 8{1} } =p rSub { size 8{2} } =1} {}. Let’s write
X ( z ) z = z ( z + 1 ) ( z − 1 ) 2 = A 0 z + 1 + A 1 z − 1 + A 2 ( z − 1 ) 2 X ( z ) z = z ( z + 1 ) ( z − 1 ) 2 = A 0 z + 1 + A 1 z − 1 + A 2 ( z − 1 ) 2 size 12{ { {X \( z \) } over {z} } = { {z} over { \( z+1 \) \( z - 1 \) rSup { size 8{2} } } } = { {A rSub { size 8{0} } } over {z+1} } + { {A rSub { size 8{1} } } over {z - 1} } + { {A rSub { size 8{2} } } over { \( z - 1 \) rSup { size 8{2} } } } } {}
The constant A0A0 size 12{A rSub { size 8{0} } } {} is given by
A 0 = ( z + 1 ) X ( z ) z ∣ z = − 1 = z ( z − 1 ) 2 ∣ z = − 1 = − 1 4 A 0 = ( z + 1 ) X ( z ) z ∣ z = − 1 = z ( z − 1 ) 2 ∣ z = − 1 = − 1 4 size 12{A rSub { size 8{0} } = \( z+1 \) { {X \( z \) } over {z} } \lline rSub { size 8{z= - 1} } = { {z} over { \( z - 1 \) rSup { size 8{2} } } } \lline rSub { size 8{z= - 1} } = - { {1} over {4} } } {}
Next we find A2A2 size 12{A rSub { size 8{2} } } {} first (the constant coresponding to the pole of highest multiplicity) by multiplying both sides by (z−1)2(z−1)2 size 12{ \( z - 1 \) rSup { size 8{2} } } {} and compute at z=p1=p2=1z=p1=p2=1 size 12{z=p rSub { size 8{1} } =p rSub { size 8{2} } =1} {}:
( z − 1 ) 2 X ( z ) z = ( z − 1 ) 2 z + 1 A 0 + ( z − 1 ) A 1 + A 2 ( z − 1 ) 2 X ( z ) z = ( z − 1 ) 2 z + 1 A 0 + ( z − 1 ) A 1 + A 2 size 12{ { { \( z - 1 \) rSup { size 8{2} } X \( z \) } over {z} } = { { \( z - 1 \) rSup { size 8{2} } } over {z+1} } A rSub { size 8{0} } + \( z - 1 \) A rSub { size 8{1} } +A rSub { size 8{2} } } {}
A 2 = ( x − 1 ) 2 X ( z ) z ∣ z = 1 = z z + 1 ∣ z = 1 = 1 2 A 2 = ( x − 1 ) 2 X ( z ) z ∣ z = 1 = z z + 1 ∣ z = 1 = 1 2 size 12{A rSub { size 8{2} } = \( x - 1 \) rSup { size 8{2} } { {X \( z \) } over {z} } \lline rSub { size 8{z=1} } = { {z} over {z+1} } \lline rSub { size 8{z=1} } = { {1} over {2} } } {}
To find A1A1 size 12{A rSub { size 8{1} } } {} we differentiate both sides of the above equation with respect to z and compute at z=p1=p2=1z=p1=p2=1 size 12{z=p rSub { size 8{1} } =p rSub { size 8{2} } =1} {}
A 1 = d dz [ ( z − 1 ) 2 X ( z ) z ] z = 1 = d dz