In the method of partial fraction expansion, after expanding the given z–transform expression into partial fractions we use the listed transform pairs (table 4.1) 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)(2z−1)
H(z)=
1.6
z(z−0.8)(2z−1)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOabaeqabaGaamisaiaacIcacaWG6bGaaiykaiabg2da9maalaaabaGaaGymaiaac6cacaaI2aaabaGaamOEaiaacIcacaWG6bGaeyOeI0IaaGimaiaac6cacaaI4aGaaiykaiaacIcacaaIYaGaamOEaiabgkHiTiaaigdacaGGPaaaaaqaaaaaaa@477A@
Solution
First rewrite H(z) as
H(z)=
0.8
z(z−0.8)(z−0.5)
H(z)=
0.8
z(z−0.8)(z−0.5)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIeacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaaicdacaGGUaGaaGioaaqaaiaadQhacaGGOaGaamOEaiabgkHiTiaaicdacaGGUaGaaGioaiaacMcacaGGOaGaamOEaiabgkHiTiaaicdacaGGUaGaaGynaiaacMcaaaaaaa@4828@
The system has 3 simple poles at z = 0 , 0.8 and 0.5 . The system is stable . Now express 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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIeacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaaicdacaGGUaGaaGioaaqaaiaadQhacaGGOaGaamOEaiabgkHiTiaaicdacaGGUaGaaGioaiaacMcacaGGOaGaamOEaiabgkHiTiaaicdacaGGUaGaaGynaiaacMcaaaGaeyypa0ZaaSaaaeaacaWGbbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamOEaaaacqGHRaWkdaWcaaqaaiaadgeadaWgaaWcbaGaaGOmaaqabaaakeaacaWG6bGaeyOeI0IaaGimaiaac6cacaaI4aaaaiabgUcaRmaalaaabaGaamyqamaaBaaaleaacaaIZaaabeaaaOqaaiaadQhacqGHsislcaaIWaGaaiOlaiaaiwdaaaaaaa@597A@
The problem is to determine the constants
A
1
,
A
2
,
A
3
A
1
,
A
2
,
A
3
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadgeadaWgaaWcbaGaaGymaaqabaGccaGGSaGaaGjbVlaadgeadaWgaaWcbaGaaGOmaaqabaGccaGGSaGaaGjbVlaadgeadaWgaaWcbaGaaG4maaqabaaaaa@3F75@
. 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
+(−1.3
A
1
−0.5
A
2
−0.8
A
3
)z+0.4
A
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
+(−1.3
A
1
−0.5
A
2
−0.8
A
3
)z+0.4
A
1
z(z−0.8)(z−0.5)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@9586@
Equating the corresponding coefficients of both sides , we get 3 simultaneours equations
A
1
+
A
2
+
A
3
=0
−1.3
A
1
−0.5
A
2
−0.8
A
3
=0
0.4
A
1
=0.8
A
1
+
A
2
+
A
3
=0
−1.3
A
1
−0.5
A
2
−0.8
A
3
=0
0.4
A
1
=0.8
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@6881@
having the roots
A
1
=2,
A
2
=−1,
A
3
=−1
A
1
=2,
A
2
=−1,
A
3
=−1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadgeadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaaIYaGaaiilaiaaywW7caWGbbWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaeyOeI0IaaGymaiaacYcacaaMf8UaamyqamaaBaaaleaacaaIZaaabeaakiabg2da9iabgkHiTiaaigdaaaa@469F@
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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIeacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaaikdaaeaacaWG6baaaiabgkHiTmaalaaabaGaaGymaaqaaiaadQhacqGHsislcaaIWaGaaiOlaiaaiIdaaaGaeyOeI0YaaSaaaeaacaaIXaaabaGaamOEaiabgkHiTiaaicdacaGGUaGaaGynaaaaaaa@4774@
Looking at the list of z transform pairs (Table .4.1 ) we see that the terms on the RHS are not ready for the inverse transform , we then proceed to rearrange H(z) :
H(z)=[
2+
z
z−0.8
−
z
z−0.5
]
z
−1
=[
2=
1
1−0.8
z
−1
−
1
1−0.5
z
−1
]
z
−1
H(z)=[
2+
z
z−0.8
−
z
z−0.5
]
z
−1
=[
2=
1
1−0.8
z
−1
−
1
1−0.5
z
−1
]
z
−1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@6328@
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)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaaikdacqaH0oazcaGGOaGaamOBaiaacMcacqGHsislcaaIWaGaaiOlaiaaiIdadaahaaWcbeqaaiaad6gaaaGccaWG1bGaaiikaiaad6gacaGGPaGaeyOeI0IaaGimaiaac6cacaaI1aWaaWbaaSqabeaacaWGUbaaaOGaamyDaiaacIcacaWGUbGaaiykaaaa@499D@
Finally , delay above function one sample due to the factor z–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)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIgacaGGOaGaamOBaiaacMcacqGH9aqpcaaIYaGaeqiTdqMaaiikaiaad6gacqGHsislcaaIXaGaaiykaiabgUcaRiaacIcacaaIWaGaaiOlaiaaiwdadaahaaWcbeqaaiaad6gacqGHsislcaaIXaaaaOGaey4kaSIaaGimaiaac6cacaaI4aWaaWbaaSqabeaacaWGUbGaeyOeI0IaaGymaaaakiaacMcacaWG1bGaaiikaiaad6gacqGHsislcaaIXaGaaiykaaaa@5279@
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) or H(z) (here we take 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−
p
1
)(z−
p
2
)(z−
p
3
)
X(z)=
N(z)
D(z)
=
N(z)
(z−
p
1
)(z−
p
2
)(z−
p
3
)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaad6eacaGGOaGaamOEaiaacMcaaeaacaWGebGaaiikaiaadQhacaGGPaaaaiabg2da9maalaaabaGaamOtaiaacIcacaWG6bGaaiykaaqaaiaacIcacaWG6bGaeyOeI0IaamiCamaaBaaaleaacaaIXaaabeaakiaacMcacaGGOaGaamOEaiabgkHiTiaadchadaWgaaWcbaGaaGOmaaqabaGccaGGPaGaaiikaiaadQhacqGHsislcaWGWbWaaSbaaSqaaiaaiodaaeqaaOGaaiykaaaaaaa@5439@
(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
+...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaadgeadaWgaaWcbaGaaGymaaqabaaakeaacaWG6bGaeyOeI0IaamiCamaaBaaaleaacaaIXaaabeaaaaGccqGHRaWkdaWcaaqaaiaadgeadaWgaaWcbaGaaGOmaaqabaaakeaacaWG6bGaeyOeI0IaamiCamaaBaaaleaacaaIYaaabeaaaaGccqGHRaWkdaWcaaqaaiaadgeadaWgaaWcbaGaaG4maaqabaaakeaacaWG6bGaeyOeI0IaamiCamaaBaaaleaacaaIZaaabeaaaaGccqGHRaWkcaGGUaGaaiOlaiaac6caaaa@4FA5@
However, on the knowledge of the transform pairs given in Equation (4.31) and (4.32) , we expand X(z)/z rather than 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
+...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@762F@
Now let’s multiply both sides of above equation with each term (z - p
ii size 12{ {} rSub { size 8{i} } } {}) , i = 1,2,…, and then evaluate the obtained expressions at the poles p
11 size 12{ {} rSub { size 8{1} } } {}, p
22 size 12{ {} rSub { size 8{2} } } {}… This will lead to the following formula for the constants :
A
i
=(z−
p
i
)
X(z)
z
|
z=
p
i
, i=0, 1, 2...
A
i
=(z−
p
i
)
X(z)
z
|
z=
p
i
, i=0, 1, 2...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadgeadaWgaaWcbaGaamyAaaqabaGccqGH9aqpcaGGOaGaamOEaiabgkHiTiaadchadaWgaaWcbaGaamyAaaqabaGccaGGPaWaaSaaaeaacaWGybGaaiikaiaadQhacaGGPaaabaGaamOEaaaacaGG8bWaaSbaaSqaaiaadQhacqGH9aqpcaWGWbWaaSbaaWqaaiaadMgaaeqaaaWcbeaakiaaysW7caGGSaGaaGzbVlaaywW7caWGPbGaeyypa0JaaGimaiaacYcacaaMe8UaaGymaiaacYcacaaMe8UaaGOmaiaac6cacaGGUaGaaiOlaaaa@57D3@
(2)
Example 2 Find the signal whose z – transform is given by
X(z)=
2
z
2
−2.05z
z
2
−2.05z+1
X(z)=
2
z
2
−2.05z
z
2
−2.05z+1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaaikdacaWG6bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaac6cacaaIWaGaaGynaiaadQhaaeaacaWG6bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaac6cacaaIWaGaaGynaiaadQhacqGHRaWkcaaIXaaaaaaa@4A0B@
Solution
The poles are
p1=0.8p1=0.8 size 12{p rSub { size 8{1} } =0 "." 8} {} and
p2p2 size 12{p rSub { size 8{2} } } {}= 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) / z , which is proper, for the expansion and write
X(z)=
2
z
2
−2.05z
z(
z
2
−2.05z+1)
=
2
z
2
−2.05z
z(z−0.8)(z−1.25)
=
A
0
z
+
A
1
z−
p
1
+
A
2
z−
p
2
X(z)=
2
z
2
−2.05z
z(
z
2
−2.05z+1)
=
2
z
2
−2.05z
z(z−0.8)(z−1.25)
=
A
0
z
+
A
1
z−
p
1
+
A
2
z−
p
2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@81E6@
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
=
2
z
2
−2.05z
z(z−1.25)
|
z=0.8
=1
A
2
=(z−1.25)
X(z)
z
|
z=1.25
=
2
z
2
−2.05z
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
=
2
z
2
−2.05z
z(z−1.25)
|
z=0.8
=1
A
2
=(z−1.25)
X(z)
z
|
z=1.25
=
2
z
2
−2.05z
z(z−0.8)
|
z=1.25
=1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@9F15@
Hence
X(z)=
z
z−0.8
+
z
z−1.25
X(z)=
z
z−0.8
+
z
z−1.25
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaadQhaaeaacaWG6bGaeyOeI0IaaGimaiaac6cacaaI4aaaaiabgUcaRmaalaaabaGaamOEaaqaaiaadQhacqGHsislcaaIXaGaaiOlaiaaikdacaaI1aaaaaaa@4606@
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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@A2C1@
Another way is to first express X(z) in terms of z
−1−1 size 12{ {} rSup { size 8{ - 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
...
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
...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@8A95@
and then evaluate the constants as
A
i
=(1−
p
i
z
−1
)X(z)
|
z=
p
i
A
i
=(1−
p
i
z
−1
)X(z)
|
z=
p
i
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadgeadaWgaaWcbaGaamyAaaqabaGccqGH9aqpcaGGOaGaaGymaiabgkHiTiaadchadaWgaaWcbaGaamyAaaqabaGccaWG6bWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaaiykaiaadIfacaGGOaGaamOEaiaacMcacaGG8bWaaSbaaSqaaiaadQhacqGH9aqpcaWGWbWaaSbaaWqaaiaadMgaaeqaaaWcbeaaaaa@4946@
(3)
Example 3 Repeat previous example but expressing X(z) in terms of z
−1−1 size 12{ {} rSup { size 8{ - 1} } } {}.
Solution
First
X(z)=
2
z
2
−2.05z
z
2
−2.05z−1
=
2−2.05
z
−1
1−2.05
z
−1
+
z
−2
X(z)=
2
z
2
−2.05z
z
2
−2.05z−1
=
2−2.05
z
−1
1−2.05
z
−1
+
z
−2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaaikdacaWG6bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaac6cacaaIWaGaaGynaiaadQhaaeaacaWG6bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaac6cacaaIWaGaaGynaiaadQhacqGHsislcaaIXaaaaiabg2da9maalaaabaGaaGOmaiabgkHiTiaaikdacaGGUaGaaGimaiaaiwdacaWG6bWaaWbaaSqabeaacqGHsislcaaIXaaaaaGcbaGaaGymaiabgkHiTiaaikdacaGGUaGaaGimaiaaiwdacaWG6bWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaey4kaSIaamOEamaaCaaaleqabaGaeyOeI0IaaGOmaaaaaaaaaa@5DBE@
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
(1−0.8
z
−1
)(1−1.25
z
−1
)
=
A
1
1−0.8
z
−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
(1−0.8
z
−1
)(1−1.25
z
−1
)
=
A
1
1−0.8
z
−1
+
A
2
1−1.25
z
−1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@8ACB@
The constants are given by
A
1
=(1−0.8
z
−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
1−0.8
z
−1
|
z=0.8
=1
A
1
=(1−0.8
z
−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
1−0.8
z
−1
|
z=0.8
=1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@8BA8@
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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaadQhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG6bGaey4kaSIaaGymaaqaaiaadQhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIZaGaamOEaiabgUcaRiaaikdaaaaaaa@45C6@
Solution
The poles of X(z) are z = -1 and z = -2 . Now let’s write
X(z)
z
=
2
z
2
+z+1
z(z+1)(z+2)
=
A
0
Z
+
A
1
z+1
+
A
2
z+2
X(z)
z
=
2
z
2
+z+1
z(z+1)(z+2)
=
A
0
Z
+
A
1
z+1
+
A
2
z+2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalaaabaGaamiwaiaacIcacaWG6bGaaiykaaqaaiaadQhaaaGaeyypa0ZaaSaaaeaacaaIYaGaamOEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadQhacqGHRaWkcaaIXaaabaGaamOEaiaacIcacaWG6bGaey4kaSIaaGymaiaacMcacaGGOaGaamOEaiabgUcaRiaaikdacaGGPaaaaiabg2da9maalaaabaGaamyqamaaBaaaleaacaaIWaaabeaaaOqaaiaadQfaaaGaey4kaSYaaSaaaeaacaWGbbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamOEaiabgUcaRiaaigdaaaGaey4kaSYaaSaaaeaacaWGbbWaaSbaaSqaaiaaikdaaeqaaaGcbaGaamOEaiabgUcaRiaaikdaaaaaaa@5884@
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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@9F28@
Hence
H(z)=0.5
z
z+1
+
1.5z
z+2
H(z)=0.5
z
z+1
+
1.5z
z+2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIeacaGGOaGaamOEaiaacMcacqGH9aqpcaaIWaGaaiOlaiaaiwdadaWcaaqaaiaadQhaaeaacaWG6bGaey4kaSIaaGymaaaacqGHRaWkdaWcaaqaaiaaigdacaGGUaGaaGynaiaadQhaaeaacaWG6bGaey4kaSIaaGOmaaaaaaa@4698@
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)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamOBaiaacMcacqGH9aqpcaaIWaGaaiOlaiaaiwdacqaH0oazcaGGOaGaamOBaiaacMcacqGHsislcaGGOaGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaad6gaaaGccaWG1bGaaiikaiaad6gacaGGPaaaaa@48A6@
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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaadQhadaahaaWcbeqaaiaaikdaaaaakeaacaWG6bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadQhacqGHRaWkcaaIXaaaaaaa@4246@
solution
The poles are given by
z
2
+2z+1=0
z
2
+2z+1=0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadQhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamOEaiabgUcaRiaaigdacqGH9aqpcaaIWaaaaa@3DC9@
whose roots are
p
1
=−
1
2
+j
3
2
=
e
j
2π
3
p
1
=−
1
2
+j
3
2
=
e
j
2π
3
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadchadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaiabgUcaRiaadQgadaWcaaqaamaakaaabaGaaG4maaWcbeaaaOqaaiaaikdaaaGaeyypa0JaamyzamaaCaaaleqabaGaamOAamaalaaabaGaaGOmaiabec8aWbqaaiaaiodaaaaaaaaa@450E@
p
2
=−
1
2
−j
3
2
=
e
−j
2π
3
=
p
1
*
p
2
=−
1
2
−j
3
2
=
e
−j
2π
3
=
p
1
*
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadchadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaiabgkHiTiaadQgadaWcaaqaamaakaaabaGaaG4maaWcbeaaaOqaaiaaikdaaaGaeyypa0JaamyzamaaCaaaleqabaGaeyOeI0IaamOAamaalaaabaGaaGOmaiabec8aWbqaaiaaiodaaaaaaOGaeyypa0JaamiCamaaDaaaleaacaaIXaaabaGaaiOkaaaaaaa@49A2@
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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@6011@
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
5π
/6
A
2
=
0.5z
z+1−j
3
|
z=−
1
2
−j
3
2
=−
1
2
+j
1
2
3
=
1
3
e
−j
5π
/6
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
5π
/6
A
2
=
0.5z
z+1−j
3
|
z=−
1
2
−j
3
2
=−
1
2
+j
1
2
3
=
1
3
e
−j
5π
/6
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@7ED5@
The time signal is
X(n)=(
A
1
p
1
n
+
A
2
p
2
n
)u(n)
=[
1
3
e
j
5π
/6
(
e
j
2π
/3
)
n
+
1
3
e
−j
5π
/6
(
e
−j
2π
/3
)
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
5π
/6
(
e
j
2π
/3
)
n
+
1
3
e
−j
5π
/6
(
e
−j
2π
/3
)
n
]u(n)
=
2
3
cos(
2π
3
n+
5π
6
)u(n)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@8B4B@
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)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaad6eacaGGOaGaamOEaiaacMcaaeaacaWGebGaaiikaiaadQhacaGGPaaaaiaaywW7cqGHshI3caaMf8UaamOtaiaacIcacaWG6bGaaiykaiabg2da9iaadgfacaGGOaGaamOEaiaacMcacaWGebGaaiikaiaadQhacaGGPaGaey4kaSIaamOuaiaacIcacaWG6bGaaiykaaaa@547E@
The order of R(z) must be smaller than that of 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)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaad6eacaGGOaGaamOEaiaacMcaaeaacaWGebGaaiikaiaadQhacaGGPaaaaiabg2da9maalaaabaGaamyuaiaacIcacaWG6bGaaiykaiaadseacaGGOaGaamOEaiaacMcacqGHRaWkcaWGsbGaaiikaiaadQhacaGGPaaabaGaamiraiaacIcacaWG6bGaaiykaaaacqGH9aqpcaWGrbGaaiikaiaadQhacaGGPaGaey4kaSYaaSaaaeaacaWGsbGaaiikaiaadQhacaGGPaaabaGaamiraiaacIcacaWG6bGaaiykaaaaaaa@5A81@
(4)
We then expand the rational polynomial term as usual.
In another way , we first temporarily forget N(z) and proceed to expand
1/D(z)1/D(z) size 12{ {1} slash {D \( z \) } } {}, afterwards multiply N(z
−1−1 size 12{ {} rSup { size 8{ - 1} } } {}) in for the full expansion of 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)
m
X(z)
z
=
N(z)
(z−p)
m
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalaaabaGaamiwaiaacIcacaWG6bGaaiykaaqaaiaadQhaaaGaeyypa0ZaaSaaaeaacaWGobGaaiikaiaadQhacaGGPaaabaGaaiikaiaadQhacqGHsislcaWGWbGaaiykamaaCaaaleqabaGaamyBaaaaaaaaaa@43BB@
(5)
The multiple pole p may be real or complex . The expansion is
X(z)
z
=
A
1
z−p
+
A
2
(z−p)
2
+...+
A
m−1
(z−p)
m−1
+
A
m
(z−p)
m
X(z)
z
=
A
1
z−p
+
A
2
(z−p)
2
+...+
A
m−1
(z−p)
m−1
+
A
m
(z−p)
m
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@5E6A@
(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 .
X(z)=
z
2
(z+1)
(z−1)
2
X(z)=
z
2
(z+1)
(z−1)
2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpdaWcaaqaaiaadQhadaahaaWcbeqaaiaaikdaaaaakeaacaGGOaGaamOEaiabgUcaRiaaigdacaGGPaGaaiikaiaadQhacqGHsislcaaIXaGaaiykamaaCaaaleqabaGaaGOmaaaaaaaaaa@44F8@
Find x(n).
Solution
X(z) has a simple pole at
p0p0 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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalaaabaGaamiwaiaacIcacaWG6bGaaiykaaqaaiaadQhaaaGaeyypa0ZaaSaaaeaacaWG6baabaGaaiikaiaadQhacqGHRaWkcaaIXaGaaiykaiaacIcacaWG6bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaacaWGbbWaaSbaaSqaaiaaicdaaeqaaaGcbaGaamOEaiabgUcaRiaaigdaaaGaey4kaSYaaSaaaeaacaWGbbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamOEaiabgkHiTiaaigdaaaGaey4kaSYaaSaaaeaacaWGbbWaaSbaaSqaaiaaikdaaeqaaaGcbaGaaiikaiaadQhacqGHsislcaaIXaGaaiykamaaCaaaleqabaGaaGOmaaaaaaaaaa@5769@
The constant A
00 size 12{ {} 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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadgeadaWgaaWcbaGaaGimaaqabaGccqGH9aqpcaGGOaGaamOEaiabgUcaRiaaigdacaGGPaWaaSaaaeaacaWGybGaaiikaiaadQhacaGGPaaabaGaamOEaaaacaGG8bWaaSbaaSqaaiaadQhacqGH9aqpcqGHsislcaaIXaaabeaakiabg2da9maalaaabaGaamOEaaqaaiaacIcacaWG6bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaaaaOGaaiiFamaaBaaaleaacaWG6bGaeyypa0JaeyOeI0IaaGymaaqabaGccqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaaI0aaaaaaa@551C@
Next we find A
22 size 12{ {} rSub { size 8{2} } } {} first (the constant coresponding to the pole of highest multiplicity ) by multiplying both sides by (z-1)
22 size 12{ {} rSup { size 8{2} } } {} and compute at z = p
11 size 12{ {} rSub { size 8{1} } } {} = p
22 size 12{ {} 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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalaaabaGaaiikaiaadQhacqGHsislcaaIXaGaaiykamaaCaaaleqabaGaaGOmaaaakiaadIfacaGGOaGaamOEaiaacMcaaeaacaWG6baaaiabg2da9maalaaabaGaaiikaiaadQhacqGHsislcaaIXaGaaiykamaaCaaaleqabaGaaGOmaaaaaOqaaiaadQhacqGHRaWkcaaIXaaaaiaadgeadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaGGOaGaamOEaiabgkHiTiaaigdacaGGPaGaamyqamaaBaaaleaacaaIXaaabeaakiabgUcaRiaadgeadaWgaaWcbaGaaGOmaaqabaaaaa@5298@
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
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadgeadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaGGOaGaamiEaiabgkHiTiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaOWaaSaaaeaacaWGybGaaiikaiaadQhacaGGPaaabaGaamOEaaaacaGG8bWaaSbaaSqaaiaadQhacqGH9aqpcaaIXaaabeaakiabg2da9maalaaabaGaamOEaaqaaiaadQhacqGHRaWkcaaIXaaaaiaacYhadaWgaaWcbaGaamOEaiabg2da9iaaigdaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaaaaa@50FA@
To find A
11 size 12{ {} rSub { size 8{1} } } {} we differentiate both sides of the above equation with respect to z and compute at z = p
11 size 12{ {} rSub { size 8{1} } } {} = p
22 size 12{ {} rSub { size 8{2} } } {}= 1
A
1
=
d
dz
[
(z−1)
2
X(z)
z
]
z=1
=
d
dz
[
z
z+1
]
|
z=1
=
1
4
A
1
=
d
dz
[
(z−1)
2
X(z)
z
]
z=1
=
d
dz
[
z
z+1
]
|
z=1
=
1
4
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadgeadaWgaaWcbaGaaGymaaqabaGccqGH9aqpdaWcaaqaaiaadsgaaeaacaWGKbGaamOEaaaadaWadaqaaiaacIcacaWG6bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaGcdaWcaaqaaiaadIfacaGGOaGaamOEaiaacMcaaeaacaWG6baaaaGaay5waiaaw2faamaaBaaaleaacaWG6bGaeyypa0JaaGymaaqabaGccqGH9aqpdaWcaaqaaiaadsgaaeaacaWGKbGaamOEaaaadaWadaqaamaalaaabaGaamOEaaqaaiaadQhacqGHRaWkcaaIXaaaaaGaay5waiaaw2faaiaacYhadaWgaaWcbaGaamOEaiabg2da9iaaigdaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGinaaaaaaa@59A3@
Thus
X(z)=−
1
4
z
z+1
+
1
4
z
z−1
+
1
2
z
(z−1)
2
X(z)=−
1
4
z
z+1
+
1
4
z
z−1
+
1
2
z
(z−1)
2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOEaiaacMcacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaaI0aaaamaalaaabaGaamOEaaqaaiaadQhacqGHRaWkcaaIXaaaaiabgUcaRmaalaaabaGaaGymaaqaaiaaisdaaaWaaSaaaeaacaWG6baabaGaamOEaiabgkHiTiaaigdaaaGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaadaWcaaqaaiaadQhaaeaacaGGOaGaamOEaiabgkHiTiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaaaaaaa@4EBB@
And for causal signal ,
X(n)=
−1
4
(
−1
)
n
u(n)+
1
2
nu(n)
=[
−
1
4
(−1)
n
+
1
2
n+
1
4
]u(n)
X(n)=
−1
4
(
−1
)
n
u(n)+
1
2
nu(n)
=[
−
1
4
(−1)
n
+
1
2
n+
1
4
]u(n)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@621C@
