1633 the inverse z-transform
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Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”)
![Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 2
The Inverse Z-Transform
• Formal inverse z-transform is based on a Cauchy integral
• Less formal ways sufficient most of the time
– Inspection method
– Partial fraction expansion
– Power series expansion
• Inspection Method
– Make use of known z-transform pairs such as
– Example: The inverse z-transform of
[ ] az
az1
1
nua 1
Zn
>
−
→← −
( ) [ ] [ ]nu
2
1
nx
2
1
z
z
2
1
1
1
zX
n
1
=→>
−
=
−](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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1633 the inverse z-transform
- 1. The Inverse z-Transform In science one tries to tell people, in such a way as to be understood by everyone, something that no one ever knew before. But in poetry, it's the exact opposite. Paul Dirac Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer, and Buck, ©1999-2000 Prentice Hall Inc.
- 2. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 2 The Inverse Z-Transform • Formal inverse z-transform is based on a Cauchy integral • Less formal ways sufficient most of the time – Inspection method – Partial fraction expansion – Power series expansion • Inspection Method – Make use of known z-transform pairs such as – Example: The inverse z-transform of [ ] az az1 1 nua 1 Zn > − →← − ( ) [ ] [ ]nu 2 1 nx 2 1 z z 2 1 1 1 zX n 1 =→> − = −
- 3. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 3 Inverse Z-Transform by Partial Fraction Expansion • Assume that a given z-transform can be expressed as • Apply partial fractional expansion • First term exist only if M>N – Br is obtained by long division • Second term represents all first order poles • Third term represents an order s pole – There will be a similar term for every high-order pole • Each term can be inverse transformed by inspection ( ) ∑ ∑ = − = − = N 0k k k M 0k k k za zb zX ( ) ( )∑∑∑ = − ≠= − − = − − + − += s 1m m1 i m N ik,1k 1 k k NM 0r r r zd1 C zd1 A zBzX
- 4. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 4 Partial Fractional Expression • Coefficients are given as • Easier to understand with examples ( ) ( )∑∑∑ = − ≠= − − = − − + − += s 1m m1 i m N ik,1k 1 k k NM 0r r r zd1 C zd1 A zBzX ( ) ( ) kdz 1 kk zXzd1A = − −= ( ) ( ) ( ) ( )[ ] 1 idw 1s ims ms ms i m wXwd1 dw d d!ms 1 C − = − − − − − −− =
- 5. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 5 Example: 2nd Order Z-Transform – Order of nominator is smaller than denominator (in terms of z-1 ) – No higher order pole ( ) 2 1 z:ROC z 2 1 1z 4 1 1 1 zX 11 > − − = −− ( ) − + − = −− 1 2 1 1 z 2 1 1 A z 4 1 1 A zX ( ) 1 4 1 2 1 1 1 zXz 4 1 1A 1 4 1 z 1 1 −= − = −= − = − ( ) 2 2 1 4 1 1 1 zXz 2 1 1A 1 2 1 z 1 2 = − = −= − = −
- 6. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 6 Example Continued • ROC extends to infinity – Indicates right sided sequence ( ) 2 1 z z 2 1 1 2 z 4 1 1 1 zX 11 > − + − − = −− [ ] [ ] [ ]nu 4 1 -nu 2 1 2nx nn =
- 7. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 7 Example #2 • Long division to obtain Bo ( ) ( ) ( ) 1z z1z 2 1 1 z1 z 2 1 z 2 3 1 zz21 zX 11 21 21 21 > − − + = +− ++ = −− − −− −− 1z5 2z3z 2 1z2z1z 2 3 z 2 1 1 12 1212 − +− +++− − −− −−−− ( ) ( )11 1 z1z 2 1 1 z51 2zX −− − − − +− += ( ) 1 2 1 1 z1 A z 2 1 1 A 2zX − − − + − += ( ) 9zXz 2 1 1A 2 1 z 1 1 −= −= = − ( ) ( ) 8zXz1A 1z 1 2 =−= = −
- 8. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 8 Example #2 Continued • ROC extends to infinity – Indicates right-sides sequence ( ) 1z z1 8 z 2 1 1 9 2zX 1 1 > − + − −= − − [ ] [ ] [ ] [ ]n8u-nu 2 1 9n2nx n −δ=
- 9. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 9 Inverse Z-Transform by Power Series Expansion • The z-transform is power series • In expanded form • Z-transforms of this form can generally be inversed easily • Especially useful for finite-length series • Example ( ) [ ]∑ ∞ −∞= − = n n znxzX ( ) [ ] [ ] [ ] [ ] [ ] ++++−+−+= −− 2112 z2xz1x0xz1xz2xzX ( ) ( )( ) 12 1112 z 2 1 1z 2 1 z z1z1z 2 1 1zzX − −−− +−−= −+ −= [ ] [ ] [ ] [ ] [ ]1n 2 1 n1n 2 1 2nnx −δ+δ−+δ−+δ= [ ] = = =− −=− −= = 2n0 1n 2 1 0n1 1n 2 1 2n1 nx
- 10. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 10 Z-Transform Properties: Linearity • Notation • Linearity – Note that the ROC of combined sequence may be larger than either ROC – This would happen if some pole/zero cancellation occurs – Example: • Both sequences are right-sided • Both sequences have a pole z=a • Both have a ROC defined as |z|>|a| • In the combined sequence the pole at z=a cancels with a zero at z=a • The combined ROC is the entire z plane except z=0 • We did make use of this property already, where? [ ] ( ) x Z RROCzXnx = →← [ ] [ ] ( ) ( ) 21 xx21 Z 21 RRROCzbXzaXnbxnax ∩=+ →←+ [ ] [ ] [ ]N-nua-nuanx nn =
- 11. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 11 Z-Transform Properties: Time Shifting • Here no is an integer – If positive the sequence is shifted right – If negative the sequence is shifted left • The ROC can change the new term may – Add or remove poles at z=0 or z=∞ • Example [ ] ( ) x nZ o RROCzXznnx o = →←− − ( ) 4 1 z z 4 1 1 1 zzX 1 1 > − = − − [ ] [ ]1-nu 4 1 nx 1-n =
- 12. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 12 Z-Transform Properties: Multiplication by Exponential • ROC is scaled by |zo| • All pole/zero locations are scaled • If zo is a positive real number: z-plane shrinks or expands • If zo is a complex number with unit magnitude it rotates • Example: We know the z-transform pair • Let’s find the z-transform of [ ] ( ) xoo Zn o RzROCz/zXnxz = →← [ ] 1z:ROC z-1 1 nu 1- Z > →← [ ] ( ) [ ] ( ) [ ] ( ) [ ]nure 2 1 nure 2 1 nuncosrnx njnj o n oo ω−ω +=ω= ( ) rz zre1 2/1 zre1 2/1 zX 1j1j oo > − + − = −ω−−ω
- 13. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 13 Z-Transform Properties: Differentiation • Example: We want the inverse z-transform of • Let’s differentiate to obtain rational expression • Making use of z-transform properties and ROC [ ] ( ) x Z RROC dz zdX znnx =− →← ( ) ( ) azaz1logzX 1 >+= − ( ) ( ) 1 1 1 2 az1 1 az dz zdX z az1 az dz zdX − − − − + =−⇒ + − = [ ] ( ) [ ]1nuaannx 1n −−= − [ ] ( ) [ ]1nu n a 1nx n 1n −−= −
- 14. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 14 Z-Transform Properties: Conjugation • Example [ ] ( ) x **Z* RROCzXnx = →← ( ) [ ] ( ) [ ] [ ] ( ) [ ] ( ) [ ] [ ]{ }nxZznxznxzX znxznxzX znxzX n n n n n n n n n n ∗ ∞ −∞= −∗ ∞ −∞= ∗∗∗∗ ∞ −∞= ∗ ∗ ∞ −∞= −∗ ∞ −∞= − === = = = ∑∑ ∑∑ ∑
- 15. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 15 Z-Transform Properties: Time Reversal • ROC is inverted • Example: • Time reversed version of [ ] ( ) x Z R 1 ROCz/1Xnx = →←− [ ] [ ]nuanx n −= − [ ]nuan ( ) 1 11- 1-1 az za-1 za- az1 1 zX − − − <= − =
- 16. Copyright (C) 2005 Güner Arslan 351M Digital Signal Processing 16 Z-Transform Properties: Convolution • Convolution in time domain is multiplication in z-domain • Example:Let’s calculate the convolution of • Multiplications of z-transforms is • ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a| • Partial fractional expansion of Y(z) [ ] [ ] ( ) ( ) 2x1x21 Z 21 RR:ROCzXzXnxnx ∩ →←∗ [ ] [ ] [ ] [ ]nunxandnuanx 2 n 1 == ( ) az:ROC az1 1 zX 11 > − = − ( ) 1z:ROC z1 1 zX 12 > − = − ( ) ( ) ( ) ( )( )1121 z1az1 1 zXzXzY −− −− == ( ) 1z:ROCasume az1 1 z1 1 a1 1 zY 11 > − − −− = −− [ ] [ ] [ ]( )nuanu a1 1 ny 1n+ − − =