the z transform - uptshannon.etc.upt.ro/teaching/sp-pi/course/4_z_transform_2013.pdf · 1 1 the z...
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The z Transform
The z transform generalizes the Discrete-time Fourier Transform for the entire complex plane.For the complex variable is used the notation:
2 2
; ,
arg
jz x j y r e x y
z r x y
z
Ω= + ⋅ = ⋅ ∈
= = +
Ω =
2
notation:
For any discrete-time LTI system:
-the eigen function:
-the eigen value:
The Discrete LTI System Response to a Complex Exponential
[ ] 00 0
j nn nx n z r e Ω= =
System
h[n]
input output
[ ] [ ] [ ] [ ] ( )0 0 0 0 0 0n n k n k n
k ky n h n z h k z z h k z z H z
∞ ∞− −
=−∞ =−∞= ∗ ⋅ = ⋅ ⋅= =∑ ∑
[ ] [ ] ( )0 0 0n ny n h n z z H z= ∗ ⋅=
( ) [ ] k
kH z h k z
∞−
=−∞= ⋅∑
( )0H z0nz
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3
fast computation of the response of any discrete-time LTI system to linear combinations of complex exponentials of the form:
transfer function, z transform of the impulse response h[n].
[ ] ( )0 0ny n z H z= ⋅
[ ] nk k
kx n c z= ∑
System
h[n]
input output[ ] ( ) n
k k kk
y n c H z z= ∑
( ) [ ] k
kH z h k z
∞−
=−∞= ⋅∑
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Bilateral z TransformThe bilateral z transform of the signal x[n] :
generalization of the discrete-time Fourier transform
Discrete-time Fourier transform = particular case of z transform on the unit circle |z|=1
[ ] ( ) ( ) [ ] , , 0jn
nZ x n z X z x n z z r e r
∞Ω−
=−∞= = ⋅ = ⋅ ≥∑
[ ] ( ) [ ]( ) [ ] ( ); 0j nn n
nZ x n z x n r e r x n r
∞− Ω− −
=−∞= ⋅ = ⋅ Ω ≥∑ F
[ ] ( ) [ ] ( )1 jz r Z x n e x nΩ= = ⇒ = ΩF|z|=1
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5
Examples
region of convergence (ROC) = set of values of z for which X(z) is convergent
rational function, zeros = roots of numerator ; poles = roots of denominator.
Pole/zero plot or constellation (PZC).
[ ] [ ]1. , 1nx n a n a= σ <
( ) ( )1
0 0
nn n
n nX z a z az
∞ ∞− −
= == =∑ ∑ 1 1az− <
( ) 11 ,
1X z z a
az−= >
−
convergent:
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convergent if:
one pole in a.
[ ] [ ]2. 1 , 1nx n a n a= − σ − − <
( ) ( )1 1
11 1 1
1 0
1 1 1n nnn n
n n n nX z a z az
az az az
− − ∞ ∞− −
− − −=−∞ =−∞ = =
⎛ ⎞ ⎛ ⎞= − = − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
∑ ∑ ∑ ∑
11 1
a z−<
( ) 11
1
1 1 ; 1 11
X z z aazaz
az
−−
−
= − = <⎛ ⎞ −−⎜ ⎟⎝ ⎠
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7
same z transform ; different ROCs
First example Second example
Region of convergence ROC : set of values of z for which theseries X(z) is convergent.
The Z transform exists where the Fourier transform of x[n]r-n
exists, r=|z|
( ) 11
1X z
az−=
−
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• ROC of a bilateral z transform can not contain any pole • for right sided signals (including causal signals), the
ROC extends outward from the outermost pole, • for left sided signals (including anticausal signals), the
ROC extends inward from the innermost pole• for infinite duration signals, ROC is a ring that doesn’t
include poles – bounded on the interior and exterior by a pole.
• for finite duration signals, the ROC is the entire z-plane, except possibly z=0 or z=∞.
• a causal and stable system’s transfer function has thepoles inside the unit circle. ROC is outside the unit circle.
Properties of the ROC of the Bilateral z Transform
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9
[ ] [ ] [ ]1 2x n x n x n= +
first component of x[n] is right sided; ROC is outside the circle with radius R-.
second component of x[n] is left sided, ROCis inside the circle with radius R+.
The ROC of X(z)is the intersectionof the ROCs of its components (ring)
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Example[ ]
[ ] [ ] [ ]1 2
3. , 0 1
nx n a a
x n x n x n
= < <
= +
[ ] [ ]
[ ] [ ]
[ ]
1 1
2
1
1 , 1
1
1 1 11 , 11
n
n
x n a n z aaz
x n a n
n za az
a
−
−
−
= σ ⎯⎯→ >−
= σ − −
−⎛ ⎞= σ − − ⎯⎯→ <⎜ ⎟⎝ ⎠ −
( )
2 1 1; 1
n a za a za az a z
a
−⎯⎯→ < <
⎛ ⎞− −⎜ ⎟⎝ ⎠
Right sided (causal) signal
Left sided (anticausal) signal
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Inverse z Transform
• Γ - counterclockwise closed path (contour) included in the region of convergence.
• Γ - encircles the origin and must encircle all poles of X(z).
• integral along a contour. • z transforms in DSP : rational functions. • partial fraction expansion and tables signal – z
transform pairs (knowing the pole/zero plot).
( ) [ ] ( ) [ ]1 11 ; 2
nZ X z n X z z dz x n ROCj
− −Γ
= = Γ ⊂π ∫
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The Bilateral z Transform computationusing its pole/zero plot
X(z) - rational function
( )( )
( )0
1
1
; N
M
kk
pkk
z zX z k z ROC
z z
=
=
−= ∈
−
∏
∏
poles and zeros ⇒ Z transform without constant k
X(z0) – also known ⇒ Z transform and constant k
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Example
unit circle ⊂ ROC ⇒ discrete-time Fourier transform exists:
A ∈ unit circle; ∠(OA, Ox) = Ψ
the spectrum of signal x[n]
0pole/zero plot 0; 0.5pz z= =( ) 1
1 for 10.5 1 0.5
zX z kz z−
= = =− −
causal signal, ROC: 0.5pz z> =
( ) ( ) 10.5 1 0.5
jj
j jeX X e
e e
ΩΩ
Ω − ΩΩ = = =− −
( ) ( ); OA
XAP
Ω = Φ Ω = ψ −ϕ
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vectors Azpk , Azok: ∠(AZ0k, Ox) = Ψk ; ∠(AZpk, Ox) = ϕk
Magnitude and phase spectrum:
The frequency Ω = length of the arc (of circle) in radians on the unit circle, between its intersection with the positive realaxis and the point A, in trigonometric sense.
General case:
( )0
1
1 1
1
; =
M
M Nkk
k kNk k
pkk
AzX k Argk
Az
=
= =
=
Ω = Φ + ψ − ϕ∏
∑ ∑∏
( )( )
( )0
1
1
; N
M
kk
pkk
z zX z k z ROC
z z
=
=
−= ∈
−
∏
∏
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The Unilateral z Transform
•For causal signals, unilateral z transform = bilateral ztransform
•ROC: entire complex plane or the outside region of a disc centered in 0.
•Useful for causal systems described by difference equations, with non zero initial conditions.
[ ] ( ) ( ) [ ]0
nu u
nZ x n z X z x n z
∞−
== = ∑
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z Transform Propertiesnotations:
1. Linearity
Proof. Directly, using the definitions. Homework - Prove it.
[ ] ( ) [ ] ( )[ ] ( ) [ ] ( )
, ;
, ;
u
u
x
y
ZZ
ZZ
x n X z z ROC x n X z
y n Y z z ROC y n Y z
←⎯→ ∈ ←⎯⎯→
←⎯→ ∈ ←⎯⎯→
[ ] [ ] ( ) ( )[ ] [ ] ( ) ( )
, at leastx y
u u
ax n by n aX z bY z z ROC ROC
ax n by n aX z bY z
+ ←⎯→ + ∈ ∩
+ ←⎯→ +
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2. Time shifting
Proof.
bilateral z transform:
If n0>0, z=0 ∉ ROC. If n0<0, z=∞ ∉ ROC.
unilateral z transform:
( )
( ) [ ]
0
0
0
0
1
0 0
,
, >0
n
n nu
n n
x n n z X z z ROC
x n n z X z x n z n
−
−− −
=−
− ←⎯→ ∈⎡ ⎤⎣ ⎦⎛ ⎞
− ←⎯→ −⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎜ ⎟⎝ ⎠
∑
[ ]
[ ] [ ]
00
0
0
0
0 00
1
00
, 0
n n m nn mu
n m n
n m m
m m n
Z x n n x n n z x m z z
z x m z x m z n
− =∞ ∞−− −
= =−
∞ −− − −
= =−
− = − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
⎛ ⎞= + >⎜ ⎟⎜ ⎟
⎝ ⎠
∑ ∑
∑ ∑
[ ] ( ) [ ]0
0 00 0
n n m m n nn m
n m mZ x n n x n n z x m z z x m z
− =∞ ∞ ∞− + −− −
=−∞ =−∞ =−∞− = − = =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦∑ ∑ ∑
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3. Modulation in time
Proof.
More generally:
Homework - Prove it.
[ ] [ ] [ ]( ) (0 0 0 0nj n j n j jn
n nZ e x n e x n z x n e z X e z
∞ ∞ −Ω Ω − Ω − Ω−
=−∞ =−∞= = =∑ ∑
[ ]
[ ]
00 0
00
, n
nu
z zz x n X ROCz z
zz x n Xz
⎛ ⎞←⎯→ ∈⎜ ⎟⎜ ⎟
⎝ ⎠⎛ ⎞
←⎯→ ⎜ ⎟⎜ ⎟⎝ ⎠
[ ] ( )0 0 , j n je x n X e z z ROCΩ − Ω←⎯→ ∈
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4. Time reversal
Proof.
[ ] [ ] [ ] 1 1mm nn
n mZ x n x n z x m X
z z
−∞ ∞=−−
=−∞ =−∞
⎛ ⎞ ⎛ ⎞− = − = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
∑ ∑
[ ] ( )1 1, x n X z ROCz
−− ←⎯→ ∈
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5. Differentiation in time
Proof. direct application of the definitions and previous properties. Homework. Prove these properties.
[ ] [ ] ( ) ( )
[ ] [ ] ( ) ( ) [ ]
1
1
1 1 ;
1 1 1u
x n x n z X z z ROC
x n x n z X z x
−
−
− − ←⎯→ − ∈
− − ←⎯→ − − −
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6. Addition in time
Proof.
[ ]( ) [ ]
1
1 ; 11
unk
k
X z x kx k z
z
−
=−∞−
=−∞
+←⎯→ >
−
∑∑
[ ] ( ) ( )1 ; 11
n
k
X z zx k X z z ROC ROCzz
∗−
=−∞←⎯→ = ∈ ∩
−−∑
[ ] [ ] [ ] ( ) ( ) ( ) [ ] [ ] [ ]1
11 1 1 ; 1uk
x n y n y n X z z Y z y y x k−
−
=−∞= − − ←⎯→ = − − − − = ∑
[ ] [ ] [ ] [ ] [ ] ( ) ( ) ( )11 1n
ky n x k x n y n y n X z z Y z−
=−∞= ⇒ = − − ←⎯→ = −∑
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7. Differentiation in z domain
Proof.
By direct application of definitions.
Homework. Prove it.
[ ] ( )
[ ] ( )
;
u
dX znx n z z ROC
dzdX z
nx n zdz
←⎯→− ∈
←⎯→−
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8. Complex conjugation in time domain
Proof.
By direct application of definitions.
Homework. Prove it.
[ ] ( )[ ] ( )
;
u
x n X z z ROC
x n X z
∗ ∗ ∗
∗ ∗ ∗
←⎯→ ∈
←⎯→
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9. Time convolution (convolution theorem)
Proof.
[ ] [ ] [ ] [ ]( ) [ ] [ ]
[ ] [ ] ( )
[ ] [ ] ( ) ( )
n n
n n k
n kk
n kn k m m k
n k
Z x n y n x n y n z x k y n k z
x k z y n k z
y m z x k z Y z X z
∞ ∞ ∞− −
=−∞ =−∞ =−∞∞ ∞ − −−
=−∞ =−∞∞ ∞− =
− −
=−∞ =−∞
∗ = ∗ = −
= −
= =
∑ ∑ ∑
∑ ∑
∑ ∑
[ ] [ ][ ] [ ] ( ) ( )[ ] [ ] ( ) ( )
,
; at leastx y
u u
x n y n
x n y n X z Y z z ROC ROC
x n y n X z Y z
∈
∗ ←⎯→ ∈ ∩
∗ ←⎯→
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10. Product theorem or convolution theorem in the complex domain
Proof.
[ ] [ ] ( )
[ ] [ ] ( )
1 , 21 ,
2 u u
z dux n y n X u Y ROCj u u
z dux n y n X u Y ROCj u u
Γ
Γ
⎛ ⎞←⎯→ Γ∈⎜ ⎟π ⎝ ⎠⎛ ⎞←⎯→ Γ ⊂⎜ ⎟π ⎝ ⎠
∫
∫
: ; : and
:
y y y yx x x xx y
y yx x
zROC R z R ROC R z R R u R R R
u
ROC R R z R R
− + − + − + − +
− − + +
⇒
⇒
< < < < < < < <
< <
[ ] [ ] [ ] [ ] ( ) [ ]
( ) [ ] ( )
112
1 1 2 2
n n n
n nn
n
Z x n y n x n y n z X u u du y n zj
z du z duX u y n X u Yj u u j u u
∞ ∞− − −
Γ=−∞ =−∞
−∞
Γ Γ=−∞
⎛ ⎞= = ⎜ ⎟π⎝ ⎠
⎡ ⎤⎛ ⎞ ⎛ ⎞= =⎢ ⎥⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
∑ ∑ ∫
∑∫ ∫
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Particular Cases
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27
11. The Initial Value Theorem
Proof.
z transform of a causal signal
At the limit :
[ ] ( ) ( )0 lim limuz zx X z X z
→∞ →∞= =
( ) ( ) [ ] [ ] [ ]0
10 1nu
nX z X z x n z x x
z
∞−
== = = + +∑ …
( ) ( ) [ ] [ ] [ ]10 1 0zuX z X z x x xz →∞= = + + ⎯⎯⎯→…
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11. The Final Value Theorem
Proof.
[ ] [ ]( ) [ ] ( ) [ ] ( ) ( )
[ ] [ ] ( ) ( ) ( ) [ ]
1 1
0 0 1
0
1 1
0 1 0
n m mn nu u
n n m
mu u
m
x n x n z x n z X z x m z X z
z x m z x X z z X z zx
∞ ∞ ∞+ = − −− −
= = =∞
−
=
+ − = + − = −
⎛ ⎞= − − = − −⎜ ⎟
⎝ ⎠
∑ ∑ ∑
∑
[ ] [ ] ( ) ( ) ( ) ( )1 1
lim lim 1 lim 1uk z zx x k z X z z X z
→∞ → →∞ = = − = −
[ ] [ ]( ) ( ) ( ) [ ]
[ ] [ ]( ) [ ] [ ]( )
[ ] [ ]( )[ ] [ ] ( ) ( ) ( ) ( )
1 10
0 0
1 1
lim 1 lim 1 0
1 lim 1
lim 1 0
lim lim 1 lim 1
nuz zn
k
kn n
k
uz zk
x n x n z z X z x
x n x n x n x n
x k x
x x k z X z z X z
∞−
→ →=∞
→∞= =
→∞
→ →→∞
+ − = − −
+ − = + −
= + −
⇒ ∞ = = − = −
∑
∑ ∑
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Relation between the z Transform and the Laplace Transform
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) [ ]
[ ] [ ] ( )
ˆIdeal sampling:
ˆ
Discrete-time signal:
a a
an
snTa a
n n
a s d
n nad d
n n
s
s s
s s s
s
x t X s
x t x nT t nT
x t x nT t nT x nT e
x nT x n
Z x n x n z x nT z
∞
=−∞∞ ∞
−
=−∞ =−∞
∞ ∞− −
=−∞ =−∞
←⎯→
= δ −
= δ − =
=
⇒ = =
∑
∑ ∑
∑ ∑
L L
( ) ( ) [ ] [ ] ( ); a a sT d dsT
sT
s s
s
x t t Z x n x n x nTe z
z e
δ = ==
=
LLaplace of thesampled analog signal
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Using the z Transform for the Study of the Discrete LTI Systems
• The z transform is useful for studying discrete-time LTI systems (theorem of convolution of discrete-time signals)
• For a causal system, with non zero initial conditions, theunilateral z transform is used.
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31
The Transfer Function for a Discrete LTI System
•transfer function H(z) - z transform of the impulse response of discrete-LTI system. •describes completely the system in the complex domain.
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Stable system: It has a frequency response. Discrete-time Fourier transform of the impulse response is convergent. The unit circle belongs to the ROC of its transfer function
|z| =1 ⊂ ROCCausal system: Hu(z) = H(z)
ROC is outside of a disc. Causal and stable system:
Unit circle belongs to the ROC: |z| =1 ⊂ ROC. All poles of H(z): inside the unit disc |zp| <1
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33
Computation of the response of a discreteLTI system using the z transform
• If we know the input signal, x[n] and the system,h[n] ↔ H (z),
• Compute the z transform of the input, X (z). • Compute the product Y(z)=H(z) X (z). • Apply inverse z transform ⇒ the response y[n]. • For a causal input signal and a causal system
with non zero initial conditions, we use theunilateral z transform
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The Computation of the Inverse z Transform
There are three methods that can be used:
1. Direct computation of the integral,2. Partial fraction expansion of the function Y(z),3. Power series expansion of function Y (z).
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35
2. Partial fraction expansion of the function Y (z)
Y(z) rational function, ratio of polynomials in z-1 or z. We use z-1, denote z-1 = x ( z transforms expressed in
function of z-1 in tables)
( ) ( ) ( )( )
( )( )
( ) ( ) ( )( )
11
1
N z N xY x Y z
D xD z
R xY x I x
D x
−−
−= = =
= +
( ) [ ]1Zk kk k k
k k kI x c x c z c n k
−−= = ←⎯⎯→ δ −∑ ∑ ∑
36
( )( ) ( )1
km ki
im ikm k
s bR x aD x x x x x=
= +− −
∑ ∑∑
( ) ( )( )m m
m
R xa x x
D xx x
⎡ ⎤= −⎢ ⎥⎣ ⎦ = ( ) ( ) ( )
( )1
!
kk
k
s is
s i kk
k
kiR xdb x xD xs i dx x x
−
−
⎧ ⎫⎡ ⎤⎪ ⎪= −⎨ ⎬⎢ ⎥− ⎪ ⎪⎣ ⎦⎩ ⎭ =
19
37
Example: 2nd order Z-Transform
( ) ( )( ) ( )
( ) ( )( ) ( )( )
( )
[ ] ( ) ( ) [ ]
[ ] ( ) ( ) [ ]
1
1 1
1 1 1 1
; ROC: 0.50.5 0.25
8 8 8 162 4 2 42 4
8 16 4 42 4 1 0.5 1 0.25
For 0.5 : 4 0.5 0.25
For 0.25 : 4 0.5 0.25 1
For 0.25
u
n n
n n
zY z Y z zz z
z xY zx x x xz z
Y zz z z z
z y n n
z y n n
−
− −
− − − −
= = >− −
= = = −− − − −− −
= − = −− − − −
⎡ ⎤⇒ > = − σ⎣ ⎦⎡ ⎤< = − − σ − −⎣ ⎦
[ ] ( ) [ ] ( ) [ ]0.5 : 4 0.5 1 4 0.25n nz y n n n< < = − σ − − − σ
Causal signal y[n]
38
3. Power series expansion of function Y (z)
Expansion of function Y(z) into a power series
Example #a
( )
[ ]
1
1 1 2
0
, : 0
1 1 1 111! 2! ! !1!
n n
n
z
z
Y z e ROC z
e z z z zn n
y nn
−
∞− − − − −
=
= >
= + + + + + =
⇒ =
∑… …
20
39
Example #b
( )
( ) ( )
[ ] [ ] ( ) [ ]( )
[ ]
2
00 1
, : 0,
1 1 1 111! 2! ! !
1 1 1! !
1 11! 1 !
m m
m
n n
n n
n
z
z
Y z e ROC z z
e z z z zm m
z zn n
y n n n nn n
∞
=−
− −
=−∞ =−∞
= ≥ ≠ ∞
= + + + + + =
= = +− −
⇒ = δ − σ − − = σ −− −
∑
∑ ∑
… …
40
( ) ( )
( ) ( ) [ ] ( )
[ ] ( )
[ ] ( ) [ ]
1
1 1
1
1
ln 1 , :
1 1; 1
initial value theorem:
0 lim ln 1 0
1 1
n nn nn
n
zn
n
Y z az ROC z a
a aY z z y n n
n n
y az
y n a nn
−
+ +∞−
=
−→∞
= + > ⇒
− −= ⎯⎯→ = ≥
= + =
−= σ −
∑
causal signal
Example #c
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41
Example #d
( ) [ ] [ ]11 ,
1nY z z a y n a n
az−= > ⇒ = σ
−
42
Example #e, same transform as #d, different ROC
( )
( ) ( ) [ ] [ ]
1
11
1
1 , 1
anticausal signal, z transform contains only powers of z
1m n n n
m n
zY z z az aaz
Y z a z a z y n a n
−
∞ −− −
= = −∞
= = <−−
⇒
⇒ = − = − ⇒ = − σ − −∑ ∑
22
43
Discrete LTI Systems Described by Linear Constant Coefficient Difference Equations
[ ] [ ] 00 0
; 0N M
k kk k
a y n k b x n k a= =
− = − ≠∑ ∑
( ) ( ) ( ) ( )( )
0
0 0
0
transfer function
Mk
N M kk k k
k k Nkk k
kk
b z N za z Y z b z X z H z
D za z
−
− − =
−= =
=
= ⇒ = =∑
∑ ∑∑
Transfer function of a discrete LTI system : rational function in z or z-1. Zeros : roots of numerator N(z); Poles : roots of denominator D(z)
44
Causal and stable system: poles inside unit circle |z|=1
The initial value theorem can be applied:
The degree of the numerator of the transfer function of a
causal and stable system is smaller or equal with the
degree of its denominator, M≤ N
1pkz <
[ ] ( ) ( ) ( )( )
0 lim lim lim - finiteuz z z
N zh H z H z
D z→∞ →∞ →∞= = =
Degree M
Degree N
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45
The Contribution of the Poles of a Causal Discrete LTI System at its
Impulse ResponseConsider two cases: simple poles and double poles.
#1 Simple complex conjugated poles
Contribution of the two poles in the impulse response :
[ ]neaear pp jnjnnp σ⋅⎟
⎠⎞⎜
⎝⎛ ⋅+⋅ Ω−Ω *
( )1 1
and
1 1
j jp p p p
j jp p
p p
p p
z r e z r e
a aH zr e z r e z
Ω − Ω∗
∗
Ω − Ω− −
= =
= + + +− −
… …
46
The pair of poles are on the unit circle.
Partial impulse response = oscillation with fixed amplitude that persists even after the end of the input signal:
System = oscillator, critically stable.
1pr < Partial impulse response decreases in timeNo instability from these poles.
1pr > Partial impulse response increases in timeInstability.
( )sin p pA nΩ +Φ
1pr =
24
47
#2. A pair of double complex conjugate poles
Contribution of the two poles in the impulse response :
( )( ) ( )
1 1 2 22 21 1 1 1
and
1 1 1 1
j jp p p p
j j j jp p p p
p p
p p p p
z r e z r e
a a a aH z
r e z r e z r e z r e z
Ω − Ω∗
∗ ∗
Ω − Ω− − Ω − Ω− −
= =
= + + + + +− − − −
… …
( ) ( ) [ ]1 2sin sinn np p p p p pA r n A nr n n⎡ ⎤Ω +Φ + Ω +Ψ σ⎣ ⎦
1pr < Partial impulse response decreases in timeNo instability from these poles.
1pr > Partial impulse response increases in timeInstability.
48
difference equation, non-zero initial conditions ⇒unilateral Z transform
Causal input signal, x[-n]=0 for n >0:
Response of a Discrete LTI System Described by a Difference Equation
[ ] [ ] [ ] [ ] 00 0 0 0
, 0 N M N M
k k k u k uk k k k
a y n k b x n k a a Z y n k b Z x n k= = = =
− = − ≠ ⇒ − = −∑ ∑ ∑ ∑
( ) [ ] ( )0 1 0
N k Mk n k
k u k uk n k
a z Y z y n z b z X z− −
= = =
⎡ ⎤+ − =⎢ ⎥⎣ ⎦
∑ ∑ ∑
( ) [ ] ( ) [ ]1 10 0
N k M kk n k n
u uk kn nk k
a z Y z y n z b z X z x n z− −
= == =
⎡ ⎤ ⎡ ⎤+ − = + −⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦∑ ∑ ∑ ∑
25
49
[ ] [ ] [ ] [ ] [ ] [ ]( ) ( ) [ ]( )
( ) ( )( )[ ]
[ ]
[ ] [ ]( )
[ ] .1 , 1 ; 1
.1
11
11
1
11
11
.1 ; 1
1
,01 , , 1
Example
0
0
0
000
0
0
0
0
111
111
111
11
><σ⎟⎟⎠
⎞⎜⎜⎝
⎛
−+
−+−=
−
−+
−−+
−−
−=
=−
−+
−−=
>−
=−+−
≠−σ==−−
Ω
+Ω
Ω
++
−−Ω−ΩΩ
Ω
−−−Ω
−Ω−
Ω
zanea
keea
kayany
azay
azeaka
zeeake
azay
azzekzY
zze
kzyzYazzY
ynkenxnxnayny
j
nj
j
nn
jjj
j
ju
juu
nj
50
First Order Systems
0pole/zero plot 0; 0.5pz z= = [ ] [ ] [ ]
( )
[ ] [ ]1
1
; 1
, 1 np
y n ay n kx nk kzH z z ROCaz z a
z a z a h n ka n
−
− − =
⇒ = = ∈− −
> = < ⇒ = σ
( ) ( ) ( )1
0 1, max. freq. response for 0, 1 0, max. for ,
Positive low-pass filterNegative high-pass filter
j jOAH e e
PA PA
aa
aa
Ψ−ϕ Ω−ϕΩ = =
< < Ω =− < < Ω = π
⇒⇒
26
51
Second Order Systems[ ] [ ] [ ] [ ] ( )
2
1 2 1 2 21 2 1 2
1 2 1
k kzy n a y n a y n kx n H za z a z z a z a− −+ − + − = ⇒ = =
+ + + +
1,2
21 2
21 2
2 21 2 1
2
1 2
4 complex conjugated poles
4 real poles . The system is considered causal and stable system.
41 magnitude
2The conditions imposed on ,
jp
a a
z e
a a
a a aa
a a
±
< ⇒
=
≥ ⇒
+ −= = <
θρ
ρ
21
2 2
21 2 2 1 2 1
for stability are :Complex conjugated poles
1 , 4
Real poles: 4 0 ; 1 ; 1.
aa a
a a a a a a
< >
− ≥ > − − > −
1.
2.
21 1 2
0 1,2
40 ,
2p
a a az z
− ± −= =
52
( ) ( ).1 : system theof responsefrequency The
pole. real double single a of existence the toscorrespond parabola The
212_____
2_____
1
ϕ−ϕ−Ω=Ω je
APAP
kH
Magnitude: even; Phase: odd
27
53
Transfer Function of Equivalent System for Serial or Parallel Interconnections of two
discrete LTI Systems
[ ] [ ] [ ]( ) ( ) ( ).zHzHzH
nhnhnh
e
e
21
21 ; +=
+=[ ] [ ] [ ]( ) ( ) ( ).zHzHzH
nhnhnh
e
e
21
21 ; =
∗=
54
Digital Filters Implementation Forms Obtained Using the z
Transform[ ] [ ] [ ] [ ] [ ]. ,1 ,
100
00∑∑∑∑====
−−−=⇔==−=−N
kk
M
kk
M
kk
N
kk knyaknxbnyNMaknxbknya
28
55
Direct form I (N+M adders). First form of implementation using z transform.
Each adder has 2 inputs. One adder with 2N inputs (M=N, a0=1).
56
Direct form II (2N adders M=N). Second form of implementation using z transform.
Each adder has 2 inputs. Two adders with N inputs each (M=N, a0=1).