1. Inverse Z-transform - Partial Fraction
Find the inverse Z-transform of
G(z) =2z2 + 2z
z2 + 2z − 3G(z)
z=
2z + 2
(z + 3)(z − 1)
=A
z + 3+
B
z − 1Multiply throughout by z+3 and let z = −3 to get
A =2z + 2
z − 1
∣∣∣∣z=−3
=−4
−4= 1
Digital Control 1 Kannan M. Moudgalya, Autumn 2007
2. Inverse Z-transform - Partial Fraction
G(z)
z=
A
z + 3+
B
z − 1
Multiply throughout by z − 1 and let z = 1 to get
B =4
4= 1
G(z)
z=
1
z + 3+
1
z − 1|z| > 3
G(z) =z
z + 3+
z
z − 1|z| > 3
↔ (−3)n1(n) + 1(n)Digital Control 2 Kannan M. Moudgalya, Autumn 2007
3. Partial Fraction - Repeated Poles
G(z) =N(z)
(z − α)pD1(z)
α not a root of N(z) and D1(z)
G(z) =A1
z − α +A2
(z − α)2+ · · ·+ Ap
(z − α)p+G1(z)
G1(z) has poles corresponding to those of D1(z).Multiply by (z − α)p
(z − α)pG(z) = A1(z − α)p−1 +A2(z − α)p−2 + · · ·+Ap−1(z − α) +Ap +G1(z)(z − α)p
Digital Control 3 Kannan M. Moudgalya, Autumn 2007
4. Partial Fraction - Repeated Poles
(z − α)pG(z) = A1(z − α)p−1 +A2(z − α)p−2 + · · ·+Ap−1(z − α) +Ap +G1(z)(z − α)p
Substituting z = α,
Ap = (z − α)pG(z)|z=α
Differentiate and let z = α:
Ap−1 =d
dz(z − α)pG(z)|z=α
Continuing, A1 =1
(p− 1)!
dp−1
dzp−1(z − α)pG(z)|z=α
Digital Control 4 Kannan M. Moudgalya, Autumn 2007
5. Repeated Poles - an Example
G(z) =11z2 − 15z + 6
(z − 2)(z − 1)2=
A1
z − 1+
A2
(z − 1)2+
B
z − 2
× (z − 2), let z = 2, to get B = 20. × (z − 1)2,
11z2 − 15z + 6
z − 2= A1(z − 1) +A2 +B
(z − 1)2
z − 2
With z = 1, get A2 = −2.Differentiating with respect to z and with z = 1,
A1 =(z − 2)(22z − 15)− (11z2 − 15z + 6)
(z − 2)2
∣∣∣∣z=1
= −9
G(z) = − 9
z − 1− 2
(z − 1)2+
20
z − 2
Digital Control 5 Kannan M. Moudgalya, Autumn 2007
6. Important Result from Differentiation
Problem 4.9 in Text: Consider
1(n)an↔ z
z − a =∞∑n=0
anz−n,∣∣az−1
∣∣ < 1
Differentiating with respect to a,
z
(z − a)2=∞∑n=0
nan−1z−n, nan−11(n)↔ z
(z − a)2
Substituting a = 1, can obtain the Z-transform of n:
n1(n)↔ z
(z − 1)2
Through further differentiation, can derive Z-transforms of n2,n3, etc.Digital Control 6 Kannan M. Moudgalya, Autumn 2007
7. Repeated Poles - an Example
G(z) = − 9
z − 1− 2
(z − 1)2+
20
z − 2
zG(z) = − 9z
z − 1− 2z
(z − 1)2+
20z
z − 2
↔ (−9− 2n+ 20× 2n)1(n)
Use shifting theorem:
G(z)↔ (−9− 2(n− 1) + 20× 2n−1)1(n− 1)
Digital Control 7 Kannan M. Moudgalya, Autumn 2007
8. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
ua(t) = A cos (2πFt+ θ), −∞ < t <∞A: amplitudeΩ: frequency in rad/sθ: phase in radF : frequency in cycles/s or Hertz
Ω = 2πF
Tp =1
FDigital Control 8 Kannan M. Moudgalya, Autumn 2007
9. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
ua(t) = A cos (2πFt+ θ), −∞ < t <∞With F fixed, periodic with period Tp
ua[t+ Tp] = A cos (2πF (t+1
F) + θ)
= A cos (2π + 2πFt+ θ)
= A cos (2πFt+ θ) = ua[t]
Digital Control 9 Kannan M. Moudgalya, Autumn 2007
10. Properties of Cont. Time Sinusoidal Signals
Cont. signals with different frequencies are different
u1 = cos
(2πt
8
), u2 = cos
(2π
7t
8
)
0 0.5 1 1.5 2 2.5 3 3.5 4−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
t
cosw
t
Different wave forms for different frequencies.Can increase f all the way to∞ or decrease to 0.Digital Control 10 Kannan M. Moudgalya, Autumn 2007
11. Discrete Time Sinusoidal Signals
u(n) = A cos (wn+ θ), −∞ < n <∞w = 2πf
n integer variable, sample numberA amplitude of the sinusoidw frequency in radians per sampleθ phase in radians.f normalized frequency, cycles/sample
Digital Control 11 Kannan M. Moudgalya, Autumn 2007
12. Discrete Time Sinusoids - Identical Signals
Discrete time sinusoids whose frequencies are sepa-rated by integer multiple of 2π are identical, i.e.,
cos ((w0 + 2π)n+ θ) = cos (w0n+ θ), ∀n• All sinusoidal sequences of the form,uk(n) = A cos (wkn+ θ),wk = w0 + 2kπ, −π < w0 < π,are indistinguishable or identical.
• Only sinusoids in −π < w0 < π are different
−π < w0 < π or − 1
2< f0 <
1
2Digital Control 12 Kannan M. Moudgalya, Autumn 2007
13. Sampling a Cont. Time Signal - Preliminaries
Let analog signal ua[t] have a frequency of F Hz
ua[t] = A cos(2πFt+ θ)
Uniform sampling rate (Ts s) or frequency (Fs Hz)
t = nTs =n
Fsu(n) = ua[nTs] −∞ < n <∞
= A cos (2πFTsn+ θ)
= A cos
(2πF
Fsn+ θ
)4= A cos(2πfn+ θ)
Digital Control 13 Kannan M. Moudgalya, Autumn 2007
14. Sampling a Cont. Time Signal - Preliminaries
In our standard notation,
u(n) = A cos (2πfn+ θ)
It follows that
f =F
Fsw =
Ω
Fs= ΩTs
Reason to call f normalized frequency, cycles/sample.Apply the uniqueness condition for sampled signals
− 1
2< f <
1
2−π < w < π
Fmax =Fs
2Ωmax = 2πFmax = πFs =
π
TsDigital Control 14 Kannan M. Moudgalya, Autumn 2007
15. Properties of Discrete Time Sinusoids - Alias
• As w0 ↑, freq. of oscillation ↑, reaches maximumat w0 = π
• What if w0 > π?
w1 = w0
w2 = 2π − w0
u1(n) = A cosw1n = A cosw0n
u2(n) = A cosw2n = A cos(2π − w0)n
= A cosw0n = u1(n)
w2 is an alias of w1
Digital Control 15 Kannan M. Moudgalya, Autumn 2007
16. Properties of Discrete Time Sinusoids - Alias
u1[t] = cos (2πt
8), u2[t] = cos (2π
7t
8), Ts = 1
u2(n) = cos (2π7n
8) = cos 2π(1− 1
8)n
= cos (2π − 2π
8)n = cos (
2πn
8) = u1(n)
0 0.5 1 1.5 2 2.5 3 3.5 4−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
t
cosw
t
Digital Control 16 Kannan M. Moudgalya, Autumn 2007
17. Fourier Transform of Aperiodic Signals
X(F ) =
∫ ∞−∞
x(t)e−j2πFtdt
x(t) =
∫ ∞−∞
X(F )ej2πFtdF
If we let radian frequency Ω = 2πF
x(t) =1
2π
∫ ∞−∞
X[Ω]ejΩtdΩ,
X[Ω] =
∫ ∞−∞
x(t)e−jΩtdt
x(t), X[Ω] are Fourier Transform PairDigital Control 17 Kannan M. Moudgalya, Autumn 2007
18. Frequency Response
Apply u(n) = ejwn to g(n) and obtain output y:
y(n) = g(n) ∗ u(n) =∞∑
k=−∞g(k)u(n− k)
=∞∑
k=−∞g(k)ejw(n−k) = ejwn
∞∑k=−∞
g(k)e−jwk
Define Discrete Time Fourier Transform
G(ejw)4=
∞∑k=−∞
g(k)e−jwk =∞∑
k=−∞g(k)z−k
∣∣∣∣∣∣z=ejw
= G(z)|z=ejw
Provided the sequence converges absolutelyDigital Control 18 Kannan M. Moudgalya, Autumn 2007
19. Frequency Response
y(n) = ejwn∞∑
k=−∞g(k)e−jwk = ejwnG(ejw)
Write in polar coordinates: G(ejw) = |G(ejw)|ejϕ - ϕ isphase angle:
y(n) = ejwn|G(ejw)|ejϕ = |G(ejwn)|ej(wn+ϕ)
1. Input is sinusoid⇒ output also is a sinusoid with followingproperties:
• Output amplitude gets multiplied by the magnitude ofG(ejw)
• Output sinusoid shifts by ϕ with respect to inputDigital Control 19 Kannan M. Moudgalya, Autumn 2007
2. At ω where |G(ejw)| is large, the sinusoid gets amplifiedand at ω where it is small, the sinusoid gets attenuated.
• The system with large gains at low frequencies and smallgains at high frequencies are called low pass filters
• Similarly high pass filters
Digital Control 20 Kannan M. Moudgalya, Autumn 2007
20. Discrete Fourier Transform
Define Discrete Time Fourier Transform
G(ejw)4=
∞∑k=−∞
g(k)e−jwk =∞∑
k=−∞g(k)z−k
∣∣∣∣∣∣z=ejw
= G(z)|z=ejw
Provided, absolute convergence∞∑
k=−∞|g(k)e−jwk| <∞⇒
∞∑k=−∞
|g(k)| <∞
For causal systems, BIBO stability. Required for DTFT.
Digital Control 21 Kannan M. Moudgalya, Autumn 2007
21. FT of Discrete Time Aperiodic Signals - Defi-nition
U(ejω)4=
∞∑n=−∞
u(n)e−jωn
u(m) =1
2π
∫ π
−πU(ejω)ejωmdω
=
∫ 1/2
−1/2
U(f)ej2πfmdf
Digital Control 22 Kannan M. Moudgalya, Autumn 2007
22. FT of a Moving Average Filter - Example
y(n) =1
3[u(n+ 1) + u(n) + u(n− 1)]
y(n) =∞∑
k=−∞g(k)u(n− k)
= g(−1)u(n+ 1) + g(0)u(n) + g(1)u(n− 1)
g(−1) = g(0) = g(1) =1
3
G(ejw)
=∞∑
n=−∞g(n)z−n|z=ejw
=1
3
(ejw + 1 + e−jw
)=
1
3(1 + 2 cosw)
Digital Control 23 Kannan M. Moudgalya, Autumn 2007
23. FT of a Moving Average Filter - Example
G(ejω)
=1
3(1 + 2 cosw), |G (ejw) | = |1
3(1 + 2 cosw)|
Arg(G) =
0 0 ≤ w < 2π
3
π 2π3≤ w < π
1 / / U p d a t e d ( 1 8 − 7 − 0 7 )
2 / / 5 . 3
3
4 w = 0 : 0 . 0 1 : %pi ;
5 subplot ( 2 , 1 , 1 ) ;
6 plot2d1 ( ” g l l ” ,w, abs(1+2∗cos (w) ) / 3 , s t y l e = 2 ) ;
7 l a b e l ( ’ ’ , 4 , ’ ’ , ’ Magnitude ’ , 4 ) ;
8 subplot ( 2 , 1 , 2 ) ;
9 plot2d1 ( ” g l n ” ,w, phasemag(1+2∗cos (w) ) , s t y l e = 2 , r e c t =[0.01 −0.5 10 2 0 0 ] ) ;
10 l a b e l ( ’ ’ , 4 , ’w ’ , ’ Phase ’ , 4 )
Digital Control 24 Kannan M. Moudgalya, Autumn 2007
24. FT of a Moving Average Filter - Example
|G (ejw) | = |13(1 + 2 cosw)|,
Arg(G) =
0 0 ≤ w < 2π
3
π 2π3≤ w < π
10−2
10−1
100
101
10−3
10−2
10−1
100
Mag
nitu
de
10−2
10−1
100
101
0
50
100
150
200
w
Pha
se
Digital Control 25 Kannan M. Moudgalya, Autumn 2007
25. Additional Properties of Fourier Transform
Symmetry of real and imaginary parts for real valued sequences
G(ejw)
=∞∑
n=−∞g(n)e−jwn
=∞∑
n=−∞g(n) coswn− j
∞∑n=−∞
g(n) sinwn
G(e−jw
)=
∞∑n=−∞
g(n)ejwn
=∞∑
n=−∞g(n) coswn+ j
∞∑n=−∞
g(n) sinwn
Re[G(ejw)]
= Re[G(e−jw
)]Im
[G(ejw)]
= −Im [G(e−jw
)]Digital Control 26 Kannan M. Moudgalya, Autumn 2007
26. Additional Properties of Fourier Transform
Recall
G(ejw)
= G∗(e−jw
)Symmetry of magnitude for real valued sequences
|G (ejw) | = [G(ejw)G∗(ejw)]1/2
=[G∗(e−jw
)G(e−jw
)]1/2= |G(e−jw)|
This shows that the magnitude is an even function. Similarly,
Arg[G(e−jw
)]= −Arg [G (ejw)]
⇒ Bode plots have to be drawn for w in [0, π] only.
Digital Control 27 Kannan M. Moudgalya, Autumn 2007
27. Sampling and Reconstruction
u(n) = ua(nTs), −∞ < n <∞Fast sampling:
Fs
2Ts
Ft
−B B
−Fs
2−32Fs
32Fs
Ua(F )
ua(nT ) = u(n) U(
FFs
)
1
Fs
ua(t)
t F
Fs
2
F0 + Fs
F0 − Fs
F0
Digital Control 28 Kannan M. Moudgalya, Autumn 2007
28. Slow Sampling Results in Aliasing
FtTs
−Fs
2Fs
2
u(n)
FtTs
−Fs
2Fs
2
t F−B B
ua(t)Ua(F )
Fs
Fs
U(
FFs
)
U(
FFs
)
Digital Control 29 Kannan M. Moudgalya, Autumn 2007
29. What to Do When Aliasing Cannot be Avoided?
Ua(F )
1Fs
U(
FFs
)
Ua(F )Ua(F + Fs) Ua(F − Fs)
U ′a(F − Fs)
U ′a(F )
1Fs
U ′(
FFs
)
Ua(F )U ′a(F + Fs)
Digital Control 30 Kannan M. Moudgalya, Autumn 2007
30. Sampling Theorem
• Suppose highest frequency contained in an analogsignal ua(t) is Fmax = B.
• It is sampled at a rate Fs > 2Fmax = 2B.
• ua(t) can be recovered from its sample values:
ua(t) =∞∑
n=−∞ua(nTs)
sinπTs
(t− nTs)
πTs
(t− nTs)• If Fs = 2Fmax, Fs is denoted by FN , the
Nyquist rate.
• Not causal: check n > 0Digital Control 31 Kannan M. Moudgalya, Autumn 2007
31. Rules for Sampling Rate Selection
• Minimum sampling rate = twice band width
• Shannon’s reconstruction cannot be implemented,have to use ZOH
• Solution: sample faster
– Number of samples in rise time = 4 to 10
– Sample 10 to 30 times bandwidth
– Use 10 times Shannon’s sampling rate
– ωcTs = 0.15 to 0.5, where,ωc = crossover frequency
Digital Control 32 Kannan M. Moudgalya, Autumn 2007
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