1. Inverse Z-transform - Partial Fraction - Moudgalya
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1. Inverse Z-transform - Partial Fraction
1. Inverse Z-transform - Partial Fraction
Find the inverse Z-transform of
G(z) =2z2 + 2z
z2 + 2z − 3
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)
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 − 1
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
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
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
2. Inverse Z-transform - Partial Fraction
G(z)
z=
A
z + 3+
B
z − 1
2. Inverse Z-transform - Partial Fraction
G(z)
z=
A
z + 3+
B
z − 1
Multiply throughout by z − 1 and let z = 1
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
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
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
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
3. Partial Fraction - Repeated Poles
G(z) =N(z)
(z − α)pD1(z)
3. Partial Fraction - Repeated Poles
G(z) =N(z)
(z − α)pD1(z)
α not a root of N(z) and D1(z)
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)
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).
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
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
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
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,
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
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:
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
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,
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
5. Repeated Poles - an Example
G(z) =11z2 − 15z + 6
(z − 2)(z − 1)2=
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
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),
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.
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,
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
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.
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,
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 =
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
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
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
6. Important Result from Differentiation
Problem 4.9 in Text:
6. Important Result from Differentiation
Problem 4.9 in Text: Consider
1(n)an↔ z
z − a
6. Important Result from Differentiation
Problem 4.9 in Text: Consider
1(n)an↔ z
z − a =∞∑n=0
anz−n,∣∣az−1
∣∣ < 1
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,
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
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
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
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:
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
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,
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
7. Repeated Poles - an Example
G(z) = − 9
z − 1− 2
(z − 1)2+
20
z − 2
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
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)
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)
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
8. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
8. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
ua(t) = A cos (2πFt+ θ),
8. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
ua(t) = A cos (2πFt+ θ), −∞ < t <∞
8. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
ua(t) = A cos (2πFt+ θ), −∞ < t <∞A: amplitude
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/s
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 rad
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
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
9. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
9. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
ua(t) = A cos (2πFt+ θ),
9. Properties of Cont. Time Sinusoidal Signals
ua(t) = A cos (Ωt+ θ)
ua(t) = A cos (2πFt+ θ), −∞ < t <∞
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
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]
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)
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)
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)
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
10. Properties of Cont. Time Sinusoidal Signals
Cont. signals with different frequencies are different
10. Properties of Cont. Time Sinusoidal Signals
Cont. signals with different frequencies are different
u1 = cos
(2πt
8
),
10. Properties of Cont. Time Sinusoidal Signals
Cont. signals with different frequencies are different
u1 = cos
(2πt
8
), u2 = cos
(2π
7t
8
)
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
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.
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
11. Discrete Time Sinusoidal Signals
u(n) = A cos (wn+ θ), −∞ < n <∞
11. Discrete Time Sinusoidal Signals
u(n) = A cos (wn+ θ), −∞ < n <∞w = 2πf
11. Discrete Time Sinusoidal Signals
u(n) = A cos (wn+ θ), −∞ < n <∞w = 2πf
n integer variable, sample number
11. Discrete Time Sinusoidal Signals
u(n) = A cos (wn+ θ), −∞ < n <∞w = 2πf
n integer variable, sample numberA amplitude of the sinusoid
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 sample
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.
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
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
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,
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),
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,
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,
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.
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
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
13. Sampling a Cont. Time Signal - Preliminaries
Let analog signal ua[t] have a frequency of F Hz
13. Sampling a Cont. Time Signal - Preliminaries
Let analog signal ua[t] have a frequency of F Hz
ua[t] = A cos(2πFt+ θ)
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)
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
Fs
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 <∞
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)
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
)
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
14. Sampling a Cont. Time Signal - Preliminaries
In our standard notation,
u(n) = A cos (2πfn+ θ)
14. Sampling a Cont. Time Signal - Preliminaries
In our standard notation,
u(n) = A cos (2πfn+ θ)
It follows that
f =F
Fs
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
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.
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
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
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
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
2
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
15. Properties of Discrete Time Sinusoids - Alias
• As w0 ↑, freq. of oscillation ↑, reaches maximumat w0 = π
15. Properties of Discrete Time Sinusoids - Alias
• As w0 ↑, freq. of oscillation ↑, reaches maximumat w0 = π
• What if w0 > π?
15. Properties of Discrete Time Sinusoids - Alias
• As w0 ↑, freq. of oscillation ↑, reaches maximumat w0 = π
• What if w0 > π?
w1 = w0
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
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
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
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
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
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
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)
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
16. Properties of Discrete Time Sinusoids - Alias
u1[t] = cos (2πt
8),
16. Properties of Discrete Time Sinusoids - Alias
u1[t] = cos (2πt
8), u2[t] = cos (2π
7t
8),
16. Properties of Discrete Time Sinusoids - Alias
u1[t] = cos (2πt
8), u2[t] = cos (2π
7t
8), Ts = 1
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)
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
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
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)
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)
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
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,
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
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
18. Frequency Response
Apply u(n) = ejwn to g(n) and obtain output y:
18. Frequency Response
Apply u(n) = ejwn to g(n) and obtain output y:
y(n) = g(n) ∗ u(n)
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)
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)
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
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
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
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
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
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
19. Frequency Response
y(n) = ejwn∞∑
k=−∞g(k)e−jwk
19. Frequency Response
y(n) = ejwn∞∑
k=−∞g(k)e−jwk = ejwnG(ejw)
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
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)
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:
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)
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 input
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 amplified
2. At ω where |G(ejw)| is large, the sinusoid gets amplifiedand at ω where it is small, the sinusoid gets attenuated.
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
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
20. Discrete Fourier Transform
Define Discrete Time Fourier Transform
20. Discrete Fourier Transform
Define Discrete Time Fourier Transform
G(ejw)4=
∞∑k=−∞
g(k)e−jwk
20. Discrete Fourier Transform
Define Discrete Time Fourier Transform
G(ejw)4=
∞∑k=−∞
g(k)e−jwk =∞∑
k=−∞g(k)z−k
∣∣∣∣∣∣z=ejw
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
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| <∞
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)| <∞
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,
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.
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
21. FT of Discrete Time Aperiodic Signals - Defi-nition
21. FT of Discrete Time Aperiodic Signals - Defi-nition
U(ejω)4=
∞∑n=−∞
u(n)e−jωn
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
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
22. FT of a Moving Average Filter - Example
y(n) =1
3[u(n+ 1) + u(n) + u(n− 1)]
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)
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)
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)
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)
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
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
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
)
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
23. FT of a Moving Average Filter - Example
G(ejω)
=1
3(1 + 2 cosw)
23. FT of a Moving Average Filter - Example
G(ejω)
=1
3(1 + 2 cosw), |G (ejw) | = |1
3(1 + 2 cosw)|
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
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
24. FT of a Moving Average Filter - Example
|G (ejw) | = |13(1 + 2 cosw)|
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
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
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
25. Additional Properties of Fourier Transform
Symmetry of real and imaginary parts for real valued sequences
25. Additional Properties of Fourier Transform
Symmetry of real and imaginary parts for real valued sequences
G(ejw)
=∞∑
n=−∞g(n)e−jwn
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
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
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
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
)]
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
26. Additional Properties of Fourier Transform
Recall
G(ejw)
= G∗(e−jw
)
26. Additional Properties of Fourier Transform
Recall
G(ejw)
= G∗(e−jw
)Symmetry of magnitude for real valued sequences
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
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
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)|
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.
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,
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)]
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
27. Sampling and Reconstruction
u(n) = ua(nTs), −∞ < n <∞
27. Sampling and Reconstruction
u(n) = ua(nTs), −∞ < n <∞Fast sampling:
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
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?
29. What to Do When Aliasing Cannot be Avoided?
Ua(F )
1Fs
U(
FFs
)
Ua(F )Ua(F + Fs) Ua(F − Fs)
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
30. Sampling Theorem
• Suppose highest frequency contained in an analogsignal ua(t) is Fmax = B.
30. Sampling Theorem
• Suppose highest frequency contained in an analogsignal ua(t) is Fmax = B.
• It is sampled at a rate Fs > 2Fmax = 2B.
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:
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)
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.
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
31. Rules for Sampling Rate Selection
• Minimum sampling rate = twice band width
31. Rules for Sampling Rate Selection
• Minimum sampling rate = twice band width
• Shannon’s reconstruction cannot be implemented,have to use ZOH
31. Rules for Sampling Rate Selection
• Minimum sampling rate = twice band width
• Shannon’s reconstruction cannot be implemented,have to use ZOH
• Solution:
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
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
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
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
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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