University of California, San Diego
Spring 2014
ECE 45 Homework 2 Solutions
2.1: Suppose we model a system by the differential equation
y(t) + 2
d x(t)
d2 y(t)
=
+ x(t)
dt
dt2
1. What is the transfer function, H(ω)
Represent x(t) and y(t) as phasors thus:
Y + 2jω X = (jω)2 Y + X
√
1 − 2jω
1 + 4ω 2 −j tan−1 (2ω)
Y
=
=
e
H(ω) =
X
1 + ω2
1 + ω2
2. What is y0 (t) when x0 (t) = 4
y0 (t) = |H(0)|x0 (t) = 4
3. What is y1 (t) when x1 (t) = cos(t/2)
y1 (t) = |H(1/2)| cos(t/2 + ∠H(1/2)) =
4√
2 cos(t/2 − π/4)
5
4. What is y2 (t) when x2 (t) = 4 + cos(t/2)
x2 (t) = x0 (t) + x1 (t) so y2 (t) = y1 (t) + y0 (t)
2.2: Are the following steady-state input-output pairs consistent with the properties of LTI systems?
1. cos(25t) → H(ω) → 99.5 sin(25t −
p
π)
Yes. The output is an amplitude-scaled and phase-shifted version of the input.
2. 2 cos(4t) → H(ω) → 1 + 4 cos(4t)
No. There is a new frequency component, ω = 0 in the output.
3. 4 → H(ω) → −8
Yes. The output is an amplitude-scaled and phase-shifted version of the input.
4. 4 → H(ω) → 8j
No, we cannot have imaginary values as output signals.
5. 4 → H(ω) → cos(3t)
No. There is a new frequency component, ω = 3 in the output.
6. sin(πt) → H(ω) → cos(πt) + 0.1 sin(πt)
Yes. The output is an amplitude-scaled and phase-shifted version of the input.
7. sin(πt) → H(ω) → cos(πt) + 0.1 sin(2πt)
No. There is a new frequency component, ω = 2π in the output.
8. sin(πt) → H(ω) → sin2 (πt)
sin2 (πt) = 21 (1 − cos(2t))
No. There are two new frequency components, ω = 0 and ω = 2
2.3: Determine the Bode Plot (magnitude and phase) of the transfer function
H(ω) = 225j
(ω 2 − 2500jω − 106 ) (1 + 10−6 jω)
(ω − 2000j) (ω − 500j) (1 + 10−6 jω)
=
225j
(ω 2 − 60000jω − 9 108 ) (25 + 5jω)
(ω − 30000j)2 (25 + 5jω)
H(ω) = 225j
ω
ω
) (1 + j 500
) (1 + 10−6 jω)
(−2000j)(−500j) (1 + j 2000
ω
(25)(−30000j)2
(1 + j 30000
)2 (1 + j ω5 )
H(ω) =
ω
ω
) (1 + j 500
) (1 + j 10ω6 )
j (1 + j 2000
ω
)2 (1 + j ω5 )
100
(1 + j 30000
Critical Points
[ω, |H(ω)|]
a = [5, −40dB]
b = [500, −40 − 20 (log(500) − log(5))] = [500, −40 − 20 (2)] = [500, −80 dB]
c = [30000, −80 + 20 (log(30000) − log(2000))] = [30000, −80 + 20 log(15)] ≈ [30000, −56 dB]
d = [106 , −56 − 20 (log(106 ) − log(30000))] = [106 , −56 − 20 log(100/3)] ≈ [106 , −86 dB]
[ω, ∠H(ω)]
e = [0.5, π/2]
f = [50, 0]
g = [200, π/4 (log(200) − log(50))] = [200, π/4 log(4)] ≈ [200, 0.15π]
h = [3000, 0.15π + π/2 (log(3000) − log(200))] = [3000, 0.15π + π/2 log(15)] ≈ [3000, 3π/4]
i = [20000, 3π/4 − π/4 (log(20000) − log(5000))] = [20000, 3π/4 − π/4 log(4)] ≈ [20000, 0.6π]
j = [105 , 0.6π − π/2 (log(105 ) − log(20000))] = [105 , 0.6π − π/2 log(5)] ≈ [105 , π/4]
k = [2 105 , π/4 − π/4(log(2 105 ) − log(105 ))] = [2 105 , π/4 − π/4 log(2)] ≈ [2 105 , π/6]
l = [107 , π/2]
2.4: Using the Bode Plot below, estimate the output for the given inputs
1. x(t) = 2 cos(4000t)
|HdB (4000)| ≈ −50dB = 20 log |H(4000)|
√
10
1000
33.75o = 3π
16
|H(4000)| ≈ 10−50/20 =
∠H(4000) ≈ 45o
√
Thus y(t) =
10
500
3
4
=
cos(4000t +
3π
)
16
2. x(t) = sin(7 104 t + π/4)
|HdB (7 104 )| ≈ −20dB = 20 log |H(7 104 )|
|H(7 104 )| ≈ 10−20/20 = 0.1
∠H(7 104 ) ≈ −45o
1
2
= −22.5o =
−π
8
Thus y(t) = 0.1 sin(7 104 t + π/8)
3. x(t) = cos2 (12500t) = 1/2 + 1/2 cos(25000 106 t)
|HdB (2.5 104 )| ≈ −30dB = 20 log |H(2.5 104 )|
|H(2.5 104 )| ≈ 10−30/20 =
∠H(2.5 104 ) ≈ −45o =
√
10
100
−π
4
|HdB (0)| ≈ −40dB = 20 log |H(0)|
|H(0)| ≈ 10−40/20 =
Thus y(t) =
1
200
√
+
10
200
1
100
cos(25000 106 t − π/4)
2.5: Determine the Bode Plot of the transfer function
√
( 2 (50 + j50) ω 2 )
100 ejπ/4 ω 2
H(ω) =
=
ω
ω
ω
ω
(1 + j 500
) (1 + j 1000
)
(1 + j 500
) (1 + j 1000
)
Critical Points:
[ω, |H(ω)|]
a = [10−1 , 0 dB]
b = [500, 40 (log(500) − log(10−1 ))] = [500, 40 log(5000)] = [500, 40 log(103 ) + 40 log(5)] =
[500, 120 + 40 log(5)] ≈ [500, 148 dB]
c = [1000, 148 + 20 (log(1000) − log(500))] = [1000, 148 + 20 log(2)] ≈ [1000, 154 dB]
[ω, ∠H(ω)]
d = [50, π/4]
e = [100, π/4 − π/4 (log(100) − log(50))] = [100, π/4 − π/4 log(2)] ≈ [100, π/6]
f = [5000, 0.175π − π/2 (log(5000) − log(100))] = [5000, π/6 − π log(50)] ≈ [5000, −2π/3]
g = [104 , −3π/4]
2.6: Determine the output of the system
H(ω) =
when the input is x(t) =
P∞
n=−∞
1 |ω| ≤ 4π
0 else
cos(π n t)+sin(2π n t+nπ)
2
|H(ω)| =
1 |ω| ≤ 4π
and ∠H(ω) = 0
0 else
∞
X
|H(π n)| cos(π n t + ∠H(π n)) + |H(2π n)| sin(2π n t + nπ + ∠H(2π n))
y(t) =
2
n=−∞
−→ H(πn) = 1 for n = ±4, ±3 ± 2, ±1, 0 and H(2πn) = 1 for n = ±2, ±1, 0 both are 0 for all other
n.
cos(πnt) = cos(−πnt) so
y(t) =
4
2
X
X
cos(π n t)
sin(2π n t + nπ)
+
2
2
n=−4
n=−2
cos(πnt)
2
+
cos(−πnt)
2
= cos(πnt), and cos(0) = 1. Thus
4
X
cos(π n t)
1
= cos(πt) + cos(2πt) + cos(3πt) + cos(4πt) +
2
2
n=−4
sin(πnt + π/n) = − sin(−(πnt + π/n)) so sin(πnt + π/n) + sin(−πnt − π/n) = 0, and sin(0) = 0
Thus
2
X
sin(2π n t + nπ)
=0
2
n=−2
so y(t) == cos(πt) + cos(2πt) + cos(3πt) + cos(4πt) +
1
2
2.7: Determine the period and fundamental frequency of the following functions:
1. f (t) = 4 + cos(3t)
T0 = 2π/3 s and ω0 = 3rad/s
2. f (t) = 8 + 4 e−j4 t + 4ej4 t = 8 + 8 cos(4t)
T0 = π/2 s and ω0 = 4rad/s
3. f (t) = 2 e−j2 t + 2 ej2 t + 2 cos(4t) = 4 cos(2t) + 2 cos(4t)
T0 = π s and ω0 = 2rad/s
2.8: Determine the Fourier coefficients Fn and the average power of the function:
1 0<t<2
f (t) =
2 2<t<4
Where f (t) is periodic with a period of T = 4
T = 4 → ω0 = π/2
1
Fn =
4
=
Z
4
jn π2 t
f (t) e
0
1
dt =
4
2
Z
jn π2 t
e
0
1
dt +
4
Z
4
π
2 ejn 2 t dt
2
1
1
(e−jnπ − 1) +
(e−jn2π − e−jnπ )
−jn2π
−jnπ
Note: e−jnπ = (−1)n and e−jn2π = 1 for all integers n
Fn =
1
1
j
((−1)n − 1) +
(1 − (−1)n ) =
(1 − (−1)n )
−jn2π
−jnπ
2πn
0
n is even and n 6= 0
Fn =
j/nπ n is odd
F0 is not well-defined, so we must calculate it separately:
Z
Z
Z
1 4
1 2
1 4
F0 =
f (t) dt =
1 dt +
2 dt = 3/2
4 0
4 0
4 2
Pavg
1
=
4
Z
0
4
1
|f (t)| dt =
4
2
Z
0
2
1
1 dt +
4
Z
4
4 dt = 5/2
2
2.9: Calculate the Fourier Series of f (t)
T0 = 1 → ω0 = 2π, and in a period from −0.5 to 0.5: f (t) = 1 − |2 t|
Z
1
2
Fn =
−jn2πt
f (t) e
0
Z
−jn2πt
dt =
(2t + 1) e
Z
0
−jn2πt
=
2t e
=e
(1 − 2t) e−jn2πt dt
0
Z
dt −
− 12
−jn2πt
dt +
− 12
− 21
1
2
Z
1
2
−jn2πt
2t e
Z
1
2
dt +
1 e−jn2πt dt
− 12
0
1/2
0
1/2
jt jt j e−2jπnt 1
1
−jn2πt
+
−e
+
+
2π 2 n2 πn −1/2
2π 2 n2 πn 0
2πn −1/2
1
1
j
1
j
1
j
jπn
−jπn
= 2 2 −e ( 2 2 −
)− e
( 2 2+
)− 2 2 +
(e−jπn − ejπn )
2π n
2π n
2πn
2π n
2πn
2π n
2πn
Note: ejnπ = e−jnπ = (−1)n
1
(−1)n
Fn = 2 2 − 2 2 =
π n
π n
0
2
π 2 n2
n is even and n 6= 0
n is odd
Not well defined for F0 so we must calculate it separately:
Z
1
2
F0 =
Z
0
f (t) dt =
− 21
Z
(2t + 1) dt +
− 21
−→ f (t) = 1/2 +
1
2
(1 − 2t) dt = 1/2
0
X
n odd
2
π 2 n2
ejn2πt
2.10: Using the results of 2.9, determine the Fourier Series of g(t). You can express g(t) as a shifted
and scaled version of f (t) and use the properties of the Fourier Series.
g(t) = 8 (f (t − 1/4) − 1/2)) = 8 f (t − 1/4) − 4
Thus by the scaling and time-shift properties of the Fourier Series:
8F0 − 4
n=0
0
n is even
=
Gn =
16
−jnπ/2
−jn2π 41
e
n is odd
n 6= 0
8Fn e
π 2 n2
Note: e−jnπ/2 = (−j)n
−→ g(t) =
X 16
(−j)n ejn2πt
2 n2
π
n odd
2.11: Determine the Fourier Series of f (t) = | sin(t)|
T0 = π, ω0 = 2. Over one period, f (t) = sin(t)
Note: By Euler’s Formula:
ejθ − e−jθ
sin(θ) =
2j
Z
Z π
1
1 π
−jn2t
sin(t) e
dt =
(ej(1−2n)t − e−j(1+2n)t ) dt
Fn =
π 0
j2π 0
j(1−2n)t
π
1
e
e−j(1+2n)t −1 ej(1−2n)π − 1 e−j(1+2n)π − 1
=
−
=
−
j2π j(1 − 2n) −j(1 + 2n) 0
2π
(1 − 2n)
(1 + 2n)
Note: e(1±2nπ) = ejπ = −1
1
Fn =
π
1
1
+
1 − 2n 1 + 2n
−→ f (t) =
=
2
1
π 1 − 4n2
∞
X
2
1
ejn2t
2
π
1
−
4n
n=−∞