Unformatted text preview: University of Illinois Spring 2018
ECE210 / ECE211  Homework 08 Solution 1. The function f (t) is periodic with period T = 4s. Between t=0 and 4s, the function is described by: −2, 0 < t < 1s
f (t) = −1, 1 < t < 3s 1,
3 < t < 4s (a) Plot f (t) between t = −5s and t = 7s.
(b) Determine the exponential Fourier coecients Fn of f (t) for n = 0, n = ±1s, and n = ±2s.
(c) Using the result of part(b), determine the compactform Fourier coecients C0 , C1 and C2 .
Solution:
(a) (b)
Fn = 1
T Z f (t)e−jnw0 t dt T
1 Z
Z
Z
π
π
1
1 3
1 4 −j π nt
−2e−j 2 nt dt +
−1e−j 2 nt dt +
1e 2 dt
4 0
4 1
4 3
π
3
π
3
2j 1
1
1
=
[ (−e−j 2 n + 1) − (e−j 2 πn − e−j 2 n ) + (e−j2πn − e−j 2 πn )]
nπ 2
4
4 = Then plug in the number for coecient
F0 =
F1 =
F−1 =
F2 =
F−2 = Z
Z
Z
1 1
1 3
1 4
3
−2dt +
−1dt +
1dt = −
4 0
4 1
4 3
4
2j 1
1
1
1 + 3j
[ (j + 1) − (j + j) + (1 − j)] =
π 2
4
4
2π
1
1
1 − 3j
2j 1
[ (−j + 1) − (−j − j) + (1 + j)] =
−π 2
4
4
2π
2j 1
1
1
3j
[ (1 + 1) − (−1 + 1) + (1 + 1)] =
2π 2
4
4
2π
2j 1
1
1
3j
[ (1 + 1) − (−1 + 1) + (1 + 1)] = −
−2π 2
4
4
2π Page 1 of 6 (c) The formula to compactform Fourier coecents is Cn = 2 Fn , where n 6= 0.
√ 10
π C1 = 2 F1  =
C2 = 2 F2  = 3
π When n = 0,
C0 = 2F0 = − 3
2 2. Consider an LTI system whose frequency response is
Z +∞ h(t) e−jωt dt = H(ω) =
−∞ sin(4 ω)
πω If the input to this system is a periodic signal
(
f (t) = +1, 0 < t < 4s
−1, 4 < t < 6s with period T = 6s.
Determine the corresponding system output y(t)
Solution:
T = 6 and w0 = 2π
T = π
3 Fn=0
Fn6=0 Since H(0) = π4 ,
y(t) = Z 4 6
1
1dt +
(−1)dt =
3
0
4
Z 4
Z 6
1
j −j 4 πn
1e−jntπ/3 dt +
(−1)e−jntπ/3 dt =
=
(e 3 − 1)
6
nπ
0
4 1
=
6 Z 3 sin( 43 πn)
π
4
j −j 4 πn
+ Σ∞
(e 3 − 1) ×
× ejn 3 t
n=−∞,n6=0
3π
nπ
nπ 2 3. Determine the Fourier series representations for the following signals:
(a) A periodic signal x(t) with period of T = 2s and
x(t) = e−t f or0 < t < 2s Solution:
1
Fn =
2 Z 1
=
2 Z =− 2 e−t e−jnπt dt 0
2 e−(jnπ+1)t dt 0 1
(e−2 − 1)
2(1 + jnπ) (b) A periodic signal x(t) with period T = 1s and
(
x(t) = cos(π t), 0 ≤ t ≤ 1s
0,
else
Page 2 of 6 Solution:
1 1 jπt
(e + e−jπt )e−jn2πt dt
2 Z 1 Z
Fn =
0 = 1
2 (ejπt(1−2n) + e−jπt(1+2n) )dt 0
1 1
1
1
jπ(1−2n)t
−jπ(1+2n)t =
e
+ (−
)e 2 jπ(1 − 2n)
jπ(1 + 2n)
0
ejπ(1−2n) − 1 e−jπ(1+2n) − 1
+
2jπ(1 − 2n)
−2jπ(1 + 2n)
4jn
=
π (1 − 4n2 ) = (c) Solution:
From the plot, we get T = 6 and w0 =
F0 = 1
6 Z 2π
6 = π
3 6 f (t)dt = 1.5
0 Z
Z
πn
1 3 −j πn t
1 6
3
Fn =
te
dt +
(6 − t) e−j 3 t dt
6 0
6 3
(−1)n (9jπn − 9(−1)n + 9) −9 + (−1)n (9 − 9jπn)
=
+
6π 2 n2
6π 2 n2
3
= − 2 2 ((−1)n − 1)2
2π n 4. Let t + 2, 1,
x(t) = −t + 2, 0, −2 < t ≤ −1,
−1 ≤ t ≤ 1,
1 < t ≤ 2,
2 < t ≤ 4. be a periodic signal with fundamental period T = 6s and Fourier coecients ak .
(a) Determine the value of a0 .
(b) Determine the Fourier series representation of dx(t)
dt Page 3 of 6 (c) Use the result of part (b) and the dierential property of the Fourier series to help determine the Fourier
series coecients of x(t).
Solution:
(a)
F0 = 1
6 Z −1 (t + 2)dt +
−2 1
6 Z 1 1dt +
−1 1
6 2 Z (2 − t) dt +
1 1
6 Z 4 0 dt =
2 1
2 a0 = 2F0 = 1 (b) Let Gn be the Fourier coecients of
Gn6=0 = 1
6 dx
dt ,
−1 Z π e−jn 3 t dt − −2 1
6 Z 2 π 1e−jn 3 t dt 1
2 π π
π
je−j 3 πn
jej 3 n
=
(−ejn 3 + 1) −
(−ej 3 n + 1)
2nπ
2nπ
j
πn
2πn
= (cos( ) − cos(
))
nπ
3
3
Z
Z
1 2
1 −1
1dt −
1dt = 0
Gn=0 =
6 −2
6 1 (c) Using the dierential property
Gn
jnωo
3
Gn
=
πnj
j
πn
2πn
3
×
(cos( ) − cos(
))
=
πnj
nπ
3
3
3
πn
2πn
= 2 2 (cos( ) − cos(
))
π n
3
3 Fn6=0 = When n = 0,
F0 = 1
2 5. Let the signal f (t) = cos3 (t) be the input of an LTI system with frequency response H(ω) = 2 e−jωπ/2 for
ω[−2, 2] rad/s and zero elsewhere. Obtain the steadystate response y(t) of the system to the input f (t).
Solution :
f (t) = cos3 (t) = (
= ejt + e−jt 3
)
2 1 j3t
(e + e−j3t + 3ejt + 3e−jt )
8 Evaluate the H(w) at w = 1, −1, we get
H(1) = −2j
H(−1) = 2j
3 jt
3
e × (−2j) + e−jt × (2j)
8
8
3
= sin(t)
2 y(t) = 6. Below is the circuit you will build and test in Lab 3. Please
Page 4 of 6 (a) using phasor method to calculate the phasor of Vo as a function of ω , assuming your input Vi (t) = cos(ωt)
(b) Please plot Vo (ω) in loglog scale, from 100Hz to 100KHz. Hint: Treating ω as a constant, calculate the
phasor of Vo in terms of ω . Then, you may get your Vo as a complex function of ω , plot the amplitute of
phase of your phasor of Vo . Solution:
(a) By applying KCL at node V2 , we have:
V1 − V2
1
jωC = V2
,
1.7kΩ By applying KCL at node V1 , we have:
V1
V1 − V2
vi − V1
V 1 − vo
= 1 +
+
,
1
5kΩ
1kΩ
jωC
jωC By applying the KCL at the node between two right 3.6kΩ resistor:
vo − V 2
V2
=
.
3.6kΩ
3.6kΩ We combine the three equations to get:
vi = 10
5vo
5
vo
1
vo + 2500vo jωC +
− vo +
+ vo ,
3.4
3400jωC
2
3400jωC
2 after the simplication, we have:
vi = ( 16
3
+ 2500jωC +
)vo .
17
1700jωC Since the phasor of vi = 1, we have:
vo = 16
17 1
+ 2500jωC + 3
1700jωC = 16
17 + 10−8 1
× 2500jω + 3×108
1700jω Page 5 of 6 (b) The loglog plot is as following: Page 6 of 6 ...
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 Spring '08
 Staff
 Fourier Series, Periodic function, dt, periodic signal, Fourier coecients

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