1. No category

Signals and Systems I
EE360: Fall 14
Homework #4
Due Th. 10/02
Note: The Basic Problems with Answers will be worth half as much as the other questions.
You must show all your work to receive credit.
1. (OW 3.21)
Solution
Given the following information about the signal
T = 8,
a1 = a∗−1 = j,
a5 = a−5 = 2,
the periodic signal x(t) can be expressed using the complex exponential form of the Fourier
Series and converted using ω0 = 2π/T = π/4.
x(t) = a−5 e−j5ω0 t + a−1 e−jω0 t + a1 ejω0 t + a5 ej5ω0 t
= 2 e−j5ω0 t + ej5ω0 t + j ejω0 t − e−jω0 t
= 4 cos(5ω0 t) + 2 sin(ω0 t)
π
5π
π
= 4 cos
t − 2 cos
t−
4
4
2
2. (OW 3.24)
Solution
(a) Find the average signal value over a period
Z
1 1
1
1
x(t)dt =
(1)(2) = .
a0 =
T T
2 2
2
(b) Define
dx(t)
g(t) =
=
dt
(
1
−1
0≤t≤1
.
1≤t≤2
Then FS coefficients can be found by noting
Z
1
g(t) −→ bk =
g(t)e−jkω0 t dt.
T T
k=0
k 6= 0
b0 = 0
bk =
1
2
1
Z
average over single period
Z 2
−jkω0 t
−jkω0 t
e
dt −
e
dt
0
1
1
2
1
1
1
1
−jkω0
−jkω0
=
e
−
e
2 −jkω0
2 −jkω0
0
1
1 1 −j2ω0
−jω0
−jω0
=
1−e
+
e
−e
2jkω0
2jkω0
h
i
1
=
1 − 2e−jkω0 + e−jk2ω0
2jkω0
"
#
1
−jkπ
−j2π
1 − 2e
+ e| {z }
=
2jkπ
=1
i
1 h
1 − e−jkπ
=
jkπ
1
Signals and Systems I
EE360: Fall 14
(c) Using the differentiation property
x(t) ←→ak
dx(t)
←→bk = jkω0 ak = jkπak
dt
1
1
1
⇒ak =
bk =
[1 − e−jπ ] = 2 2 [e−jπ − 1]
2
jkπ
(jkπ)
k π
3. (OW 3.25)
Solution
(a)
1
x(t) = [ej4πt + e−j4πt ]
2
1
⇒ ak −→ a1 = a−1 =
2
(b)
1 j4πt
[e
− e−j4πt ]
2j
1
1
⇒ bk −→ b1 = , b−1 = −
2j
2j
y(t) =
(c) The multiplicative property states the product in the time domain has convolution of
the FS coefficients.
∞
X
z(t) = x(t)y(t) ←→ ck =
al bk−l .
l=−∞
1
1
1
1
ck = [ δ(k + 1) + δ(k − 1)] ∗ [− δ(k + 1) + δ(k − 1)]
2
2
2j
2j
1
1
1
1
= − δ(k + 2) + δ(k) − δ(k) + δ(k − 2)
4j
4j
4j
4j
1
1
= − δ(k + 2) +
δ(k − 2)
4j
4j
|{z}
|{z}
c−2
c2
(d) Using pure trigonometric expansion
z(t) = cos (4πt) sin (4πt)
1
1
= [ej4πt + e−j4πt ] · [ej4πt − e−j4πt ]
2
2j
1
= [ej2(4π)t − e0 + e0 − e−j2(4π)t ]
4j
1 j2(4π)t 1 −j2(4π)t
=
e
− e
4j
4j
|{z}
|{z}
c2
c−2
2
Signals and Systems I
EE360: Fall 14
4. (OW 3.27)
Solution
N = 5,
a2 = a∗−2 = 2ejπ/6 ,
a0 = 2,
a4 = a∗−4 = ejπ/3 ,
ω0 =
x[n] = 2 + a2 ej2ω0 n + a−2 e−j2ω0 n + a4 ej4ω0 n + a−4 e−j4ω0 n
= 2 + 2ejπ/6 ej2ω0 n + 2e−jπ/6 e−j2ω0 n + ejπ/3 ej4ω0 n + e−jπ/3 e−j4ω0 n
h
i h
i
= 2 + 2 ej(2ω0 n+π/6) + e−j(2ω0 n+π/6) + ej(4ω0 n+π/3) + e−j(4ω0 n+π/3)
= 2 + 4 cos (2ω0 n + π/6) + 2 cos (4ω0 n + π/3)
= 2 + 4 sin (2ω0 n + π/6 + π/2) + 2 sin (4ω0 n + π/3 + π/2)
= 2 + 4 sin (2ω0 n + 2π/3) + 2 sin (4ω0 n + 5π/6)
5. (OW 3.33)
Solution
∂
y(t) + 4y(t) = x(t)
∂t
assume input of form ejωt and find frequency response
ejωt −→ H(jω)ejωt for LTI system
⇒
∂
y(t) = jωH(jω)ejωt
∂t
⇒ jωH(jω)ejωt + 4H(jω)ejωt = ejωt
⇒ H(jω) [jω + 4] = 1
1
H(jω) =
4 + jω
←− use this for y(t) =
inf
X
ak H(jkω0 )ejkω0 t
k=− inf
(a) ω0 = 2π, T = 1
x(t) = cos(2πt) =
1 j2πt
e
+ e−j2πt
2
1
H(j2π)ej2πt + H(−j2π)e−j2πt
2
1
1
1
j2πt
−j2πt
e
+
e
=
2 4 + j2π
4 − j2π
y(t) =
1
⇒ b1 =
2
1
4 + j2π
3
, b−1
1
=
2
1
4 − j2π
2π
5
Signals and Systems I
EE360: Fall 14
(b)
x(t) = sin( |{z}
4πt ) + cos( |{z}
6πt +π/4)
T1 =1/2
T2 =1/3
⇒ T = 1, ω0 = 2π
1
1 j2ω0 t
1
=
e
− e−j2ω0 t + ejπ/4 ej3ω0 t + e−jπ/4 e−j3ω0 t
2j
2
2
y(t) =
1
1
1
[H(j4π) − H(−j4π)] + ejπ/4 H(j6π)ej3ω0 t + e−jπ/4 H(−j6π)e−j3ω0 t
2j
2
2
1
1
⇒ b2 =
,
2j 4 + j4π
1
1
,
b3 = ejπ/4
2j
4 + j6π
b−2
b−3
1
1
=−
,
2j 4 − j4π
1
1
= e−jπ/4
2j
4 − j6π
6. (OW 3.36)
Solution
Similar to 3.33
1
y[n] − y[n − 1] = x[n], y[n] = H(ejω )ejωn , y[n − 1] = H(ejω )ejω(n−1)
4
1 −jω jωn
jω
H(e ) 1 − e
e
= ejωn
4
H(ejω ) =
1
1−
1 −jω
4e
(a)
3π
2π
π
x[n] = sin( n), N0 = 8 ⇒ ω0 =
=
4
8
4
i
3π
1 h j 3π n
=
e 4 − e−j 4 n
2j
i
3π
3π
1 h j 3π j 3π n
H e 4 e 4 − H e−j 4 e−j 4 n
2j
"
#
3π
1
1
1
j 3π
n
−j
n
4
=
e 4 −
3π e
2j 1 − 1 e−j 3π
4
1 − 1 ej 4
y[n] =
4
1
⇒ b3 =
2j
4
!
1
1 − 41 e−j
4
3π
4
, b−3 =
1
1
1
2j 1 − ej 3π
4
4
!
Signals and Systems I
EE360: Fall 14
(b)
π π x[n] = cos
n + 2 cos
n .
| {z4 } | {z 2 }
N1 =8
Therefore the period of x[n] is N = 8 ⇒ w0 =
Following the same procedure as above
N2 =4
2π
8
= π4 .
1
x[n] = [ejπ/4n + e−jπ/4n ] + [ej2π/4n + e−j2π/4n ]
2
1
y[n] = [H(ejπ/4 ) ejπ/4n + H(e−jπ/4 ) e−jπ/4n ] + [H(ej2π/4 ) ej2π/4n + H(ej2π/4 ) e−j2π/4n ]
| {z }
| {z }
| {z }
2 | {z }
k=1
k=−1
1
⇒ b1 =
2
b−1 =
1
2
k=2
!
1
,
1 − 41 e−jπ/4
!
1
,
1 − 41 ejπ/4
5
b2 =
b−2 =
k=−2
!
1
,
1 − 14 e−jπ/2
!
1
1 − 14 ejπ/2