EECE 340 Homework, Spring 2008
American University of Beirut · Signals and Systems · 4 pages
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American University of Beirut
Department of Electrical and Computer Engineering
EECE 440 Signals and Systems
Homework 4 – Solution
Problem 1
Signal at b
)t( cos(10,000 )t
2m
)t(
m
)t(
b
2
1
+
=
Signal at c
)t( cos(30,000 )t
2m
)t( cos(10,000 )t
2m
)t( cos(20,000 )t
2m
)t(c
2
2
1
+
+
=
C(t) is a band-pass signal that rum from 9500 rad/s to 30500 rad/s.
Therefore, its bandwidth is 21000 rad/s.
Problem 2
Y(t) is a low pass signal of bandwidth 12 rad/s.
Problem 3
The signal at the input of the filter is:
)t
T cos(3
2 rect t
1
t
T cos
2 rect t
1
)t(
x
0
0
ω
⎟⎠
⎞
⎜⎝
⎛
+
ω
⎟⎠
⎞
⎜⎝
⎛
=
The signal at the output of the filter is:
)t(
2 f
1
t
T cos
2 rect t
1
)t(
g
ω0 =
⎟⎠
⎞
⎜⎝
⎛
=
Problem 4
2
1
H( )
ω
+
ω
ω =
Plotting
( )
H ω vs ω, we can easily conclude that the filter is a high-pass
filter
Problem 5
At the input of the filter
[
]
[
220 ])
(
220)
(
20
160)
(
160)
(
12
S( )
+ δ ω +
π δ ω −
+
+ δ ω +
π δ ω −
ω =
At the output of the filter
[
220 ])
(
220)
(
20
G( )
+ δ ω +
π δ ω −
ω =
20cos(220t
g )t(
=
200
2
(20)
P
2
g
=
=
Watts
Problem 6
From your notes
)t
fˆ )t( sin(
)t
)t( cos(
f
)t(
g
0
0
ω
+
ω
=
Problem 7
The sinusoidal signal has a frequency of 1 Hz or a period of 1 s. The
maximum allowable time between intervals should be T=
2
1
2B
1 =
s.
Problem 8
Discussed in class
Problem 9
2.5 cos(1600 t)
)t
cos(2400
5.2
)t
cos(1800
5.2
)t
cos(2200
5.2
)t
10cos(2000
)t(s
π
+
π
+
π
+
π
+
π
=
b. Total power =
2
( 5.2 )
2
( 5.2 )
2
( 5.2 )
2
( 5.2 )
2
(10)
2
2
2
2
2
+
+
+
+
= 50+2(6.25)=62.5 Watts
c. The total sideband power = total power – Carrier power = 62.5 -50
=12.5 Watts
d. Modulation index =
min
max
min
max
A
A
A
A
+
−
μ =
Using MATLAB: Amax = 1, and Amin= -0.5
Therefore,
μ = .0 64
Problem 10
a. Carrier Power =
50
2
A
P
2
c
c
=
=
watts, implies that
c
A =10 Volts
b.
10 rad s/
2
6
c
ω = π×
c.
μ = amka
=0.8
d.
510 rad s/
2
3
m
= π×
ω
[
]
)t
cos(10,000 t cos( ,2 000,000
8.0
101
)t(s
π
π
+
=
Problem 11
a. Modulation index =
min
max
min
max
A
A
A
A
+
−
μ =
Using Plot: Amax = 12, and Amin= 4
Therefore,
8.0
μ =
b.
]
A 1[
A
c
max
+ μ
=
, implies that
Ac = 8Volts
Problem 12
a. The AM wave
)t
k m t( )]cos(
A 1[
)t(s
c
a
c
ω
+
=
for m=0.5,
k a =1/16
Therefore, the AM wave is written as:
(2/16)cos(60 t)]cos200 t
)t
(6/16)cos(20
100 1[
)t(s
π
π
−
π
−
=
b.
)t
)t( sin(200
A k mˆ
)t
1( /16)m t( )]cos(200
100 1[
)t(s
c a
π
−
π
+
=