ECE 453: Homework Assignment 2 Solution
111 Solution
a.
s
(
t
) = 10 cos[2
π
10
8
t
−
2 cos(2
π
10
4
t
+
π/
4)]
.
This signal is in the form
A
cos(
ω
c
t
+
θ
(
t
))
,
where
θ
(
t
)=
−
2 cos(2
π
10
4
t
+
4)
. Hence, the signal is purely angle modulated. The message signal consists of a single
tone with frequency
f
m
= 10
4
Hz =
10
kHz. The peak phase deviation is 2 radians. The instantaneous
frequency deviation
∆
ω
(
t
d
dt
θ
(
t
)=4
π
10
4
sin(2
π
10
4
t
+
4)
. The peak frequency deviation,
∆
f
max
, is
20
kHz, so using Carson’s rule, the bandwidth is
BW
= 2(
f
m
+∆
f
max
) = 2(10 + 20)
kHz
= 60
kHz.
b.
s
(
t
) = 50 cos[2
π
10
8
t
] cos[2
π
10
4
t
+
4]
.
This signal is in the form
Am
(
t
) cos[2
π
10
8
t
]
, where
m
(
t
cos(2
π
10
4
t
+
4)
. Hence, the signal is amplitude modulated using DSBSC. As in part a., the message signal,
m
(
t
)
, is a tone with frequency
f
m
= 10
kHz. Hence, the bandwidth occupied by
s
(
t
)
is
=2
f
m
= 20
kHz.
112 Solution
s
(
t
) = 50 cos(
ω
c
t
+ 5 sin(2
π
1000
t
))
.
This signal is in the form
A
cos(
ω
c
t
+
θ
(
t
))
,
where
θ
(
t
) = 5 sin(2
π
1000
t
)
.
Hence, the signal is purely angle modulated. The message signal consists of a single tone with frequency
f
m
= 1000
Hz =
1
kHz.
a. The peak phase deviation is 5 radians.
b. The instantaneous frequency deviation
∆
ω
(
t
d
dt
θ
(
t
) = 10
4
π
cos(2
π
1000
t
)
s
−
1
.
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 Spring '08
 Staff
 Frequency, Hertz, usb

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