11 Solution
Signal 1 contains components at
f
1
=0
.
5
kHz and
f
2
.
9
kHz and signal 2 contains components at
f
3
=3
.
5
kHz and
f
4
= 10
kHz. The output of the multiplier is
[cos
ω
1
t
+ cos
ω
2
t
][cos
ω
3
t
+ cos
ω
4
t
]
.
When expanded, the product includes 4 terms:
cos
ω
i
t
cos
ω
j
t
, where
(
i, j
)
∈{
(1
,
3)
,
(1
,
4)
,
(2
,
3)
,
(2
,
4)
}
.
Using the identify
cos
ω
i
t
cos
ω
j
t
=
1
2
cos(
ω
i
+
ω
j
)
t
+
1
2
cos(
ω
i
−
ω
j
)
t
it is clear that the output of the multiplier contains the following frequencies:
f
1
+
f
3
=4
kHz

f
1
−
f
3

kHz
f
1
+
f
4
= 10
.
5
kHz

f
4
−
f
1

=9
.
5
kHz
f
2
+
f
3
.
4
kHz

f
2
−
f
3

=2
.
6
kHz
f
2
+
f
4
= 10
.
9
kHz

f
2
−
f
4

.
1
kHz
The components that pass through the Flter are 4 kHz, 4.4 kHz, 9.1 kHz, and 9.5 kHz. Note that when
di±erence frequencies are being calculated the absolute value is used to ensure that the result is positive. In
other words, since we are discussing realvalued signals, every negativefrequency component is accompanied
by a positivefrequency counterpart. When asked for “the frequency” of a sinusoidal signal, we report the
positive number.
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 Spring '08
 Staff
 Inductance, RL circuit, LC circuit, F3

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