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THE STATE UNIVERSITY OF NEW JERSEY
RUTGERS
College of Engineering
Department of Electrical and Computer Engineering
332:322
Principles of Communications Systems
Spring 2004
Problem Set 4
1.
Phase Locked Loop Demodulator (quickie):
(a) Using what you know about phase locked loops (PLLs), show that the PLL can be used
as a FM demodulation circuit.
SOLUTION:
Just use the
˙
φ
t
output – it’s the derivative of whatever was inside the
argument of that incoming
cos2
π
f
c
t
ϖ
t
. That is,
˙
φ
t
2
π
f
c
˙
ϖ
t
. So you filter
the D.C., and throw in a few multiplicative constants and you’re done.
(b) How can you modify your PLL to obtain a PM demodulation circuit?
SOLUTION:
Integrate the output of your demodulator above. But note, you’ve GOT
to get rid of the DC offset, otherwise you’ll have a lawsuit on your hands when
2
π
f
c
t
exceeds about ten kilovolts – how fast would that happen if the listener were tuned to
99.9FM? :)
2.
Hilbert Transform Stuff for Culture:
The Hilberttransform of a Fourier transformable
signal
m
t
, denoted by
m
H
t
, is defined by
m
H
t
1
π
∞
∞
m
t
t
τ
d
τ
In the frequency domain, we have
M
H
f
j sgn
f
M
f
where
m
t
,
M
f
and
m
H
t
,
M
H
f
are Fourier transform pairs and
sgn
f
is the signum
function. Using the definition of the Hilbert transform, show that a singlesideband modu
lated signal resulting from the message signal
m
t
and carrier cos
2
π
f
c
t
of unit amplitude
is given by
s
t
1
2
m
t
cos
2
π
f
c
t
m
H
t
sin
2
π
f
c
t
where the minus sign corresponds to the transmission of upper sideband and plus to the
transmission of lower sideband.
SOLUTION:
Consider first the modulated signal
s
t
1
2
m
t
cos
2
π
f
c
t
m
H
t
sin
2
π
f
c
t
Applying Fourier transform to the above equation, we get
S
f
1
4
M
f
f
c
M
f
f
c
1
4
j
M
H
f
f
c
M
H
f
f
c
1
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View Full DocumentFrom the defenition of Hilbert Transform, we have
M
H
f
j sgn
f
M
f
where sgn
f
is the signum function. Equivalently, we may write
1
j
M
H
f
f
c
sgn
f
f
c
M
f
f
c
1
j
M
H
f
f
c
sgn
f
f
c
M
f
f
c
From the definition of signum function, we note the following for f
0
and f
f
c
sgn
f
f
c
sgn
f
f
c
1
Correspondingly, the equation for S
f
reduces to,
S
f
1
4
M
f
f
c
M
f
f
c
1
4
M
f
f
c
M
f
f
c
1
2
M
f
f
c
In words, the above result means that, except for a scaling factor, the spectrum of the modu
lated signal s
t
is the same as that of a DSBSC modulated signal for f
f
c
.
For f
0
and f
f
c
, we have
sgn
f
f
c
1
sgn
f
f
c
1
Correspondingly, S
f
reduces to,
S
f
1
4
M
f
f
c
M
f
f
c
1
4
M
f
f
c
M
f
f
c
0
In words, for f
f
c
, the modulated signal s
t
is zero.
Combining the two results, the modulated signal s
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 Spring '08
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