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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 5
Reading: Haykin 3.1–3.4
1.
Nyquist 101:
Specify the Nqyusit rate and Nyquist interval for each of the following signals.
Note that
sinc
(
x
)
≡
sin(
πx
)
πx
.
(a)
g
(
t
) =
sinc
(200
t
)
SOLUTION:
This sinc pulse corresponds to a bandwidth of
W
= 100
Hz. Hence, the
Nyquist rate is
200
Hz, and the Nyquist interval is
1
/
200
seconds.
(b)
g
(
t
) =
sinc
2
(200
t
)
SOLUTION:
This signal may be viewed as the product of the sinc pulse
sinc
(200
t
)
with itself. Since multiplication in time domain corresponds to convolution in frequency
domain, we find that the signal
g
(
t
)
has a bandwidth equal to twice that of the sinc pulse
sinc
(200
t
)
, that is
200
Hz. The Nyquist rate of
g
(
t
)
is therefore
400
Hz, and the Nyquist
interval is
1
/
400
seconds.
(c)
g
(
t
) =
sinc
(200
t
) +
sinc
2
(200
t
)
SOLUTION:
The bandwidth of
g
(
t
)
is determined by the highest frequency content
of either
sinc
(200
t
)
or
sinc
2
(200
t
)
. From earlier parts, we know that
sinc
2
(200
t
)
has
the higher bandwidth equal to
200
Hz. Correspondingly, the Nyquist rate is
400
Hz
and the Nyquist interval is
1
/
400
seconds.
2.
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