Digital Signal Processing Reference
In-Depth Information
ω ω 0
2 W
ω + ω 0
2 W
(c) X 3 ( ω ) = rect
+ rect
;
(d) X 4 ( ω ) = u ( ω − ω 0 ) u ( ω − 2 ω 0 ) .
9.3 The converse of the uncertainty principle, explained in Problem 9.2, is
also true. In other words, a time-limited signal cannot be band-limited. By
calculating the CTFT of the following time-limited signals, show that the
converse of the uncertainty principle is indeed true (assume that τ, T , and
α are real, positive constants):
(a) x 1 ( t ) = cos( ω 0 t )[ u ( t + T ) u ( t T )];
t
τ
t
τ
(b) x 2 ( t ) = rect
rect
( denotes the CT convolution operator);
t
τ
(c) x 3 ( t ) = e −α t
rect
;
(d) x 4 ( t ) = δ ( t 5) + δ ( t + 5).
9.4 The CT signal x ( t ) = v 1 ( t ) v 2 ( t ) is sampled with an ideal impulse train:
s ( t ) =
δ ( t kT s ) .
k =−∞
(a) Assuming that v 1 ( t ) and v 2 ( t ) are two baseband signals band-limited
to 200 Hz and 450 Hz, respectively, compute the minimum value of
the sampling rate f s that does not introduce any aliasing.
(b) Repeat part (a) if the waveforms for v 1 ( t ) and v 2 ( t ) are given by the
following expression:
v 1 ( t ) = sinc(600 t )
and
v 2 ( t ) = sinc(1000 t ) .
(c) Assuming that a sampling interval of T s = 2 ms is used to sample
x ( t ) = v 1 ( t ) v 2 ( t ) specified in part (b), sketch the spectrum of the sam-
pled signal. Can x ( t ) be accurately recovered from the sampled signal?
(d) Repeat part (c) for a sampling interval of T s = 0 . 1 ms.
9.5 The CT signal x ( t ) = sin(400 π t ) + 2 cos(150 π t ) is sampled with an ideal
impulse train. Sketch the CTFT of the sampled signal for the following
values of the sampling rate:
(a) f s = 100 samples/s;
(b) f s = 200 samples/s;
(c) f s = 400 samples/s;
(d) f s = 500 samples/s.
In each case, calculate the reconstructed signal using an ideal LPF with the
transfer function given in Eq. (9.7) and a cut-off frequency of ω s / 2 = π f s .
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