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P18.1
Consider the system in Figure P18.1-1 for discrete-time processing of a continuoustime
signal using sampling period T, where the C/D operation is as shown in Figure P18.1-2
and the D/C operation is as shown in Figure P18.1-3.
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The filter G(Q) is the lowpass filter shown in Figure P18.1-4.
The Fourier transform of xc(t), Xc(w) is given in Figure P18.1-5.

The sampling frequency is 8 kHz. Sketch accurately the following transforms.
(a) X,(w)
(b) X(Q)
(c) Y(Q)
(d) Ye(w)
Consider the continuous-time frequency response in Figure P18.2.

We want to implement this continuous-time filter using discrete-time processing.
(a) What is the maximum value of the sampling period T required?
(b) What is the required discrete-time filter G(Q) for T found in part (a)?
(c) Sketch the total system.
P18.3
The system in Figure P18.3 is similar to that demonstrated in the lecture. Note
that, as in the lecture, there is no anti-aliasing filter.

P18.4 Suppose we want to design a variable-bandwidth, continuous-time filter
using the structure in Figure P18.4-1.
Find, in terms of wc, the value of the sampling period To and the corresponding
value co, such that the total continuous-time filter has the frequency response
shown in Figure P18.4-2.

P18.5 Consider the system in Figure P18.5-1.
Let H(Q) be as given in Figure P18.5-2 and X(co) as given in Figure P18.5-3.

(a) Sketch X(Q) and Y(Q).
(b) Suppose we replace the system in Figure P18.5-1 by the P18.5-4. Find
G(w) such that y[n] = z[n].
P18.6
Suppose we are given the system in Figure P18.6-1.
(a) Find the appropriate values of the sampling period To to avoid aliasing. Also
find the proper value for K so that the overall system has a gain of unity at w = 0
(i.e., no overall dc gain).

(b) Suppose To is halved, but the anti-aliasing and reconstruction filters are not
modified.
(i) If X(w) is as given in Figure P18.6-2, find Y(Q).
(ii) If Y(Q) is as given in Figure P18.6-3, find Y(w).

P18.7
Figure P18.7 shows a system that processes continuous-time signals using a
digital filter. The digital filter h[n] is linear and causal with difference equation
For input signals that are bandlimited so that Xe(w) = 0 for I l > ir/T, the system
is equivalent to a continuous-time LTI system. Determine the frequency
response He(w) of the equivalent overall system with input xc(t) and output
yc(t).

P18.8 Figure P18.8-1 depicts a system for which the input and output are discrete-time
signals. The discrete-time input x[n] is converted to a continuous-time impulse train
x,(t). The continuous-time signal x,(t) is then filtered by an LTI system to produce the
output yc(t), which is then converted to the discrete-time signal y[n]. The LTI system
with input xc(t) and output yc(t) is causal and is characterized by the linear constant-
coefficient difference equation

The overall system is equivalent to a causal discrete-time LTI system, as indicated
in Figure P18.8-2. Determine the frequency response H(Q) of the equivalent LTI
system.
P18.9
We wish to design a continuous-time sinusoidal signal generator that is capable of
producing sinusoidal signals at any frequency satisfying wi : W5 W2, where w,
and W2 are positive numbers.
Our design is to take the following form. We have stored a discrete-time
cosine wave of period N; that is, we have stored x[O], . . . , x[N - 1], where
Every T seconds we output an impulse weighted by a value of x[k], where we
proceed through the values of k = 0, 1, ... , N - 1 in a cyclic fashion. That is,

(a) Show that by adjusting T we can adjust the frequency of the cosine signal
being sampled. Specifically, show that
where wo = 21r/NT. Determine a range of values for T so that y,(t) can
represent samples of a cosine signal with a frequency that is variable over the
full range
(b) Sketch Y,(w).
The overall system for generating a continuous-time sinusoid is depicted in
Figure P18.9-1. H(w) is an ideal lowpass filter with unity gain in its passband:

The parameter we is to be determined such that y(t) is a continuous-time cosine signal
in the desired frequency band.
(c) Consider any value of T in the range determined in part (a). Determine the
minimum value of N and some value for w,such that y(t) is a cosine signal in the
range wi : O o02.
(d) The amplitude of y(t) will vary depending on the value of wchosen between wi
and W2. Determine the amplitude of y(t) as a function of w and as a function of N.

P18.10
In many practical situations, a signal is recorded in the presence of an echo, which
we would like to remove by appropriate processing. For example, Figure P18.10-1
illustrates a system in which a receiver receives simultaneously a signal x(t) and an
echo represented by an attenuated delayed replication of x(t). Thus, the receiver
output is s(t) = x(t) + ax(t - TO), where jal < 1. The receiver output is to be processed
to recover x(t) by first converting to a sequence and using an appropriate digital filter
h[n] as indicated in Figure P18.10-2.

Assume that x(t) is bandlimited, i.e., X(w) = 0 for IwI > wm, and that al < 1.
(a) If To < lr/M and the sampling period is taken equal to To (i.e., T = TO), determine the
difference equation for the digital filter h[n] so that yc(t) is proportional to x(t).
(b) With the assumptions of part (a), specify the gain A of the ideal lowpass filter so that
yc(t) = x(t).
(c) Now suppose that 2r/wm < To < 27r/WM. Determine a choice for the sampling period
T, the lowpass filter gain A, and the frequency response for the digital filter h[n] such
that yc(t) is equal to x(t).

S18.1 (a) Since x,(t) = xc(t)p(t), then X,(w) is just a replication of Xe(w) centered at
multiples of the sampling frequency, namely 8 kHz or 27r8 X 10' rad/s. The sampling
period is T = 1/8000.
(b) X(Q) is just a rescaling of the frequency axis, where 21r8 X 103 becomes 2 1r.
X(Q) is shown in Figure S18.1-2.

(c) Y(Q) is the product G(Q)X(Q). Therefore, Y(Q) appears as in Figure S18.1-3.
(d) Y,(w) is a frequency-scaled version of Y(w) but only in the range 0 = --w to 7,
as shown in Figure S18.1-4. Also note the gain of T.

S18.2 (a) The maximum nonzero frequency component of H(w) is 5 00 w. Therefore,
this frequency can correspond to, at most, the maximum digital frequency before
folding, i.e., Q = x. From the relation wT = Q,we get
(b) Since w = 500w maps to Q = 7r, the discrete-time filter G(Q) is as shown in
Figure S18.2-1.
(c) The complete system is given by Figure S18.2-2. Note the need for an anti-
aliasing filter.

S18.3
(a) Recall that Xc(w) is as given by Figure S18.3-1.

Ye(w) is given by eq. (S18.3-2) and Figure S18.3-3.
Thus x(t) = y(t) in this case.
(b) Xc(w) is as given in Figure S18.3-4.

We now use eq. (S18.3-1), shown in Figure S18.3-5.
Thus, in the range ± r, X(Q) = 20000 E",,U X,[20000(9 - 2irn)] is given as in
Figure S18.3-6.

Using eq. (S18.3-2), we find Ye(w) as in Figure S18.3-7.
Note aliasing since 27000 Hz is above half the sampling rate of 20000 Hz.
(c) Xe(w) is as given in Figure S18.3-8.

Again we use eq. (S18.3-1), shown in Figure S18.3-9.

Thus X(Q) is given as in Figure S18.3-10.
Finally, from eq. (S18.3-2) we have Y,(w) shown in Figure S18.3-11.
S18.4 It is required that we sample at a rate such that the discrete-time frequency
7r/ 2 will correspond to c. The relation between 9, and cis Q, = ocTo. Thus, we
require

As wc increases, demanding a wider filter, To decreases, and consequently the sampling
frequency must be increased. There are two ways to calculate Wa. First, since we are
sampling at a rate of
we need an anti-aliasing filter that will remove power at frequencies higher than half
the sampling rate; therefore wa = 2wc. Alternatively, we note that the "folding
frequency," or the frequency at which aliasing begins, is 0 = 7r. Since 9 = r/2
corresponds to wc, then r must correspond to 2w,.
S18.5 (a) We sketch X(Q) by stretching the frequency axis so that 2 7 corresponds to
the sampling frequency with a gain of 1/TO. We then repeat the spectrum, as shown
in Figure S18.5-1.

After filtering, Y(Q) is given as in Figure S18.5-2.
(b) We see that Y(Q) looks like X(w) filtered and then sampled. The discrete-time
frequency is -/3. Again, 2-x corresponds to 27r/To, so 7/3 corresponds to ir/3T. Thus, if
x(t) is filtered by G(w) as given in Figure S18.5-3, then y[n] = z[n].

Solutions to Optional Problems
S18.6 (a) Since we are allowing all frequencies less than 1007r through the anti-
aliasing filter, we need to sample at least twice 100r, or 200r. Thus, 2 007r = 2-
x/T or To = 10 ms. To find K, recall that impulse sampling introduces a gain of
1/TO. To account for this, K must equal To, or K = 0.01. (b) (i) Since X(w) is
bandlimited to 100-r, the anti-aliasing filter has no effect. The Fourier transform
of x,(t), the modulated pulse train, is given in Figure S18.6-1.
Since To = 0.005, the sampling frequency is 4007r. After conversion to a
discrete-time signal, X(Q) appears as in Figure S18.6-2.

(ii) There are three effects to note in D/C conversion: (1) a gain of To, (2) a frequency
scaling by a factor of T,, and (3) the removal of repeated spectra. Thus, Y(w) is as
shown in Figure S18.6-4.

S18.7 After the initial shock, you should realize that this problem is not as difficult
as it seems. If instead of h[n] we had been given the frequency response H(Q), then
He(w) would be just a scaled version of H(Q) bandlimited to ir/T. Let us find, then,
H(Q). Using properties of the Fourier transform, we have

Therefore, the magnitude and phase of He(c) are as shown in Figure S18.7.

From our previous study, we know that Xe(o) in the range ±ir/Tlooks just like X(Q) in
the range ±ir. Similarly, Y,(w) between -- r/T and +r/T looks like Y(Q) in the range -ir
to r. Although there is a factor of T, we can disregard it in analyzing this system
because it is accounted for in the H(w) filter. The transformation of xc(t) to yc(t) will
correspond to filtering x[n], yielding y[n]. In fact, the equivalent system will have a
system function H(Q) given by

where He(w) is the Fourier transform of h(t). Thus, we need to find Hc(w). The relation
between yc(t) and xc(t) is governed by the following differential equation:
Using the properties of the Fourier transform, we have
S18.9
(a) It is instructive to sketch a typical y,(t), which we have done in Figure S18.9-1.

Let us suppose that T is changed by being reduced. Then the envelope of y,(t)
seems to correspond to a higher-frequency cosine. At time kt,
where we use the sampling property of the impulse function. Thus,
(b) Recall that sampling with an impulse train repeats the spectrum with a period
of 2r/T and a gain factor of 1/T. Since 5([cos(2t/NT)J is as given by Figure S18.9-
2, Y,(w) is then given by Figure S18.9-3.

(c) The minimum value of N is 2, corresponding to the impulses at o and (27r/T
wo) being superimposed at w/T. The lowpass filter cutoff frequency must be such
that the (superimposed) impulses at 7r/T are in the passband and those at 37r/T are
outside the passband. Consequently,

(d) Comparing Y(w) and Y,(w) in Figures S18.9-2 and S18.9-3 respectively, we
see that for N > 2 the cosine output will have an amplitude of 1/T = w/21r. If N =
2, then the output amplitude will be 2/T = w/7r.
S18.10
(a) By sampling sc(t), we get
This is a first-order difference equation, so given s[n], we can find x[n]. Since x(t)
is appropriately bandlimited, we can then set

(b) From part (a) we see that T = A will make y(t) = x(t).
(c) Since we do not want to alias, we still need T < 7r/wM. Now
Taking the continuousFourier transform, we see that
Thus, the continuous-time inverse system has frequency response
We want to implement this in discrete time. Therefore, using the relation, we obtain

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