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13 Continuous-Time Modulation Recommended
Problems
P13.1
In the amplitude modulation system in Figure P13.1-1, the input x(t) has the Fourier
transform shown in Figure P13.1-2.
For each choice of carrier c(t) in the following list, draw the magnitude and phase of
Y(w), the Fourier transform of y(t).
Problem
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The Fourier transform of s(t) is given by Figure P13.2-2.
The Fourier transform of x(t) is given by Figure P13.2-3.

For which of the following choices for m(t) and d(t) is y(t) nonzero?
P13.3
In Section 7.1 of the text, we discussed the effect of a loss in synchronization in phase
between the carrier signals in the modulator and demodulator for sinusoidal amplitude
modulation. Specifically, we showed that the output of the demodulator is attenuated by
the cosine of the phase difference; in particular, when the modulator and demodulator
have a phase difference of ir/2, the demodulator output is zero. As we demonstrate in this
problem, it is also important to havefrequency synchronization between the modulator and
demodulator.

Consider the amplitude modulation and demodulation systems in Figure P13.3-1, with
0e = 0 and with a change in frequency of the demodulator carrier such that
w(t) = y(t) cos wAdt
where y(t) = x(t) cos Wct

Let us denote the frequency difference between the modulator and demodulator as Aw [i.e.,
(Wd - Wc) = Aw]. Also, assume that x(t) is bandlimited with X(W) = 0 for Iwl > wm and
assume that the cutoff frequency W of the lowpass filter in the demodulator satisfies the
inequality
(wM + |IA|) < W < (2we + Awl -Mo)
(a) Show that the output of the demodulator lowpass filter is proportional to x(t)cos(AWt).
(b) (b) If the spectrum of x(t) is that shown in Figure P13.3-2, sketch the spectrum of the
output of the demodulator.
You may find it useful to use the trigonometric identity
P13.4
As discussed in Section 7.1.1 of the text, asynchronous modulation-demodulation requires
the injection of the carrier signal so that the modulated signal is of the form

where [A + x(t)] > 0 for all t. The presence of the carrier means that more transmitter
power is required, representing an inefficiency.
(a) Let x(t) be given by x(t) = cos omt with wm < we and [A + x(t)] > 0. For a periodic
signal y(t) with period T, the time average power P, is defined as P, = (1/T) fTy 2 (t)dt.
Determine and sketch Py for y(t) in eq. (P13.4-1). Express your answer as a function of
the modulation index m, defined as the maximum absolute value of x(t) divided by A.
(b) The efficiency of transmission of an amplitude-modulated signal is defined to be the
ratio of the power in the sidebands of the signal to the total power in the signal. With
x(t) = cos wMt and with wM < wc and [A + x(t)] > 0, determine and sketch the efficiency
E of the modulated signal as a function of the modulation index m.
Optional
Problems
P13.5
Consider the modulated signal z(t) = A(t)cos(wc + 0,), where we, is known but 0c is
unknown. We would like to recover A(t) from z(t).
(a) Show that z(t) = x(t)cos wet + y(t)sin wet and express x(t) and y(t) in terms of
A(t) and 0e,.
(b) (b) Show how to recover x(t) from z(t) by modulation followed by filtering.
(c) Show how to recover y(t) from z(t) by modulation followed by filtering.
(d) Express A(t) in terms of x(t) and y(t) with no reference to 0c and show in a block
diagram how to recover A(t) from z(t). The following trigonometric identities may
be useful:

P13.6
A single-sideband modulation system with carrier frequency we, is shown in Figure
P13.6-1.
Sketch the Fourier transform of si(t), s2(t), sA(t), s4(t), sr(t), s,(t), and y(t), thus showing
that y(t) is x(t) single-sideband-modulated on the carrier We. Assume that x(t) has the
real Fourier transform shown in Figure P13.6-2 and that H(w) is a lowpass filter as
shown in Figure P13.6-3.

P13.7
Consider the system in Figure P13.7, which can be used to transmit two real signals
over a single transmission channel.
For y 1(t) to be the same as s,(t), and y 2(t) to be the same as s2(t), choose the
proper filter H(w) and place the proper restrictions on the bandwidth of si(t) and s2(t).
P13.8
A commonly used system to maintain privacy in voice communications is a speech
scrambler. As illustrated in Figure P13.8-1, the input to the system is a normal
speech signal x(t) and the output is the scrambled version y(t). The signal y(t) is
transmitted and then unscrambled at the receiver.

We assume that all inputs to the scrambler are real and bandlimited to frequency wM;
that is, X(w) = 0 for IwI > WM. Given any such input, our proposed scrambler permutes
different bands of the input signal spectrum. In addition, the output signal is real and
bandlimited to the same frequency band; that is, Y(w) = 0 for IwI > wM. The specific
permuting algorithm for our scrambler is
(a) If X(w) is given by the spectrum shown in Figure P13.8-2, sketch the spectrum of the
scrambled signal y(t).
(b) Using amplifiers, multipliers, adders, oscillators, and whatever ideal filters you find
necessary, draw the block diagram for such an ideal scrambler. (c) Again, using
amplifiers, multipliers, adders, oscillators, and ideal filters, draw a block diagram for the
associated unscrambler.

13 Continuous-Time Modulation
Solutions to Recommended Problems
S13.1
(a) By the shifting property,
The magnitude and phase of Y(w) are given in Figure S13.1-
1.
(b) Since ej 3wc+ji2 eer/2e 3 wt, we are modulating the same carrier as in part (a) except
that we multiply the result by eij/ 2 . Thus
Note in Figure S13.1-2 that the magnitude of Y(w) is unaffected and that the phase is
shifted by ir/2.
Solution

(c) Since
we can think of modulation by cos 3wet as the sum of
modulation by
Thus, the magnitude and phase of Y(o) are as shown in Figure
S13.1-3. Note the scaling in the magnitude.

(d) We can think of modulation by sin 3wet as the sum of modulation by
Thus, the magnitude and phase of Y(w) are as given in Figure S13.1-4. Note the
scaling by 1in the magnitude.
(e) Since the phase terms are different in parts (c) and (d), we cannot just add
spectra. We need to convert cos 3wct + sin 3wet into the form A cos(3wet + 0). Note

Modulating by each exponential separately and then adding yields the magnitude and
phase given in Figure S13.1-5. (Note the scaling in the magnitude.)

S13.2
In Figure S13.2-1 we redraw the system with some auxiliary signals labeled
By the modulation property, Ri(w), the Fourier transform of ri(t), is
Since S(w) is composed of impulses, Ri(w) is a repetition of X(w) centered at -2c, 0, and
2we, and scaled by 1/(27r). See Figure S13.2-2.
(a) Since m(t) = d(t) = 1, y(t) is ri(t) filtered twice by the same ideal lowpass filter with
cutoff at we. Thus, comparing the resulting Fourier transform of y(t), shown in Figure
S13.2-3, we see that y(t) = 1/(27r)x(t), which is nonzero.

(b) Modulating ri(t) by ei-c' yields R1(w - we) as shown in Figure S13.2-4.
Similarly, modulating by e ~j'Wyields Ri(o + coc) as shown in Figure S13.2-
5.
Since cos wot = (ewct + e -iwct)/ 2 , modulating ri(t) by cos wct yields a Fourier
transform of r2(t) given by

Thus, R 2 (W)is as given in Figure S13.2-6.
After filtering, R3(o) is given as in Figure S13.2-7.
R 4(w) is given by shifting R3(W)up and down by we and dividing by 2. See Figure
S13.2-8.

After filtering, Y(w) is as shown in Figure S13.2-9.
Comparing Y(w) and X(w) yields
which is drawn in Figure S13.2-10.

After filtering, R3(x) = 0. Therefore, y(t) = 0.
(d) In this case, it is not necessary to know r3(t) exactly. Suppose r 3 (t) is nonzero, with
RA(W) given as in Figure S13.2-11.
After modulating by d(t) = cos 2cot, R 4 (W) is given as in Figure
S13.2-12.

After filtering, y(t) = 0 since R 4(w) has no energy from -we, to we,.
(e) For this part, let us calculate R2(w) explicitly.
which is drawn in Figure S13.2-13.

After filtering, R3 (w) is as shown in Figure S13.2-14.
Modulating again yields R 4(w) as shown in Figure S13.2-15.
Finally, filtering R 4 (w) gives the Fourier transform of y(t), shown in Figure
S13.2-16.

Thus,
S13.3
(a) The demodulator signal w(t) is related to x(t) via
W(t) = (cos wdt) (cos Wt)x(t)
The first term is bandlimited to ± (wM + IAwl), while the second term is
bandlimited from Aw + 2wc - wM to Aw + 2wc + wM. Thus after filtering, only the
first term remains. Therefore, the output of the demodulator lowpass filter is
given by .x(t)cos AWt.
(b) Consider first IAw I > wM Then for X(w) as given, ix(t)cos Awt has a Fourier
transform as shown in Figure S13.3-1.

S13.4
(a) In this case,

We recognize that the preceding expression is a Fourier series expansion. Using
Parseval's theorem for the Fourier series, we have

(b) The power in the sidebands is found from P, when A = 0. Thus, P,y =
and the efficiency is
which is sketched in Figure S13.4-2.
Solutions to Optional Problems
S13.5

If we use an ideal lowpass filter with cutoff wc and if A(t), and thus x(t), is bandlimited
to _oc, then we recover the term x(t)/2. Thus the processing is as shown in Figure
S13.5-1.
(c) Similarly, consider
Filtering z(t) sin wct with the same filter as in part (b) yields y(t), as shown in
Figure S13.5-2.

(d) We can readily see that
Note that to be able to recover A(t) in this way, the Fourier transform of A(t)
must be zero for o > oc| I and A(t) > 0. Also note that we are implicitly
assuming that A(t) is a real signal.
S13.6
From Figures P13.6-1 to P13.6-3, we can relate the Fourier
transforms of all the signals concerned.
Thus, Si(w) and S2(O) appear as in Figure S13.6-1.

After filtering, S3(w) andS 4(w) are given as in Figure S13.6-2.

S5(U)is as follows (see Figure S13.6-3):
Note that the amplitude is reversed since (1/2j)(1/2j) = -I.
S6(w) is as follows and as shown in Figure S13.6-4.
Finally, Y(w) = S5(w) + S6(w), as shown in Figure S13.6-5.

Thus, y(t) is a single-sideband modulation of x(t).
S13.7
Note that
Using trigonometric identities, we have
Thus, if s1(t) is bandlimited to ± wo and we use the filter H(w) as given in Figure S13.7, y
1(t) will then equal s1(t).

Similarly,
Using the same filter and imposing the same restrictions on s 2(t), we obtain y 2(t) = s2
(t).
S13.8
(a) X(w) is given as in Figure S13.8-1.
For Y(w), the spectrum of the scrambled signal is as shown in Figure
S13.8-2.
Thus, X(w) is reversed for w > 0 and w < 0.

(b) Suppose we multiply x(t) by cos Wot. Denoting z(t) = x(t)cos wmt, we find that Z(w) is
composed of scaled versions of X(w) centered at ± wM. See Figure S13.8-3.
Filtering z(t) with an ideal lowpass filter with a gain of 2 yields y(t), as shown in
Figure S13.8-4.
(c) Suppose we use the same system to recover x(t). Let y(t)cos Wmt = r(t). Then
R(w) is as given in Figure S13.8-5.

Filtering with the same lowpass filter yields x(t).

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