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2. 14 Demonstration of Amplitude Modulation
Recommended Problems
P14.
1 Consider the AM modulation system in Figure P14.1-1.
K/A is called the modulation index, where K is the maximum amplitude of x(t).
Parts (a)-(c) contain plots of y(t) versus t for several different modulation indices,
with x(t) = B cos o0t. Find the modulation index for each signal.
Problems
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3. (a) Consider the signal x(t) in Figure P14.2-1.
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5. (b) Suppose that x(t) has the Fourier transform shown in Figure P14.2-5. Find
Y(w) for each case in part (a).
P14.3
For each of the time waveforms (a)-(j) (Figures P14.3-1 to P14.3-10), match its
possible spectrum (i)-(x) (Figures P14.3-11 to P14.3-20).
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10. P14.4
The spectrum analyzer discussed in the lecture computed the estimate of the
magnitude of the Fourier transform of x,(t) by taking samples of x,(t) at equally
spaced intervals T, stopping after N samples, and computing the discrete-time
Fourier transform of the N-point sequence.
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11. Thus,
(a) Suppose x,(t) = cos w0t. Find and sketch IX(Q) 1.
(b) In any practical system, X(Q) can be explicitly calculated only at
a finite set of Q. A common choice is
For the following situations
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12. 14 Demonstration of Amplitude Modulation
Solutions to Recommended Problems
S14.1
(a) We see in Figure S 14.1-1 that the modulating cosine wave has a peak
amplitude of 2K = 2, so that K = 1. At the point in time when the modulating
cosine wave is zero, the total signal is A = 2, so K/A = 0.5. Therefore, the signal
has 50% modulation. See Figure S14.1-1.
(b) 2K = 2, K = 1, A = 1, so K/A = 1, and the signal has 100% modulation. See
Figure S14.1-2.
Solutions
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16. where cos irt 5 H(w), and H(w) is as shown in Figure S14.2-5. Therefore, Y(w) is as
given in Figure S14.2-6.
P(w) is an impulsive spectrum, as shown in Figure S14.2-7, because the
corresponding p(t) is periodic. (Note that only odd harmonics are present.)
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17. S14-6
Therefore Y(w) is as shown in Figure S14.2-8.
S14.3
(a) ii
(b) i
(c) iii
(d) vi
(e) v
(f) iv
(g) vii
(h) x
(i) ix
(ii) (j) viii
S14.4
(a) We are considering
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18. which is effectively the Fourier transform of a signal of infinite duration multiplied
by a window of length N:
From the convolution theorem we can compute the Fourier transform of the
product of these two sequences:
Therefore,
as shown in Figure S14.4-1. (Note that the spectrum is periodic
with period 27r.)
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19. However, when k = 2 we have the ratio of
and we treat the limit as k - 0. Using L'H6pital's rule, we have i(5) = 2.5.
Similarly, the second term is zero except when k = ... -2, 3, 8, . . . . Taking the
limit yields 2.5. So X(27rk/5) is as shown in Figure S14.4-2.
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20. Note that X(27rk/5) is periodic in k with period 5 since X(Q) is periodic in 9 with
period 27r.
Now woT = 2r-3, and the numerator and denominator are nonzero for all k.
Evaluating the preceding expression yields X(k) as shown in Figure S14.4-3.
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