14 Demonstration of
Amplitude Modulation
Solutions to
Recommended Problems
S14.1
(a) We see in Figure S14.1-1 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.
S14-1
Signals and Systems
S14-2
2K
A
0
.2
.4
.6
.8
Figure S14.1-2
(c) 2K = 2, K = 1, A = 0.5, so K/A = 2, and the signal has 200% modulation.
2
2K
T-0
(till 11 zm
irri~irur~v
UW
Figure S14.1-3
"V
TU
Demonstration of Amplitude Modulation / Solutions
S14-3
S14.2
(a) (i)
y(t)
32-1-10
12
Figure S14.2-1
(ii)
J, (t0
/1
3\
t
0--
1
2
3
4
5
6
-2
-3 Figure S14.2-2
Signals and Systems
S14-4
(iii)
(b) (i)
Y(w) =
=
=
y(t)e -j"' dt
x (-t) e j' tdt,
x(t')e -jw"'
2
t'
=
2'
,
dt' = 1
dt'
= 2 X(2w)
Therefore, Y(w) is a compressed version of X(w). See Figure S14.2-4.
(ii)
From the convolution theorem,
Y(W) = 27r-
-.
X(Q)H(w - Q)dQ,
Demonstration of Amplitude Modulation / Solutions
S14-5
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.
H(w)
iT
7T
-7T
7T
Figure S14.2-5
Y(W)
2
37r
-n
3n
_7
2
2
2
2
Figure S14.2-6
(iii)
Y(W)
=
X(Q)P(w - Q)dQ
--
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.)
P(w)
4
-31r
-5 7
I
-7r
i
3
T
Figure S14.2-7
S5.
Signals and Systems
S14-6
Therefore Y(w) is as shown in Figure S14.2-8.
S14.3
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
ii
i
iii
vi
v
iv
vii
x
ix
(j) viii
S14.4
(a) We are considering
N-I
X(Q) =
Z x[n]e'-j,
n=O
which is effectively the Fourier transform of a signal of infinite duration multiplied by a window of length N:
X(Q)
cos wonT(u[n] - u[n - N])e -jn
=
From the convolution theorem we can compute the Fourier transform of the
product of these two sequences:
cos wonT -
u[n]-u[n-]
x[b(Q - woT) + b(O + woT)],
9r
1 - e -jfN
1 - e-
ei-j(N-1)/2
sin N/2
sin 0/2
-w < 0 < i
Demonstration of Amplitude Modulation / Solutions
S14-7
Therefore,
X(
=
2
e -j(--woT)(N-1)/2 sin[N(Q - woT)/21 + 1 e
sin[(Q - wOT)/21
2
-j(g+w 0T)(N-1)/2
sin[N(O + woT)/2]
sin[(Q + woT)/2]
as shown in Figure S14.4-1. (Note that the spectrum is periodic with period
27r.)
IX(G2)I
-o
0
0
T
007
7
Figure S14.4-1
N-I
(b)
x[ne -j"k*
X(Qk) =
n=O
cos wonTe -j(2rk/N)n
X (2;k
N)
N-1
n=O
1 jwonTe
(i)
2~
2
S(
2
N-1
-j(2ik/N)n
j(oT-
1
-
ej(woT-
-JwonTe -j(2ik/N)n
1 j( ~w1oT-
2k/N)N)
2wk/N)
2 1
2k/N)N
2k/N)
-ej(-woT-
For woT = 27r(2) and N = 5, the first term is zero for
k =. . . -3,
2, 7, . . .
However, when k = 2 we have the ratio of
2
1 ei 21(2 / 5 -k/ 5 ) 5
1 e j21(2/5--k/5)
0
0
Signals and Systems
S14-8
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.
X( 27rk)
5
2.5 --
k
0
1
2
3
4
Figure S14.4-2
Note that X(27rk/5) is periodic in k with period 5 since X(Q) is periodic in
9 with period 27r.
(
( 27k)
(i)N )
1 (1 - e j(WOT 2wk/N)N
2 1 - e itw0T-21rk/N)
(
2
-
eji(w0T -2xk/N)N
j(- w0T- 2xk/N)
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.
IX(k)|
3--
k
0
1
2
3
Figure S14.4-3
4
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Resource: Signals and Systems
Professor Alan V. Oppenheim
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