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