13 Continuous-Time Modulation

Transcription

1 3 Continuous-Time Modulation Solutions to Recommended Problems S3. (a) By the shifting property, x(t)ejawc t - X(w - 3we) = Y(w) The magnitude and phase of Y(w) are given in Figure S3.-. IY(W)I 4Y(W) w.. 3wc 4wc 4w. w~ 3w~ Figure S3.- (b) Since ej 3 wc+ji er/e 3 wt, we are modulating the same carrier as in part (a) except that we multiply the result by eij/. Thus Y(w) = ej' X(w - 3c) Note in Figure S3.- that the magnitude of Y(w) is unaffected and that the phase is shifted by ir/. Y(w) I w() 4 Y(w) 7r - wc Figure S3.- 4wc (c) Since ej3wct cos 3 xeot = + ' g -j3wct S3-

2 Signals and Systems S3- we can think of modulation by cos 3wet as the sum of modulation by Sj3eet, -j3wc t and Thus, the magnitude and phase of Y(o) are as shown in Figure S3.-3. Note the scaling in the magnitude. jy(w)i -3wc 3ic 4 Y(w) -3w 3w c Figure S3.-3 (d) We can think of modulation by sin 3wet as the sum of modulation by e -j3wct -fr/ ej3wct -jr/ and Thus, the magnitude and phase of Y(w) are as given in Figure S3.-4. Note the scaling by in the magnitude. IY(w)I -3oic 4 Y(W) 3oc -4c -wc c 4 Figure S3.-4 (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

3 Continuous-Time Modulation / Solutions S3-3 that cos(a - #) = cos a cos # + sin a sin # Let a = 3wet and # = 7r/ 4. Then cos (3xet - = (cos 3cet + sin 3wet) Thus cos 3Uet + sin 3wet = \/ cos (3wet -r Now we write c(t) as e 3 wct - ("!) + e j3wt -(r/4)j Modulating by each exponential separately and then adding yields the magnitude and phase given in Figure S3.-5. (Note the scaling in the magnitude.) Y(W) -3co~ -3wc 4 Y(W) 3co~ 3 Ct wc Figure S3.-5 S3. In Figure S3.- we redraw the system with some auxiliary signals labeled. r, t) r(t) ir3(t) r4(t) x(t) X X y(t) s(t) m(t) d(t) Figure S3.-

4 Signals and Systems S3-4 By the modulation property, Ri(w), the Fourier transform of ri(t), is R(w) = - [X(w) * S(w)] 7r Since S(w) is composed of impulses, Ri(w) is a repetition of X(w) centered at -c, 0, and we, and scaled by /(7r). See Figure S3.-. R (w) ir Figure S3.- (a) Since m(t) = d(t) =, 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 S3.-3, we see that y(t) = /(7r)x(t), which is nonzero. ir Y(W) Figure S3.-3 WCo (b) Modulating ri(t) by ei-c' yields R (w - we) as shown in Figure S3.-4. R (o - 7 wc) S. Figure S3.-4

5 Continuous-Time Modulation / Solutions S3-5 Similarly, modulating by e ~j'wyields Ri(o + coc) as shown in Figure S3.-5. R I(w + wc) n -wc -c Figure S3.-5 Since cos wot = (ewct + e -iwct)/, modulating ri(t) by cos wct yields a Fourier transform of r (t) given by Ri(w - wc) + Ri(w + wo) Thus, R (W)is as given in Figure S3.-6. R (CJ) -t -wc II co -Fc -- oc Wc 3c. Figure S3.-6 After filtering, R3(o) is given as in Figure S3.-7. R 3 (W) -- c Wc Figure S3.-7

6 Signals and Systems S3-6 R 4 (w) is given by shifting R 3 (W)up and down by we and dividing by. See Figure S3.-8. R 4 (W) 4w -c -wc Wc.c Figure S3.-8 After filtering, Y(w) is as shown in Figure S3.-9. Y(W) 4w Figure S3.-9 Comparing Y(w) and X(w) yields y(t) = -- x(t) 4-r (c) Since then sin wet = which is drawn in Figure S3.-0. g jwct _ -jwct R (w) = Ri(w - oc) - Ri(w + we) j j R(W) I/ -4I 4w Figure S3.-0 After filtering, R 3 (x) = 0. Therefore, y(t) = 0.

7 Continuous-Time Modulation / Solutions S3-7 (d) In this case, it is not necessary to know r 3 (t) exactly. Suppose r 3 (t) is nonzero, with RA(W) given as in Figure S3.-. After modulating by d(t) = cos cot, R 4 (W) is given as in Figure S3.-. R 4 (w) -wc -~oc c o Figure S3.- After filtering, y(t) = 0 since R 4 (w) has no energy from -we, (e) For this part, let us calculate R (w) explicitly. R(W) = RI(w - wc) + R,(w + wc) n to we,. which is drawn in Figure S3.-3. R (W) 4ff -wc Wc Wc wc Figure S3.-3

8 Signals and Systems S3-8 After filtering, R 3 (w) is as shown in Figure S3.-4. R 3 (w) 47r -coc Wc Figure S3.-4 Modulating again yields R 4 (w) as shown in Figure S3.-5. R 4 (W) c S c Figure S3.-5 Finally, filtering R 4 (w) S3.-6. gives the Fourier transform of y(t), shown in Figure Y(o) 8ir Figure S3.-6 Thus, y(t) = -x(t) 87r S3.3 (a) The demodulator signal w(t) is related to x(t) via W(t) = (cos wdt) (cos Wt)x(t)

9 Continuous-Time Modulation / Solutions S3-9 Since cos A cos B = [cos(a - B) + cos(a + B)], w(t) = i[cos(aw)t + cos(aw + we)t]x(t) - i[cos(a)t]x(t) + -[cos(aw + wc)t]x(t) The first term is bandlimited to ± (wm + IAwl), while the second term is bandlimited from Aw + wc - wm to Aw + wc + 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 S3.3-. _ I -lawi Figure S3.3- Iol For IAwl <wm, there is some overlap. See Figure S3.3-. M O <IAWl<wJ M M O<IAWK. TM 4 -awi lwl -JAWI AW Figure S3.3- S3.4 (a) In this case, But y(t) = [A + cos wl)mt cos(wct + Oc) Thus, Cos omt cos(wct + O') = [cos((wm ~ cjt ~ Oc)+ cos((wm + Oc)t + cj] y(t) = A cos(wct + Oc) + cos((wm w)t O + cos((wm + W')t + Oc = eiwct + e -Jwct + -e ijcej(wm- C)t 4 + ejoce -i(wm-wc)t + ejoce(wm+)dt +! e -joce -j(wm±(cc) t 4 4 4

10 Signals and Systems S3-0 We recognize that the preceding expression is a Fourier series expansion. Using Parseval's theorem for the Fourier series, we have Thus, Since then as shown in Figure S3.4-. TO k= A P,=(A) + 4 max Ix(t) A A 9 M 4, P y 3 4M overmodulation Figure S3.4- (b) The power in the sidebands is found from P, when A = 0. Thus, P,y = efficiency is and the E= = /(M ) + } + M > which is sketched in Figure S3.4-.

11 Continuous-Time Modulation / Solutions S3- Solutions to Optional Problems S3.5 (a) Using the identity for cos(a + B), we have Thus, we see that Therefore, A(t)cos(oct + Oc) = A(t) (cos 0e cos oct - sin 0c sin oct) x(t) = A(t) cos 0c, y(t) = -A(t) sin 0, z(t) = A(t)cos(oct + 0c) = x(t)cos wct + y(t)sin wot (b) Consider modulating z(t) by cos oc. Then z(t)cos wot = x(t)cos wet + y(t)sin oct cos wct Using trigonometric identities, we have z(t)cos oct = + cos et + sin wet 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)/. Thus the processing is as shown in Figure S3.5-. z (t) X: x (t) -WC C Cos coct (c) Similarly, consider Figure S3.5- z(t)sin oct = x(t)cos oct sin wct + y(t)sinswct = sin wet + y) y) cos wct Filtering z(t) sin wct with the same filter as in part (b) yields y(t), as shown in Figure S3.5-.

12 Signals and Systems S3- z(t) rx eyt sin coct Figure S3.5- (d) We can readily see that x (t) + y (t) = A (t) (cos 0e + sin0') = A (t) Therefore, A(t) = x (t) + y (t). The block diagram in Figure S3.5-3 summarizes how to recover A(t) from z(t). z(t) A (t) sin ct Figure S3.5-3 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. S3.6 From Figures P3.6- to P3.6-3, we can relate the Fourier transforms of all the signals concerned. SI(M) = - [X (W 00-- X( j ) S(W) = [ Xx ' -O + X o Thus, Si(w) and S (O) appear as in Figure S3.6-.

13 Continuous-Time Modulation / Solutions S3-3 SI (w) S(w) 0 0 Figure S3.6- After filtering, S 3 (w) ands 4 (w) are given as in Figure S3.6-. S 3 (w) S 4 (Co) S 5 (U)is as follows (see Figure S3.6-3): Figure S3.6- S 5 (W)= [S3 -) ( W - We - S3 (W + W,+ Note that the amplitude is reversed since (/j)(/j) = -I.

14 Signals and Systems S3-4 Ss (C) 4 -- oc c 4 Figure S3.6-3 S 6 (w) is as follows and as shown in Figure S S 6 (w)= S 4 -cw- + S4 x + We + )] Finally, Y(w) = S 5 (w) + S 6 (w), as shown in Figure S Y(W) - ~-oc I CO oc Figure S3.6-5 Thus, y(t) is a single-sideband modulation of x(t). S3.7 Note that Using trigonometric identities, we have qi(t) = [si(t)cos wot + s (t)sin wot]cos wot = s (t)cos xot + s (t)sin wot cos wot qi(t) = isi(t) + isi(t)cos wet + is (t)sin wot Thus, if s (t) is bandlimited to ± wo and we use the filter H(w) as given in Figure S3.7, y (t) will then equal s (t).

15 Continuous-Time Modulation / Solutions S3-5 H(w) Figure S3.7 Similarly, q (t) = si(t)cos wet sin wot + s (t)sin Wot SIM sin w t + s (t) s (t) wot Using the same filter and imposing the same restrictions on s (t), we obtain y (t) = s (t). S3.8 (a) X(w) is given as in Figure S3.8-. X(w) - -ojm WM Figure S3.8- For Y(w), the spectrum of the scrambled signal is as shown in Figure S3.8-. Y(W) Figure S3.8- Thus, X(w) is reversed for w > 0 and w < 0.

16 Signals and Systems S3-6 (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 S Z(W) -wm M WM WM Figure S3.8-3 Filtering z(t) with an ideal lowpass filter with a gain of yields y(t), as shown in Figure S x (t) M X. iy(t) Cos WMt _M - IjMJ Figure S3.8-4 (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 S R(o) WM WM -wm -- "M ljm wm Figure S3.8-5 Filtering with the same lowpass filter yields x(t).

17 MIT OpenCourseWare Resource: Signals and Systems Professor Alan V. Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our Terms of Use, visit: