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Chapter 1 Solutions 111. (a) E=ÊMPdt+ fitaPa=3 (b) E=-IPa+ PPA =3 (o) E= fi(Pdt+ f(-2Pdt = 12 (d) E= J(At+ JX-1at=3 “omments: Changing the sign of a signal does not change its energy. Doubling a signal quadruples its energy. Shifting a signal does not change its energy. Multiplying a signal by a constant K increases its energy by a factor K? 112. + 1 1 E 1 1 = 2dt= > =, En = —tdt= =+, E. fa á > , Ff paro 1 2 1 ae lepol [ “92 [aos IK Wdt= 5 Cho = 3» Pa = | (t-1d Pde= 5, 113. (a) 2 1 2 - 2d = 244 4 AVdi = B.= [fa 2, fas [aa 2, 1 2 Ee [ (22dt=4, Evo [ (2)2dt = 4 o 1 Therefore Er+y = Br + Ey. (b) o 1 1 E, -[ sinZtdt = > (Udt — E) cos(2t)dt= 7 +0=7 o 2 o 2 o am E, [ (1)2 dt = 27 b am | a am 2 Emo | Gintripato [ smé(gama / sin(ear+ [ (1)2dt = 1+0+2% = 37 b 6 E o In both cases (a) and (b), Es+y = Es + Ey. Similarly we can show that for both cases E, y = Es + Ey. 23 (c) As seen in part (a), E, =[ sinZtdt=7/2 o Furthermore, x m=[, dt=a o Thus, Em | Gintetat= [ sinê(gatsa singde+ / (dt = m/2+2(2) india o o o o Additionally, Esy -[ (Sint-1Pdi=m/2-44+71= Ei -4 o In this case, Es+y £ Es-y £ Es + Ey. Hence, we cannot generalize the conclu- sions observed in parts (a) and (b). LI4. P=1 [2 (1)dt = 64/7 (a) Ps = PP M-EPdt = 64/7 (b) Pa = 1 [2 (2)dt = 4(64/7) = 256/7 (0) Pes = 3 JP, (ct2dt = 6402/7 Comments: Changing the sign of a signal does not affect its power. Multiplying a signal by a constant c increases the power by a factor c2. 11-5. (a) Power of a sinusoid of amplitude C is C2/2 [Eq. (1.43)) regardless of its frequency (w 4 0) and phase. Therefore, in this case P = 52 + (10)2/2 = 75. (b) Power of a sum of sinusoids is equal to the sum of the powers [Eq. (1.45). Therefore, im this case P = L9Ê + GO — 78, (c) (10 4 2 sin 3t) cos 10t = 10cos 10t + sin 13t — sin 3t. Hence from Eq. (1.4b) = 09) 1410 P= ea +5+5=5. of the sinusoids SÊ -as. (d) 10cos St cos 10t = 5(cos 5t-+cos 15t. Hence from Eq. (1.4b) P = SÊ (e) 10sin 5tcos 10t = 5(sin 15t-sin 5t. Hence from Eq. (1.4b) P = SÊ, caí =25. (f) cit coswot = 3 [e!tetwo)t 4 ee-wot. Using the result in Prob. 1.1-5, we obtain P=(1/4)+(1/4)=1/2. 116. First, a(t) = 34 (TP pa = 340 infinite. Thus, 24 1.21 1.2-2 3. integrands are periodic signals (made up of sinusoids). These terms, when divided by T > 00, yield zero. The remaining terms (k = 7) yield xp E dm, [a DO Vou dt= DO Da T/2 p=m (d) à H 5 + 10cos(100t + 7/3) 5 + 5ei(100+5) 4 5e-I0100+5) = 54 5eim/3ç5100t 4 som im/3p-S100€ ] Hence, P=5 + [574 [pe irBp=25+25+25=75. Thought of another way, note that Do = 5, Dx) = 5 and thus P, = 52452 + 2=75. ER e(t) = 10cos(100t+ 7/3) + 16sin(150t + 1/5) 5eJ7/3/100t 4 50-57/86-5100t — jgçin/5ç5150t 4 jge-in/5p-5150t Hence, P, = [57 P use TP -g8e7/SPrljse IP = 25425+64+64 = 178. Thought of another way, note that Da; = 5 and Dx, = 8. Hence, P, = 5245248248? = 178. iii. (10 +2 sin 3t) cos 10t = 10 cos 10t + sin 13t — sin 3t. In this case, Da, = 5 : D+2 = 0.5 and Ds = 0.5. Hence, P=52 +52 +(0.5)24 (0.5)2 “+ (05)2+ (0.5)2 = 51 iv. 10cos St cos 10t = (cos | u + cos 154). In this case, Dx, = 2.5 and Dyo = 2.5. Hence, P = (2.5)2 4 (2.5)2 + (2.5)2 4 (2. 25 v. 10sin 5tcos 10t = 5(sin 15t-sin 5t). In this case, Dx = 2.5 and Do = 2.5. Hence, P = (2.52 + (2.52 + (2.5)? + (2.5)2 = 25 cit coswgt = 3 [elarwolt 4 edta-wolt In this case, Ds, = 0.5 . Hence, P=(1/22+(1/22 = 1/2 vi. . First, notice that x(t) = «?(t) and that the area of each pulse is one. Since x(t) has an infinite number of pulses, the corresponding energy must also be infinite. To compute the power, notice that N pulses requires an interval of width 2.9 2(i+1) = N2+3N As N — 09, power is computed by the ratio of area to width, or P = limy-oo yEGy = O. Thus, P=0and E =00. Refer to Figure S1.2-). Refer to Figure S1.2-2. (a) a1(t) can be formed by shifting «(t) to the left by 1 plus a time-inverted version of (t) shifted to left by 1. Thus, a(=(tED)+ra(t+D=2(t+D) +20 —0. 26 1.21 1.2-2 3. integrands are periodic signals (made up of sinusoids). These terms, when divided by T > 00, yield zero. The remaining terms (k = 7) yield xp E dm, [a DO Vou dt= DO Da T/2 p=m (d) à H 5 + 10cos(100t + 7/3) 5 + 5ei(100+5) 4 5e-I0100+5) = 54 5eim/3ç5100t 4 som im/3p-S100€ ] Hence, P=5 + [574 [pe irBp=25+25+25=75. Thought of another way, note that Do = 5, Dx) = 5 and thus P, = 52452 + 2=75. ER e(t) = 10cos(100t+ 7/3) + 16sin(150t + 1/5) 5eJ7/3/100t 4 50-57/86-5100t — jgçin/5ç5150t 4 jge-in/5p-5150t Hence, P, = [57 P use TP -g8e7/SPrljse IP = 25425+64+64 = 178. Thought of another way, note that Da; = 5 and Dx, = 8. Hence, P, = 5245248248? = 178. iii. (10 +2 sin 3t) cos 10t = 10 cos 10t + sin 13t — sin 3t. In this case, Da, = 5 : D+2 = 0.5 and Ds = 0.5. Hence, P=52 +52 +(0.5)24 (0.5)2 “+ (05)2+ (0.5)2 = 51 iv. 10cos St cos 10t = (cos | u + cos 154). In this case, Dx, = 2.5 and Dyo = 2.5. Hence, P = (2.5)2 4 (2.5)2 + (2.5)2 4 (2. 25 v. 10sin 5tcos 10t = 5(sin 15t-sin 5t). In this case, Dx = 2.5 and Do = 2.5. Hence, P = (2.52 + (2.52 + (2.5)? + (2.5)2 = 25 cit coswgt = 3 [elarwolt 4 edta-wolt In this case, Ds, = 0.5 . Hence, P=(1/22+(1/22 = 1/2 vi. . First, notice that x(t) = «?(t) and that the area of each pulse is one. Since x(t) has an infinite number of pulses, the corresponding energy must also be infinite. To compute the power, notice that N pulses requires an interval of width 2.9 2(i+1) = N2+3N As N — 09, power is computed by the ratio of area to width, or P = limy-oo yEGy = O. Thus, P=0and E =00. Refer to Figure S1.2-). Refer to Figure S1.2-2. (a) a1(t) can be formed by shifting «(t) to the left by 1 plus a time-inverted version of (t) shifted to left by 1. Thus, a(=(tED)+ra(t+D=2(t+D) +20 —0. 26 tiplying it by 4/3 to obtain dzi(t/2) = à [n(2) + 2(258)]. From this, we subtract a rectangular pedestal of height 1/3 and width 4. This is obtained by time-expanding z2(t) by 2 and multiplying it by 1/3 to yield Iza(t/2) = 1 [e(H2) + e(258)). Hence, 3 ug = 5 [Bras] su atSA (e) z5(t) is a sum of three components: (1) z2(t) time-compressed by a factor 2, (ii) x(t) left-shifted by 1.5, and (iii) a(t) time-inverted and then night shifted by 1.5 Hence, cs(t) = 2(t + 0.5) + 2(0.5 — t) + (t+ 1.5) + o(1.5 — 2). Es- ft-cor de= foco dt= Es Ecs) = [teor di= f. 22(e) do = E, Ex = [ tett-mpa= [todo = E, Exu = FÉ teca ar - f 222) de = Es a uu) = feto -opat= :F. 2a) da = Bo/a, Esujo = fteterop dt= af 2?(a) di = 0B, East) = [E lesorta =a f “(0 dt = E, Comment: Multiplying a signal by constant a increases the signal energy by a factor a2. 1.25. (a) Calling y(t) = 2u(=3t+1) = t(u(=t-1) = u(=t+1)), MATLAB is used to sketch ul). >> t = [-1.5:.001:1.5]; y = inline('t.*((t<=-1)-(t<=1))"); >> plot(t,y(t),'k-'); axis([-1.5 1.5 -1.1 1.1]); >> xlabel('t'); ylabel('2x(-3t+1)º); (b) Since y(t) = 27(=3t+1), 0.5+y(=t/3+1/3) = 0.5(2)z(=3(=t/3+1/3)+1) = a(8). MATLAB is used to sketch z(t). >> y = inline('t.*((t<=-1)-(t<=1))º); >> t [-3:.001:5); x = 0.5+y(-t/3+1/3); >> plot(t,x,'k->); axis([-3 5 -0.6 0.6]); >> xlabel(?t'); ylabel('x(t)º); " ) 1.2-6. MATLAB is used to compute each sketch. Notice that the unit step is in the exponent of the function z(t). 28 1,3-2. (a) (b ) Figure $1.2-5a: Plot of 2r(—3t + 1). Figure $1.2-5b: Plot of x(t) = 0.5y(=t/3 + 1/3) >> t = [-1:.001:1]; >> x = inline(?2."(-t.+(t>=0))"); >> plot(t,x(t),?k'); axis([-1 10 1.1)); >> xlabel(?t”); ylabel('x(t)'); >> plot(t,0.5+x(1-24t),ºk); axis([-1 10 1.1]); >> xlabel(?t?); ylabel(?y(t)'); False. Figure 1.11b is an example of a signal that is continuous-time but digital. False. Figure 1.1Ic is discrete-time but analog. False. e-* is neither an energy nor a power signal. False. e-tu(t) has infinite duration but is an energy signal. False. u(t) is a power signal that is causal. True. A periodic signal, by definition, exists for all t. True. Every bounded periodic signal is a power signal. False. Signals with bounded power are not necessarily periodic. For example, =(t) = cos(t)u(t) is non-periodic but has a bounded power of P, = 0.25 29 Pa, A JP” cost(e)dt (0.5(t + sin(t) cos(t)fi£o) dllor=i mm"? Ps 3 Jo sin?(mt)dt : 2 1 ( E(mt — sin(nt) cos(mt)) Eco) = lly=-l 22m 2 r 2 Pr = lim ap [q (cos(t) + sin(rt))? dt = limenco 55 JE, (cos2(t) + sin?(t) + cos(t) sin(xt)) dt = Po, + Protlimco de J2,.0.5 (sin(mt — t) + sin(nt + 8) dt = Pa +Pr+0= Thus, 1 Pa =Pa=5 and P=1 1.34. No, f(t) = sin(uwt) is not guaranteed to be a periodic function for an arbitrary constant w. Specifically, if w is purely imaginary then f(t) is in the form of hyperbolic sine, which is not a periodic function. For example, if w = 9 then f(t) = ysinh(t). Only when w is constrained to be real will f(t) be periodic. 135. (a) Ey = [Sovi(dt = [No dz?(2t)dt. Performing the change of variable ! = 2t —oco 5 yields [O da?(1) SE = Ex. Thus, E, 1.0417 em En = TE = = 0.0579 (b) Since ya(t) is just a (Ty, = 4)-periodic replication of z(t), the power is easily obtained as E. Es . P, = — == 0.2604 Po To 4 (e) Notice, Ty = Tu/2 = 2 Thus, Pg = Ely, MO! = 3 fr, SUe(2Odt Performing the change of variable = 2t yields Py = 3 fr, Ult) = E Jo e(t)ar = Ez. Thus, = Es = 0.0289. Pos 36 1.3-6. For all parts, y(t) = y(b)=t? over 0
t = [-3:.001:3]; mt = mod(t,2); >> y.1 = (mt<=1).*(mt.2) + (mt>1).*((mt-2).72); >> plot(t,y 1,'k?); xlabel(?t'); ylabel(?y 1(t)?); axis tight; 31 na IVAVAVI Figure $1.3-6a: Plot of qn (t) (b) Let k 1(!) is odd and (T> = 3)-periodic. The constant k is determined dy constraining the power to be unity, P =1 = 1 (x? + Elio). Solving for k yields k2 = 3 2/5 = 13/5 or k = 3/5. Thus, vBh a t = [-3:.001:3]; mt = mod(t,3); >> y.2 = (mt=2) .*((mt-3).72)+... ((mt>=1)&(mt<1.5))+sgrt(13/5)-... C(mt>=1.5)&(mt<2))+sgrt(13/5); >> plot(t,y.2,ºk'); xlabel('t”); ylabel('y 2(t)?); axis([-3 3 -2 2]); (c Define ya(t) = qn (t) + 3ya(t). To be periodic, ys(t) must equal ys(t-+ 3) for some value T3. This implies that qa (t) = qn(t + T3) and qo = ya(t+ T3). Since qn (t) is (Ty = 2)-periodie, Ty must De an integer multiple of 71. Similarly, since ya(t) is (To = 3)-periodic, T3 must be an integer multiple of To. Thus, periodicity of va(t) requires Ty = Tiky = 2k = Toky = 3ko, which is satisfied letting ky = 3 and k = 2. Thus, ya(t) is periodic with Ty = 6. Noting va(Dys (1) = vi(t) + 9308), Pos = 75 Srs (0) + 08(D) dt = Po, + Pos. Thus, (d 1 5 1.41, Refer to Figure S1.4-1 32 (a) 'The impulse is located at 7 = t and a(7) at 7 = t is e(t). Therefore oo f z(r)ó(t — 1) dr = «(0). (b) The impulse 5(7) is at 7 = 0 and a(t — 7) at 7 = 0 is a(t). Therefore [ o(r)z(t = 1) dr = ab). “o Using similar arguments, we obtain (Ja (a) 0 (e) é (1) 5 (g) z(-1) (1) -e 14-5. For sketches, refer to Figure S1.4-5. (a) Recall that the derivative of a function at the jump discontinuity is equal to en impulse of strength equal to the amount of discontinuity. Hence, dz/dt contains impulses 45(t+4) and 25(t—2). In addition, the derivative is —1 over the interval (=4, 0), and is 1 over the interval (0, 2). The derivative is zero for t < —4 and t>2. The result is sketched in Figure $1.4-5(a). (b) Using the procedure in part (a), Figure S1.4-5(b) depicts d?x/dt? for the signal in Figure P1.4-2a. 18 1 e Ml o st SO =1 0" 2 E-= (s [apa +) Figure $1.4-5 1.4-6. For sketches, refer to Figure $1.4-6. (a) Recall that the area under an impulse of strength k is k. Over the interval 0O3 0 1 3-€ 34 (b) lt) = [-sta-n-se-m-sa-9+-3 de = tu(b)-u(t-1)-u(t-2)-u(t-3)-... Figure $1.4-6 1.4-7. Changing the variable t to —z, we obtain Fsosna=- [ Cacaitoda= [” o(-a(0) de = 640) This shows that PE sooma= [Pao dt = 460) Therefore ó(t) = 6(-t) 1.4-8. Letting at = x, we obtain (for a > 0) Fesosaga = 1 f (2) de = 56(0) Similarly for a < 0, we show that this integral is —19(0). Therefore Pe Had = Toto) f * g(s(o) dt Therefore 1 ó(at) = —o(t (et) = 13510 1.4-9. Posmoma = swmor- [Tava u 0- [ótmsma = -5(0) 1.4-10. For sketches, refer to Figure S1.4-10. (a) s12 = 73 (b) e-*tcos 3t = 0.5[e-+i8)t + e-(8-33)]. Therefore the frequencies are s,2 = -3+53. (c) Using the argument in (b), we find the frequencies 5,2 =2+j3 (d) s=-2 35 (b) Ex. = [So z2(t)dt. Because etu(t) and etu(-t) are disjoint in time, the E cross-product term in z2(t) is zero. Hence, oo oo 0 Es. - [ a2(t)dt = 1 TA ettdt +[ esa] =L o alh no 8 Using a similar argument, we have Ba =L 8 Also, ad 1 E; -[ etdt=—. o 4 Hence, E, = Ex + Ex, (c) To generalize this result, we first consider causal e(t). In this case, z(t) and a(=t) are disjoint. Moreover, energy or z(t) is identical to that of z(-t). Hence, E =5 TA to(t)Pde f. iet-opal =5E. Using a similar argument, it follows that E,, = 1 E,. Hence, for causal signals, E, = Es + Es, Identical arguments hold for anti-causal signals. Thus, for anti-causal signal (t) E, = Es + Es, Now, every signal can be expressed as a sum of a causal and an anti-causal signal. Also, the signal energy is equal to the sum of energies of the causal and the anti-causal components. Hence, it follows that for a general case Es — Es + Ex, sto) = Ale(o) + 20-00 = 2(-0) = Aet)P -tet-of?) Since the areas under |r(t)2 and |z(-t)f2 are identical, it follows that f. ze(Bxo(t)dt = 0 (6) f. eddt = 5 sta + 5 Panda Because the areas under z(t) and z(-t) are identical, it follows that FÊ. ze(t)dt = FÊ. e(t)dt. 154. volt) = 0.5(2(t) — (=) = 0.5(sin(mt)u(t) — sin(>mt)Ju(=)) = 0.5sin(mt)(u(t) + u(-t)). Since sin(0) = O, this reduces to zo(t) = 0.5sin(xt), which is a (T = 2)- periodic signal. Therefore, xo(t) = 0.5sin(rt) is a periodic signal. 1.5.5. vet) = 0.5(2() + 2(=0)) = 0.5(cos(rt)u(t) + cos(=mtJu(=8)) = 0.5 cos(me)(u(t) + u(=t)). Written another way, Ze() = ( 0.5 cos(t) o TO such that e.(t+ T) = 2d), . Since there exists no xe(t) is not à periodic function. It is worth pointing out that sometimes the unit step is defined as u(t) = 1 t>0 0.5 t=0 . Using this alternate definition, ze(t) is periodic. o t<0 1.5-6. (a) Using the figure, 2(1) = (+ D)(u(t+ 1) =u() +(=t+1)(u(t)—u(t— 1)). MATLAB is used to plot v(t) = 32 (=3(t+1)) >> x = ânline( (t+1).+((t>=-1)&(t<0))+(-L+1) .+((t>=0)&(t<1))2); >> t = [-5:.001:5]; v = 3+x(-0.5+(t+1)); >> plot(t,v,'k-'); xlabel('t?); ylabelCv(t)'); GOuT a 4 A 0 1 2 3 48 Figure S1.5-6a: Plot of v(t) = 3x (=1(t + 1)). (b) Since v(t) is finite duration, P, = 0. Signal is unaffected by shifting, so v(t) is shifted to start at t = 0. By symmetry, the energy of the first half is equal to the energy of the second half. Thus, E, = 2 J; (Bt) dt = 23 Thus, E, =12and P,=0. 38 s 4 a a su a O ' t n a d Fa a 2 E? o 4 42 õ Figure S1.5-6d: Plots of v(2t + 3), v(2t) + 3, 2u(t + 3), and 2u(t) + 3. >> subplot (221); plot(t,v(axt+b),?k-'); grid; >> xlabel('t'); ylabel(ºv(at+b)'); axis(ax); >> subplot(222); plot(t,v(a+*t)+b,?k-'); grid; >> xlabel(?t?); ylabel(ºv(at)+b”); axis(ax); >> subplot(223); plot(t,atv(t+b),ºk-'); grid; >> xlabel(ºt?); ylabel('av(t+b)?); axis(ax); >> subplot (224); plot(t,arv(t)+b,'k-?); grid; >> xlabel(?t'); ylabel(av(t)+b”); axis(ax); EN EVAN vin) may E] Figure $1.5-6e: Plots of v(-3t — 2), v(=3t) — 2, —3u(t — 2), and —3u(t) — 2. 1.57. (a) Using the figure, v(t) = t(u(t) — u(t— 1)) + (u(t= 1) — u(t — 2)). MATLAB is used to plot yo(t) = “-u9 >> y = inline(?t.*((t>=0)&(t<1))+((t>=1)&(t<2))"); > t= [-5:.001:5]; y.o = (y(t)-y(-t))/2; >> plot(t,y.0,'k-?); xlabel(ºt?); ylabel('y.o(t)'); axis([-5 5 -.6 .6]); 40 E) Figure S1.5-7a: Plot of yo(t) = “U-ulco, Thus, -1/2 -2> y = inline('t.*((t>=0)&(t<1))+((t>=1)&(t<2))"); >> t = [-8:.001:0]; x = 5+y(-0.5+*t-1.5); >> plot(t,x,'k-'); xlabel('t'); ylabel('x(t)'); axis([-8 0 -.5 5.5)); Figure $1.5-7b: Plot of a(t) = 5y(-0.5t — 1.5) Thus, 5 -T
