Baixe Solução Lathi 2a ed - Sistemas Lineares - chapter 01 e outras Resumos em PDF para Engenharia Elétrica, somente na Docsity! Chapter 1 Solutions 111. (a) E=POPdt+ fiaPat=3 () E=i-Ia+ PAPA =3 (o) E= fi(Pat+ [(-2Pdt = 12 (d) E= JR(At+ JM-1)at=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= = = 0, Es = —tdt = =+, E, fa á > , fo paro E 1 2 1 a lepol f “9 [oao=s fe Wdt= 5 Cho= 3» Pa = | (1d Pde= 5, 113. (a) 2 1 2 - 2d = 20 4 AVdi = B= [fa 2, fas [ram 2, 1 2 Eu [ (2Pdt=4, Esy= [ (2)2dt = 4 o 1 Therefore Er+y = Er + Ey. (b) a 1 1 E; =[ sinZtdt => (Udt — E) cos(2t)dt= 7 +0=7 o 2 o 2 o a E, f (1)2 dt = 27 h am | ao, à a Emo | Gintripato [ smé(gama / sin(iar+ [ (1)2dt = 1+0+2% = 37 À h A b In both cases (a) and (b), Es+y = Es + Ey. Similarly we can show that for both cases E, y = Er + Ey. 23 (c) As seen in part (a), E,= [ simZtdt =7/2 o Furthermore, x m=[ dt=a o Thus, Exey -[ (sint+1)2 dt =[ sinê(gatea sin(gde+ / (dt = m/2+2(2) tn ÉL 44 o o o o Additionally, . 2 37 Esy= | (Sint-1di=a/2-44n=0-4 o In this case, Ersy £ Es-y £ Es + E, Hence, we cannot gencralize the conclu- sions observed in parts (a) and (b). LI-4. P=1 [2 (1)dt = 64/7 (a) Ps = 4 PME dt =64/7 (b) Pa = 1 [2 (20)dt = 4(64/7) = 256/7 2 (ct Pat = 64c2/7 (0) Pr =1 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. 1.15. (a) Power of a sinusoid of amplitude C is C2/2 [Eq. (1.43)) regardless of its frequency (w £ 0) and phase. Therefore, in this case P = 52 + (10)2/2 = 75. of the sinusoids (b) Power of a sum of sinusoids is equal to the sum of the powers [Ea. (1.45). Therefore, in this case P = L9Ê + GO — 78, (c) (10 + 2 sin 3t) cos 10t = 10cos 10t + sin 13t — sin 3t. Hence from Eq. (L.4b) P= 8, 141=5]. d) 10cos 5t cos 10t = 5(cos 5t+cos 15. Hence from Eq. (1.4b) P = SÊ + SÊ = 95, ( q 2 (e) 10sin 5tcos 10t = 5(sin 15t-sin 5t. Hence from Eq. (1.4b) P = SÊ, CÊ =925. (f) cit coswot = 3 [e!etwolt 4 ee-wo. Using the result in Prob. 1.1-5, we obtain P=(1/4)+(1/4)=1/2. 2 o<t<T 11-6. First, z(t) = T<t<T Next, P. = + = vt 4a? sA2 7º TP pm — 34 ho Pat= SA infinite. Thus, 2 E 36) = 4. Since power is finite, energy must be 2 p= and E, = 00. 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 rp n dim, (a DO Vou dt= Do Da T/2 p=m (db) à NH 5 + 10cos(100t + 7/3) 5 + 5ei(100+5) 4 5e-i(100+5) = 5 + 5eim/3ç5100t 4 som im/3p-S100€ 1 Hence, P=5 + [574 [pe irBp =25425+25=75. Thought of another way, note that Do = 5, Dx, = 5 and thus P, = 52452 + 2=7 5. H a(t) = 10cos(100t+ 7/3) + 16sin(150t + 1/5) 5ei7/36/100t 4 50-57/86-5100t — jgçin/5ç5150t 4 jge-in/5p-5150t Hence, P, = [5 P use ir Pa |-g8e7/SPlj8e IP = 25425+64+64 = 178. Thought of another way, note that Day = 5 and Dx, = 8. Hence, P, = 5245248248? = 178. iii. (10 +2 sin 3t) cos 10t = 10 cos 10t + sin 13 — sin 3t. In this case, Da, = 5 : D+2 = 0.5 and Dss = 0.5. Hence, P=52452+(0.5)24 (0.5)2 + (05)2+ (0.5)2 = 51 iv. 10cos Stcos 10t = (cos | a + 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 Ds = 2.5. Hence, P = (2.52 + (2.52 + (2.5)? + (2.5)2 = 25 eictcoswgt = 3 [east 4 eda-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 yEG = O. Thus, P=0and E=00. Refer to Figure S1.2-1. Refer to Figure S1.2-2. (a) a1(t) can be formed by shifting 2(t) to the left by 1 plus a time-inverted version of x(t) shifted to left by 1. Thus, ab) = (ED) ra t+D=2(t+D+20—0. 26 Ae 1 z â À aértreo a Xt3t) + X(&) A] 7 tt» + | 24 30 34 48 +t> -á Figure $1.2-1 a -2 z | 4 bb GESTO SI -2 6 4 4> al xezta) 4- XD 2 2 . o| 23 t= o. ÇG t+ Figure $1.2-2 (b) z2(t) can be formed by time-expanding (t) by factor 2 to obtain 2(t/2) Now, lefi-shift 2(t/2) by unity to obtain «(“$2). We now add to this a time-inverted version of 2(15!) to obtain za(t). Thus, (e (c) Observe that xa(t) is composed of two parts: First, a rectangular pulse to form the base is constructed by time-expanding za(t) by a factor of 2. This is obtained by replacing t with t/2 in 22(t). Thus, we obtain aa(t/2) = a(82) + (25º) Second, the two triangles on top of the rectangular base are constructed by time- expanded (factor of 2) and shifted versions of z(t) according to z(t/2) + z(=1/2) Thus, est) = (EB) sa? TS) + 2(t/2) + a(-1/2) (d) za(t) can be obtained by time-expanding 21(t) by a factor 2 and then mul- 27 tiplying it by 4/3 to obtain dai(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, ao [055 +a (e) z5(t) is a sum of three components: (1) z2(t) time-compressed by a factor 2, (ii) a(t) left-shifted by 1.5, and (iii) a(t) time-inverted and then right shifted by 1.5 Hence, cs(t) = 2(t+ 0.5) + 2(0.5 — t) + (t+ 1.5) + a(1.5 — 2). Es- ftp dt= Fem dt=Es Ecs) o di= f. 22(2) do = E, Exen= [ tett-mpa= [" eto)do= E Exu= FE teca as - Ef 222) de = Exa Eua) = feto -opdt= :F. 2a) da = Boa, Esujo = Ftevrop di= af 2?(a) dt = 0B, East) = [testa =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(ºt'); 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) = «(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?); ylabelC'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 Pa, A JP” cost(e)dt & (0.5(t + sin(t) cos(t)fi£o) dlor=i =" 2 Pro 3 Jo sin?(mt)dt : 2 1 ( &(mt — sin(nt) cos(mt)) [a] = lly-l 22m 2 r 2 Pr = lim ap [q (cos(t) + sin(mt))? dt = limenco 55 JE, (cos2(t) + sin?(t) + cos(t) sin(xt)) dt = Po, + Protlimco de J2,.0.5 (sin(nt — t) + sin(nt + 8) dt = Pa +Pr+0= Thus, 1 Pa = Pa = 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) = sinh(t). Only when w is constrained to be real will f(t) be periodic. 135. (a) Ey = [Sovi(dt = [No dv?(2t)dt. Performing the change of variable ! = 2t —oco 5 yields [O da?(1) SE = Ez. Thus, E, 1.0417 em En= TE => = 00579 (b) Since ya(t) is just a (Ty, = 4)-periodic replication of 2(t), the power is easily obtained as E. E; . P, = — == 0.2604 “To 4 (e) Notice, Ty = Tu/2 = 2 Thus, Pg = Ely, MO! = 3 fr, Sue(2Od Performing the change of variable = 2t yields Py = 3 fr, Aut) = 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<1. (a) To ensure yi(t) is even, yi(t) = over -1 <t <0. Since yn(t) is (Ti = 2)- a DA periodic, q1(t) = gn(t + 2) for all t. Thus, qn(t) = ( nte +2) a : Past Ji(tPa = 055 =1/5. Thus, é Pu =1/5 A sketch of 3 (t) over 3 <t <3 is created using MATLAB. > 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)º); axis tight; 31 na | | U Figure $1.3-6a: Plot of qn (t) (b) Let k 1<t<15 t? 0<t<l vO= mm) o w va(t+3) vt With this form, y2(!) is odd and (T> = 3)-periodic. The constant k is determined by constraining the power to be unity, P = 1 = 1 (x? + Elio). Solving for k yields k2 = 3 2/5 = 13/5 0r k = 3/5. Thus, vB5 a<t<15 nt) = t O<t<1 ? =ya(=t) vt va(t+ 3) vt A sketch of ya(t) over -3 <t < 3 is created using MATLAB. > t = [-3:.001:3]; mt = mod(t,3); >> y.2 = (mt<t).+(mt."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 periodie, ys(t) must equal ys(t-+ 3) for some value T. This implies that qa (t) = qa(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 = 3kp, which is satisfied letting ky = 3 and kp = 2. Thus, ya(t) is periodic with Ty = 6. Noting va(Dys(1) = vi(t) + 9300), Pos = 55 Jrs (HO) + 08(D) dt = Po, + Poa. Thus, (d 1.41, Refer to Figure S1.4-1 32 Figure $1.3-6b: Plot of y>(t) z 5 Â (ao í e o! 5 7 t> 2 (d) o t> 2 Figure S1.4-1 14.2. alt) = (4t+Dfu(t+1) — u(t)) + (-2t+ 4)fu(t) — u(t — 2)) = (A+ Du(t+1) — Gtu(t) + 3u(t) + (24 — 4)u(t — 2) z(t) = Plut)-u(t-2)] + (2t— B)fu(t— 2) — u(t — 4) = Put) -(P-2t+8)u(t-2)-(2t- B)u(t— 4) 1.4-3. Using the fact that J(z)ó(x) = f(0)ó(x), we have (a) 0 (b) 55(0) (e) 35(8) (d) —56(t-1) (e) sigó(v +3) (E) kó(w) (use Lº Hôpital's rule) 1.44. Im these problems remember that impulse ó(z) is located at 2 = 0. Thus, an impulse ó(t — 7) is located at 7 = t, and so on. 33 () s=2 (1) 5 = 5eºt so that s = 0. fe *-|53 J3 à x A, o. -3 [o o| 2 -a |O 2 pos bla afã a | 0 Figure S1.4-10 1.5-1. For sketches, refer to Figure S1.5-1. (a) xe(t) = 0.5[u(t) + u(=8)] = 0.5 and co(t) = 0.5fu(t) — u(= 0]. (b) 2e(t) = 0.5[tu(t) — tu(=0)] = 0.5]t] and xo(t) = O.5ftu(t) + tu(=)] = 0.5€. (0) ze(t) = 05[sinwot+sin(=wot)] = O and zo(t) = 0.5[sinwot = sin(=wot)] = sincoot. (d) xe(t) = 0.5[coswot + cos(=wot)] = coswot and xo(t) = 0.5[coswot — cos(=wot)] = 0. (e) cos(wot +89) = coswpt cos O — sin wpt sin O. Hence ze(t) = coswpt cos O and co(t) = = sinwotsin O O.Ssinwot (8) xe(t) = 0.5[sinwotu(t) + sin(=wot)u(=)] = 0.5[sinwot u(t) — sinwotu(=t)] and zolt) = 0.5[sin wot u(t) — sin(=wot)u(=)] = 0.5[sinwotu(t) + sinwotu(=t)) = (8) zelt) = 0.5fcoswotult) + cos(=wot)u(-t)] = 0.5fcoswotu(t) + coswotu(=)] = 0.5coswot and zo(t) = 0.5[coswotu(t) — cos(=wot)u(—t)] = O.5[coswotu(t) — coswotu(-b)]. Xo (95 ) (45 X, x, — huge ya % ro 6 11X Ao o — 30 na4s e = dy (E) x (23x x 1.5-2. (a) 36 (b) Ex. = [So z2(t)dt. Because etu(t) and etu(-t) are disjoint in time, the 2 cross-product term in z2(t) is zero. Hence, oo oo 0 Es. -[ a2(t)dt = 1 / ettdt +[ esa] =L o alh no 8 Using a similar argument, we have Bo =L “8 Also, o 1 E; -[ etidt=—. » 4 Hence, E, = Es + 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 [ te(t)Pde +f. iet-opal =2B. 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, seo) = Ale) + 2(-0]( = 2(-0) = Aet)P -tet-of?) Since the areas under |r(t)]2 and |z(-t)]2 are identical, it follows that F ze xo(t)dt = 0 (6) f. edtdt = 5) sta + 5 Piso Because the areas under z(t) and z(-t) are identical, it follows that Ff. 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) = 0, this reduces to xo(t) = 0.5sin(xt), which is a (T = 2)- periodic signal. Therefore, £o(t) = 0.5sin(rt) is a periodic signal. 1.55. 2e(1) = 0.5(2(0) + 2(=0)) = 0.5(cos(me)u(t) + cos(=mt)Ju(=8)) = 0.5cos(mt)(u(t) + u(=8)). Written another way, Ze() = ( 0.5 cos(t) o TO such that ze(t+ T) = ze(t), - Since there exists no xe(t) is not a 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. 0 t<0 1.5-6. (a) Using the figure, 2(t) = (+ 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))?); >> t = [-5:.001:5]; v = 3+x(-0.5+(t+1)); >> plot(t,v,'k-'); xlabel('t?); ylabelCv(t)'); “o “O a 4 A 0 1 2 34 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 Figure S1.5-7a: Plot of yo(t) = “U-ulco, Thus, -1/2 -2<t< Sar a<e<a v(D=4 172 át<2 0 otherwise (b) Since u(t) = .2a(—2t— 3), 5y(—0. 5t—1.5) = 5(0.2)x(—2(—0. 5t—1.5) -3)= a(t) MATLAB is used to sketch c(t). >> 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<t<-5 a()=4 -5(t+3)/2 -D<t<-3 0 otherwise 1.5-8. Let the graphed signal be named y(t). 41 (a) (5) Since y(t) = —0.52(-3t+2), —2y(-t/3+2/3) = —2(-0.5)u(-3(>t/3+2/3)+2) = a(t). MATLAB is used to sketch v(t) >> y = inline(? ((t>=-1)&(t<0))+(-t+1) .+(Ct>=0)&(t<1)) 7); >> t = [-2:.001:6]; x = -2+y(-t/3+2/3); >> plot(t,x,'k-?); xlabel('t?); ylabel('x(t)); axis([-2 6 -2.5 0.5)); os «os Figure S1.5 : Plot of z(t) = —2y(-t/3+ 2/3) Thus, -2(t+1)/3 I<t<2 at) = -2 2<t<5 0 otherwise The even portion of a(t) is xe(t) = 0.5(2(1) + 2(-8)). >> x = inline(?-2*(t+1)/3.*((t>=-1)&(t<2))-2*((t>=2)&(1<5))º); >> t = [-6:.001:6]; x e = 0.5+*(x(t)+x(-t)); >> plot(t,x.e,'k-'); xlabel(?t'); ylabel('x.e(t)?); axis([-6 6 -1.5 0.5]); [o Figure $1.5-8b: Plot of ze(t) Thus, a a<tkt<s =) lt-D/3 1<li<2 Te(t) = -2/3 kej<1 0 otherwise (c) The odd portion of z(t) is xo(t) = 0.5(2(t) — a(=1)). >> x = inline(?-2+(t+1)/3.*((t>=-1)&(t<2))-2+((t>=2)&(t<5))?); >> t = [-6:.001:6]; x. 0 = 0.5*(x(t)-x(-t)); >> plot(t,x.0,'k-?); xlabel('t'); ylabel(?x.e(t)'); axis([-6 6 -1.5 1.5]); & | Figure $1.5-8c: Plot of xo(t). Thus, 1 —-b<t<-2 (-t+1)/3 -2<t<— eott) = 2/3 -1<t<1 MUSA (+13 I<t<2 a 2<t<5 0 otherwise 1.5-9. Notice, wi(=t) = 0.5(w(=t) + w"(0))" = 0.5(w*(=t) + w(t)) = wcs(t). In Cartesian form, this becomes wji(—-t) = a(=t) — y(-t) = x(t) + gu(t) = wes(t). Equating the real portions yields z(-t) = z(t), and equating the imaginary portions yields t) = y(t). Thus, by definition, the real portion of w.s(t) is even and the imaginary portion of wes(t) is odd. 1.5-10. Notice, —wz(=t) = —0.5(w(=t) — w'(t)* = 0.5(-w'(=t) + w(b) = wca(t). In Cartesian form, this becomes —wt,(—t) (8) + gu(=t) = a(t) 4 3u(t) = wea(t). Equating the real portions yields —z(=t) = z(t), and equating the imaginary portions yields y(-t) = y(t). Thus, by definition, the real portion of wca(t) is odd and the imaginary portion of wca(t) is even. 15-11. Complex signal tw(t) is defined over (0 << 1). (a) Assigning certain properties to w(t) allows us to plot w(t) over (-1 <t<1) à TE o(t) is even, w(t) = w(=t), it is even in both the real and imaginary components. Thus, the graph folds back on itself and appears unchanged. 43 and . o . . vo=[ alojar = [ ardr+ [ atér= 40) +f a(1)dr zeroinput Se Put ero-state 1.6-2. From Newton's law a o D=M— (= and vo) = Tl. a(r)dr = z/ d+ alodr = O) + | atra 1.7-1. Only (b), (f), and (h) are linear. Al the remaining are nonlinear. This can be verified by using the procedure discussed in Example 1.9. 1.7-2. (a) The system is time-invariant because the input <(t) yields the output y(t) = e(t— 2). Hence, if the input is e(t — T), the outputis a(t-T-2) = y(t-T), which makes the system time-invariant. The system is time-varying. The input z(t) yields the output v(t) = a(=8). Thus, the output is obtained by changing the sign of t in (t). Therefore, when the input is z(t — T), the output is a(=t— T) = =(—[t+ T]) = y(t + T), which represents the original output advanced by T (not delayed by T). (c) The system is time-varying. The input z(t) yields the output v(t) = z(at), which is a scaled version of the input. Thus, the output is obtained by replacing t in the input with at. Thus, if the input is z(t — T) (z(t) delayed by T), the output is e(at — T) = a(aft — £]), which is «(at) delayed by T/a (not T). Hence the system is time-varying. (d) (d) The system is time-varying. The input z(t) yields the output y(t) = tz(t). For the input a(t- T), the output is te(t— T), which is not te(t) delayed by T. Hence the system is time-varying. L The system is time-varying. The output is a constant, given by the area under z(t) over the interval [t| < 5. Now, if z(t) is delayed by T, the output, which is the area under the delayed z(t), is another constant. But this ontput is not the same as the original output delayed by T. Hence the system is time-varying. The system is time-invariant. The input c(t) yields the output y(t), which is the square of the second derivative of z(t). If the input is delayed by T, the output is also delayed by T. Hence the system is time-invariant. 1.7-3. We construct the table below from the first three rows of data. Let r; denote the jth row. a(t) qn(0) qr(o) v(t) o 1 1 ETult) o 2 1 e(3 + 2)u(t) ut) 1 q 2u(t) 0 1 0 (t+ 1)e-tu(t) EO) 0 q Get + Du(t) re=(ra+rs) qu(t) Lo 1 (5etatet+Du(t) r=Aretm) cult) 0 0 (et+2tet + Du(t) 46 In our case, the input a(t) = u(t+ 5) — u(t — 5). From 77 and the superposition and time-invariance properties, we have vb) = m(t+5)-r(t—5) [9 + 2(r + se 49 2) u(t 5) — [e + aqi jet 42) u(e 5) 1.7-4. 1H the input is ke(1), the new output y(t) is u(t) = 6*2*(0)7 (15) =KX(0/ (E Hence the homogeneity is satisfied. If the input-output pair is denoted by x; > yi, then ) and oa = (232/(65) But mtm>(m+)/Mintio)tn+y mom = (me 1.7-5. From the hint it is clear that when vc(0) = 0, the capacitor may be removed, and the circuit behaves as shown in Figure $1.7-5. It is clearly zero-state linear. To show that it is zero-input nonlinear, consider the circuit with z(t) = O (zero-input). The current v(t) has the same direction (shown by arrow) regardless of the polarity of vc (because the input branch is a short). Thus the system is zero-input nonlinear. tata gidco + Figure $1.7-5 1.7-6. The solution is trivial. The input is a current source, which has infinite impedance. Hence, as far as the output y(t) is concerned, the circuit behaves as shown in Figure S1.7-6. The nonlinear elements are irrelevant in computing the output y(t), and the output y(t) satisfies the linearity conditions. Yet, the circuit is nonlinear because it contains nonlinear elements. o-1H À AVE + te) a 22 VO Figure $1.7-6 177. (a) y(t) = a(t— 2). Thus, the output y(t) always starts after the input by 2 seconds (see Figure S1.7-7a). Clearly, the system is causal 47 (b) y(t) = z(=t). The output g(t) is obtained by time inversion in the input. Thus, if the input starts at t = 0, the output starts before t = O (see Figure S1.7-7b). Hence, the system is not causal. (c) y(t) = z(at), a > 1. The output y(t) is obtained by time compression of the input by factor a. Hence, the output can start before the input (see Figure S1.7-7c), amd the system is not causal. (d) v(t) = e(at), a < 1. 'The output y(t) is obtained by time expansion of the input by factor 1/a. Hence, the output can start before the input (see Figure S1.7-7d), and the system is not causal. YO xt “ y6 ) E cd A, X) a e Y é > o +-> o|JTTÃ > - -T |º Es bs RC x | (43 Figure S1.7-7 1.7-8. (a) Invertible because the input can be obtained by taking the derivative of the output. Hence, the inverse system equation is y(t) = dx/dt. (d Not invertible for even values of n, because the sign information is lost. However, the system is invertible for odd values of n. The inverse system equation is u(t) = feto”. Not invertible because differentiation operation irretrievable loses the constant part of 2(t). (c (d) The system y(t) = z(3t — 6) = =(3[t — 2]) represents an operation of signal compression by factor 3, and then time delay by 2 seconds. Hence, the input can be obtained from the output by first advancing the output by 2 seconds, and then time-expanding by factor 3. Hence, the inverse system equation is v(t) = «(é + 2). Although the system is invertible, it is not realizable because it involves the operation of signal compression and signal advancing (which makes it noncausal). However, if we can accept time delay, we can realize a noncausal system. (e) Not invertible because cosine is a multiple valued function, and cos“1[z(t)] is not unique. (£) Invertible. z(t) = Iny(b) 179. (a) Yes, the system is linear. Begin assuming qa (t) = r(t)xi(t) and yo(t) = r(Bxa(t). Applying azi(t) + bzo(t) to the system yields y(t) = r(t) (ami (0) + bxa()) = ar(t)mi(t) + br(t)xa(t) = ant) + bya(t). Yes, the system is memoryless. By inspection, it is clear that the system only depends on the current input (b 1.8-1. The loop equation for the circuit is 3) + Dial) =2(8) or (D+3)n(t) =2(8) (1) Also 1 Dino) = velo) => a(o) = Sua(t) (2) Substitution of (2) in (1) yields (D+3) D w(b=a(t) or (D+3)ya(t) = Da(t) 1.8-2. The currents in the resistor, capacitor and inductor are 2y2(t), Dya(t) and (2/D)ya(t), respectively. Therefore (D+24+ Soto) ==) or (D? +2D + 9ya(t) = De(t) (1) Also ' an(t) = Dya(t) or volt) = sun(t) (2) Substituting of (2) in (1) yields D24+2D+2 5 ul) = Da(t) or (D? +2D +29) (t) = D2a(t) 1.8-3. The freebody diagram for the mass M is shown in Figure 1.8-3. From this diagram it follows that Mj=B(t-)+K(z-9) A (MD? + BD + K)y(t) = (BD + KJa(t) M -. B(k-Y) KOWY) Figure 81.8-3 1.8-4. The loop equation for the field coil is (DL; + RyJis(t) = «(t) (1) If T(t) is the torque generated, then T(t) = Kyir(t) = (JD? + BDJo(t) (2) 51 1.8-6. Substituting of (1) in (2) yields K dn= ESA = (JD? + BDJo(t) a (JD? + BDDL, + Ry)o(t) = Kyz(t) Ig:(t) — go(t)JAt = AMh or | ig) = alalt) — ao(b] 0) But Go(t) = Rh(t) (2) Differentiation of (2) yields dot) = Rb(e) = Elas) — qo(0) and A A (» + 5) go(t) = qa:(t) or A (D + a)ao(t) = aqu(t) a=5 (3) and a vo) = sq ut) substituting this in (1) yields no=5(1-55)u0= guto or , (D+ ajh(o = Tato) (a) The order of the system is zero; there are no energy storage components such as capacitors or inductors (b) Using KVL on the left loop yields a(t) = Ray (t) + Ra(gn (t) — va(t)) = 3ya(t) — 2ya(t). KVL on the middle loop yields 0 = Ro(ya(t) —3n (t)) + Raya(8) + Ra(ya(t)— va(8) = 2 (8)+9yo(t)—4ys(t). Finally, KVL on the right loop yields Ra(ys(t) — v2(t) + (Rs + Re)ya(t) = —4ya(t) + 15ya(t). Combining together yields | 3 20 | [| [| -2 9 —4 wlt) |=| 0 |. 0 -4 15 ya(t) 0 52 (c) Cramer's rule suggests 3-2 «(t) 290 0 40 v(D = 0 -2 9 4 0 4 15 MATLAB computes the denominator determinant. >> det([3 -2 0;-2 9 -4;0 -4 15]) ans = 297 The numerator determinant is easy computed by hand as 0 + 0 + Su(t) = 8e(t). Thus, 8 8 va(t) = soze(t) = 397 (2 Icos(b) ult — 1). 1.10-1. From Figure P1.8-2, we obtain c()=0/2+àh +40 Moreover, the capacitor voltage q) (t) equals the voltage across the inductor, which is io. Hence, the state equations are h=-0/2-qg-2 amd qp=20 1.10-2. The capacitor current Cgs = ás is qy — qo. Therefore da = 2h — 292 (1) The two loop equations are nthte=r=h=-2n-g+a (2) 1. . “grita =0=> do =—3qo +3q5 (3) 3 Equations (1), (2) and (3) are the state equations. For the 29 resistor: current is q1, voltage is 23. For the 1H inductor: current is qy, voltage is Gy = z(t) — 2qy — q3- For the capacitor: current is q1 — q2, voltage is qa. For the 3H inductor: current is q2, voltage is lo = —q» + q3- For the 19 resistor: current is q2 and voltage is q2. At the instant t, q = 5, qo = 1, q3 = 2 and 2 = 10. Substituting these values in the above results yields 20 resistor: current 5A, voltage 104. 1H capacitor: current 5A, voltage 10 — 10 — 2 = —2V. The capacitor: current 5 - 1=4A, voltage 2V. The 3H inductor: current 1A, voltage -1+2 = 1V. The 19) resistor: current 1A, voltage 1V. 53