Baixe chapter 07 e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! Chapter 7 Solutions TAL. TI. oo oo o x / atestar f alt) cosatdt — 5 / 2(t) Sinwt dt JE <(t) is an even function of t, z(t) sinwt is an odd function of t, and the second integral vanishes. Moreover, z(t) coswt is am even function of t, and the first integral is twice the integral over the interval 0 to oo. Thus when z(t) is even oo x(t) coswt dt a) ho Similar argument shows that when s(t) is odd + X(w)= -2 [ e(t) sincot dt (2) h If a(t) is also real (in addition to being even), the integral (1) is real. Moreover from (1) - xX(-u)= :/ a(t) coswt dt = Xu) o Hence X(w) is real and even function of w. Similar arguments can be used to prove the rest of the properties. qt) = =. X (we! dy = =. |X (00) [e 2X põvt day = ã [E |X(w)I costot + “Xu do 5 [” |X (wo) sinfot + ZX (w)] do) Since |X (w)| is an even function and LX (w) is an odd function of w, the integrand in the second integral is an odd function of w, and therefore vanishes. Moreover the integrand in the first integral is an even function of w, and therefore 1 po z() = j |X (eo)] cosfiot + 2X (ww) duo (a) Because z(t) = xo(t) + ze(t) and ei“! = cos wt + jsin wt X(w) = Ff Izo(t) + rele ivtdt 324 = f Est) + coli] cosatat — 5 [” olt) + ze(t)) sinwtdt Because ze(t) coswt and xo(t) sinwt are even functions and xo(t) coswt and ze(t) sinwt are odd functions of t, these integrals [properties in Eqs. (B.43), p. 38) reduce to oo oo X(v)= 2/ met) coswt dt — of colf) sinwwt dt a) k b Akso, from the results of Prob. 7.1-1, we have oo oo Flo) =2 xe(t) coswtdt and Flvo()y = 25 [ 2o(b) sinvotdt (2) b k From Egs. (1) and (2), the desired result follows. (b) We can express u(t) in terms of its even and odd components as follows u(t) Slu(o) + u(=0] + Gfu(t) = u(-0] 1 1 = 5 +osgn(t) 2 2 Ls Te(t) Tolt) * and Xew)=76(w) — and Xou) = 5 Clearly, Xe(w) is the real part and Xo(w) is the odd part of X (wo). =Stu(t) We follow the same procedure for z(t) = Etu(t) = Me Stu(t) + e ul] 4 lotto) — e-Stu(-0] Qrtaloot zelt) zo) Also , ipa a x. > - +. (eo) sm wi Fa? and afoa 1 2ju Xodw)=> = (e, iate wipa? Clearly, X.(w) is the real part and Xo(w) is the odd part of X(w). TA. (a) T I-eiu+ar ePeivtdt = dretrlcl eita gg — jura 0 (b) T T =(dw-a)T x)= eectetar= [ ela gp= "20 + J ju-a 325 + > a| tectCE) A SB) c 1 Es Jo = . snelE)reet (ção) Cf) (e ST Eos Figure $7.2-1 723. 104 ot (104 at) = ES tdo = = festrtme — nom] 2x ho-x 2m(jw) lo, S2mo jr0t = = = [25 sin gt] = sinc(at)e710t Fm * T24. (a) Lipo 1 qu , O) = [esmo du = sã [O tio o) da 1 esto to) |? sinwo(t—-to) wo. = eo ram) O e smelelt- o) Em (b) H 1 º ut “o iojwt a(t) =|/ je? do + [ jet dy m —wo o = gu A ql? (1 coswot 2mt 2mt o mt 7.25. (a) When a > 0, we cannot find the Fourier transform of eStu(t) by setting s = ju in the Laplace transform of e“tu(t) because the ROC is Res > a, which does not include the jw-axis (b) The Laplace transform of z(t) is T T x9= [Cetera [Petosras 2 poco] Interestingly, because z(t) has a finite width, the ROC of its X(s) is the entire ju-axis. Hence, the Fourier transform s-plane, which include: Xu) = X(5) lacju = [1 e-ts-or] 328 7.31. To verify this, we find the Fourier transform of z(t) T T xuj= [ eteimao [Cestas Lo fo seen] o o JW a Which agrees with X(juw) (a) 1 u(t) > mô(w) + — (8) (tw) jo =(8) X(w) Application of duality property yields 1 nó(t) + > > 27u(-w) Jt O 2me(-w) x(8 or 5 [o+ al + u(-w) Application of Eq. (4.35) yields 5 e 9- | += u(w) But ó(t) is an even function, that is d(—t) = ó(t), and 1 5 5150) + 57] ++ u(o) (b) cos wot += m[ó(w + wo) + ó(w — wo)] cos tuoi = 1 Tt, a(t) X(w) Application of duality property yields m(6(t + wo) + ô(t — wo)| <=> 2m cos (—wow) = 2 cos (wow) CTT To Toto, x(t) 2ra(—w) Setting wo = T yields t+ T)+6(t- T) > 2c0s Tw (e) sin wot <=> jrló(w + wo) — ó(w — wo) a(t) X(w) Application of duality property yields jm[6(t+ wo) — 6(t — wo)] <=> 2m sin(-wouw) = —27 sin(wpw) ATA A o) O Otto too) tes Gu int oo) x(t) 2x7(—w) 329 Setting wo = T yields MU+T)-S(t— T) + 2jsin Tw 7.3-2. Refer to the solution of Prob. 1.2-3 for description of these signals. (a) c(t=a(t+1)+=(1-t) (t+ 1) > X (we Time inversion of z(t + 1) yields z(—t + 1), Hence (1-1) + X(-wje-ie Hence (ft) > X (we + X(-wjeis (db) ga) vo(15) 2 (e + 5) + Xu and & (=) =2 (5 + 3) es 2X (Pujeiis Therefore ( I- 5 += 2X(-Qw)er it Hence — xa(t) <> 2 [X(2w)eie + X(-2w)e 1] (c) ga(t)=a (2) +2 (=) +e (5) +z (5) z (2) + 4X (du)eizs 2 (9) + 4X (-4ujerio : (5) ES 2X) and & (-5) + 2X (2) Hence — za(t) 5 4 [X(de)e2e + X(-dw)e-52e + 2X(20) + 2X(-20)] (a) 4 t+2 4 2-t 1 t+2 1 2-t c()=3e (2) tar (35) (Ss (5) 2 á -—t z (=) es 2X and 4 (5) +» 2X(-2w)e- 32 330 Also T=3. The signal in Figure S7.3-4a is a(t+ 3) + a(t— 3), and a(t+3) + a(t— 3) <> 4sinc(w) cos 3w (b) Here a(2) is a triangular pulse shown in Figure $7.3-4b. From the Table 4.1 (pair 19) ft a(o =()=A (5) + sine! (5) Also T=3. The signal in Figure P7.3-4b is a(t+ 3) + 2(t— 3), and a(t+ 3) + o(t— 3) &> 2sine? (5) cos Ju | 4 1208 [ (as -f ( £t> Figure S7.3-4 7.3-5. Frequency-shifting property states that , a(tjefivot > X(w F wo) Therefore e(b) sinwnt = astettjeie pele io] = agbtto “w)+X(o wo) Time-shifting property states that e(t&T) <=> X (uw)etie7 Therefore ct+T)-e(t- T) > X (we — X (we de7 = 25X (w) sinwT and stett ED -elt- T) > X(v)sinTo The signal in Pigure P7.3-5 is z(t + 3) — a(t — 3) where t : a(t) = rect (5) + 2sinc(w) Therefore 2(t43) — e(t— 3) <> 2j[2sinc(w) sin 3] = 4j sinc(tw) sin 3 7.3-6. (a) The signal z(t) in this case is a triangle pulse A(5É) (Figure S7.3-6) multiplied by cos 10t. «()=4 () cos 10% 27 333 Also from Table 4.1 (pair 19) A(£) <=> x sinc?(Z£) From the modulation prop- erty (4.41), it follows that (tw + 10) 2 The Fourier transform in this case is a real function and we need only the amplitude spectrum in this case as shown in Figure S7.3-6a. m(w — 10) 2 =a(L Z Jsinc? c(t)=A (5) cos 10t <=> 3 (enc [tm (b) The signal z(t) here is the same as the signal in (a) delayed by 2x. From time shifting property, its Fourier transform is the same as in part (a) multiplied by eiv(2n), Therefore xtoj=5 (one [C=10) sine [ELO cosmo 2 The Fourier transform in this case is the same as that in part (a) multiplied by e-i27». “This multiplying factor represents a linear phase spectrum —27w. Thus we have an amplitude spectrum [same as in part (a) as well as a linear phase spectrum ZX(w) = —2mw as shown in Pigure S7.3-6b. the amplitude spectrum in this case as shown in Figure S7.3-6b. Note: In the above solution, we first multiplied the triangle pulse A() by cos 10t and then delayed the result by 27. This means the signal in (b)gis expressed as A(S2Z) cos 10(t — 27). We could have interchanged the operation in this particular case, that is, the triangle pulse A(É) is first delayed by 27 and then the result is multiplied by cos 10t. In this altenate procedure, the signal in (b) is expressed as A(S2r) cos 10t. This interchange of operation is permissible here only because the sinusoid cos 10t executes integral number of cycles in the interval 27. Because of this both the expressions are equivalent since cos 10(t — 2x) = cos 10 (c) In this case the signal is identical to that in (b), except that the basic pulse is rect(£) instead of a triangle pulse A(£). Now t rect (x) <=> 2x sinc(mw) 2x Using the same argument as for part (b), we obtain X(wo) = m(sincfm(w + 10)] + sincfr(w — 10)]je- 32" w = rect ( Esinc(t) +» rect (5) 737. (a) X(w Also Therefore » a(t) = =sinc(t) cos 4t 334 6 G 40 12 my cl -—» 1X 1 ==. Jo W— LX03- -2mos Figure S7.3-6 E (b) X(w)=A (25) +4 (25) Also | > sinc? o =sincê(t) +=» A E) Therefore » a(t) = +sinc?(t) cos 4t 738. (a) 1 xt no u(t) and u(t) > nó(w) + ; ju = If x(t) = eXtu(t) + u(t), then ves = (55) (me) = nô(w) 1 É qua [go - a E 1 = —Sô(u) + E + 5 because g(x)ó(x) = 9(0)5(2) il (eo) 335 (b) ut) es = -a d 1 . cite ui) + Eli) gos 1 tu) + aa T41. 1 Ho) = 511 (a) Xu) = =p 1 1 1 Vo) = Go)GarD mai aa vi) = (tuo) (b) 1 Xo) = = 1 “e = qu vt) = tertu(t) (e) X(w) = “5 o q Ao ap Vw) = Go+DGw-1) jo+1l jul vt) = Stu) + eta) (a) x(w) = nóto) + 1 1 Yu) = a (nto +] - o) +) [pecauseg(z)ô(2) = 9(0)5(2)] = tó(w) + ma o) = 1-uo) 338 743. 744 and H(w)= ju-—2 and — afiada -DGw+1) 3ljwu+l ju-2 Therefore E (60) = 3 [etu(t) + Stu(-8)) (b) and Therefore u(t) = [e — Ju(>)] Xi(w)=sinc (sê) and X(0)=1 Figure $7.4-3 shows Xi (1), Xo(w), Hi (w) and Ha(w). Now Xi(w)H(w) Yo(w) = Xo(w)Ho(w) The spectra Yi(w) and Y5(w) are also shown in Figure S7.4-3. Because y(t) = in (B)ya(t), the frequency convolution property yields Y(w) = Yi(w) + Yo(w). From the width property of convolution, it follows that the bandwidth of Y (w) is the sum of bandwidths of Yi(wo) and Yo(w). Because the bandwidths of Yi (w) and Ya(w) are 10 kHz, 5 kHz, respectively, the bandwidth of Y (w) is 15 kHz. H(w) = 10"? sinc (585) and P(w) = 0.5 x 10-ºsinc? (qt) The two spectra are sketched in Figure 87.4-4. It is clear that H (w) is much narrower than P(w), and we may consider P(w) to be nearly constant of value P(0) = 10-5/2 over the entire band of H(w). Hence, Y(w) = P()H(wo) = P(O)H(w) = 0.5 x 10"ºH(w) =» y(t)=0.5x 10-9h(t) Recall that A(t) is the unit impulse response of the system. Hence, the output y(t) is equal to the system response to an input 0.5 x 10-96(w) = Aó(w) H(w) = 10"? sinc (és) and P(w) = 0.5sinc?(%) The two spectra are sketched in Figure S7.4-5. It is clear that P(w) is much narrower than H(w), and we may consider H(w) to be nearly constant of value H(0) = 10"? 339 -3 10 A Note that the dc gain of the system is k = H(0) = 1073. Hence, the output is nearly kP(8). Dê dv) P/D) el 4 X to) =— í -. — wo -20000E | &9000 ti , 4 H te) BD) —20000 2000] = “187 dr w> x (o) E XD (0 DR, lo) Hole -aosoom | 2oompT to-> CocidT dt de Figure 87.4-3 . 6 P Ho) sro (O — | con > ué TO o O > ox Figure S7.4-4 over the entire band of P(w). Hence, Y(w) = P(w)H(w) = P(w)H(0) = 103P(w) => y(t) = 10"2p(t) OST a É | | T CO —> Figure 87.45 1900TT Hence, h(t) can be expressed as à sum of its even and odd components à h(t) = he(t) + ho(t) 340 7.46. Every signal can be expressed as a sum of even and odd components (see Sec. 1.5-2). [wo - E, ==. [X(w)P do = > f. Su? dy “dal O Letting ow = 5 and consequently dw = ade 11 o 2 E=>—s E do = - Xrovã Le “E amovã ZovT 7.6.2. Consider a signal z(t) = sine(kt) oo p= sinc?(kt) dt : a mf, de= 7.63. 1f 22(t) <=> A(w), then the output Y(w) = A(w)H(w), where H(w) is the lowpass filter transfer function (Figure S7.6-3). Because this filter band Af — 0, we may express it as an impulse function of area 474. Thus, xa H(w) = [4x fJó(w) and Y(w) = [47 A()A f]ó(w) = [47A(0)A f]ó(w) Here we used the property g(z)ô(x) = 9(0)5(2) [Ea- (1.23a)]. This yields u(t) = 24(0)Af Next, because 22(t) <= A(w), we have A(w) = / E erberietat so that A(0)= / “Cdi= E. Hence, y(t) = 2E:Af. Figure S7.6-3 7.6-4. Recall that 1 (o . oo = [ Xo(w)et du and f ri(teivt dt = Xi(=0w) Therefore f (dana = =D a(otf” Xolu ei! do] dt 343 1 o Re o =L f Xatol/ (eis! dt) do = + [esaxa(o) do da Jo o Zr Interchanging the roles of z1(t) and x>(t) in the above development, we can show that PE storage = So (xau) Xal-u) do Im the generalized Parseval's theorem in Prob. 7.6-4, if we identify g1(t) = sine(Wt — 7.65. mx) and ga(t) = sinc(Wt — nx), then irge we, and Ga( rea (Je Gi(w) = rec (5) e Therefore PE sumantyar= fo (5) [O free (5) sr ao But rect (54) = 1 for lw| < W, and is 0 otherwise. Hence o é suis 0 ngm Festa sã; [ (ee qo= [80 nÉn * &j2rk — 1 when k is an integer. In evaluating the integral, we used the fact that e 7.6-6. Application of duality property [Eq. (4.31)] to pair 3 (Table 4.1) yields da some atol t+a? The signal energy is given by :[ Pere" do = 4m [ e dy o The energy contained within the band (0 to W ) is W Ew = 4m [ eau dy = 0 — o a 1 Ew = 0.99Es, then e2eW = 0,01 w = 2 naajs = CS, a a TIA. (i) For m(t) = cos 1000t Posa-selt) = m(t) cos 10,000t = cos 1000t cos 10, 000t 1 =Icos 90004 + cos 11, 000%] 2 = LSB USB (ii) For m(t) = 2cos 1000t + cos 2000t m(t) cos 10, 000t = [2 cos 1000t + cos 2000] cos 10, 000t Posa-selt) 344 cos 9000t + cos 11, 000t + áleos 8000t + cos 1 =D LSB USB (iii) For m(t) = cos 1000 cos 3000t Ppsa-so(t) 1 álcos 8000 + cos 12, 0004] + áloos 6000t + c 2,000] 1 1 [cos 9000t + cos 80008] + [cos 11,000t + + cos 12, 0004] 2 mt) cos 10, 000t = leos 2000t + cos 4000€] cos 10, 000t os 14, 0008] 5 (cos 8000t + cos 6000] + 5 cos 12, 000t + cos 14, 000% LSB USB This information is summarized in a table below. Figure 87.7-1 shows various spectra. “Mw Modulated signal speetrim 15 1 1 ! ( | 0 tt ol v> 414% =4C00" | 1600 —Ito? -qroo QlOC cod | - * ; tt, tt tt. 41 -Z000 Op 2ro0. > ok “9K-8K o O» o gra TIK RE, ve = - “ 1 L Lille 144 41 tt. tt -AK Ol 2x 4 -i96 “I2K -8 EIS o WU MK SK I2K I4K Figure S7.7-1 [ case | Baseband frequency [ DSB frequency | LSB frequency | USB frequency i 1000 9000 and 11,000 | 9000 11,000 j 1000 9000 and 11,000 9000 11,000 2000 8000 and 12,000 8000 12,000 — Hi 2000 8000 and 12,000 8000 12,000 4000 6000 and 14,000 6000 14,000 7.7-2. (a) The signal at point b is êml The term 2 za(t) m(t) cos? wet l m(t) fi coswet + i cos 304] (t) coswet is the desired modulated signal, whose spectrum is centered at +wç. The remaining term im(t) cos3wct is the unwanted term, which represents the modulated signal with carrier frequency 3w, with spectrum É 345 TIA. TM. ubhressed i x s uy “158 |oists + H3—> Figure $7.7-5 is centered at +2w.. Hence the output of the lowpass filter is vt) = A+ m(t) When this signal is passed through a de block, the de term A is suppressed yielding the output m(t). This shows that the system can demodulate AM signal regardless of the value of A. This is a synchronous or coherent demodulation. (a) u=05=72=12 > 4=20 (b)u=10=Z:=2 > 4=10 ()u=20=Z2=1 54=5 ()u=0=2Z=1 >4=0 + This means that y = oo represents the DSB-SC case. Figure S7.7-7 shows various waveforms. Figure S7.7-7 The signal z(t) = e-“tu(t) has Fourier Transform given by X (w) = A and energy Es = 3. Using this information, MATLAB program MS7P2 is modified function [W,E W] = MS7P2modi(a,beta,tol) % MS7P2modi.m % Function M-file computes essential bandwidth W for exp(-at)u(t). Y% INPUTS: a = decay parameter of x(t) 4% beta = fraction of signal energy desired in W % tol = tolerance of relative energy error % OUTPUTS: W = essential bandwidth [rad/s] 348 % E.W = Energy contained in bandwidth W W=0; step=a; Y Initial guess and step values X. squared = inline('1./(omega."2ta."2)”,'omega”,'a?); E = beta/(2%a); % Desired energy in W relerr = (E-0)/E; % Initial relative error is 100 percent while(abs(relerr) > tol), if (relerr>0), % W too small W=W+step; Y Increase W by step elseif (relerr<0), Y W too large step = step/2; W = W-step; Y Decrease step size and then W. end EM = 1/(2+pi)*quad(X squared,-W,W, [],[],a); relerr = (E - E W)/E; end (a Setting a = 1 and using 95% signal energy results in >> [W,E W]=MS7P2modi(1,.95,1e-9) W = 12.7062 W = 0.4750 Thus, W = 12.7062. + From the text example, “the essential bandwidth corresponding to 95% signal energy is derived as W = 12.706a radians per second. For a = 1, this corresponds nicely with the computed value of W = 12.7062. (b) Setting a = 2 and using 90% signal energy results in >> [W,E W)=MS7P2modi(2,.90,1e-9) W= 12.6275 W = 0.2250 Thus, Wo = 12.6275. (c) Setting a = 3 and using 75% signal energy results in >> [W,E W]=MS7P2modi(3, .75,1e-9) W=7.2426 W = 0.1250 Thus, Was = 7.2426. 7.M-2. To solve this problem, program MS7P2 is modified to solve for the pulse width to achieve a desired essential bandwidth, rather than solving for the essential bandwidth that corresponds to a desired pulse. function [tau,E W] = MS7P2mod2(W,beta,tol) 4 MSTP2mod2.m 4 Function M-file computes essential bandwidth W for square pulse. % INPUTS: W = essential bandwidth [rad/s] % beta = fraction of signal energy desired in W u tol = tolerance of relative energy error % OUTPUTS: tau = pulse width 349 7.M3. % E W = Energy contained in bandwidth W tau = 1; step = 1; % Initial guess and step values x. squared = inline(” (tau+MS7P1 (omega*+tau/2)).72”,?omega”,?tau”); EM = 1/(2+pi)*quad(X squared,-W,W, [), [J, tau); E = betartau; % Desired energy in W relerr = (E - E W)/E; while(abs(relerr) > tol), if (relerr>0), Y% tau too small taustau+step; Y Increase tau by step elseif (relerr<0), Y tau too large step = step/2; tau = tau-step; ) Decrease step size and then tau. end E W = 1/(2+pi)+quad(X squared,-W,W, [), [) tau); E = betattau; Desired energy in W relerr = (E - E W)/E; end (a) Set W = 275 and select 95% signal energy. >> [tau,E W] = MS7P2mod2(2xpi*5, .95,1e-9) tau = 0.4146 E.W = 0.3939 Thus, m = 0.4146 (b) Set W = 2710 and select 90% signal energy. >> [tau,E W) = MS7P2mod2(2+*pi+10,.90,1e-9) tau = 0.0849 EM = 0.0764 Thus, 1.2 = 0.0849. (c) Set W = 2720 and select 75% signal energy. >> [tau,E W] = MS7P2mod2(2+pi+20, .75,1e-9) tau = 0.0236 E W=0.0177 Thus, 73 = 0.0236. To solve this problem, program MS7P2 is modified to solve for the decay parameter a to achieve a desired essential bandwidth, rather than solving for the essential bandwidth that corresponds to a desired decay parameter. function [a,E.W] = MS7P2mod3(W, beta, tol) % MSTP2mod3.m Y% Function M-file computes decay parameter a needed to % achieve a given essential bandwidth. % INPUTS: W = essential bandwidth [rad/s] % beta = fraction of signal energy desired in W % tol = tolerance of relative energy error 350 7.M-6. MATLAB is used to evaluate and plot the first ten coefficients. >> tau = 2*pi/3; T.O = 2*pi; n = [0:10]; >> Dn = tau/T 0+MS7P1i(n+pi*tau/T 0); >> stem(n,D n,'k”); xlabel(?n”); ylabel(ºD nº); >> axis([-0.5 10.5 -0.2 0.55]); CSA AI E E E FB ES Figure S7.M-5a: Fourier series coefficients Dy for a(t). (b) Setting To = 7 and 7 = 7/3 yields Da= Fsinc (5) . Notice, the coefficients D, depend only on the duty-cycle of the signal, not the period. Since the duty cycle is fixed, the coefficients D, are identical to those determined in 7.M-5a. Refer to solution 7.M-5a for the MATLAB code and plot. X(w) JS u(t)er set dt oo T| > SS Penta JS, et rota (gu /2)? (gu /2)?) dp = USP [o eita a Substituting t'/v2 = t and dt'/V2 = dt yields -w2/4 pos o X(w) = f ea apr, (D=5 Le =(t-a2 AA? However, L= [2 e gy =1 tor any a, so [E e gy = V2r. Thus, X(w) = re PA, MATLAB is used to plot x(t) and X(w). >> t = linspace(-10,10,1001); x = exp(-t.72); >> omega = linspace(-10,10,1001); X = sgrt(pi)+exp(-omega."2/4); 353 >> >> >> subplot (211); plot(t,x,'k); xlabel('t'); ylabel('x(t)”); subplot(212); plot(t,X,ºk'); xlabel("Nomega”); ylabel('X(Nomega)”); A 4 4 4 29 2 4 88 xo Figure S7.M-6: a(t) = e” and X( Figure S7.M-6 confirms that X (w) is just a scaled and stretched version of 2(t). ThiSis something remarkable; the Fourier Transform of a Gaussian pulse is itself a Gaussian pulse! 354