Problem Set Solution 7: Continuous-Time Fourier Series

7 Continuous-Time Fourier Series Solutions to Recommended Problems S7.1 (a) For the LTI system indicated in Figure S7.1, the output y(t) is expressed as h(r)x(t - r) dr, y(t) =f where h(t) is the impulse response and x(t) is the input. LTI x(t) ON h(t) y (t) Figure S7.1 For x(t) = ew', y(t) = f h()ew(- { = ejwt T) dr h(r)e-j' dr = e+iwtH(w) (b) We are given that the first-order differential equation is of the form dy(t) + ay(t) = x(t) dt From part (a), when x(t) = eiw', then y(t) = ej"'H(w). Also, by differentiating y(t), we have dy(t) = jcoej"'H(co) dt) Substituting, we get ewt jwewtH(w) + aejw'H(w) Hence, jwH(o) + aH(w) = 1, H(w) = or 1 a + jw. S7.2 (a) The output of a discrete-time LTI system is given by the discrete-time convo­ lution sum h[k]x[n - k] y[n] = ( k S7-1 Signals and Systems S7-2 If x[n] = z", then y[n] = ( h[k]z"-k k h[k]z -k = z" k = z"H(z) (b) We are given that the first-order difference equation is of the form y[n] + ay[n From part (a), if x[n] = z", then y[n] - 1] = x[n] = z"H(z). Hence, y[n - 1] = z" -1H(z). By substitution, z"H(z) + az"- H(z) =z" which implies (1 + az-')H(z) = 1, = H(z) 1 1 + az-1 _ S7.3 (a) x(t) sin (lort + 6 = j2xt5 _ = -j21t5 2j 2j We choose wo, the fundamental frequency, to be 21r. x(t) = a eik-ot, k where ejr/6 a5 =a_5 __e -jr/6 2j 2j Otherwise ak = 0. (b) x(t) = 1 + cos(27rt) ej2xt =1+ 2 For wo = 2 e-j2wt +ejit 2 7r, a_ = ai = -, and ao = 1. All other ak's = 0. (c) x(t) = [1 + cos(27rt)] [sin(lort + = sin(10it + /6 e = -_ . _}15_ 2j + ej2rt5 e j21rt4 4j + cos(2irt)sin( 10irt + e -j2t5) e 2 2j e ee -j2xts -j2t6 4j + j2+rt 4j ej 2 + le -j2+ t eji( 6 e t6 _ 4j e-j2xt4 2 t5 r __e 2* e -j27rt5 Continuous-Time Fourier Series / Solutions S7-3 Therefore, x(t) = where wo = 2 ake*4' 1r. a4 ejr/6 . =- a5 = a6 All other ( a-4 43j e jr/6 -, 2j e jr/6 4j. _ a-5, /6 e-ir 2j I _e-jr16 . a_6 = 4j 4j ak's = 0. S7.4 (a) Note that the period is To = 6. Fourier coefficients are given by x(t)e -jk" 0 t dt ak = - We take wo = 27r/To = r/3. Choosing the period of integration as -3 have ak = _ 6 2 e - jk( /3)t e 1 _ 1 6 f1 2 [e +j(7/3)k _ -jk(r/3)t 1 -jk(r/3)t 6 -jk(7r/3) 1 F2 1 1 dt 1 -jk(ir/3)t 6 -jk(7r/3) e +j(2r/3)k - e-j(21/3)k -j27rk cos(2ir/3)k cos(7r/3)k jrk jrk Therefore, x(t) akejkwot = k 7r 3 and ak = cos(27r/3)k - cos(r/3)k jrk 2 1 + e j(-/3)kI to 3, we Signals and Systems S7-4 Note that ao = 0, as can be determined either by applying L'H6pital's rule or by noting that f ao = (1|T0 ) x(t)dt. (b) x(t) Figure S7.4-2 The period is To = 2, with wo = 2-x/2 = r. The Fourier coefficients are ak = fTO x(t)e -iko'dt - To o Choosing the period of integration as k= a 3/2 2 -1/2 -i2 t o we have -2, dt -­ x(t)e ikot 3/2 2 1/2 1 2 [6(t) 26(t - . e-k-o 1)]eikO' - dt 1. = - 2 (e-i-)k Therefore, a o - 2, ak = - ()k It is instructive to plot ak, which we have done in Figure S7.4-3. Continuous-Time Fourier Series / Solutions S7-5 S7.5 (a) (i) and (ii) From Problem 4.12 of the text (page 260), we have x {t - T)= -x(t), which means odd harmonics. Since x(t) is real and even, the waveform has real coefficients. (b) (i) and (iii) -x(t) = x t ­ which means odd harmonics. Since x(t) is real and odd, the waveform has imag­ inary coefficients. (c) (i) W~ T) -x(t) = x t 4 )' which means odd harmonics. Also, x(t) is neither even nor odd. Solutions to Optional Problems S7.6 x(t) is specified in the interval 0 < t < T/4, as shown in Figure S7.6-1. x(t) | T 8 t T 4 Figure S7.6-1 (a) Since x(t) is even, we can extend Figure S7.6-1 as indicated in Figure S7.6-2. x (t) t _T _T 4 8 T T 8 4 Figure S7.6-2 Signals and Systems S7-6 Since x(t) has only odd harmonics, it must have the property that x(t - T/2) = -x(t), as shown in Figure S7.6-3. x(t) in Figure S7.6-2 shifted x(t - T) = -x(t) 2 T 4 -x(t) in T 4 T 2 37 4 / Figure S7.6-2 Figure S7.6-3 So we have x(t) as in Figure S7.6-4. x(t) 0 _T 4 T 4 T 2 3T 4 T -1-­ Figure S7.6-4 (b) In the interval from t = 0 to t = T/4, x(t) is given as in Figure S7.6-5. Continuous-Time Fourier Series / Solutions S7-7 Since x(t) is odd, for - T/4 < t < T/4 it must be as indicated in Figure S7.6-6. x(t) T 8 T 4 T 4 Figure S7.6-6 Since x(t) has odd harmonics, x[t - (T/2)] = -x(t). Consequently x(t) is as shown in Figure S7.6-7. x(t) T T T 4 T 8 -1 Figure S7.6-7 S7.7 1 x(t)e -jkw 0t dt ak = - TO (a) alk TO = - T0 to)e -jk" 0 t dt x(t fTO Substituting r = t - to, we obtain ak = - 1F I x(r)eikwor dr - e -jkoto OTO = ake -jk"0'0o (b) ak = 1 - T0Of17O x(- t)e jkwot dt Substituting r = - t, we have 1 ak = - T0OfT 0 x(T)e kwOdr = a-k Signals and Systems S7-8 (c) x*(t)e -jk 0t dt = 50TO 0 = T0 - x(t)ekwot dt = ak, 6k = ak (d) a(d) = x(at)eik( 2 ra/T)t fi 0 To dt la Let r = at. Then ak = - TO x(r)e -jk(2"/To)' dr = ak To Therefore, a S7.8 (a) Since O4(t) are eigenfunctions and the system is linear, the output is y(t) L = XkCk'k(t). k=-00 (b) y(t)=t 2 d 2x(t) 2 dt dt2 dx(t) +t dt kk(t) = tk, d0k(t) ktk1_ dt d2dtk(t) dt 2 "k-Jt2 ' ktk = So if k(t) 1)tk-2 - = x(t), then y(t) = t 2k(k - 2 1)tk- + tktk 1)tk + ktk = k(k = kotk = k 4k(t) - The eigenfunction 4k(t) has eigenvalue Xk I = k2 S7.9 (a) 9(t) = 2 1(t) * 2 2(t) = fT I 1(r)A2 (t - r) dr The Fourier coefficients for P(t) are given by Ck = -- = 1 -- 1(r) I 50TO J Toakbk 2 (t - r) dre )t -jk(2wTot TOL 1 (r)e - TO/2, where TO/2 4T. S7.10 (a) The approximation is N N(t)= akk(t k=-N with the corresponding error signal eN(t) = x(t) ­ Nt) N = x(t) - ( k=-N akk(t) Signals and Systems S7-10 Hence, |eN(t) I = = x(t) a*t) x*() kk - z ( akx(t)4k(t) - |x(t)| 2 - k If we integrate, akx(04k() + k ( (I k aka14k(M 1() eN(t) 12 dt, and use the property that f k a we get 1, 1, = otherwise, b= E = fb Ix(t) I'dt Za* - T - x(t)kp(t) cit k a X*(tk(t) dt ak + T a k |akI k Since ai = bi + jc, =E an d OC r x(t)k7(t) dt x*(t)Oi(t) dt + 2bi ­ and -_ aci = x(t)(t )dt - j j f x*(t)i(t) dt + 2c, Setting -= 0 and = 0, we can multiply the second equation byj and add the two equations to get (E aE - ab + = 0 By substitution, we get bi + jci = = ax(t)O(t) dt a, (b) If {4(t)} are orthogonal but not orthonormal, then the only thing that changes from the result of part (a) is a Tab ab qk*~c(t) dt Z Iak IAk k k1 It is easy to see that we will now get a = fbx(t)4it) dt 2 k Continuous-Time Fourier Series / Solutions S7-11 (c) Since fTo+a einwot e-jnwot dt = for all values of a, using parts (a) and (b) we can write ai = a _ x(t)e-"nwo' 1 =- x(t)e-nwot dt T0 fTo dt