8 Continuous-Time Fourier Transform Solutions to Recommended Problems S8.1 (a) x(t) t Tj 2 Tj 2 Figure S8.1-1 Note that the total width is T,. (b) i(t) t 3T 1 2 -- T1 To 2 T1 2 T1 2 To 2 Tl 3 T 1 =O 2 Figure S8.1-2 (c) Using the definition of the Fourier transform, we have ox(t)e -j X(w) = dt Til e = le-j" dt since x(t) -T1/2 = 0 for ItI> sin wTi 12 (e -jwTI1/2 - JW Ti/2 = _ 2 eT1/ ) = 2 (e See Figure S8.1-3. S8-1 Signals and Systems S8-2 (d) Using the analysis formula, we have T f ak = X(t)e ~jk 0 t dt, where we integrate over any period. ak _ 1 -T f -T To12 1 (t)e -jk(2w/T)t dt = TO f T1 /2 /2 -T e -jk(2/T )t dt O 1 To ak = TO (e jkirToT 0 -- eikir1/TO) = -jk 2, sin kr(T1 /T 0 ) ,rk _ sin ,r(2k/3) 7rk Note that ak = 0 whenever (27rk)/3 = irm for m a nonzero integer. (e) Substituting (21rk)/To for o, we obtain T_1X( TO =2)T I"=(21rk)/ To 1 2 sin(7rkT1 /T 0 ) To 2,rk/To sin rk(T 1 /T 0 ) ,rk = ak (f) From the result of part (e), we sample the Fourier transform of x( t), X(w), at w = 2irk/To and then scale by 1/To to get ak. Continuous-Time Fourier Transform / Solutions S8-3 S8.2 (a) X(w) = fx(t)e 5)e -j' dt = e ~j (t - -j4t dt = = cos 5w - j sin 5w, by the sifting property of the unit impulse. IX(w)| = |ej 5wI = 1 0, e -at goes to zero as t goes to infinity. Therefore, X(W) = -1 1(0 a + jw - 1) = IX(W)I = [X(W)X*(w)]l/ Re{X(w)} = Im{X(W)} -X(O) = 2 = X(w) + X*(O) 1 a +jw a +ja a + X(W) - X*(W) 2 a2 + = tan- Im{X(W)) \/ 2 a 2 2 -jw)] (2 i -W w 2 '> = -tan _ w a The magnitude and angle of X(w) are shown in Figure S8.2-2. 1 +o2 > Signals and Systems S8-4 IX(w)I a -a 4X(W) 2 Tro -aa 2T Figure S8.2-2 (c) X(o) = e 1+12)tU(t )e 1 1 + j(2 [ 1-+j(2-)t - e(-1+ 2)te -jct dt -''' dt = 0 )0 Since Re{-1 + j(2 - w)} < 0, lim, e[- 1+j(2-w) = 0. Therefore, 1 1 + j(o - 2) 1 IX(wO)| = [X(x)X*(W)]11 2 = Re{X()} = Im{X(W)} = X() = 1 + (w 2 X*(co) - 2 + (ow- 2) 2 1 X(w) + X*(W) X(w) \/1 1 tan-' ImX(W) LRe{X(w)} - 2)2 -(o - 2) + ((A -2)2 -tan-'(w - 2) The magnitude and angle of X(w) are shown in Figure S8.2-3. Continuous-Time Fourier Transform / Solutions S8-5 X(w)I 1 1 3 2 4X(W) Figure S8.2-3 Note that there is no symmetry about w = 0 since x(t) is not real. S8.3 X3(t)e -' dt (a) X3(w) = Substituting for x 3(t), we obtain X 3(() = J ax 1(t)e -wt dt + =f = [ax 1 (t) + bx 2(t0)e 1-'dt af x 1(t)e -j dt + b f bx 2 (t)e -jwt dt x 2(t)e -- ' dt = aX1 (w) + bX 2 (w) (b) Recall the sifting property of the unit impulse function: f h(t)b(t - to) dt = h(t o) Therefore, 21rb(w - wo)ej-' do = 2rewot Signals and Systems S8-6 Thus, 2 1- 2,r 7rb(w - wo)ei-' dw = ei-o' -"oo Note that the integral relating 2-7rS(w - wo) and e-o' is exactly of the form X(w)ej- t dw, - x(t) = where x(t) = e jot and X(w) = 2-7rb(co - wo). Thus, we can think of ei-o' as the inverse Fourier transform of 2 7rb(w - wo). Therefore, 27rb(o - wo) is the Fourier transform of eiwo'. (c) Using the result of part (a), we have X(M) = 1 ake 7{x(t )} = 51 2r j = T, a, From part (b), = 2,irb ( W 5{eik(2,rT)t -21rk) Therefore, Z(w) 27rab = (W k = -00 (d) X(w) 2 sin 2 sin 2 3 To 3 3 Figure S8.3 S8.4 (a) We see that the new transform is Xa(f) = X(w) W=2,wf We know that x(t) =- f X(w)e'j t do 5T {eik( 2,rs/t Continuous-Time Fourier Transform / Solutions S8-7 Let w = 2irf. Then dw = 21rdf, and x(t) = 1 X(21rf)ej 2"2ir df = Xa(f )ei2 f= -o 27 ,"f df Thus, there is no factor of 21r in the inverse relation. IXa(f)I TI 2 1 1 2 TI TI TF T, Figure S8.4 (b) Comparing Xb(V) x(t)e = -1 jv' and dt x(t X(w) = -j~" dt, we see that Xb(V) = V2_ 7r or X(W) X(w) = \7 X(w) -V The inverse transform relation for X(w) is x(t) - 1 *0 X(w)e =1 j Xb(v)e X2r(w)ej- dw dw = 1 jvt dv, where we have substituted v for w. Thus, the factor of 1/2-x has been distributed among the forward and inverse transforms. S8.5 (a) By inspection, To = 6. (b) ak = 1 TO J TO T0 )e 2 /To)t -k( We integrate from -3 to 3: 1 1 ak= = 1 J- dt 2 (ti + 1)+ +l 1 + 'je 1 6(t) + -6(t 2 - (1 -Jlk6.. + c 6 Cos 2k Signals and Systems S8-8 (t) (c) (i) X 1(w) = - X 2(w) = (ii) + Cos 27rk 6( = x 1 (t)e -"' dt ie * + - ie -"1 1 + eik(2 / 6 )t [6(t + 1) + b(t) + -1(t - 1)]e -j't dt + cos w x 2(t)e -''' dt = [3( t) + it(t - 1) + -3(t - 5)]e - dt = 1 + ie -j + le -i5 w (d) We see that by periodically repeating x 1 ( t) with period Ti = 6, we get t(t), as shown in Figure S8.5-1. x 1 (t) 2 -1 0 1 t x 1 (t - 6) 0 5 6 7 5 6 7 xl(t + 6) 2 -7 t -6 t t |tt) -5 0 x(t) -7 -6 -5 -1 0 1 Figure S8.5-1 Similarly, we can periodically repeat x 2(t) to get 1(t). Thus S8.5-2. T2 = 6. See Figure Continuous-Time Fourier Transform / Solutions S8-9 x 2 (t) tilt t 01 X2 (t -6) 2 t 0 6 t2~ x 2 (t ± 6) -6-5 t 1 1 7 t 2 0 7 X(t) IL -6 -5 -1 0 t2 t It 1 5 6 t 7 Figure S8.5-2 (e) Since I(t) is a periodic repetition of x 1 (t) or X 2 (0), the Fourier series coefficients of t(t) should be expressible as scaled samples of X,(co). Evaluate X 1 ( 0 ) at w= 21rk/6. Then 6 co-= Similarly, we can get X2 - ak X1(x) But e as a scaled sample of X 2(w). Consider X 2(21rk/6): 1 j2k/6 5e + e =irk1 +2o 62 njlOrk6 = e j(lOrk/62rk) X2 (27rk\ 1 k*k2rk A e =6,rk 2 2 n 6 rkI . Thus, = 1 + cos X6 - Although X,(w) # X 2(w), they are equal for w = 6 a. =27rk/6. X x Signals and Systems S8-10 S8.6 (a) By inspection, e -atu(t) 7 1 9 . a + jw Thus, 1 7I e- 7tu(t) 1 7 + jw Direct inversion using the inverse Fourier transform formula is very difficult. (b) Xb(w) = 26(w + 7) + 26(o - Xb(w)ej do = - = Xb(t) 1 7), 1 e- It + e . 2 = 2 - 7) + 6(w - 7)]ei-' dw [6(w + cos 7t (c) From Example 4.8 of the text (page 191), we see that 37 e alti 9 2a _2a a 2 + W2 However, note that ax(t) aX(w) since x(t)e -jw dt = aX(w) ax(t)e -j" dt = a Thus, 1 -" -e 2a (d) 1 l 1 or a2 + (02 = Xa(w)[26(o + 7) + 26(w - Xa(w)Xb(W) = 2X(-7)(co 2 - XdCO)= 7 - j7 xaOt = 1f*[2 1 2 b(o7) 7 +j7 ±7+ 2 1 . e 7' ir7-j7 1 1 + b(co - 7) 7 +j7 7 -37 = - XA(t) 7)] .7 6(o + 7) + f- 1 - e -31t 6 + 7) + 2Xa(7)6(w - 7) (w +7) + 27r-- 7 9 + .2 . ej_ xr7+j7 ej" do ' Note that 1 7 +j7 7 2 -jr/4 1 7 -j7 =_ 1=( , 7 2 +jir/4 Thus XA(t) = - r7 - - 2 [e -j(7t -/4) + ei( 7 t-/4)] = -77r cos 7t - -) 4 Continuous-Time Fourier Transform / Solutions S8-11 we ~j3 w, 0 5 w < 1, 3 we -j w, (e) Xe(w) = -1 elsewhere, t0, f x(t) = 0, (O ! X(w)e-' dw = i [J' we -jamei' we ~i3-ejwl dw do - Note that xe" f dx = e ax -1) ax , -(ax Substituting a = j(t - 3) into the integrals, we obtain 1 x(t) = L( (j(t j(t - 21r egj(t I - 3)w (j(t - - 3)) 3)w - 1) 0 which can be simplified to yield 1 [ cos (t - 3) - 1 x(t) = (t - 7r 3)2 - 0 3)w (j )) (j(t - 3)w - 1) (j(t - 3)) 2 0 sin (t - 3) (t - + 3)_ Solutions to Optional Problems S8.7 (a) y(t)e -j Y(w) = S (b) Let r t - ft -) dr e -j'' dt x(r)h( t - dt x(,r)h(t - 1)e -j"' dr dt T and integrate for all r and r. Then Y(w) = x(r)e -' = = x(r)h(r)e -jw( r-+*T) .fro TT=- d-r dr dr h(r)e {r=-0 ~'j dr X(w)H(w) S8.8 (a) Using the analysis equation, we obtain 1 rT/2_ Ilk = T I T/ -T/2 e(t)e -jk(21r/T) t dt-T 1 T Thus all the Fourier series coefficients are equal to 1/T. (b) For periodic signals, the Fourier transform can be calculated from ak as X(w) = 21 ak k=-00 w-T J Signals and Systems S8-12 In this case, 2rk 21 PM=T 6 T P(wj) 21Tr T _47 2_ 0 _ 2n 4 T T T T Figure S8.8 (c) We are required to show that 2(t) = x(t )* p(t) Substituting for p(t), we have x(t)* p(t) = xt kT) * k Ot Using the associative property of convolution, we obtain x(t) * p(t) = [x(t) * b(t T, kT)] - k= -o From the sifting property of b(t), it follows that x(t - kT) =t(t) x(t)* pX)O k= Thus, x(t) * p(t) is a periodic repetition of x(t) with period T. (d) From Problem P8.7, we have X(w) = X()P(x) = 27rk - X(-) -T k= =jI 2xk -- X(w)5(c Since each summation term is nonzero only at w = 27rk/T, *0 i = y=) 2xk 2rk 27r T X( T B From this expression we see that the Fourier series coefficients of t(t) are ak = - X Ti T , which is consistent with our previous discussions.

Problem Set Solution 8: Continuous-Time Fourier Transform

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock