8 Continuous-Time Fourier Transform

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1 8 Continuous-Time Fourier Transform Solutions to Recommended Problems S8. (a) x(t) Tj Tj t Figure S8.- (b) Note that the total width is T,. i(t) t 3T -- T To T T To Tl 3 T =O Figure S8.- (c) Using the definition of the Fourier transform, we have X(w) = o x(t)e -j dt = Ti/ le-j" dt since x(t) = 0 for ItI> Til sin wti = e - (e -jwti/ _ et/ ) = JW -T/ (e See Figure S8.-3. S8-

i(t) t 3T -- T To T T To Tl 3 T =O Figure S8.

2 Signals and Systems S8- (d) Using the analysis formula, we have ak = T f X(t)e ~jk 0 t dt, where we integrate over any period. _ f T (t)e -jk(w/t)t dt = f T / O ak -T e -jk(/t )t dt - To TO -T / ak = To sin kr(t /T 0 ) _ sin,r(k/3) (e jkirtot 0 -- eikir/to) = TO,rk 7rk -jk, Note that ak = 0 whenever (7rk)/3 = irm for m a nonzero integer. (e) Substituting (rk)/to for o, we obtain T _ sin(7rkt X( =)T /T 0 ) sin rk(t /T 0 ) TO I"=(rk)/ To To,rk/To,rk = ak (f) From the result of part (e), we sample the Fourier transform of x( t), X(w), at w = irk/to and then scale by /To to get ak.

-jk, Note that ak = 0 whenever (7rk)/3 = irm for m a nonzero integer.

3 Continuous-Time Fourier Transform / Solutions S8-3 S8. (a) X(w) = fx(t)e -j4t dt = (t - 5)e -j' dt = e ~j = cos 5w - j sin 5w, by the sifting property of the unit impulse. IX(w) = ej 5 wi = for all w, LIM{X(xc)} <tx(w) = tan-' tan -sin5wo) = -5w Re{X(co)} cos 5w (b) X(w) = e -a t u(t)e -i"' dt = e -ate -j't dt 0 - (a+jw)t dt - a+jw Since Re{a} > 0, e -at e - (a+jo)t o goes to zero as t goes to infinity. Therefore, - X(W) = (0 - ) = a +jw a + jw IX(W)I = [X(W)X*(w)]l/ = a +ja X(w) + X*(O) a Re{X(w)} = a + ( i -jw)] \/ +o > Im{X(W)} = X(W) - X*(W) -W a + w '> Im{X(W)) _ w -X(O) = tan- = -tan a The magnitude and angle of X(w) are shown in Figure S8.-.

IX(w) = ej 5 wi = for all w, LIM{X(xc)} <tx(w) = tan-' tan -sin5wo) = -5w Re{X(co)} cos 5w (b) X(w) = e -a t u(t)e -i"' dt = e -ate -j't dt 0 - (a+jw)t dt - a+jw

4 Signals and Systems S8-4 IX(w)I -a a 4X(W) Tro -aa T Figure S8.- (c) X(o) = e + )tu(t )e -'''dt = e(-+ )te -jct dt + j( - )0 [ -+j(-)t 0 Since Re{- + j( - w)} < 0, lim, e[- +j(-w) = 0. Therefore, + j(o - ) IX(wO) = [X(x)X*(W)] = \/ + (ow- ) Re{X()} = Im{X(W)} = X(w) + X*(W) + (w - ) X(w) - X*(co) -(o - ) + ((A -) X() = tan-' ImX(W) -tan-'(w - ) LRe{X(w)} The magnitude and angle of X(w) are shown in Figure S8.-3.

$Therefore, + j(o - ) IX(wO) = [X(x)X*(W)] = \/ + (ow- ) Re{X()} = Im{X(W)} = X(w) + X*(W) + (w - ) X(w)$

5 Continuous-Time Fourier Transform / Solutions S8-5 X(w)I 3 4X(W) Figure S8.-3 Note that there is no symmetry about w = 0 since x(t) is not real. S8.3 (a) X 3 (w) = X 3 (t)e -' dt Substituting for x 3 (t), we obtain X 3 (() = J [ax (t) + bx (t0)e -'dt =f ax (t)e -wt dt + bx (t)e -jwt dt = af x (t)e -j dt + b f x (t)e -- ' dt = ax (w) + bx (w) (b) Recall the sifting property of the unit impulse function: Therefore, f h(t)b(t - to) dt = h(t o ) rb(w - wo)ej-' do = rewot

3 (a) X 3 (w) = X 3 (t)e -' dt Substituting for x 3 (t), we obtain X 3 (() = J [ax (t) + bx (t0)e -'dt =f ax (t)e -wt

6 Signals and Systems S8-6 Thus, - 7rb(w - wo)ei-' dw = ei-o',r -"oo Note that the integral relating -7rS(w - x(t) = - wo) and e-o' is exactly of the form X(w)ej- t dw, where x(t) = e jot and X(w) = -7rb(co - wo). Thus, we can think of ei-o' as the inverse Fourier transform of 7rb(w - wo). Therefore, 7rb(o - wo) is the Fourier transform of eiwo'. (c) Using the result of part (a), we have X(M) = 7{x(t )} = 5 ake r j = T, a, 5T {eik(,rs/t From part (b), 5{eik(,rT)t =,irb ( W -rk) Therefore, Z(w) = k = -00 7rab ( W (d) X(w) sin 3 sin T o 3 3 Figure S8.3 S8.4 (a) We see that the new transform is We know that Xa(f) = X(w) W=,wf x(t) = - f X(w)e'j t do

Therefore, 7rb(o - wo) is the Fourier transform of eiwo'.

7 Continuous-Time Fourier Transform / Solutions S8-7 Let w = irf. Then dw = rdf, and x(t) = X(rf)ej "ir df = Xa(f )ei,"f df 7 f= -o Thus, there is no factor of r in the inverse relation. IXa(f)I TI (b) Comparing TI TI TF T, Figure S8.4 Xb(V) = - x(t)e jv' dt and X(w) = x(t -j~" dt, we see that Xb(V) = X(W) or X(w) = \7 X(w) V_ 7r -V The inverse transform relation for X(w) is x(t) - *0 X(w)e j dw = Xr(w)ej- dw = Xb(v)e jvt dv, where we have substituted v for w. Thus, the factor of /-x has been distributed among the forward and inverse transforms. S8.5 (a) By inspection, To = 6. (b) ak = TO )e -k( /To)t dt TO J T0 We integrate from -3 to 3: ak= (ti + )+ 6(t) + -6(t - + c 6 J- = +l + 'je -Jlk6.. ( Cos k

$4 Xb(V) = - x(t)e jv' dt and X(w) = x(t -j~" dt, we see that Xb(V) = X(W) or X(w) = \7 X(w) V_ 7r -V The inverse transform relation for X(w) is x(t) - *0 X(w)e j dw = Xr(w)ej-$

8 Signals and Systems S8-8 (t) = 6( + Cos 7rk eik( / 6 )t (c) (i) X (w) = x (t)e -"' dt - [6(t + ) + b(t) + -(t - )]e -j't dt - ie * + + ie -" + cos w (ii) X (w) = x (t)e -''' dt = [3( t) + it(t - ) + -3(t - 5)]e - dt = + ie -j + le -i 5 w (d) We see that by periodically repeating x ( t) with period Ti = 6, we get t(t), as shown in Figure S8.5-. x (t) - 0 t x (t - 6) xl(t + 6) t t tt) t x(t) Figure S8.5- Similarly, we can periodically repeat x (t) to get (t). Thus T = 6. See Figure S8.5-.

periodically repeating x ( t) with period Ti = 6, we get t(t), as shown in Figure S8.5-.

9 Continuous-Time Fourier Transform / Solutions S8-9 x (t) tilt t 0 X (t -6) t t t t~ x (t ± 6) X(t) IL t tit t Figure S8.5- (e) Since I(t) is a periodic repetition of x (t) or X (0), the Fourier series coefficients of t(t) should be expressible as scaled samples of X,(co). Evaluate X ( 0 ) at w= rk/6. Then 6 co-= k*krk (7rk\ Similarly, we can get ak as a scaled sample of X (w). Consider X (rk/6): X - jk/6 5e + e X(x) =irk +o =6,rk n X x 6 But e njlork6 = e j(lork/6rk) e rki 6. Thus, X = + cos - = A X6 6 a. Although X,(w) # X (w), they are equal for w =7rk/6.

$X,(co). Evaluate X ( 0 ) at w= rk/6. Then 6 co-= k*krk (7rk\ Similarly, we can get ak as a scaled sample of X (w).$

10 Signals and Systems S8-0 S8.6 (a) By inspection, Thus, e -atu(t) 7 9. a + jw (b) 7I e- 7tu(t) 7 + jw Direct inversion using the inverse Fourier transform formula is very difficult. Xb(w) = 6(w + 7) + 6(o - 7), Xb(t) = - Xb(w)ej do = -. e- It + e = cos 7t [6(w + 7) + 6 (w - 7)]ei-' dw (c) From Example 4.8 of the text (page 9), we see that However, note that 37 a e alti 9 _a a + W since ax(t) ax(t)e -j" dt = a ax(w) x(t)e -jw dt = ax(w) (d) Thus, - -e -" l a 7 a + (0 or 9 +. Xa(w)Xb(W) = Xa(w)[6(o + 7) + 6(w - 7)] = X(-7)(co + 7) + Xa(7)6(w - 7) XdCO)= - (w +7) j7 7 +j7 b(o7) f*[ ±7+ xaot = f-.7 6(o + 7) + 7r j7 b(co - 7) ej" do.. XA(t) = - e 7 ' + - ej_ ' ir7-j7 xr7+j7 Note that Thus 7 +j7 7 -jr/4 7 -j7 = =_ ( 7 - e -3t 6, +jir/4 XA(t) = - - r7 - [e -j(7t -/4) + ei( 7 t-/4)] = -- 77r cos 7t - -) 4

8 of the text (page 9), we see that However, note that 37 a e alti 9 _a a + W since ax(t) ax(t)e -j" dt = a ax(w) x(t)e -jw dt = ax(w) (d) Thus, - -e -" l a 7 a + (0 or 9 +.

11 Continuous-Time Fourier Transform / Solutions S8- we ~j 3 w, 0 5 w <, (e) Xe(w) = we -j 3 w, -! (O 0, t0, elsewhere, x(t) = f X(w)e-' dw = i [J' we -jamei' do - we ~i 3 -ejwl dw Note that Substituting a = j(t - x(t) e ax xe" f dx =, -(ax -) 3) into the integrals, we obtain j(t - 3)w I egj(t - 3)w 0 = - L( (j(t - 3)w - ) 0 (j )) (j(t - 3)w - ) J r (j(t - 3)) 0 (j(t - 3)) which can be simplified to yield [ cos (t - 3) - sin (t - 3) x(t) = 7r (t - 3) + (t - 3)_ Solutions to Optional Problems S8.7 (a) Y(w) = y(t)e -j dt x(r)h(t - -) dr e -j'' dt S ft x(,r)h(t - )e -j"' dr dt (b) Let r t - T and integrate for all r and r. Then Y(w) = TT=-.fro x(r)h(r)e -jw( r-+*t) dr dr = x(r)e -' d-r {r=-0 h(r)e ~'j dr = X(w)H(w) S8.8 (a) Using the analysis equation, we obtain rt/_ Ilk = I T/ e(t)e -jk(r/t) t dt-t T -T/ T Thus all the Fourier series coefficients are equal to /T. (b) For periodic signals, the Fourier transform can be calculated from ak as X(w) = ak w-t k=-00

3)w 0 = - L( (j(t - 3)w - ) 0 (j )) (j(t - 3)w - ) J r (j(t - 3)) 0 (j(t - 3)) which can be simplified to yield [ cos (t - 3) - sin (t - 3) x(t) = 7r (t - 3) + (t - 3)_ Solutions to Optional Problems

12 Signals and Systems S8- In this case, rk PM=T 6 T P(wj) Tr T _47 _ n 0 _ 4 T T T T (c) We are required to show that Substituting for p(t), we have Figure S8.8 (t) = x(t )* p(t) x(t)* p(t) = xt * kt) k Ot Using the associative property of convolution, we obtain x(t) *p(t) = T, [x(t) * b(t - kt)] k= -o From the sifting property of b(t), it follows that x(t)* px)o x(t - kt) =t(t) 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) = X(-) - 7rk -Tk= =ji -- X(w)5(c xk Since each summation term is nonzero only at w = 7rk/T, *0 7r rk xk i = y=) T X( T B From this expression we see that the Fourier series coefficients of t(t) are ak = - X, T i which is consistent with our previous discussions. T

follows that x(t)* px)o x(t - kt) =t(t) k= Thus, x(t) * p(t) is a periodic repetition of x(t) with period T. (d) From Problem P8.

13 MIT OpenCourseWare Resource: Signals and Systems Professor Alan V. Oppenheim The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our Terms of Use, visit:

OpenCourseWare, but has been provided by the author as an individual learning resource.

9 Fourier Transform Properties

9 Fourier Transform Properties The Fourier transform is a major cornerstone in the analysis and representation of signals and linear, time-invariant systems, and its elegance and importance cannot be overemphasized.