Question

Problem 3. 0 Figure 2 Given: f(t) = { 2.5, -1.5 <=<= 1.5 f(t) = { 0 otherwise See figure(2) above. A) Find the Fourier transform for f( (see figure 2) and sketch its waveform. B) Determine the values of the first three frequency terms (w1, W2, W3) where F(w) = 0. C) Given x(t) = 1.58(-0.8) edt Determine whether or not Fourier transform exists for x(t). If yes, find the Fourier transfe not explain why it does not. Problem 4. Compute numerical solutions for the following equations: 3 2 1) (t = -0 to +00) 8(t) (t - 2t + 10t + 1) dt 2)S (t = 0 to 1/2) sin(t) 8(t-12) dt 3) S (t = 0 to +00) 8( t + 1/2 ) sin( t - 1)dt 10 2t 4)(t = 0 to 0) 10 e dt 5S (t = -0 to +00) cos( t/4) 8( t - 2) dt

Answer 1

Giveni DC) 125 1. Sti= 205 rect ( +/- 2) -1.5 106 T Arect (th] ET AT Sa (WI) 2.55eck (+13) > (2.4)(3) sa (web) ETA 7.5 sa ca Et to sa (15w) LETA 15 Sein Closw) . 1050 fah 6 . . . 115 wa Fo tos sin €166) 7.6 sin ( to 5 kw) ...1.5,800 f(t) 25 sinc (1.5 W)

. -67-45ar o anuk GÃ 87. . 75 sin (1.50) = F(W) F(W) = 0 => sin (1.5w] = 0 doy w = 624,40, =l for ws*,341 - Fluor=0 for wo = 20 W2 = 45 is = 6* © xati = f 15 sct-o.g) e just dit a CA) f (t-böldt: alto); teto Etz 0.3 Otherwise! X(+) = 1,5 ēju Co.B) 20+) = 1.52 10.8w - For x64) Fourier kans form in de redefined but it contains impulse for because 264 is power signal. 105 As an fcil 1.5 210.90 FT, ens (w+0.8)

C+3 - 2++100+1) det 03-2.02 +1000)+12=1 ſ sint 8 (+-)dt = undefined because at boundaries integral value is under ched $ 8C++^]2) sin(t-1.) dt = 0. Because t==12 is outside the limit. f roest dt zo area at a point is zero o. ☺ si cos Cat $64-2.)ikan cos (4.2) ::..cos (A/2)