Question

7. The signal x(t) shown below is modulated (multiplied) by cos(10nt). Find the Fourier transform of x(t)cos(10nt) and neatly sketch the magnitude? Useful transform pairs. rect (9) = t sinc (); «(t)cos (Wgt) }(x(w+wo) + X(w – wo)); «(t – to) ~X(w)e-juto (10 points) x(+) 1 t

Answer 1

xH) it -1 Let two function alt=2, 4)+22 (4) H2 (4) X,(+) ro + t 1 Step 1 4,4) H, (++%) [ Left Shift by / I ti (let a new signally, (f) - % 1². We know, At sac( we 2/2 2/ A=1, ?= 1 [ by compare with 924] "Y, (W) sac ( We know, xict) X, (W) 2, (t-to) - X, (W) e gwo to time shifting CS Scanned with CamScanner

X, (+) = 4,1+-£) il tus) to X, (W) = Y/(w).e è i wo to XL (W) = Y We -wat X, (W) - 0 saclü Step ② H2H) 1 -1 s --X(+) 122 H 1 -1 -- Hz (+-+) ya (t) = -1/2 xt 112 CS Scanned with CamScanner

(3 We know Arect (the) to - 22 2% * Alsace we by compare, A = 1) て=1 1 sac( we Y2(W) = sacl we -X2(W) sac (w Step jwet zyk)= 424 +b] e X(W) = x1(W) + X2 (W) vot i wote we wel Sac ( Sac X (W) = e Step If delt is multiply by cus Lott then in frequency domain it is convalulated, Hi(t). Hg(t) X;(W) * Xa(w) $(W-101) + f (w +501) $(w=107) Cuslent) 2 CS Scanned with CamScanner

then XH.cos lott מן X(W) * f(w-167) +$(w+101 2 wa ] 2(W)*fw-wo) = x (w wa) x (w+lowy = X (W-101) + 20 2 +j (WHOA! e = 3 [4ac (uzun) - Pleak - sec (women 2 ANS CS Scanned with CamScanner