EECE 340 Exam, Spring 2008

Problem | AMERICAN UNIVERSITY OF BEIRUT EECE440 - Sample problems 3 (Karameh) (i) We have . 1 AX(e™) = : OO Tee ve = ee —_ “lle =| Taking the inverse Fourier transform, we obtain yin] =3 (3) * ln) -2 (6) u{n]. nay 1 1 me wy 1 1 res) = fea =| 4 2 Therefore, (ii) We have Therefore, 3 i-de [ote flea ‘Taking the inverse Fourier transform, we obtain win) =4 (5) um -2 ( i) ula] —3(n +1) (3) * un X(e™) = ar > Sl ~ (2k + 1). ke-oo (ili) We have Therefore, ¥{e) aa > dtu - (2k + ) ; - | sited - nS Sw — (2k + 1x) 3 ‘Taking the inverse Fourier transform, we obtain a[n) = xy", Problem 2 (2) Since the two systema are cascaded, the frequency response of the overall system Ae) = Bye Hy (ety 2— evi 1+ pee Therefore, the Foutier transforms of the input and output of the overall Telated by syatem ¥{e™*) _ 8-e X(e™) 14 Tei Cross-multiplying and taking the inverse Fourier transform, we get 1 vin] + Sul — 3] = 2rfn} - 2fp — 1], b) We may rewrite the overall frequency response as ate) = 48, G+ivays | U- iva ry rl letee Tyo a pense ju Taking the inverse Fourier transform we get An] = : (-3) um +3 tiv (5) ty + inva (Get) at Problem 3 - iy x > 202) yw. (rviewy s ¥f< [e y alr) < fn) 2), Yah tn C Ly Cry Yad ere wl 4 hed cow ve Wee b-0F VY oVld<[2lol2lor} ac: fr fy. Dery ots Oe a a C es ¥,(e) - {eo IiLoo a2) m{n) 2 ~9,[r) S) ii ~ >be 2 + YO Oy Ye ) 2b 2 lar Y(E-r3 s) Pri. (210 12} 01) Yy@-r,\. [01202821] Ya(Orns\ - (\ ol 2eost) a BOY clQ ars ye & - 2 7 an) 4 Ar 7 y,{h-r} + [oorlolto] > yey. \- > _— A 3 4 Problem 4 (a) x[n] = (4)" un + 2] 1 n z[n] = (3) u[n + 2] 1» 1, = (3) *(g) Puln + 2] 1 n DTFT 1 (3) un] _ 2e-i0 s[n + 2] et e922. 5(¢59) . QeI22 x(e™) ~ 1- ze I (b) x[n] = (nm — 2)(u[n + 4] — uf[n — 5]) 92 DIFT i uln + 4] — u[n — 5] ——> sin( sin($) ns[n] aw 1-552) [nl] _ j d sin(%2) 7 yin dQ sin(2) sin(#) Problem 5 (a) X(ei%) = jsin(4Q) — 2 _ so" _ fesin 9 1 1 a[n] = zn +4) - 56ln — 4] — 26[n] (b) X(ei%) = [eee] ® [2a] sin(2) sin(2) Let the first part be A((e3®)), and the second be B(e®). { 1 |n-2|<7 a[n] = : 0, otherwise < Yn] = {| nse 0, otherwise x((e%)) = A((e))@ Bie) 2{n] = 2ra{n]b[n] (e In| <3 a(n] 0, otherwise