Question

NULE. (TJLIUSE UUUL, JUNI 2"Sue ulleal SHCCL No electronic device. 14 I Ime: 2:00-3:1 spil, 3/13/2018, RIN! NAL-S/TIU 1. Consider that the signal m(t) = cos(it) is naturally sampled by s(t) = P(t-n) where the sampling period is 1 second and the pulse is rectangular p(t) = 1 for t E (0,0.5) and (t) = 0 otherwise, (a) (18%) Draw s(t), m(t), and qt). (b) (7%) Determine the Fourier series form III for (C). m(t) - -→ 4t) (c) (5%) Determine the math expression of natural-top sampled signal (0) (d) (12%) Determine the Fourier transforms of m(t), s(t), and qt). (e) (5%) Draw the Fourier transform of qt) forf € (-3,3). s(t) (f) (8%) If (t) passes through a low-pass filter with frequency response U() = 1 for ISIS 1 Hz and U ) = 0 for | > 1 Hz, determine the output signal of U )

Answer 1

somn lit considera menage Signal met) = (os (At) sampled by signal scto = Pct-n) such that 91(t) = S(t) mct). ne-a ~ S(+). Amit). MAI 10 160 2-19 ---f L atino OOSS, - - As sct) is periodic with period T=1, Wo = 27 = 27 DE 3(t) = 2 an einwati n=-00 an= + , Plt) e inust ot = +11 e-jowot dt = -1 [e-jowot 71 jnwoT

an the [e-in yoo -603 [e-aj-] an = (Conson) - Jj , sct) = É un minut 8 wor=2x d) Fourer tranx form of meth'= cos(at) mit) = etjt te-ajt MCF) = + [36-1/2) +SC++)]-O SE+) = an ezanit. 5ff) = Ê an. SC-n). - - 2 n- in=-60 auct) = do's(t) x mit) n=- qJCt) = mF) * SC4), = mg) * Žan $cF=n) 96) = Žan mcm) = an(scensov :)) C) as q/lt) = mas set) = m(t) žame zaint 1/(t)= an cosfat) e znjnt n=-20 n= 0 - n=-ao n=-40

946) = { an [8cn-+) + scn++)]. as his integer away. scn-) = 0 8(n ++) = 0 QUCt). = 0 -3,56% 3 P QVCF) tow pass filter. &). When we pan qit) through hct). then we get output uit) Ult) = n(t) * VLt). TUGA) = HCf) QVCF) = 0 hence Signal is not recovered.