UNIVERSITY OF AKRON DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING 4400: 34 INTRODUCTION TO COMMUNICATION SYSTEMS - Spring 07 SAMPLE FINAL EXAM TIME: hour 30 minutes INSTRUCTIONS: () Write your nme on ech pge () Clerly show ll your work, in steps, in the spce ssigned for ech question. (3) Answers without resoning will not receive ny credit. (4) With some thinking, there re lwys short cuts to the nswer. If you re tking too much time to nswer question it is dvisble to leve it nd return to it fter you hve ttempted the rest of the problems. (5) If you feel tht you do not understnd question or it is mbiguous, plese sk bout it. (6) The list of the needed equtions nd formule is ttched to the end of this booklet. (7) Relx nd Perform. GOOD LUCK Question Totl Grde Your Score 0 0 3 30 4 0 Totl 80
Nme: Problem (0 points) Consider the signl x(t) = e t u(t). ) Clculte its energy in time nd frequency domins nd verify the Prsevl s theorem for it. b) Clculte its utocorreltion nd ESD functions. c) Clculte the essentil bndwidth of this signl where the essentil bndwidth must contin 95% of the signl energy. d) If we pss x(t) through system whose impulse response is h(t) = δ(t 4), find the output signl ESD nd its energy.
Nme: Problem (0 points) ) The signl shown in the figure is supposed to be trnsmitted using DSB+C modultion. Clculte the required modultion index nd mximum power efficiency of this modultion. b) We hve superheterodyne AM receiver with chep non-idel locl oscilltor. In ddition to f LO, this locl oscilltor lwys genertes nother frequency tone t f LO -0KHz. The intermedite frequency f IF is set to be t 00KHz. Plot the structure of the superheterodyne receiver. Assuming tht the receiver is tuned to receive chnnel t center frequency f c = 350KHz, clculte f LO. Clculte possible imge frequencies.
Nme: Problem 3 (30 points) An ngle modulted signl with frequency f c = 50KHz is described by the eqution, ϕ FM (t) = 0 cos(π f c t + 0.0sin 000πt) ) Find the power of the modulted signl nd its frequency devition. Estimte the bndwidth of the modulted signl. b) Design nd sketch the block digrm of n Armstrong indirect FM modultor tht generte WBFM signl with crrier 96.3MHz nd frequency devition 0.48KHz using the bove signl. Only frequency doublers re vilble s frequency multipliers. In ddition, n oscilltor with djustble frequency from 3MHz to 4MHz is lso vilble for mixing, long with bndpss filters of ny specifictions. c) To demodulte the WBFM signl generted in prt (b), we use slope detector s shown in the figure which is in fct high pss filter. We know tht the 3dB frequency of this filter is t /RC. Assuming R = KW, find suitble vlue for the cpcitor, C.
Nme: Problem 4 (0 points) Consider the signl x t = cos 0πt + cos (30πt). ) Find out the Nyquist smpling rte for this signl. b) If we smple this signl t 5Hz, nd then from the smpled signl try to reconstruct the originl signl by pssing in through LPF with bndwidth 8Hz, find out the signl in time domin t the output of the LPF. Sketch the spectrum of this signl. Is it perfect reconstruction? Explin.
e ± jx e jx + e jx cos x = = cos x ± j sin x sin x = ( cos x ) cos x = ( + cos x ) Signl Power: Pg = Tlim Signl Energy: Eg = e jx e jx sin x = j T T / T / x(t)x * (t)dt = x(t)x * (t)dt = dx x = ( ) tn ( ) +x Sg ( f )df = Rg (0) Eg ( f )df = Rg (0) Fourier Trnsform nd Inverse Fourier Trnsform: X( f ) = x(t) = x(t)e j πft dt X( f )e j πft df mmx mmin A + # mmx + mmin 4400:34 Exm 5/6 usefulpower Ps AM power efficiency: η = = totlpower Pc + Ps ± jx e = cos x ± j sin x cos x cos y = 0.5[cos(x + Δfy) + cos(x y)] jx jx = FM nde jxpm index: where B is the messge bndwidth + e modultion e jx e β B cos x = sin x = AM modultion index: µ = sin x =k f mp cos x ) Δf = ( for FM π j p x ) cos x = k( +mcos nd Δf = p for π cos x = cos x sin x PM x x xe dx = e x x sin(x)dx = cos(x) + sin(x) FM nd PM bndwidth: B = B(β + ) x FM T / Signl Power: P = lim T / x(t)x * (t)dt Signl Energy: E g = x(t)x * (t)dt Nyquist Rteg ist BT(B is the nlog signl bndwidth). mp Fourier Trnsform nd Inverse Fourier Trnsform: j πft X( f ) = N Quntiztion Noise Power: =x(t)e 3L dt x(t) = X( f )e j πft df