University of Illinois Spring 2004ECE 361: First MidSemester ExamThursday February 26, 2004, 8:30 a.m. – 9:50 a.m.This is a closed-book closed-notes examination except that one 8.5” × 11” sheet of notes is per-mitted: both sides may be used. Tables of integrals, calculators, laptop computers, PDAs, iPods,cellphones, e-mail pagers, etc. are neither needed nor permitted.This Examination contains three problemsThroughout this exam, an “AWGN channel” means an additive white Gaussian noise channel inwhich the noise process is independent of the signals being transmitted, and has two-sided powerspectral density N0/2 volts2/Hz.1. Consider a binary digital communication syste m operating over an AWGN channel usingsignalss0(t) =A, 0 ≤ t < T,0, otherwise,and s1(t) =At/T, 0 ≤ t < T,0, otherwise.(a) Suppose that the receiver consists of a linear filter with impulse response h(t) = s0(t)followed by a sampler at time t = T , and that the minimax threshold is used. What isthe minimax error probability achieved by this receiver?(b) Repeat part (a) for the case where h(t) = s1(t). The sampling time remains T and theminimax threshold appropriate for this system is used.(c) State the optimum minimax error probability achievable with these signals and provethat it is smaller than your answers to parts (a) and (b).2. (a) The signals s0(t) and s1(t) are used to communicate over an AWGN channel. The signalenergies E0and E1respectively are such that E0> E1> 0. You are allowed to choosethe signals subject to the above energy constraint. How would you choose the signals soas to achieve minimax error probability that is as small as possible? State clearly whatthis minimum error probability is.(b) Now suppose that signals s2(t) and s3(t) of energies E0and E1respectively togetherwith signals s0(t) and s1(t) occupy a two-dimensional signal space. Notice that even-subscripted s ignals have energy E0and odd-subscripted signals have energy E1. Onceagain, you are allowed to choose signals subject to the stated energy constraint. What4-ary signal constellation minimizes the symbol error probability? You can re-choosesignals s0(t) and s1(t) if you want to in this part. Explain your reasoning in arrivingat your answer (a formal proof that the symbol error probability is minimized is notrequired.)(c) For your constellation of part (b), how would you assign bits b0b1to the four signals soas to minimize the average bit error probability? Explain your reasoning.(d) Now consider four additional signals s4(t) and s6(t) (of energy E0) and s5(t) and s7(t)(of energy E1) in the same two-dimensional space as in part (b). What 8-ary signalconstellation minimizes the symbol error probability? Once again, you can re-choosesignals s0(t)-s3(t) if you want to in this part. Explain your reasoning in arriving at youranswer (a formal proof that the symbol error probability is minimized is not required.)(e) For any given value of E1, for what value(s) of E0can the eight signals of part (d) bechosen to so as to lie at the intersections of uniformly spaced gridlines? Note that thesignal set thus resembles a QAM signal set except that not every gridline intersection isrequired to have a signal point. What is the minimum distance dminof this QAM-likesignal set?3. Consider a c ommunication system that uses a 22m-ary rectangular QAM signal constellationwith nearest neighbors d apart to communicate over an AWGN channel.(a) What is the average energy per bit?(b) What is the average symbol error probability?(c) Now consider a receiver that operates as follows. If the sample (X0, X1) is at distance< d/2 from a signal point sj, then the receiver decides that sjwas the transmittedsignal. (Note that the sample cannot be within d/2 of two different signal points.) Ifthe sample is at distance ≥ d/2 from all 22msignal points, the receiver outputs a specialsymbol ? indicating that it is unable to make a decision. (A re-transmission is usuallyrequested in such cases but that’s a matter for another course).Show that the probability of a correct decision by such a bounded-distance demodulatoris 1 − exp(−d2/4N0) and compare this to the probability of a correct decision by theusual optimum