EE160. Spring 2003. San Jos´e State UniversitySolution of sample final examPROBLEM 1.(a)f2(t)1-1-1s1s2f1(t)R2R1(b) Since f1(t)andf2(t) ha ve disjoint domains, they are orthogonal. The signals are normalizedto unit energy if10f21(t)dt = c21=1 −→ c1=1,and21f22(t)dt = c22=1 −→ c2=1.Note that ¯s1=(0, 1) and ¯s2=(−1, −1). It follows thats1(t)=f2(t), and s2(t)=−f1(t) − f2(t).s1(t)s2(t)t112t12-1(c) d212=[0−(−1)]2+[1− (−1)]2=5. Consequently, with N0/2=5/2,Pe= Q 52N0= Q 12=0.241PROBLEM 2.Rb=1Tb=1× 106. Therefore, Tb=1× 10−6. In addition, we are given A =10−3volt. It followsthat the average energy of the transmitted pulses is E = A2Tb/2=0.5 × 10−12W.(a) The probabilit y of error isPe= Q EN0≤ 3.17 × 10−5.Using Table 4.1, p. 152, of textbook, we find that EN0≥ 4.0 −→ N0≤E16=132× 10−12.(b) Evidently, N = 1. The basis function is given in the figure b elow:f1(t)1TbtTb(c)R1R2s1s2f1A Tb00.5x10-6= 10-6PROBLEM 3.(a) E =12(A)2+12(−A)2= A2. It follows that A =√E.(b) Conditional PDF of Y :fY(y|U =0) =1√πN0e−1N0(y−A)2=1√πe−(y −√E)2,i.e., given that U = 0 is sent, the decision variable Y is Gaussian distributed with mean√E andvariance12.2(c) The threshold is set to 0.1. (Note: This is not the optimal setting.) The correspondingprobability of error isPe=12Pr {Y ≥ 0.1|U =1} +12Pr {Y<0.1|U =0}=12∞0.11√πe−(y +√E)2dy +120.1−∞1√πe−(y −√E)2dy=12Q√E +0.11/2+ Q√E − 0.11/2s1s2f1-E
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