1 4 a n 1 2 with prob 1 4 a n with prob 1 2 by

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14a+n+12with Prob.14a+nwith Prob.12By symmetry,Pe=P(e|a=1)=P(e|a= −1), hence,Pe=P(e|a= −1)=12P(n1>0)+14Pparenleftbiggn32>0parenrightbigg+14Pparenleftbiggn12>0parenrightbigg=12Qparenleftbigg1σnparenrightbigg+14Qparenleftbigg32σnparenrightbigg+14Qparenleftbigg12σnparenrightbiggProblem 10.31)If the transmitted signal isr(t)=summationdisplayn=−∞anh(tnT)+n(t)then the output of the receiving filter isy(t)=summationdisplayn=−∞anx(tnT)+ν(t)wherex(t)=h(t)⋆h(t)andν(t)=n(t)⋆h(t). If the sampling time is off by 10%, then the samplesat the output of the correlator are taken att=(m±110)T. Assuming thatt=(m110)Twithoutloss of generality, then the sampled sequence isym=summationdisplayn=−∞anx((m110TnT)+ν((m110)T)If the signal pulse is rectangular with amplitudeAand durationT, thenn=−∞anx((m110TnT)is nonzero only forn=mandn=m1 and therefore, the sampled sequence is given byym=amx(110T)+am1x(T110T)+ν((m110)T)=910amA2T+am1110A2T+ν((m110)T)The power spectral density of the noise at the output of the correlator isSν(f)= Sn(f)|H(f)|2=N02A2T2sinc2(fT)428
Thus, the variance of the noise isσnu2=integraldisplay−∞N02A2T2sinc2(fT)df=N02A2T21T=N02A2Tand therefore, the SNR isSNR=parenleftbigg910parenrightbigg22(A2T)2N0A2T=811002A2TN0As it is observed, there is a loss of 10 log1081100= −0.9151 dB due to the mistiming.2)Recall from part a) that the sampled sequence isym=910amA2T+am1110A2T+νmThe termam1A2T10expresses the ISI introduced to the system. Ifam=1 is transmitted, then theprobability of error isP(e|am=1)=12P(e|am=1,am1=1)+12P(e|am=1,am1= −1)=12radicalbigπN0A2TintegraldisplayA2T−∞eν2N0A2T+12radicalbigπN0A2Tintegraldisplay810A2T−∞eν2N0A2T=12QradicalBigg2A2TN0+12Qradicaltpradicalvertexradicalbtparenleftbigg810parenrightbigg22A2TN0Since the symbols of the binary PAM system are equiprobable the previous derived expression isthe probability of error when a symbol by symbol detector is employed. Comparing this with theprobability of error of a system with no ISI, we observe that there is an increase of the probabilityof error byPdiff(e)=12Qradicaltpradicalvertexradicalbtparenleftbigg810parenrightbigg22A2TN012QradicalBigg2A2TN0Problem 10.41)Taking the inverse Fourier transform ofH(f), we obtainh(t)= F1[H(f)]=δ(t)+α2δ(tt0)+α2δ(t+t0)Hence,y(t)=s(t)⋆h(t)=s(t)+α2s(tt0)+α2s(t+t0)2)If the signals(t)is used to modulate the sequence{an}, then the transmitted signal isu(t)=summationdisplayn=−∞ans(tnT)429
The received signal is the convolution ofu(t)withh(t). Hence,y(t)=u(t)⋆h(t)=

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