Problem 5.12 :The correlation of the two signals in binary FSK is:ρ =sin(2π∆fT)2π∆fTTo find the minimum value of the correlation, we set the derivative of ρ with respect to ∆fequal to zero. Thus:ϑρϑ∆f=0=cos(2π∆fT)2πT2π∆fT−sin(2π∆fT)(2π∆fT)22πTand therefore :2π∆fT =tan(2π∆fT)Solving numerically (or graphically) the equation x =tan(x), we obtain x =4.4934. Thus,2π∆fT =4.4934 =⇒ ∆f =0.7151Tand the value of ρ is −0.2172.We know that the probability of error can be expressed in terms of the distance d12between thesignal points, as :Pe= Qd2122N0where the distance between the two signal points is :d212=2Eb(1 − ρ)and therefore :Pe= Q2Eb(1 − ρ)2N0= Q1.2172EbN0Problem 5.13 :(a) It is straightforward to see that :Set I : Four − level PAMSet II : OrthogonalSet III : Biorthogonal(b) The transmitted waveforms in the first set have energy :12A2or129A2. Hence for the firstset the average energy is :E1=14212A2+2129A2=2.5A2All the waveforms in the second and third sets have the same energy :12A2.Hence :E2= E3= A2/2(c) The average probability of a symbol error for M-PAM is (5-2-45) :P4,P AM=2(M − 1)MQ 6Eav(M2− 1)N0=32QA2N0(d) For coherent detection, a union bound can be given by (5-2-25) :P4,orth< (M −1) QEs/N0=3QA22N0while for non-coherent detection :P4,orth,nc≤ (M − 1) P2,nc=312e−Es/2N0=32e−A 2/4N0 (e) It is not possible to use non-coherent detection for a biorthogonal signal set : e.g. withoutphase knowledge, we cannot distinguish between the signals u1(t)andu3(t)(oru2(t)/u4(t)).(f) The bit rate to bandwidth ratio for M-PAM is given by (5-2-85) :RW1=2log2M =2log24=4For orthogonal signals we can use the expression given by (5-2-86) or notice that we use a symbolinterval 4 times larger than the one used in set I, resulting in a bit rate 4 times smaller :RW2=2log2MM=1Finally, the biorthogonal set has double the bandwidth efficiency of the orthogonal set :RW3=2Hence, set I is the most bandwidth efficient (at the expense of larger average power), but set IIIwill also be satisfactory.P r oblem 5.13 :Problem 5.14 :The following graph shows the decision regions for the four signals :✻✲ ✲✻❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅ABCDU1U2A = U1 > +|U2|B = U1 < −|U2|C = U2 > +|U1|D = U2 < −|U1|W2W1ABDCæAs we see, using the transformation W1= U1+ U2,W2= U1−U2alters the decision regions to :(W1> 0,W2> 0 → s1(t); W1> 0,W2< 0 → s2(t); etc.) . Assuming that s1(t) was transmitted,the outputs of the matched filters will be :U1=2E + N1rU2= N2rwhere N1r,N2rare uncorrelated (Prob. 5.7) Gaussian-distributed terms with zero mean andvariance 2EN0. Then :W1=2E +(N1r+ N2r)W2=2E +(N1r− N2r)will be Gaussian distributed with means : E [W1]=E [W2]=2E, and variances : E [W21]=E [W22]=4EN0. Since U1,U2are independent, it is straightforward to prove that W1,W2areindependent, too. Hence, the probability that a correct decision is made, assuming that s1(t)was transmitted is :Pc|s1= P [W1> 0] P [W2> 0] = (P [W1> 0])2=(1− P [W1< 0])2=1 − Q2E√4EN02=1 − QEN02=1 − Q2EbN02where Eb= E/2 is the transmitted energy per bit. Then :Pe|s1=1− Pc|s1=1− 1 − Q 2EbN02=2Q 2EbN01 −12Q 2EbN0This is the exact symbol error probability for the 4-PSK signal, which is expected since thevector space representations of the 4-biorthogonal and 4-PSK signals are identical.Problem 5.15 :(a) The output of the matched filter can be expressed as :y(t)=Rev(t)ej2πfctwhere v(t) is the lowpass equivalent of the output :v(t)=t0s0(τ)h(t − τ)dτ =t0Ae−(t−τ)/Tdτ = AT1 − e−t/T, 0 ≤ t ≤ TT0Ae−(t−τ)/Tdτ = AT (e − 1)e−t/T,T≤ t(b) Asketchofv(t) is given in the following figure :0 T tv(t)(c) y(t)=v(t)cos2πfct, where fc>> 1/T. Hence the maximum value of y corresponds to themaximum value of v, or ymax= y(T )=vmax= v(T )=AT (1 − e−1).(d) Working with lowpass equivalent signals, the noise term at the sampling instant will be :vN(T )=T0z(τ)h(T − τ)dτThe mean is : E [vN(T )] =T0E [z(τ)] h(T − τ)dτ =0, and the second moment :E|vN(T )|2= ET0z(τ)h(T − τ)dτT0z∗(w)h(T − w)dw=2N0T0h2(T − τ)dτ= N0T (1 − e−2)The variance of the real-valued noise component can be obtained using the relationship Re[N]=12(N + N∗)toobtain:σ2Nr=12E|vN(T )|2=12N0T (1 − e−2)(e) The SNR is defined as :γ =|vmax|2E|vN(T )|2=A2TN0e − 1e +1(the same result is obtained if we consider the real bandpass signal, when the energy termhas the additional factor 1/2 compared to the lowpass energy term, and the noise term isσ2Nr=12E|vN(T )|2)(f) If we have a filter matched to s0(t), then the output of the noise-free matched filter will be :vmax= v(T )=T0s2o(t)=A2Tand the noise term will have second moment :E|vN(T )|2= ET0z(τ)s0(T − τ)dτT0z∗(w)s0(T − w)dw=2N0T0s20(T − τ)dτ=2N0A2Tgiving an SNR of :γ =|vmax|2E|vN(T )|2=A2T2N0Compared with the result we obtained in (e), using a sub-optimum filter, the loss in SNR isequal to :e−1e+112−1=0.925 or approximately 0.35 dBProblem 5.16 :(a) Consider the QAM constellation of Fig. P5-16. Using the Pythagorean theorem we can findthe radius of the inner circle as:a2+ a2= A2=⇒ a =1√2AThe radius of the outer circle can be found using the cosine rule. Since b is the third side of atriangle with a and A the two other sides and angle between then equal to θ =75o, we obtain:b2= a2+ A2− 2aA cos 75o=⇒ b =1+√32A(b) If we denote by r the radius of the circle, then using the cosine theorem we obtain:A2= r2+ r2− 2r cos 45o=⇒ r =A2 −√2(c) The average transmitted power of the PSK constellation is:PPSK=8×18×A2 −√22=⇒ PPSK=A22 −√2whereas the average transmitted power of the QAM constellation:PQAM=18 4A22+4(1 +√3)24A2=⇒ PQAM=2+(1+√3)28A2The relative power advantage of the PSK constellation over the QAM constellation is:gain =PPSKPQAM=8(2 + (1 +√3)2)(2 −√2)=1.5927 dBProblem 5.18 :For binary phase modulation, the error probability isP2= Q2EbN0= QA2TN0With P2=10−6we find from tables thatA2TN0=4.74 =⇒ A2T =44.9352 × 10−10If the data rate is 10 Kbps, then