Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống
1
/ 20 trang
THÔNG TIN TÀI LIỆU
Nội dung
The structure of this optimal receiver is shown in the next figure. The optimal receivers, derived in this problem, are more costly than those derived in the text, since N is usually less than M, the number of signal waveforms. For example, in an M-ary PAM system, N = 1 always less than M. ☛ ✂ ❘ ☛ ✂ ❘ ☛ ✂ ❘ ❦ ❦ ❦ ❅ ❅ ❅ ❅ ❅ ❅ ✲ ✲ ✲ ❄ ❄ ❄ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ t = T t = T t = T h M (t)=s m (T −t) h 2 (t)=s 2 (T −t) h 1 (t)=s 1 (T −t) . largest the Select c M c 2 c 1 r ·s 1 r ·s M r ·s 2 r(t) . . . . Problem 7.21 1) The optimal receiver (see Problem 7.20) computes the metrics C(r, s m )= ∞ −∞ r(t)s m (t)dt − 1 2 ∞ −∞ |s m (t)| 2 dt + N 0 2 ln P(s m ) and decides in favor of the signal with the largest C(r, s m ). Since s 1 (t)=−s 2 (t), the energy of the two message signals is the same, and therefore the detection rule is written as ∞ −∞ r(t)s 1 (t)dt s 1 > < s 2 N 0 4 ln P (s 2 ) P (s 1 ) = N 0 4 ln p 2 p 1 2) If s 1 (t) is transmitted, then the output of the correlator is ∞ −∞ r(t)s 1 (t)dt = T 0 (s 1 (t)) 2 dt + T 0 n(t)s 1 (t)dt = E s + n where E s is the energy of the signal and n is a zero-mean Gaussian random variable with variance σ 2 n = E T 0 T 0 n(τ)n(v)s 1 (τ)s 1 (v)dτdv = T 0 T 0 s 1 (τ)s 1 (v)E[n(τ)n(v)]dτ dv = N 0 2 T 0 T 0 s 1 (τ)s 1 (v)δ(τ − v)dτdv = N 0 2 T 0 |s 1 (τ)| 2 dτ = N 0 2 E s Hence, the probability of error P (e|s 1 )is P (e|s 1 )= N 0 4 ln p 2 p 1 −E s −∞ 1 √ πN 0 E s e − x 2 N 0 E s dx = Q 2E s N 0 − 1 4 2N 0 E s ln p 2 p 1 178 Similarly we find that P (e|s 2 )=Q 2E s N 0 + 1 4 2N 0 E s ln p 2 p 1 The average probability of error is P (e)=p 1 P (e|s 1 )+p 2 P (e|s 2 ) = p 1 Q 2E s N 0 − 1 4 2N 0 E s ln 1 −p 1 p 1 +(1− p 1 )Q 2E s N 0 + 1 4 2N 0 E s ln 1 −p 1 p 1 3) In the next figure we plot the probability of error as a function of p 1 , for two values of the SNR = 2E s N 0 . As it is observed the probability of error attains its maximum for equiprobable signals. 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 P(e) Probability p SNR=1 0 db 0 1 2 3 4 5 6 7 8 x10 -24 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 P(e) Probability p SNR=100 20 db Problem 7.22 1) The two equiprobable signals have the same energy and therefore the optimal receiver bases its decisions on the rule ∞ −∞ r(t)s 1 (t)dt s 1 > < s 2 ∞ −∞ r(t)s 2 (t)dt 2) If the message signal s 1 (t) is transmitted, then r(t)=s 1 (t)+n(t) and the decision rule becomes ∞ −∞ (s 1 (t)+n(t))(s 1 (t) −s 2 (t))dt = ∞ −∞ s 1 (t)(s 1 (t) −s 2 (t))dt + ∞ −∞ n(t)(s 1 (t) −s 2 (t))dt = ∞ −∞ s 1 (t)(s 1 (t) −s 2 (t))dt + n s 1 > < s 2 0 where n is a zero mean Gaussian random variable with variance σ 2 n = ∞ −∞ ∞ −∞ (s 1 (τ) −s 2 (τ))(s 1 (v) −s 2 (v))E[n(τ)n(v)]dτ dv 179 = T 0 T 0 (s 1 (τ) −s 2 (τ))(s 1 (v) −s 2 (v)) N 0 2 δ(τ − v)dτ dv = N 0 2 T 0 (s 1 (τ) −s 2 (τ)) 2 dτ = N 0 2 T 0 T 0 2Aτ T − A 2 dτ = N 0 2 A 2 T 3 Since ∞ −∞ s 1 (t)(s 1 (t) −s 2 (t))dt = T 0 At T 2At T − A dt = A 2 T 6 the probability of error P (e|s 1 ) is given by P (e|s 1 )=P ( A 2 T 6 + n<0) = 1 2π A 2 TN 0 6 − A 2 T 6 −∞ exp − x 2 2 A 2 TN 0 6 dx = Q A 2 T 6N 0 Similarly we find that P (e|s 2 )=Q A 2 T 6N 0 and since the two signals are equiprobable, the average probability of error is given by P (e)= 1 2 P (e|s 1 )+ 1 2 P (e|s 2 ) = Q A 2 T 6N 0 = Q E s 2N 0 where E s is the energy of the transmitted signals. Problem 7.23 a) The PDF of the noise n is f(n)= λ 2 e −λ|n| The optimal receiver uses the criterion f(r|A) f(r|−A) = e −λ[|r−A|−|r+A|] A > < −A 1=⇒ r A > < −A 0 180 The average probability of error is P (e)= 1 2 P (e|A)+ 1 2 P (e|−A) = 1 2 0 −∞ f(r|A)dr + 1 2 ∞ 0 f(r|−A)dr = 1 2 0 −∞ λ 2 e −λ|r−A| dr + 1 2 ∞ 0 λ 2 e −λ|r+A| dr = λ 4 −A −∞ e −λ|x| dx + λ 4 ∞ A e −λ|x| dx = λ 4 1 λ e λx −A −∞ + λ 4 − 1 λ e −λx ∞ A = 1 2 e −λA b) The variance of the noise is σ 2 n = λ 2 ∞ −∞ e −λ|x| x 2 dx = λ ∞ 0 e −λx x 2 dx = λ 2! λ 3 = 2 λ 2 Hence, the SNR is SNR = A 2 2 λ 2 = A 2 λ 2 2 and the probability of error is given by P (e)= 1 2 e − √ λ 2 A 2 = 1 2 e − √ 2SNR For P (e)=10 −5 we obtain ln(2 ×10 −5 )=− √ 2SNR =⇒ SNR = 58.534=17.6741 dB If the noise was Gaussian, then P (e)=Q 2E b N 0 = Q √ SNR where SNR is the signal to noise ratio at the output of the matched filter. With P (e)=10 −5 we find √ SNR=4.26 and therefore SNR = 18.1476 = 12.594 dB. Thus the required signal to noise ratio is 5 dB less when the additive noise is Gaussian. Problem 7.24 The energy of the two signals s 1 (t) and s 2 (t)is E b = A 2 T The dimensionality of the signal space is one, and by choosing the basis function as ψ(t)= 1 √ T 0 ≤ t< T 2 − 1 √ T T 2 ≤ t ≤ T 181 we find the vector representation of the signals as s 1,2 = ±A √ T + n with n a zero-mean Gaussian random variable of variance N 0 2 . The probability of error for antipodal signals is given by, where E b = A 2 T . Hence, P (e)=Q 2E b N 0 = Q 2A 2 T N 0 Problem 7.25 The three symbols A, 0 and −A are used with equal probability. Hence, the optimal detector uses two thresholds, which are A 2 and − A 2 , and it bases its decisions on the criterion A : r> A 2 0: − A 2 <r< A 2 −A : r<− A 2 If the variance of the AWG noise is σ 2 n , then the average probability of error is P (e)= 1 3 A 2 −∞ 1 2πσ 2 n e − (r−A) 2 2σ 2 n dr + 1 3 1 − A 2 − A 2 1 2πσ 2 n e − r 2 2σ 2 n dr + 1 3 ∞ − A 2 1 2πσ 2 n e − (r+A) 2 2σ 2 n dr = 1 3 Q A 2σ n + 1 3 2Q A 2σ n + 1 3 Q A 2σ n = 4 3 Q A 2σ n Problem 7.26 The biorthogonal signal set has the form s 1 =[ E s , 0, 0, 0] s 5 =[− E s , 0, 0, 0] s 2 =[0, E s , 0, 0] s 6 =[0, − E s , 0, 0] s 3 =[0, 0, E s , 0] s 7 =[0, 0, − E s , 0] s 4 =[0, 0, 0, E s ] s 8 =[0, 0, 0, − E s ] For each point s i , there are M − 2 = 6 points at a distance d i,k = |s i − s k | = 2E s and one vector (−s i ) at a distance d i,m =2 √ E s . Hence, the union bound on the probability of error P (e|s i ) is given by P UB (e|s i )= M k=1,k=i Q d i,k √ 2N 0 =6Q E s N 0 + Q 2E s N 0 182 Since all the signals are equiprobable, we find that P UB (e)=6Q E s N 0 + Q 2E s N 0 With M =8=2 3 , E s =3E b and therefore, P UB (e)=6Q 3E b N 0 + Q 6E b N 0 Problem 7.27 It is convenient to find first the probability of a correct decision. Since all signals are equiprobable P (C)= M i=1 1 M P (C|s i ) All the P (C|s i ), i =1, ,M are identical because of the symmetry of the constellation. By translating the vector s i to the origin we can find the probability of a correct decision, given that s i was transmitted, as P (C|s i )= ∞ − d 2 f(n 1 )dn 1 ∞ − d 2 f(n 2 )dn 2 ∞ − d 2 f(n N )dn N where the number of the integrals on the right side of the equation is N, d is the minimum distance between the points and f(n i )= 1 πN 0 e − n 2 i N 0 Hence, P (C|s i )= ∞ − d 2 f(n)dn N = 1 − − d 2 −∞ f(n)dn N = 1 −Q d √ 2N 0 N and therefore, the probability of error is given by P (e)=1− P(C)=1− M i=1 1 M 1 −Q d √ 2N 0 N =1− 1 −Q d √ 2N 0 N Note that since E s = N i=1 s 2 m,i = N i=1 ( d 2 ) 2 = N d 2 4 the probability of error can be written as P (e)=1− 1 −Q 2E s NN 0 N 183 Problem 7.28 Consider first the signal y(t)= n k=1 c k δ(t −kT c ) The signal y(t) has duration T = nT c and its matched filter is g(t)=y(T − t)=y(nT c − t)= n k=1 c k δ(nT c − kT c − t) = n i=1 c n−i+1 δ((i −1)T c − t)= n i=1 c n−i+1 δ(t −(i −1)T c ) that is, a sequence of impulses starting at t = 0 and weighted by the mirror image sequence of {c i }. Since, s(t)= n k=1 c k p(t −kT c )=p(t) n k=1 c k δ(t −kT c ) the Fourier transform of the signal s(t)is S(f)=P (f) n k=1 c k e −j2πfkT c and therefore, the Fourier transform of the signal matched to s(t)is H(f )=S ∗ (f)e −j2πfT = S ∗ (f)e −j2πfnT c = P ∗ (f) n k=1 c k e j2πfkT c e −j2πfnT c = P ∗ (f) n i=1 c n−i+1 e −j2πf(i−1)T −c = P ∗ (f)F[g(t)] Thus, the matched filter H(f) can be considered as the cascade of a filter,with impulse response p(−t), matched to the pulse p(t) and a filter, with impulse response g(t), matched to the signal y(t)= n k=1 c k δ(t −kT c ). The output of the matched filter at t = nT c is ∞ −∞ |s(t)| 2 = n k=1 c 2 k ∞ −∞ p 2 (t −kT c )dt = T c n k=1 c 2 k where we have used the fact that p(t) is a rectangular pulse of unit amplitude and duration T c . Problem 7.29 The bandwidth required for transmission of an M-ary PAM signal is W = R b 2 log 2 M Hz Since, R b =8×10 3 samples sec × 8 bits sample =64×10 3 bits sec 184 we obtain W = 16 KHz M =4 10.667 KHz M =8 8 KHz M =16 Problem 7.30 The vector r =[r 1 ,r 2 ] at the output of the integrators is r =[r 1 ,r 2 ]=[ 1.5 0 r(t)dt, 2 1 r(t)dt] If s 1 (t) is transmitted, then 1.5 0 r(t)dt = 1.5 0 [s 1 (t)+n(t)]dt =1+ 1.5 0 n(t)dt =1+n 1 2 1 r(t)dt = 2 1 [s 1 (t)+n(t)]dt = 2 1 n(t)dt = n 2 where n 1 is a zero-mean Gaussian random variable with variance σ 2 n 1 = E 1.5 0 1.5 0 n(τ)n(v)dτ dv = N 0 2 1.5 0 dτ =1.5 and n 2 is is a zero-mean Gaussian random variable with variance σ 2 n 2 = E 2 1 2 1 n(τ)n(v)dτ dv = N 0 2 2 1 dτ =1 Thus, the vector representation of the received signal (at the output of the integrators) is r =[1+n 1 ,n 2 ] Similarly we find that if s 2 (t) is transmitted, then r =[0.5+n 1 , 1+n 2 ] Suppose now that the detector bases its decisions on the rule r 1 − r 2 s 1 > < s 2 T The probability of error P (e|s 1 ) is obtained as P (e|s 1 )=P (r1 −r 2 <T|s 1 ) = P (1 + n 1 − n 2 <T)=P (n 1 − n 2 <T−1) = P (n<T) where the random variable n = n 1 − n 2 is zero-mean Gaussian with variance σ 2 n = σ 2 n 1 + σ 2 n 2 − 2E[n 1 n 2 ] = σ 2 n 1 + σ 2 n 2 − 2 1.5 1 N 0 2 dτ =1.5+1− 2 × 0.5=1.5 185 Hence, P (e|s 1 )= 1 2πσ 2 n T −1 −∞ e − x 2 2σ 2 n dx Similarly we find that P (e|s 2 )=P (0.5+n 1 − 1 − n 2 >T) = P (n 1 − n 2 >T+0.5) = 1 2πσ 2 n ∞ T +0.5 e − x 2 2σ 2 n dx The average probability of error is P (e)= 1 2 P (e|s 1 )+ 1 2 P (e|s 2 ) = 1 2 2πσ 2 n T −1 −∞ e − x 2 2σ 2 n dx + 1 2 2πσ 2 n ∞ T +0.5 e − x 2 2σ 2 n dx To find the value of T that minimizes the probability of error, we set the derivative of P (e) with respect to T equal to zero. Using the Leibnitz rule for the differentiation of definite integrals, we obtain ϑP (e) ϑT = 1 2 2πσ 2 n e − (T −1) 2 2σ 2 n − e − (T +0.5) 2 2σ 2 n =0 or (T − 1) 2 =(T +0.5) 2 =⇒ T =0.25 Thus, the optimal decision rule is r 1 − r 2 s 1 > < s 2 0.25 Problem 7.31 a) The inner product of s i (t) and s j (t)is ∞ −∞ s i (t)s j (t)dt = ∞ −∞ n k=1 c ik p(t −kT c ) n l=1 c jl p(t −lT c )dt = n k=1 n l=1 c ik c jl ∞ −∞ p(t −kT c )p(t −lT c )dt = n k=1 n l=1 c ik c jl E p δ kl = E p n k=1 c ik c jk The quantity n k=1 c ik c jk is the inner product of the row vectors C i and C j . Since the rows of the matrix H n are orthogonal by construction, we obtain ∞ −∞ s i (t)s j (t)dt = E p n k=1 c 2 ik δ ij = nE p δ ij Thus, the waveforms s i (t) and s j (t) are orthogonal. 186 b) Using the results of Problem 7.28, we obtain that the filter matched to the waveform s i (t)= n k=1 c ik p(t −kT c ) can be realized as the cascade of a filter matched to p(t) followed by a discrete-time filter matched to the vector C i =[c i1 , ,c in ]. Since the pulse p(t) is common to all the signal waveforms s i (t), we conclude that the n matched filters can be realized by a filter matched to p(t) followed by n discrete-time filters matched to the vectors C i , i =1, ,n. Problem 7.32 a) The optimal ML detector selects the sequence C i that minimizes the quantity D(r,C i )= n k=1 (r k − E b C ik ) 2 The metrics of the two possible transmitted sequences are D(r,C 1 )= w k=1 (r k − E b ) 2 + n k=w+1 (r k − E b ) 2 and D(r,C 2 )= w k=1 (r k − E b ) 2 + n k=w+1 (r k + E b ) 2 Since the first term of the right side is common for the two equations, we conclude that the optimal ML detector can base its decisions only on the last n −w received elements of r. That is n k=w+1 (r k − E b ) 2 − n k=w+1 (r k + E b ) 2 C 2 > < C 1 0 or equivalently n k=w+1 r k C 1 > < C 2 0 b) Since r k = √ E b C ik + n k , the probability of error P (e|C 1 )is P (e|C 1 )=P E b (n −w)+ n k=w+1 n k < 0 = P n k=w+1 n k < −(n −w) E b The random variable u = n k=w+1 n k is zero-mean Gaussian with variance σ 2 u =(n − w)σ 2 . Hence P (e|C 1 )= 1 2π(n −w)σ 2 − √ E b (n−w) −∞ exp(− x 2 2π(n −w)σ 2 )dx = Q E b (n −w) σ 2 187 [...]... following figure depicts a 4-cube and the way that one can traverse it in Gray-code order (see John F Wakerly, Digital Design Principles and Practices, Prentice Hall, 1990) Adjacent points are connected with solid or dashed lines 1 110 0 110 1 010 1011 0111 0 010 1111 1100 1101 0011 100 0 0101 0100 0000 0001 196 100 1 One way to label the points of the V.29 constellation using the Gray-code is depicted in the... Q N0 A2 T N0 With P2 = 10 6 we find from tables that A2 T = 4.74 =⇒ A2 T = 44.9352 × 10 10 N0 If the data rate is 10 Kbps, then the bit interval is T = 10 4 and therefore, the signal amplitude is A= 44.9352 × 10 10 × 104 = 6.7034 × 10 3 Similarly we find that when the rate is 105 bps and 106 bps, the required amplitude of the signal is A = 2.12 × 10 2 and A = 6.703 × 10 2 respectively Problem 7.35... as 3 is only 2 Having labeled the innermost points, all the adjacent nodes can be found using the previous figure 100 0 1 100 1 1 101 1 0011 2 1 2 1 1 1 2 0101 0111 1111 1 0001 1 1 2 1 1 1 0000 0 010 0100 2 0 110 1 1 2 1 1100 1 110 1 010 1 1101 Problem 7.46 1) Consider the QAM constellation of Fig P-7.46 Using the Pythagorean theorem we can find the radius of the inner circle as 1 a2 + a2 = A2 =⇒ a = √ A 2 The... for the four-phase constellation, we find d 2 2 r1 + r1 = d2 =⇒ r1 = √ 2 The radius of the 8-PSK constellation is found using the cosine rule Thus, 2 2 2 d2 = r2 + r2 − 2r2 cos(45o ) =⇒ r2 = d 2− √ 2 The average transmitted power of the 4-PSK and the 8-PSK constellation is given by P4,av = d2 , 2 P8,av = d2 √ 2− 2 Thus, the additional transmitted power needed by the 8-PSK signal is P = 10 log10 2d2 √ =... where Es = Eg /2 is the energy of the transmitted signal As it is observed the phase error φ − φ reduces the SNR by a factor ˆ SNRloss = 10 log10 cos2 (φ − φ) ˆ 2) When φ − φ = 45o , then the loss due to the phase error is SNRloss = 10 log10 cos2 (45o ) = 10 log10 1 = 3.01 dB 2 Problem 7.38 1) The closed loop transfer function is H(s) = G(s) 1 G(s)/s = = 2 √ 1 + G(s)/s s + G(s) s + 2s + 1 The poles... the power of the involved signals The probability of error is given by P (error) = Q 2PT N0 1− Pc PT The loss due to the allocation of power to the pilot signal is SNRloss = 10 log10 1 − When Pc /PT = 0.1, then Pc PT SNRloss = 10 log10 (0.9) = −0.4576 dB The negative sign indicates that the SNR is decreased by 0.4576 dB 192 Problem 7.37 1) If the received signal is r(t) = ±gT (t) cos(2πfc t + φ) + n(t)... 2ρs sin π M where ρs is the SNR per symbol In this case, equal error probability for the two signaling schemes, implies that sin π π ρ8,s 2 π 4 =⇒ 10 log10 ρ4,s sin = 20 log10 = 5.3329 dB = ρ8,s sin 4 8 ρ4,s sin π 8 2 Problem 7.43 The constellation of Fig P-7.43(a) has four points at a distance 2A from the origin and four points √ at a distance 2 2A Thus, the average transmitted power of the constellation... where θn = an cos−1 Pc PT Any method used to extract the carrier phase from the received signal can be employed at the receiver The following figure shows the structure of a receiver that employs a decision-feedback PLL The operation of the PLL is described in the next part t = Tb v(t) E× n E E T Tb 0 (·)dt E Threshold E DFPLL cos(2πfc t + φ) 2) At the receiver the signal is demodulated by crosscorrelating... equality we identify the time constants as τ2 = R2 C, τ1 = (R1 + R2 )C Problem 7.41 Assuming that the input resistance of the operational amplifier is high so that no current flows through it, then the voltage-current equations of the circuit are V2 = −AV1 V1 − V2 = R1 + 1 i Cs V1 − V0 = iR where, V1 , V2 is the input and output voltage of the amplifier respectively, and V0 is the signal at the input of the... of the correlators are r1 = r2 = 1 2 1 2 1 ˆ ˆ 2PT − ns (t) sin(φ − φ + θn ) + nc (t) cos(φ − φ + θn ) 2 1 ˆ ˆ 2PT − ns (t) cos(φ − φ + θn ) + nc (t) sin(φ − φ − θn ) 2 where nc (t), ns (t) are the in-phase and quadrature components of the noise n(t) If the detector has made the correct decision on the transmitted point, then by multiplying r1 by cos(θn ) and r2 by sin(θn ) and subtracting the results, . dashed lines. 0000 0001 0011 0 010 0 110 0111 0101 0100 1100 1101 1111 1 110 1 010 1011 100 0 100 1 196 One way to label the points of the V.29 constellation using the Gray-code is depicted in the next figure the innermost points, all the adjacent nodes can be found using the previous figure. 0000 0001 0101 0100 1100 1101 101 0 0 110 0 010 0011 100 1 100 0 101 1 0111 1111 1 110 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 21 2 2 1 2 1 Problem. is A = 44.9352 10 10 × 10 4 =6.7034 10 −3 Similarly we find that when the rate is 10 5 bps and 10 6 bps, the required amplitude of the signal is A =2.12 × 10 −2 and A =6.703 × 10 −2 respectively. Problem