Es e j4φ +4N n. 16 KE s /N 0. σ 2ˆφ4 1 γ s. p(φ e )= exp 1 ( 2πσ φ b cos N 2 φ e 0

Transcription

1 Problem 6.15 : he received signal-plus-noise vector at the output of the matched filter may be represented as (see (5-2-63) for example) : r n = E s e j(θn φ) + N n where θ n =0,π/2,π,3π/2 for QPSK, and φ is the carrier phase. By raising r n to the fourth power and neglecting all products of noise terms, we obtain : r 4 n ( Es ) 4 e j4(θ n φ) +4 ( Es ) 3 Nn ( Es ) 3 [ Es e j4φ +4N n ] If the estimate is formed by averaging the received vectors {rn 4 } over K signal intervals, we have the resultant vector U = K E s e jφ +4 K n=1 N n. Let φ 4 4φ. hen, the estimate of φ 4 is : 1 Im(U) ˆφ 4 = tan Re(U) N n is a complex-valued Gaussian noise component with zero mean and variance σ 2 = N 0 /2. Hence, the pdf of ˆφ 4 is given by (5-2-55) where : ( ) 2 K Es γ s = 16 (2Kσ 2 ) = K2 E s 16KN 0 = KE s 16N 0 o a first approximation, the variance of the estimate is : σ 2ˆφ4 1 γ s = 16 KE s /N 0 Problem 6.16 : he PDF of the carrier phase error φ e,isgivenby: p(φ e )= 2 e 2σ φ 2 1 e φ 2πσφ hus the average probability of error is : P 2 = = = P 2 (φ e )p(φ e )dφ e [ ] 2Eb Q cos N 2 φ e p(φ e )dφ e 0 1 [ exp 1 ( 2E 2πσ φ b cos N 2 φ e x 2 + φ2 e σ 2 φ )] dxdφ e

2 Problem 6.17: he log-likelihood function of the symbol timing may be expressed in terms of the equivalent low-pass signals as Λ L (τ) = R [ 1 N 0 0 r(t)s l (t; τ)dt ] = R [ 1 N 0 0 r(t) n I n g (t n τ)dt ] = R [ 1 N 0 n I n y n (τ) ] where y n (τ) = 0 r(t)g (t n τ)dt. Anecessary condition for ˆτ to be the ML estimate of τ is dλ L (τ) = 0 τ d dτ [ n I n y n (τ)+ n I n y n (τ)] = 0 d n I n y dτ n(τ)+ d n I n y dτ n (τ) = 0 If we express y n (τ) into its real and imaginary parts : y n (τ) = a n (τ) +jb n (τ), the above expression simplifies to the following condition for the ML estimate of the timing ˆτ n R[I n ] d dτ a n(τ)+ n I[I n ] d dτ b n(τ) =0 Problem 6.18: We follow the exact same steps of the derivation found in Sec For a PAM signal I n = I n and J n = 0. Since the pulse g(t) is real, it follows that B(τ) in expression (6.4-6) is zero, therefore (6.4-7) can be rewritten as where Λ L (φ, τ) =A(τ)cosφ A(τ) = 1 N 0 In y n (τ) hen the necessary conditions for the estimates of φ and τ to be the ML estimates (6.4-8) and (6.4-9) give ˆφ ML =0 and n I n d dτ [y n(τ)] τ=ˆτml =0 133

3 Problem 9.15 : he SNR at the detector is : E b = P b = P b(1 + β) N 0 N 0 N 0 W =30dB

4 Since it is desired to expand the bandwidth by a factor of 10 while maintaining the same SNR, 3 the received power P b should increase by the same factor. hus the additional power needed is P a =10log Hence, the required transmitted power is : = db P S = = dbw P r oblem 9.16 : he pulse x(t) having the raised cosine spectrum given by (9-2-26/27) is : x(t) =sinc(t/ ) cos(πβt/) 1 4β 2 t 2 / 2 he function sinc(t/ ) is1whent = 0and0whent = n. herefore, the Nyquist criterion will be satisfied as long as the function g(t) is: g(t) = cos(πβt/) { 1 4β 2 t 2 / = 1 t =0 2 bounded t 0 he function g(t) needs to be checked only for those values of t such that 4β 2 t 2 / 2 =1or βt =. However : 2 cos(πβt/) 1 4β 2 t 2 / = lim cos( π x) 2 2 x 1 1 x and by using L Hospital s rule : Hence : lim βt 2 cos( π lim x) 2 x 1 1 x = lim π x 1 2 sin(π 2 x)=π 2 < x(n )= { 1 n =0 0 n 0 meaning that the pulse x(t) satisfies the Nyquist criterion. P r oblem 9.17 : Substituting the expression of X rc (f) given by (8.2.22) in the desired integral, we obtain : 1 β [ X rc (f)df = 1+cos π 2 β ( f 1 β ] 1 β ) df + df 1+β 1 β

5 = 1+β + 1 β 1 β + 1+β = 1+ = 1+ 1 β 1+β 0 β β β [ 1+cos π 2 β (f 1 β ) ( ) 1 β 1+β 2 df β ] df 2 df cos π β (f + 1 β 1+β )df + cos π β xdx + β 0 1 β cos π β xdx cos π xdx =1+0=1 β cos π β (f 1 β )df

6 Problem 9.19 : he bandwidth of the channel is : W = = 2700 Hz

7 Since the minimum transmission bandwidth required for bandpass signaling is R, where R is the rate of transmission, we conclude that the maximum value of the symbol rate for the given channel is R max = If an M-ary PAM modulation is used for transmission, then in order to achieve a bit-rate of 9600 bps, with maximum rate of R max, the minimum size of the constellation is M =2 k = 16. In this case, the symbol rate is : R = 9600 k = 2400 symbols/sec and the symbol interval = 1 = 1 sec. he roll-off factor β of the raised cosine pulse used R 2400 for transmission is is determined by noting that 1200(1 + β) = 1350, and hence, β = herefore, the squared root raised cosine pulse can have a roll-off of β =0.125.

8 Problem 9.23 : he roll-off factor β is related to the bandwidth by the expression 1+β =2W,orequivalently R(1 + β) =2W. he following table shows the symbol rate for the various values of the excess bandwidth and for W = 1500 Hz. β R he above results were obtained with the assumption that double-sideband PAM is employed, so the available lowpass bandwidth will be from W = 3000 to W Hz. If single-sideband 2 transmission is used, then the spectral efficiency is doubled, and the above symbol rates R are doubled. Problem 9.24 : he following table shows the precoded sequence, the transmitted amplitude levels, the received signal levels and the decoded sequence, when the data sequence modulates a duobinary transmitting filter. Data seq. D n : Precoded seq. P n : ransmitted seq. I n : Received seq. B n : Decoded seq. D n : Problem 9.25 : he following table shows the precoded sequence, the transmitted amplitude levels, the received signal levels and the decoded sequence, when the data sequence modulates a modified duobinary transmitting filter. Data seq. D n : Precoded seq. P n : ransmitted seq. I n : Received seq. B n : Decoded seq. D n :

9 Problem 10.1 : Suppose that a m = +1 is the transmitted signal. hen the probability of error will be : P e 1 = P (y m < 0 a m =+1) = P (1 + n m + i m < 0) = 1 4 P (1/2+n m < 0) P (3/2+n m < 0) P (1 + n m < 0) = 1 [ ] 1 4 Q + 1 [ ] 3 2σ n 4 Q + 1 [ ] 1 2σ n 2 Q σn Due to the symmetry of the intersymbol interference, the probability of error, when a m = 1 is transmitted, is the same. hus, the above result is the average probability of error.

10 Problem : (a) he equivalent discrete-time impulse response of the channel is : 1 h(t) = h n δ(t n )=0.3δ(t + )+0.9δ(t)+0.3δ(t ) n= 1 If by {c n } we denote the coefficients of the FIR equalizer, then the equalized signal is : 1 q m = c n h m n n= 1 which in matrix notation is written as : c 1 c 0 c 1 = 0 1 0

11 he coefficients of the zero-force equalizer can be found by solving the previous matrix equation. hus, c c 0 = c (b) he values of q m for m = ±2, ±3are given by q 2 = q 2 = q 3 = q 3 = 1 c n h 2 n = c 1 h 1 = n= 1 1 c n h 2 n = c 1 h 1 = n= 1 1 c n h 3 n =0 n= 1 1 c n h 3 n =0 n= 1 Problem : (a) he output of the zero-force equalizer is : q m = 1 n= 1 c n x mn With q 0 =1andq m =0form 0, we obtain the system : c c c 1 = Solving the previous system in terms of the equalizer s coefficients, we obtain : c 1 c 0 c 1 =

12 (b) he output of the equalizer is : q m = Hence, the residual ISI sequence is and its span is 6 symbols. 0 m 4 c 1 x 2 =0 m = 3 c 1 x 1 + c 0 x 2 = 0.49 m = 2 0 m = 1 1 m =0 0 m =1 c 0 x 2 + x 1 c 1 = m =2 c 1 x 2 = m =3 0 m 4 residual ISI = {...,0, 0.49, 0, 0, 0, , , 0,...}

13 Problem : (a) F (z) = z 1 X(z) F (z)f (z 1 )=( z 1 )( z) =1 0.48z z hus, x 0 =1,x 1 = x 1 = (b) 1 ( H n= ω + 2πn ) 2 = X ( e jω) =1 0.48e jω 0.48e jω = cos ω (c) For the linear equalizer base on the mean-square-error criterion we have : J min = 2π π/ π/ = 1 π 2π π N 0 dω 1+N cos ω N 0 1+N cos θ dθ ( ) = 1 N0 π π 1+N 0 π dθ, a = 1 a cos θ 1+N 0 But : herefore : 1 π 1 2π π 1 a cos θ dθ = 1 1 a 2,a2 < 1 J min = N N 0 1 ( ) = N 0 N 0 (1 + N 0 ) 2 (0.96) 2 (d) For the decision-feedback equalizer : J min = 2N 0 1+N 0 + (1 + N 0 ) 2 (0.96) 2 which follows from the result in example Note that for N 0 << 1, J min 2N N (0.96) 2 0 In contrast, for the linear equalizer we have : J min N 0 1 (0.96) N 0

14 (d) he metrics are (y 1 0.8I 1 ) 2,i=1 and i (y i 0.8I i +0.6I i 1 ) 2,i 2 µ 1 (I 1 =3)=[ ] 2 =3.61 µ 1 (I 1 =1)=[ ] 2 =0.09 µ 1 (I 1 = 1) = [ ] 2 =1.69 µ 1 (I 1 = 3 )=[ ] 2 =8.41 µ 2 (I 2 =3,I 1 =3)=µ 1 (3) + [ ] 2 =5.57 µ 2 (3, 1) = µ 1 (1) + [ ] 2 =0.13 µ 2 (3, 1) = µ 1 ( 1) + [ ] 2 =6.53 µ 2 (3, 3 )=µ 1 ( 3)+[ ] 2 =13.25 µ 2 (1, 3)=µ 1 (3) + [ ] 2 =12.61 µ 2 (1, 1) = µ 1 (1) + [ ] 2 =3.33 µ 2 (1, 1) = µ 1 ( 1) + [ ] 2 =2.05 µ 2 (1, 3 )=µ 1 ( 3)+[ ] 2 =8.77 µ 2 ( 1, 3)=µ 1 (3) + [ ] 2 =24.77 µ 2 ( 1, 1) = µ 1 (1) + [ ] 2 =11.65 µ 2 ( 1, 1) = µ 1 ( 1) + [ ] 2 =6.53 µ 2 ( 1, 3 )=µ 1 ( 3)+[ ] 2 =9.41 µ 2 ( 3, 3)=µ 1 (3) + [ ] 2 =42.05 µ 2 ( 3, 1) = µ 1 (1) + [ ] 2 =25.09 µ 2 ( 3, 1) = µ 1 ( 1) + [ ] 2 =16.13 µ 2 ( 3, 3 )=µ 1 ( 3)+[ ] 2 =15.17 he four surviving paths at this stage are min I1 [µ 2 (x, I 1 )],x=3, 1, 1, 3or : I 2 =3,I 1 = 1 with metric µ 2 (3, 1) = 0.13 I 2 =1,I 1 = 1 with metric µ 2 (1, 1) = 2.05 I 2 = 1,I 1 = 1 with metric µ 2 ( 1, 1) = 6.53 I 2 = 3,I 1 = 3with metric µ 2 ( 3, 3) = Now we compute the metrics for the next stage : µ 3 (I 3 =3,I 2 =3,I 1 =1)=µ 2 (3, 1) + [ ] 2 =2.69 µ 3 (3, 1, 1) = µ 2 (1, 1) + [ ] 2 =9.89 µ 3 (3, 1, 1) = µ 2 ( 1, 1) + [ ] 2 =22.53 µ 3 (3, 3, 3 )=µ 2 ( 3, 3)+[ ] 2 =42.21

15 µ 3 (1, 3, 1) = µ 2 (3, 1) + [ ] 2 =0.13 µ 3 (1, 1, 1) = µ 2 (1, 1) + [ ] 2 =7.81 µ 3 (1, 1, 1) = µ 2 ( 1, 1) + [ ] 2 =12.29 µ 3 (1, 3, 3 )=µ 2 ( 3, 3)+[ ] 2 =28.13 µ 3 ( 1, 3, 1) = µ 2 (3, 1) + [ ] 2 =2.69 µ 3 ( 1, 1, 1) = µ 2 (1, 1) + [ ] 2 =2.69 µ 3 ( 1, 1, 1) = µ 2 ( 1, 1) + [ ] 2 =7.17 µ 3 ( 1, 3, 3 )=µ 2 ( 3, 3)+[ ] 2 =19.17 µ 3 ( 3, 3, 1) = µ 2 (3, 1) + [ ] 2 =10.37 µ 3 ( 3, 1, 1) = µ 2 (1, 1) + [ ] 2 =2.69 µ 3 ( 3, 1, 1) = µ 2 ( 1, 1) + [ ] 2 =7.17 µ 3 ( 3, 3, 3 )=µ 2 ( 3, 3)+[ ] 2 =15.33 he four surviving sequences at this stage are min I2,I 1 [µ 3 (x, I 2,I 1 )],x=3, 1, 1, 3or : I 3 =3,I 2 =3,I 1 = 1 with metric µ 3 (3, 3, 1) = 2.69 I 3 =1,I 2 =3,I 1 = 1 with metric µ 3 (1, 3, 1) = 0.13 I 3 = 1,I 2 =3,I 1 = 1 with metric µ 3 ( 1, 3, 1) = 2.69 I 3 = 3,I 2 =1,I 1 = 1 with metric µ 3 ( 3, 1, 1) = 2.69 (e) For the channel, δmin 2 = 1 and hence : P 4 =8Q 6 15 γ av