# 数字通信习题解答

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19. Page 609, 6 lines above Equation(10.1-34) should be k+1L- k+-L-1 20 Page 646, Figure 10.3-5 delete the"hat"from I(z) 21. Page 651, 4 lines from the top replace“ over with“ about 22. Page 651, 2 lines above Section 10.6 Turob” should be“ Turbo 23. Page 673, Figure 11.1-6 Lower delay line elements: z should be z 24. Page 750, Figure 13.2-8 Replace adders”with“ multipliers 25. Page 752, figure 13 2-9 Replace"adders' with multipliers 26. Page 856, Equation (146-5) Replace k with k 27. Page 885, Figure 14.7-7 The"Input "should bc 02310 28. Page 894, Problem 14.16 ri=hiST h2s2 t n hIS2 hsi t 29.Page895 Delete 2 from the expression on the error probability 30. Page 915, top of page (1547) should bc(153-47) 31. Page 925, 6 lines form top To should be Tp 32.Page935 a) top of page r=b sqrt(C1)+ b2p sqrt(82)+ n1 r2=bip sqrt(E1)+ b2 sqrt(E2)+n2 b Problem 15.8, last equation delete factor of 12 c)Problem 15.9, first equation delete comma after b=1 33. Page 936, first equation at top of page, second term should be coshi [rz sqrt(E2)-blp sgrt(61 2)]/No 34. Page 936, second equation from top of page divide each of the arguments in the cosh function by No 35. Page 936, Problem 15.10 should be 36. Page 936 Problem 15.11 the last term in the equation should be 8,+8-2 pl sqrt(,8 (1/2)Q了sqrt No/2 C五 APTER2 Prob 2.1 P(A)=∑P(A,B),2=1,2,3,4 Hence P(A1)=∑P(A1,B)=0.1+0.08+0.13=0.31 (42)=∑P(A2,B)=0.05+0.03+0.09=0.17 (A3)=∑P(A3,B)=0.05+0.12+014=031 3 (A)=∑P(A4,B)=0.11+0.04+0.06=0.21 (B1)=∑P(A,B)=0.10+0.05+0.05+0.11=0 P(B2)=∑P(A1,B2)=0.08+0.03+0.12+004=0.27 P(B3)=∑P(A,B3)=0.13+009+0.14+0.06=0.42 Problem 2.2 The relationship holds for m=2(2-1-34): p( 21, C 2)=p(r2 C1p(a1) Suppose it holds for n-k,ie:p(x1,x2,…,xk)-p(xk|xk-1,…,x1)p(xk-1{xk-2,…,x1)…p(x1) Then for n=k+1 )=p(xk+1xk,xk-1,…,x1)p(xk,xk-1…x1) p(ak+laCk, Tk-1,., 1p(ak Tk 1p)(k-1k-2 Hence the relationship holds for n=k+l, and by induction it holds for any n Problem 2.3 Following the same procedure as in example 2-1-1, we prove Problem 2.4 Relationship(2-1-44) gives 1/3 pX 3[(y-b)/ X is a gaussian r.v. with zero mean and unit variance: px() Hence 32(y-b/mc5(2)23 df of y 05 0.45 0.35 0.25 02 0.15 0.05 Problem 2.5 (a) Since(Xr, Xi) are statistically independent px(r, i)=px(C TO tjYi=(X+ Xi) cos+Yisin +j(Y sin +Yi cos o) Yr cos o +Yisin o Yr sin + cos The jacobian of the above transformation is aX aX COS O sin p OXr sin g Hence, by(2-1-55) PYyr,i)= px((Y cos o+Yisin ) (Y sin Yi cos o)) (b)Y=AX and X=A-Y O xx/2o(the covariance matrix M of the random variables C1, ..., n is M=OI, since they are i i d )and J=1/ det(a). Hence py(y (27a2)m/2|det(A川 For the pdf's of X and y to be identical we require that I det(a)=1 and(A'A=I=A=A Hence, A must be a unitary(orthogonal)matrix Problem 2.6 E E X But po(d ()=1+p+pe1 tp+ pe 3 E(Y) duly (ju d in(1-p+pet)"-jpelu aI E(Y2) d2m lu=o dv ln( (1-p+pein)m-lpeiunmp+np(n →E(Y Problem 2.7 (733- (v11+02x2+03x3+04x4) E(X1X2X3X4)=(-j) 01(01,102,3,y3) duna2au3 an 1=02=03=04=0 From(2-1-151)of the text, and the zero-mean property of the given rv's reV三01,02,03,0 We obtain the desired result by bringing the exponent to a scalar form and then performing quadruple differentiation. We can simplify the procedure by noting that 0 where u=[uil, i2, P:3, Pig. Also note that Hij=l, Hence 04(1v1,jvy2,jv3,jv4) 0v020304 12/34+p123/14+/24/13 Problem 2.8 For the central chi-square with n degress of freedom 1-12 /2 O d(0) 10 /2-1 E(Y dv(u) )200 (1v)2ma4(m/2+1) (1-23)/+2→E(Y2)= d-n(3 do2-|=0=m(n+2)2 The variance is o=E(Y)-E(Y For the non-central chi-square with n degrees of freedom 1-2072 2σ where by definition: S-=iiM d (iv) (1-202 272)2+/em(12my Hence, E(Y) dwb(i du U=0 d2(jv) 14(n+2)-82(n+4)a2 1-120 1-120 /2+3 (1-2 n/2+4 Hence ElY ( =0=2mn04+422+(no2+ d ane =E(Y2)-E(Y)2=2m04+422 Problem 2.9 The Cauchy r.v. has p(a) 2⊥a2 ∞<x<∞(a) E(X p(a) c=0 since pla) is an even function C E(X Note that for large w,2+ +ar-1(i. e non-zero value). Hence, E(X

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