Table Of ContentCorrections to Digital Communications, 4th Edition
1. Page 31, Equation (2.1-54)
First line: y instead of y
l 2
Second line: g instead of g
n 1
2. Page 163, Equation (4.2-30)
∞
Σ
s hou ld be: s(t) = ao/2 +
k=1
3. Page 163, Equation (4.2-31)
T
should be: a = (2/T) ∫ s(t) cos 2πkt/T dt , k>0
k
0
T
b = (2/T) ∫ s(t) sin 2πkt/T dt , k>1
k
0
4. Page 178, 7 lines from the top
ε) ε
should be: sqrt (2 instead of sqrt (2)
5. Page 238, Equation (5.1-19)
should be: h(T-τ ) instead of h(t-τ )
6. Page 238, two lines below Equation (5.1-20)
should be: y2 (T) instead of y2(t)
n n
7. Page 244, Equation (5.1 (cid:150) 45)
should be: m = 1,2,(cid:133)M
8. Page 245, Equation (5.1-48)
ε ε
should be: : sqrt ( ) instead of sqrt ( )
b n
9. Page 309, Equation (5.4-39)
R1 sqrt (2εs/N0) instead of sqrt (2εsR1/N0)
10. Page 318, Equation (5.5-17)
add the term: (cid:150) (N )
0 dBW/Hz
11. Page 366, Equation (6.4-3)
Replace + sign with (cid:150) sign in the second term of the summation
12. Page 367, Equation (6.4-6)
Replace + sign with (cid:150) sign in the second term of the summation
13. Page 367, Equations (6.4-8) and (6.4-9)
add the subscript L to the log-likelihood function
14. Page 422, lines 2 and 3 above Equation (8.1-14)
delete the phrase (cid:147)no more than(cid:148)
15. Page 468, 12 lines from the top and 5 lines from the bottom
should be: b < instead of b <
16. Page 491, Figure 8.2-15
solid line corresponds to soft-decision decoding
broken line corresponds to hard-decision decoding
17. Page 500, Equation (8.2-41)
In the denominator, M should be M and M should be M
k j j J
18. Page 591, Figure P9.9
The lower shaping filter in the modulator and demodulator,
q(t) should have a (cid:147)hat(cid:148) on it
19. Page 609, 6 lines above Equation (10.1-34)
ε should be ε
k+1(cid:150) L-1 k+l (cid:150) L-1
20. Page 646, Figure 10.3-5
delete the (cid:147)hat(cid:148) from I(z)
21. Page 651, 4 lines from the top
replace (cid:147)over(cid:148) with (cid:147)about(cid:148)
22. Page 651, 2 lines above Section 10.6
(cid:147)Turob(cid:148) should be (cid:147)Turbo(cid:148)
23. Page 673, Figure 11.1-6
Lower delay line elements: z1 should be z-1
24. Page 750, Figure 13.2-8
Replace (cid:147)adders(cid:148) with (cid:147)multipliers(cid:148)
25. Page 752, Figure 13.2-9
Replace (cid:147)adders(cid:148) with multipliers(cid:148)
26. Page 856, Equation (14.6-5)
Replace K with k
27. Page 885, Figure 14.7-7
The (cid:147)Input(cid:148) should be 02310
28. Page 894, Problem 14.16
r = h s + h s + n
1 1 1 2 2 1
r = h s * + h s * + n
2 1 2 2 1 2
29. Page 895
Delete 2k from the expression on the error probability
30. Page 915, top of page
(15.47) should be (15.3-47)
31. Page 925, 6 lines form top
T should be T
0 p
32. Page 935
a) top of page:
ε ε
r = b sqrt( ) + b ρ sqrt ( ) + n
1 1 1 2 2 1
ε ε
r = b ρ sqrt( ) + b sqrt ( ) + n
2 1 1 2 2 2
b) Problem 15.8, last equation
delete factor of l/2
c) Problem 15.9, first equation
delete comma after b =1
2
33. Page 936, first equation at top of page, second term
should be:
l { [ ε ε ε ]/ }
n cosh r sqrt ( ) (cid:150) b ρ sqrt ( ) N
2 2 1 1 2 0
34. Page 936, second equation from top of page
divide each of the arguments in the cosh function by N
0
35. Page 936, Problem 15.10
2
[ ]
should be η =
k
36. Page 936, Problem 15.11
the last term in the equation should be:
ε ε ε ε
+ -2׀ρ׀ sqrt ( )
(1/2) Q {sqrt[ 1 2 1 2 ]}
N /2
0
CHAPTER 2
Problem 2.1 :
(cid:1)3
P(A ) = P(A ,B ),i = 1,2,3,4
i i j
j=1
Hence :
(cid:1)3
P(A ) = P(A ,B ) = 0.1+0.08+0.13 = 0.31
1 1 j
j=1
(cid:1)3
P(A ) = P(A ,B ) = 0.05+0.03+0.09 = 0.17
2 2 j
j=1
(cid:1)3
P(A ) = P(A ,B ) = 0.05+0.12+0.14 = 0.31
3 3 j
j=1
(cid:1)3
P(A ) = P(A ,B ) = 0.11+0.04+0.06 = 0.21
4 4 j
j=1
Similarly :
(cid:1)4
P(B ) = P(A ,B ) = 0.10+0.05+0.05+0.11 = 0.31
1 i 1
i=1
(cid:1)4
P(B ) = P(A ,B ) = 0.08+0.03+0.12+0.04 = 0.27
2 i 2
i=1
(cid:1)4
P(B ) = P(A ,B ) = 0.13+0.09+0.14+0.06 = 0.42
3 i 3
i=1
Problem 2.2 :
The relationship holds for n = 2 (2-1-34) : p(x ,x ) = p(x |x )p(x )
1 2 2 1 1
Suppose it holds for n = k,i.e : p(x ,x ,...,x ) = p(x |x ,...,x )p(x |x ,...,x ) ...p(x )
1 2 k k k−1 1 k−1 k−2 1 1
Then for n = k +1 :
p(x ,x ,...,x ,x ) = p(x |x ,x ,...,x )p(x ,x ...,x )
1 2 k k+1 k+1 k k−1 1 k k−1 1
= p(x |x ,x ,...,x )p(x |x ,...,x )p(x |x ,...,x ) ...p(x )
k+1 k k−1 1 k k−1 1 k−1 k−2 1 1
Hence the relationship holds for n = k +1, and by induction it holds for any n.
1
Problem 2.3 :
Following the same procedure as in example 2-1-1, we prove :
(cid:2) (cid:3)
1 y −b
p (y) = p
Y |a| X a
Problem 2.4 :
Relationship (2-1-44) gives :
(cid:2) (cid:3)
1 y −b 1/3
p (y) = p
Y 3a[(y −b)/a]2/3 X a
X is a gaussian r.v. with zero mean and unit variance : p (x) = √1 e−x2/2
X
2π
Hence :
pY(y) = 3a√2π[(y1−b)/a]2/3e−12(y−ab)2/3
pdf of Y
0.5
0.45
0.4 a=2
0.35 b=3
0.3
0.25
0.2
0.15
0.1
0.05
0
−10 −8 −6 −4 −2 0 2 4 6 8 10
y
Problem 2.5 :
(a) Since (X ,X ) are statistically independent :
r i
1
pX(xr,xi) = pX(xr)pX(xi) = 2πσ2e−(x2r+x2i)/2σ2
2
Also :
Y +jY = (X +X )ejφ ⇒
r i r i
X +X = (Y +jY )e−jφ = Y cosφ+Y sinφ+j(−Y sinφ+Y cosφ) ⇒
r i r i r i r i
(cid:8) (cid:9)
X = Y cosφ+Y sinφ
r r i
X = −Y sinφ+Y cosφ
i r i
The Jacobian of the above transformation is :
(cid:10) (cid:10) (cid:10) (cid:10)
(cid:10)(cid:10) ∂Xr ∂Xi (cid:10)(cid:10) (cid:10)(cid:10) cosφ −sinφ (cid:10)(cid:10)
J = (cid:10) ∂Yr ∂Yr (cid:10) = (cid:10) (cid:10) = 1
(cid:10) ∂Xr ∂Xi (cid:10) (cid:10) sinφ cosφ (cid:10)
∂Y ∂Y
i i
Hence, by (2-1-55) :
p (y ,y ) = p ((Y cosφ+Y sinφ),(−Y sinφ+Y cosφ))
Y r i X r i r i
= 1 e−(yr2+yi2)/2σ2
2πσ2
(b) Y = AX and X = A−1Y
Now, p (x) = 1 e−x(cid:2)x/2σ2 (the covariance matrix M of the random variables x ,...,x is
X (2πσ2)n/2 1 n
M = σ2I, since they are i.i.d) and J = 1/|det(A)|. Hence :
1 1
p (y) = e−y(cid:2)(A−1)(cid:2)A−1y/2σ2
Y (2πσ2)n/2|det(A)|
For the pdf’s of X and Y to be identical we require that :
|det(A)| = 1 and (A−1)(cid:3)A−1 = I =⇒ A−1 = A(cid:3)
Hence, A must be a unitary (orthogonal) matrix .
Problem 2.6 :
(a)
(cid:14) (cid:16)
(cid:11) (cid:12) (cid:11) (cid:13) (cid:12) (cid:15)n (cid:15)n (cid:11) (cid:12) (cid:17) (cid:18)
ψY(jv) = E ejvY = E ejv ni=1xi = E ejvxi = E ejvX = ψX(ejv) n
i=1 i=1
But,
p (x) = pδ(x−1)+(1−p)δ(x) ⇒ ψ (ejv) = 1+p+pejv
X X
(cid:17) (cid:18)
n
⇒ ψ (jv) = 1+p+pejv
Y
3
(b)
dψ (jv)
E(Y) = −j Y | = −jn(1−p+pejv)n−1jpejv| = np
dv v=0 v=0
and
d2ψ (jv) d (cid:11) (cid:12)
E(Y2) = − Y | = − jn(1−p+pejv)n−1pejv = np+np(n−1)p
d2v v=0 dv v=0
⇒ E(Y2) = n2p2 +np(1−p)
Problem 2.7 :
(cid:11) (cid:12)
ψ(jv ,jv ,jv ,jv ) = E ej(v1x1+v2x2+v3x3+v4x4)
1 2 3 4
∂4ψ(jv ,jv ,jv ,jv )
E(X X X X ) = (−j)4 1 2 3 4 |
1 2 3 4 ∂v ∂v ∂v ∂v v1=v2=v3=v4=0
1 2 3 4
From (2-1-151) of the text, and the zero-mean property of the given rv’s :
ψ(jv) = e−12v(cid:2)Mv
where v = [v ,v ,v ,v ](cid:3),M = [µ ].
1 2 3 4 ij
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 :
∂ψ(jv)
∂v = −µ(cid:3)ive−12v(cid:2)Mv
i
where µ(cid:3) =[µ ,µ ,µ ,µ ] . Also note that :
i i1 i2 i3 i4
∂µ(cid:3)v
j = µ = µ
∂v ij ji
i
Hence :
∂4ψ(jv ,jv ,jv ,jv )
1 2 3 4 | = µ µ +µ µ +µ µ
∂v ∂v ∂v ∂v V=0 12 34 23 14 24 13
1 2 3 4
Problem 2.8 :
For the central chi-square with n degress of freedom :
1
ψ(jv) =
(1−j2vσ2)n/2
4
Now :
dψ(jv) jnσ2 dψ(jv)
= ⇒ E(Y) = −j | = nσ2
dv (1−j2vσ2)n/2+1 dv v=0
d2ψ(jv) −2nσ4(n/2+1) (cid:17) (cid:18) d2ψ(jv)
= ⇒ E Y2 = − | = n(n+2)σ2
dv2 (1−j2vσ2)n/2+2 dv2 v=0
The variance is σ2 = E(Y2)−[E(Y)]2 = 2nσ4
Y
For the non-central chi-square with n degrees of freedom :
1
ψ(jv) = ejvs2/(1−j2vσ2)
(1−j2vσ2)n/2
(cid:13)
where by definition : s2 = n m2 .
i=1 i
(cid:14) (cid:16)
dψ(jv) jnσ2 js2
= + ejvs2/(1−j2vσ2)
dv (1−j2vσ2)n/2+1 (1−j2vσ2)n/2+2
Hence, E(Y) = −jdψ(jv)| = nσ2 +s2
dv v=0
(cid:14) (cid:16)
d2ψ(jv) −nσ4(n+2) −s2(n+4)σ2 −ns2σ2 −s4
= + + ejvs2/(1−j2vσ2)
dv2 (1−j2vσ2)n/2+2 (1−j2vσ2)n/2+3 (1−j2vσ2)n/2+4
Hence,
(cid:17) (cid:18) d2ψ(jv) (cid:17) (cid:18)
E Y2 = − | = 2nσ4 +4s2σ2 + nσ2 +s2
dv2 v=0
and (cid:17) (cid:18)
σ2 = E Y2 −[E(Y)]2 = 2nσ4 +4σ2s2
Y
Problem 2.9 :
The Cauchy r.v. has : p(x) = a/π ,−∞ < x < ∞ (a)
x2+a2
(cid:19)
∞
E(X) = xp(x)dx = 0
−∞
since p(x) is an even function.
(cid:17) (cid:18) (cid:19) ∞ a (cid:19) ∞ x2
E X2 = x2p(x)dx = dx
−∞ π −∞ x2 +a2
Note that for large x, x2 → 1 (i.e non-zero value). Hence,
x2+a2
(cid:17) (cid:18)
E X2 = ∞,σ2 = ∞
5
