Solutions Manual
for
Digital Communications, 5th Edition
(Chapter 2) 1
Prepared by
Kostas Stamatiou
January 11, 2008
2
Problem 2.1
a.
ˆx(t) = 1
x(a)
b. In exactly the same way as in part (a) we prove :
c. x(t) = cos ω0t, so its Fourier transform is : X(f) = 1
2[δ(f−f0) + δ(f+f0)] , f0= 2πω0.
Exploiting the phase-shifting property (2-1-4) of the Hilbert transform :
d. In a similar way to part (c) :
e. The positive frequency content of the new signal will be : (−j)(−j)X(f) = −X(f), f > 0,while
ˆ
f. Since the magnitude response of the Hilbert transformer is characterized by : |H(f)|= 1,we
have that : ˆ
X(f)=|H(f)||X(f)|=|X(f)|.Hence :
3
and using Parseval’s relationship :
−∞
−∞
g. From parts (a) and (b) above, we note that if x(t) is even, ˆx(t) is odd and vice-versa. Therefore,
Problem 2.2
1. Using relations
X(f) = 1
and Parseval’s relation, we have
Z∞
−∞
x(t)y(t)dt =Z∞
−∞
X(f)Y∗(f)dt
=Z∞
−∞ 1
2Xl(f−f0) + 1
2Xl(−f−f0)1
2Yl(f−f0) + 1
2Yl(−f−f0)∗df
2Re Z∞
−∞
where we have used the fact that since Xl(f−f0) and Yl(−f−f0) do not overlap, Xl(f−
2. Putting y(t) = x(t) we get the desired result from the result of part 1.
4
Problem 2.3
A well-known result in estimation theory based on the minimum mean-squared-error criterion states
that the minimum of Eeis obtained when the error is orthogonal to each of the functions in the
series expansion. Hence :
Z∞
−∞ “s(t)−
K
X
k=1
skfk(t)#f∗
n(t)dt = 0, n = 1,2, …, K (1)
−∞
The corresponding residual error Eeis :
Emin =R∞
−∞ hs(t)−PK
k=1 skfk(t)ihs(t)−PK
n=1 snfn(t)i∗dt
where we have exploited relationship (1) to go from the second to the third step in the above
calculation.
Note : Relationship (1) can also be obtained by simple differentiation of the residual error with
d
danEe=d
danR∞
−∞ hs(t)−PK
k=1 skfk(t)ihs(t)−PK
n=1 snfn(t)i∗dt = 0
Problem 2.4
The procedure is very similar to the one for the real-valued signals described in the book (pages
33-37). The only difference is that the projections should conform to the complex-valued vector
space :
c12= Z∞
−∞
s2(t)f∗
1(t)dt
−∞
Problem 2.5
The first basis function is :
g4(t) = s4(t)
√E4
=s4(t)
√3=
−1/√3,0≤t≤3
0,o.w.
Then, for the second basis function :
2/3,0≤t≤2
0,o.w
where E3denotes the energy of g′
3(t) : E3=R3
0(g′
3(t))2dt = 8/3.
For the third basis function :
6
Hence :
Finally for the fourth basis function :
c41 =Z∞
−∞
s1(t)g4(t)dt =−2/√3,c31 =Z∞
−∞
s1(t)g3(t)dt = 2/√6, c21 = 0
Hence :
The last result is expected, since the dimensionality of the vector space generated by these signals
is 3. Based on the basis functions (g2(t), g3(t), g4(t)) the basis representation of the signals is :
s4=0,0,√3⇒ E4= 3
Problem 2.6
Consider the set of signals e
φnl(t) = jφnl(t),1≤n≤N, then by definition of lowpass equivalent
signals and by Equations 2.2-49 and 2.2-54, we see that φn(t)’s are √2 times the lowpass equivalents
and using the result of problem 2.2 we have
hφn(t),e
φm(t)i= 0 for all n, m
7
and
he
φn(t),e
φm(t)i= 0 for all n6=m
Using the fact that the energy in lowpass equivalent signal is twice the energy in the bandpass
Problem 2.7
Let x(t) = m(t) cos 2πf0twhere m(t) is real and lowpass with bandwidth less than f0. Then
Problem 2.8
For real-valued signals the correlation coefficients are given by : ρkm =1
√EkEmR∞
−∞ sk(t)sm(t)dt and
and:
d(e)
12 = 2 d(e)
13 =q2 + 3 −2√62
√6= 1 d(e)
14 =q2 + 3 + 2√62
√6= 3
Problem 2.9
We know from Fourier transform properties that if a signal x(t) is real-valued then its Fourier
transform satisfies : X(−f) = X∗(f) (Hermitian property). Hence the condition under which sl(t)
Problem 2.10
a. To show that the waveforms fn(t), n= 1,…,3 are orthogonal we have to prove that:
Z∞
−∞
fm(t)fn(t)dt = 0, m 6=n
Clearly:
Similarly:
c13 =Z∞
−∞
f1(t)f3(t)dt =Z4
0
f1(t)f3(t)dt
and :
c23 =Z∞
−∞
f2(t)f3(t)dt =Z4
0
f2(t)f3(t)dt
9
Thus, the signals fn(t) are orthogonal. It is also straightforward to prove that the signals have unit
energy : Z∞
b. We first determine the weighting coefficients
xn=Z∞
−∞
x(t)fn(t)dt, n = 1,2,3
Problem 2.11
a. As an orthonormal set of basis functions we consider the set
f1(t) =
1 0 ≤t < 1
0 o.w f2(t) =
1 1 ≤t < 2
0 o.w
1 2 ≤t < 3
1 3 ≤t < 4
10
b. The representation vectors are
s1=h2−1−1−1i
c. The distance between the first and the second vector is:
d1,2=p|s1−s2|2=rh4−2−2−1i
2=√25
Similarly we find that :
2=√5
Problem 2.12
As a set of orthonormal functions we consider the waveforms
1 0 ≤t < 1
1 1 ≤t < 2
1 2 ≤t < 3
The vector representation of the signals is
s1=h2 2 2 i
s2=h2 0 0 i
Problem 2.13
1. P(E2) = P(R2, R3, R4) = 3/7.
2. P(E3|E2) = P(E3E2)
P(E2)=P(R2)
3/7=1
3.
Problem 2.14
1. P(R) = P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C) = 0.2×0.05 + 0.3×0.1 + 0.5×0.15 =
Problem 2.15
The relationship holds for n= 2 (2-1-34) : p(x1, x2) = p(x2|x1)p(x1)
Problem 2.16
1. Let Tand Rdenote channel input and outputs respectively. Using Bayes rule we have
p(T= 0|R=A) = p(T= 0)p(R=A|T= 0)
and therefore p(T= 1|R=A) = 3
4, obviously if R=Ais observed, the best decision would
2. Here we know that a 0 is transmitted, therefore we are looking for p(error|T= 0), this is
the probability that the receiver declares a 1 was sent when actually a 0 was transmitted.
3. We have p(error|T= 0) = 1
3, and p(error|T= 1) = p(R=B|T= 1) = 1
3. Therefore, by the
total probability theorem
Problem 2.17
Following the same procedure as in example 2-1-1, we prove :
Problem 2.18
Relationship (2-1-44) gives :
pY(y) = 1
3a[(y−b)/a]2/3pX“y−b
a1/3#
3a√2π[(y−b)/a]2/3e−1
−10 −8 −6 −4 −2 0 2 4 6 8 10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
y
pdf of Y
a=2
b=3
Problem 2.19
1) The random variable Xis Gaussian with zero mean and variance σ2= 10−8. Thus p(X > x) =
Q(x
σ) and
p(X > 10−4X > 0) = p(X > 10−4, X > 0)
Problem 2.20
1) y=g(x) = ax2. Assume without loss of generality that a > 0. Then, if y < 0 the equation
y=ax2has no real solutions and fY(y) = 0. If y > 0 there are two solutions to the system, namely
x1,2=py/a. Hence,
fY(y) = fX(x1)
|g′(x1)|+fX(x2)
|g′(x2)|
√ay√2πσ2e−y
2) The equation y=g(x) has no solutions if y < −b. Thus FY(y) and fY(y) are zero for y < −b. If
equal to
FY(−b+)−FY(−b−) = FX(−b)
and
fY(y) = FX(−b)δ(y+b) + (1 −FX(b))δ(y−b) + fX(y)[u−1(y+b)−u−1(y−b)]
σ(δ(y+b) + δ(y−b)) + 1
3) In the case of the hard limiter
Thus FY(y) is a staircase function and
4) The random variable y=g(x) takes the values yn=xnwith probability
Thus, FY(y) is a staircase function with FY(y) = 0 if y < x1and FY(y) = 1 if y > xN. The PDF
is a sequence of impulse functions, that is
N
X
i=1 hQai
Problem 2.21
For nodd, xnis odd and since the zero-mean Gaussian PDF is even their product is odd. Since
the integral of an odd function over the interval [−∞,∞] is zero, we obtain E[Xn] = 0 for nodd.
This is true for all n. Now let n= 2k−1, we will have I2k= (2k−1)σ2I2k−2, with the initial
condition I0=√2πσ2. Substituting we have
I2=σ2√2πσ2
and in general if I2k= (2k−1)(2k−3)(2k−5) ×···×3×1σ2k√2πσ2, then I2k+2 = (2k+ 1)σ2I2k=
for neven.
Problem 2.22
a. Since (Xr, Xi) are statistically independent :
pX(xr, xi) = pX(xr)pX(xi) = 1
2πσ2e−(x2
r+x2
i)/2σ2
Also :
Yr+jYi= (Xr+Xi)ejφ ⇒
Xi=−Yrsin φ+Yicos φ
The Jacobian of the above transformation is :
∂Xr
∂Xi
=
cos φ−sin φ
Hence, by (2-1-55) :
pY(yr, yi) = pX((Yrcos φ+Yisin φ),(−Yrsin φ+Yicos φ))
2πσ2e−(y2
r+y2
b. Y =AX and X=A−1Y
Now, pX(x) = 1
(2πσ2)n/2e−x′x/2σ2(the covariance matrix Mof the random variables x1, …, xnis
Problem 2.23
Since we are dealing with linear combinations of jointly Gaussian random variables, it is clear
that Yis jointly Gaussian. We clearly have mY=E[AX] = AmX. This means that Y−mY=
Problem 2.24
a.
ψY(jv) = EejvY =Ehejv Pn
i=1 xii=E“n
Y
i=1
ejvxi#=
n
Y
i=1
EejvX =ψX(ejv)n
b.
E(Y) = −jdψY(jv)
dv |v=0 =−jn(1 −p+pejv)n−1jpejv|v=0 =np
1. In the figure shown below
18
x
x
u
v
let us consider the region u > x, v > x shown as the colored region extending to infinity, call
this region R, and let us integrate e−u2+v2
2over this region. We have
ZZ
R
e−u2+v2
2du dv =ZZ
R
e−r2
2r dr dθ
2
where we have used the fact that region Ris included in the region outside the quarter circle
as shown in the figure. On the other hand we have
ZZ
e−u2+v2
2du dv =Z∞
x
e−u2
2du Z∞
x
e−v2
2dv
From the above relations we conclude that
2π(Q(x))2≤π
2e−x2
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2. In R∞
2dy
2and dv =dy
yand du =−ye−y2
Z∞
x
e−y2
2dy
y2=
−e−y2
2
y
∞
x
−Z∞
x
e−y2
2dy =e−x2
2
x−√2πQ(x)
x−√2πQ(x)>0⇒Q(x)<1
On the other hand, note that
Z∞
x
e−y2
2dy
y2<1
x2Z∞
x
e−y2
2dy =√2π
x2Q(x)
3. From x
√2π(1 + x2)e−x2
2< Q(x)<1
√2πx e−x2
2
Problem 2.26
20
1. FYn(y) = P[Yn≤y] = 1 −P[Yn> y] = 1 −P[x1> y, X2> y, . . . , Xn> y] = 1 −(P[X > y])n
1
2.
f(y) = n
A1−y
An−1
1−y
A1−λy
nn
Problem 2.27
ψ(jv1, jv2, jv3, jv4) = Ehej(v1x1+v2x2+v3x3+v4x4)i
where v= [v1, v2, v3, v4]′,M= [µij].
We obtain the desired result by bringing the exponent to a scalar form and then performing
∂vi
ive−1
where µ′
i= [µi1, µi2, µi3, µi4] . Also note that :
∂µ′
jv
∂vi
=µij =µji