Also, from Problem 12.15, H (X, Y ) H (X) + H (Y ). Combining the two relations, we obtain
H (Y ) + H (X|Y ) H (X) + H (Y ) H (X|Y ) H (X)
Suppose now that the previous relation holds with equality. Then,
p(x) log p(x|y) =
p(x) log p(x)
x
x
p(x) log(
x

E
J0
=
W/R
Nu 1
=
106 /104
14
=
100
14
= 7.14 (8.54 dB)
1. The processing gain is 100.
2. With Nu = 30 and Eb /J0 = 7.14, the processing gain should be increased to
W/R = (7.14) (29) = 207
Hence, the bandwidth must be increased to W = 2.07MH z.
Problem 11

Solution to Chapter 2 Problems
Problem 2.1
1.
2 t+
(2t + 5) =
plot is shown below:
5
2
. This indicates rst we have to plot
5
(2t) and then shift it to left by 2 . A
T
(2t + 5)
1
E t
9
11 4
4
2.
(2t + 8) =
(2(t 4). This operation combines a scaling, ippi

Since Am = max[|m(t)|] we conclude that the modulation index is
2bAm
a
=
Problem 3.20
1) When USSB is employed the bandwidth of the modulated signal is the same with the bandwidth of the
message signal. Hence,
WUSSB = W = 104 Hz
2) When DSB is used, then

then,
H (Y |X) = pH (Y |X = 0) + (1 p)H (Y |X = 1)
= ph( ) + (1 p)h( ) = h( )
where h( ) is the binary entropy function. As it is seen H (Y |X) is independent on p and therefore I (X; Y )
is maximized when H (Y ) is maximized. To nd the distribution p(x)

1)
=
Ex
T
2
lim
T
T
2
2
x1 (t)dx
T
2
= lim
T
e2t cos2 tdt
0
T
2
1
(2 cos2 t + sin 2t) 1 e2t
= lim
T 8
0
3
1
T
= lim
(2 cos2 + sin T 1)eT + 3 =
T 8
2
8
Thus x1 (t) is an energy-type signal and the energy content is 3/8
2)
=
Ex
=
T
2
lim
T
T
2
2
x2 (t)dx

Problem 11.32
Since T = 2900 + 150 = 3050 K, it follows that
N0 = kT = 1.38 1023 305 = 4.21 1021 W/Hz
The transmitting wavelength is
=
c
3 108
= 0.130 m
=
f
2.3 109
Hence, the gain of the receiving antenna is
GR =
2
D
= 0.55
3.14159 64
0.130
2
= 1.3156 1

t = T is
T
y(T ) =
r( )s( )d
0
T
=
(s( ) + s( T ) + n( )s( )d
0
T
=
T
s 2 ( )d +
0
n( )s( )d
0
= Es + n
where Es is the energy of the signal pulse and n is a zero-mean Gaussian random variable with variance
2
n = N02Es . Similarly, the output of the match

Problem 9.19
The frequency response of the RC lter is
C(f ) =
R
1
j 2RCf
1
+ j 2RCf
=
1
1 + j 2 RCf
The amplitude and the phase spectrum of the lter are
1
1 + 4 2 (RC)2 f 2
|C(f )| =
1
2
,
c (f )
= arctan(2 RCf )
The envelope delay is
1 d c (f )
RC
1
2 RC

and the probability of error is
P (e|a) =
1
1
P (e|s1 , a) + P (e|s2 , a)
2
2
For a given a, the optimal value of V (a) is found by setting
V (a) =
a
4
P (e|a)
V (a)
equal to zero. By doing so we nd that
A2 T
2
The mean square estimation of V (a) is
1
V

where n is a zero-mean Gaussian random variable with variance
2
n = E
=
=
N0
2
N0
4
Tb
Tb
0
n(t)n( ) cos(2fc t + ) cos(2fc + )dtd
0
Tb
cos2 (2fc t + )dt
0
Note that Tb has been normalized to 1 since the problem has been stated in terms of the power of the

1110
0110
1010
1011
0111
0010
1111
1100
1101
0011
1000
0100
0000
1001
0101
0001
One way to label the points of the V.29 constellation using the Gray-code is depicted in the next gure. Note
that the maximum Hamming distance between points with distance bet

In order to transmit 9600 bps with a symbol rate R = 2400 symbols per second, the number of information
bits per symbol should be
k=
9600
=4
2400
Hence, a 24 = 16 QAM signal constellation is needed. The carrier frequency fc is set to 1800 Hz, which is
the

Solution to Chapter 9 Problems
Problem 9.1
1) The following table shows the values of Eh (W )/T obtained using an adaptive recursive Newton-Cotes
numerical integration rule.
WT
0.5
1.0
1.5
2.0
2.5
3.0
Eh (W )/T
0.2253
0.3442
0.3730
0.3748
0.3479
0.3750
A

Type
M=4
PAM
2 log2 M
4
QAM
log2 M
2
FSK (coherent)
To achieve a ratio
(QAM).
R/W
2 log2 M
M
log2 M
M
1
FSK (noncoherent)
0.5
of at least 2, we have to select either the rst signal set (PAM) or the second signal set
R
W
Problem 10.17
1) If the transmitted

The received signal is the convolution of u(t) with h(t). Hence,
an s(t nT )
y(t) = u(t) h(t) =
(t) +
n=
an s(t nT ) +
=
n=
2
(t t0 ) + (t + t0 )
2
2
an s(t t0 nT ) +
n=
2
an s(t + t0 nT )
n=
Thus, the output of the matched lter s(t) at the time instant t

1
1
However, V (f ) is odd with respect to 0 and since V (f + 2T ) and V (f 2T ) are even, the translated spectra
satisfy
1
2T
1
2T
1 j 2f t
)e
V (f
df =
2T
1
2T
1
2T
V (f +
1 j 2f t
)e
df
2T
U (f +
1 j 2f t
df
)e
2T
Hence,
x(t) = sinc(t/T ) + 2j sin(

At time t = T we have
T
y(T ) =
T
si ( )d +
0
n( )d =
0
Eb
T +
T
T
n( )d
0
The signal energy at the output of the integrator at t = T is
Eb
Es =
T
T
2
= Eb T
whereas the noise power
T
Pn = E
n( )n(v)d dv
0
T
=
0
T
E[n( )n(v)]d dv
0
=
T
N0
2
0
T
0
T
( v)

Problem 8.13
1) The optimum threshold is given by
N0
1p
N0
= ln 2
= ln
p
4 Eb
4 Eb
2) The average probability of error is ( =
N
0
4 Eb
ln 2)
2
1
e(r+ Eb ) /N0 dr
N0
2
1
e(r Eb ) /N0 dr
+p(am = 1)
N0
+ Eb
2
Eb
1
Q
=
+ Q
3
3
N0 /2
N0 /2
1
2N0 /Eb l

If t 0 or t 2T , then y2 (t) = 0. If 0 < t
t T
2
y2 (t) =
T
,
2
T
2
4d +
t T
2
0
If T < t
3T
2
T
2
4d +
tT
3T
2
t
(2)d +
T
2
= 2t. If
9
d = 7t T
2
, then
y2 (t) =
For,
t
0 (2)d
then y2 (t) =
t T
2
(2)d +
T
2
T
t T
2
d =
19T
7t
2
< t 2T , we obtain
T
y2

corresponding to that point, we have two signals with energy d 2 , two signals with energy 9d 2 , two signals
with energy 25d 2 ,., and two signals with energy (M 1)2 d 2 . The average energy is
Eav =
1
M
Ei =
i
2d 2
1 + 9 + 25 + + (M 1)2
M
Using the well

R2
R1
0d
d
d
d
R3
d
d
5) If the signals are equiprobable then,
P (e|m1 ) = P (|r m1 |2 > |r m2 |2 m1 ) + P (|r m1 |2 > |r m3 |2 m1 )
When m1 is transmitted then r = [ T + n1 , n2 ] and therefore, P (e|m1 ) is written as
P (e|m1 ) = P (n2 n1 > T ) + P (n1

where n is a zero mean Gaussian random variable with variance
=
T
2
n =
T
0
=
=
=
(s1 ( ) s2 ( )(s1 (v) s2 (v)E[n( )n(v)]d dv
(s1 ( ) s2 ( )(s1 (v) s2 (v)
0
T
N0
( v)d dv
2
N0
(s1 ( ) s2 ( )2 d
2 0
2
N0 T T 2A
A d
2 0 0
T
N 0 A2 T
2 3
Since
s1 (t)(s1 (t)

and
T
y(t) = etT
e2 d
tT
T
1
= etT e2
2
tT
1 t+T
1 t3T
=
e
e
2
2
Therefore
y(t) =
1 T
et et
e
2
1 t+T
1
e
2 et3T
2
T <t 2
0
otherwise
0<t T
Problem 8.19
We have Pav = REb = 2 106 Eb , hence
2Eb
N0
Pb = Q
=Q
2Pav
RN0
= 106
Using the Q-function table (pag

and therefore, the Fourier transform of the signal matched to s(t) is
H (f ) = S (f )ej 2f T = S (f )ej 2f nTc
n
= P (f )
ck ej 2f kTc ej 2f nTc
k=1
n
= P (f )
cni+1 ej 2f (i1)T c
i=1
= P (f )F[g(t)]
Thus, the matched lter H (f ) can be considered as the

The quantity n cik cj k is the inner product of the row vectors C i and C j . Since the rows of the matrix Hn
k=1
are orthogonal by construction, we obtain
n
si (t)sj (t)dt = Ep
2
cik ij = nEp ij
k=1
Thus, the waveforms si (t) and sj (t) are orthogonal.
2

E .
.
. k
E
E
c1
c
r s1 E k E
()dt
()dt
T
s1 (t)
.
E .k
. .
c2
Select
r s2 E c E
k
the
E
.
.
.
largest
T
r(t)
E
.
.
.
s2 (t)
.
E .k
. .
E
cM
r sM E c E
k
()dt
T
sM (t)
Parallel to the development of the optimal receiver using N lters matched to the orthon