then the matched waveform is
h(t) = gT (T t) = sinc
2
T
T
t
2
= gT (t)
where the last equality follows from the fact that gT (t) is even with respect to the t =
output of the matched lter is
=
F 1
=
T2
T
f ej2 f T
4
2
T
T
T
2
sinc
(t T ) = gT t
2
T
2
2
T

1) The envelope delay of the RC lter is (see Problem 9.19)
Tc (f ) =
RC
1 + 4 2 (RC)2 f 2
A plot of T (f ) with RC = 106 is shown in the next gure
10
x -7
10
9.999
9.998
9.997
Tc(f)
9.996
9.995
9.994
9.993
9.992
9.991
9.99
0
500 1000 1500 2000 2500 3000 3

Thus,
Q
3 Eb
15 N0
=
Eb
1
105
= 25.3688
3
N0
3 If the bit rate of transmission is 19200 bps, then
k=
19200
=8
2400
In this case a 256-QAM constellation is used and the probability of error is
PM
1
Q
=1 12 1
16
With PM = 105 we obtain
3 8 Eb
255 N0
2
Eb

With
Y (z) = 0.3z + 0.9 + 0.3z1 = (f0 + f1 z1 )(f0 + f1 z)
we obtain the parameters f0 and f1 as
0.7854
f0 =
,
0.1146
0.1146
f1 =
0.7854
The parameters f0 and f1 should have the same sign since f0 f1 = 0.3. However, the sign itself does
not play a

then its lth element is
gk,l
J(ck )
=
2ck,l
=
=
Thus, the vector gk is
gk =
N
1
E 2
ck,n ykn ak ykl
2
n=N
E ek ykl = E ek ykl
E[ek yk+N ]
.
.
.
E[ek ykN ]
= E[ek yk ]
where yk is the vector yk = [yk+N ykN ]T . Since k = ek yk , its expected value is

6) Let a be xed to some value between 0 and 1. Then, if we argue as in part 5) we obtain
A2 T
V (a)
8
P (e|s1 , a)
=
P (n2 n1 > 2
P (e|s2 , a)
=
P (n1 n2 > (a + 2)
A2 T
+ V (a)
8
and the probability of error is
P (e|a) =
1
1
P (e|s1 , a) + P (e|s2 , a)
2

4) The variance of the noise at the output of the generalized matched lter is
2 =
Sn (f )|G(f )|2 df =
|S(f )|2
df
Sn (f )
At the sampling instant t = t0 = T , the signal component at the output of the matched lter is
y(T )
=
=
Y (f )ej2 f T df =
S(f )
S

9
8.5
8
PAR
7.5
7
6.5
6
5.5
5
5
10
Realization
15
20
The MATLAB script for the problem is given below.
T = 1;
Fs = 200;
t = 0 : 1/(Fs*T) : T1/(Fs*T);
K = 32;
k = 1 : K1;
rlz = 20;
% No. of realizations
PAR = zeros(1,rlz);
% Initialization for speed
echo o

Problem 12.5
H(X, Y )
=
=
H(X, g(X) = H(X) + H(g(X)|X)
H(g(X) + H(X|g(X)
But, H(g(X)|X) = 0, since g() is deterministic. Therefore,
H(X) = H(g(X) + H(X|g(X)
Since each term in the previous equation is non-negative we obtain
H(X) H(g(X)
Equality holds when

Therefore, there is a signicant amount of spectral overlap among the signals transmitted on different subcarriers. Nevertheless, these signals are orthogonal when transmitted synchronously in
time.
20
0
20
40
|Uk(f)| (dB)
60
80
100
120
140
160
180
0
0.5
1

a=rand(1,36);
a=sign(a0.5);
b=reshape(a,9,4);
% Generate the 16QAM points
XXX=2*b(:,1)+b(:,2)+j*(2*b(:,3)+b(:,4);
XX=XXX;
X=[0 XX 0 conj(XX(9:1:1)];
xt=zeros(1,101);
for t=0:100
for k=0:N1
xt(1,t+1)=xt(1,t+1)+1/sqrt(N)*X(k+1)*exp(j*2*pi*k*t/T);
echo o
end

0
10
1
Pb
10
2
10
3
10
Adaptive LMS equalizer
No ISI
4
10
0
2
4
6
8
1
10log 2
10
12
14
16
Figure 10.24: Symbol error probability of LMS adaptive equalizer for 2 = 0.01, 2 = 0.1 and 2 = 1
487
Chapter 11
Problem 11.1
The analog signal is
1
x(t) =
N
N1
Xk e

1
Ray
5
1
2
= 1
1
2
1
5
The equalizer coecients obtained by inverting the matrix Ry and the results are as follows
0.0956
0.7347
1 q =
copt = X
1.6761
0.7347
0.0956
Figure 10.21 presents the equalized pulse. Note the small amount of residua

X(m) = (1/W)*cos(pi*f/(2*W);
else
X(m)=0;
end
end;
X=X/max(abs(X);
X= 20.* log10(abs(X);
% impulse response of the cascade of the transmitter and the receiver lters
g R=g T;
imp resp of cascade=conv(g R,g T);
% Plotting commands follow.
30
40
Computer Pro

The function g(t) needs to be checked only for those values of t such that 42 t 2 /T 2 = 1 or t =
However,
cos( x)
cos(t/T )
2
= lim
lim
2 2
2
x1 1 x
t T 1 4 t /T
2
and by using LHospitals rule
lim
cos( 2 x)
1x
x1
Hence,
sin( x) =
<
x1 2
2
2
= lim
1
x(nT

semilogy(SNRindB1,smld err prb,*);
hold
semilogy(SNRindB2,theo err prb);
function [p]=smldPe83(snr in dB)
% [p]=smldPe83(snr in dB)
%
SMLDPE83 simulates the probability of error for a given
%
snr in dB, signal-to-noise ratio in dB.
E=1;
alpha opt=1/2;
% s

the signal corresponding to that point, we have two signals with energy d2 , two signals with energy
9d2 , two signals with energy 25d2 ,., and two signals with energy (M 1)2 d2 . The average energy
is
2d2
1
Ei =
Eav =
1 + 9 + 25 + + (M 1)2
M i
M
Using th

where the random variable n = n1 n2 is zero-mean Gaussian with variance
2
n
=
2
2
n1 + n2 2E[n1 n2 ]
=
2
2
n1 + n2 2
1.5
N0
d
2
1
1.5 + 1 2 0.5 = 1.5
=
Hence,
P (e|s1 ) =
T 1
1
2
2n
e
x2
2
2n
dx
Similarly we nd that
P (e|s2 )
=
=
=
P (0.5 + n1 1 n2 > T )

3) In the next gure we plot the probability of error as a function of p1 , for two values of the
2E
SNR = N0s . As it is observed the probability of error attains its maximum for equiprobable signals.
0.12
6
0.1
x -24
10
7
5
P(e)
8
0.14
P(e)
0.16
0.08
0.0

Problem 8.31
1) Consider the QAM constellation of Fig. P-10.12. Using the Pythagorean theorem we can nd the
radius of the inner circle as
1
a2 + a2 = A2 a = A
2
The radius of the outer circle can be found using the cosine rule. Since b is the third side o

Note that Tb has been normalized to 1 since the problem has been stated in terms of the power of
the involved signals. The probability of error is given by
P (error) = Q
2PT
N0
1
Pc
PT
The loss due to the allocation of power to the pilot signal is
SNRloss

5T
2
T
2T
T . . . . . . . .
3) The signal waveform matched to s3 (t) is
h3 (t) =
The output of the matched lter is
2
0
y3 (t) = h3 (t) s3 (t) =
In the next gure we have plotted y3 (t).
0t
T
2
T
2
<tT
4t 2T
4t + 6T
T
2
t<T
T t
3T
2
2T . . . . . . . . . .

-1
-0.5
0.5
1
-3
x 10
Figure 8.17: The autocorrelation function of the raised-cosine signal
391
Chapter 9
Problem 9.1
1) The rst set represents a 4-PAM signal constellation. The points of the constellation are cfw_A, 3A.
The second set consists of four or