NATIONAL AUSTRALIAN UNIVERSITY
Department of Engineering
ENGN3226 Digital Communications
Tutorial #6 Solutions
1. (Proakis and Salehi 7.11) In a binary PAM system for which two signals occur with unequal probabilities p and 1 p the optimum detector is specied by b z
s1 s2 2 N0 1 p n 1 p ln = ln . 2 p 4 p
Thus, the optimum threshold is =
N0 4 b
ln
1p p
. (Note if p = 0.5, = 0.)
(a) The probability of error is found, as follows. P (e) = P (e|s1 )p(s1 ) + P (e|s2 )p(s2 )
=p
f (z|s1 )dz + (1 p)
f (z|s2 )dz
=p
1 2N0 /2
e
(zs1 )2 2N0 /2
dz + (1 p)
1 2N0 /2 z s2 N0 /2
e
(zs2 )2 2N0 /2
dz
change of variables,
(s1 )/ N0 /2
x=
z s1 N0 /2
and x =
=p
x2 1 e 2 dx + (1 p) 2
(s2 )/
N0 /2
x2 1 e 2 dx 2
= pQ = pQ
+ s1 N0 /2 + b N0 /2
+ (1 p)Q + (1 p)Q
s2 N0 /2 + b N0 /2
(b) Evaluate the probability of error for p = 0.3 and p = 0.5 with b /N0 = 10. For p = 0.3, is given by 5 7 . = N0 ln 2 3 And, so the probability of error is given by P (e) = 0.3Q 5 7 ln + 2 5 + 0.7Q 3 2 2 = 0.3Q(2.9743) + 0.7Q(5.97) = 0.0004 = 4e4 7 5 ln + 2 5 3 2 2
1
2. (Proakis and Salehi 7.12) A binary PAM communication system is used to transmit data over an AWGN channel. The prior probabilities for the bits are P (am = 1) = 1/3 and P (am = 1) = 2/3. (a) To determine the optimum threshold, start with the MAP criterion, as we have signals with unequal probabilities. That is, us the probability metrics PM(sm ) = f (z|sm )p(sm ) So, for 2 signals, that is, m 1, 2 we write, knowing that z|sm is Gaussian with mean sm and variance N0 /2. PM(s1 ) s1 1 PM(s2 ) s2 f (z|s1 )p(s1 ) s1 1 f (z|s2 )p(s2 ) s2
1 e 2N0 /2 1 e 2N0 /2
(zs1 )2 2N0 /2 (zs2 )2 2N0 /2
1 3 s1 2 3 s2
1
e Take logs.
(zs2 )2 (zs1 )2 2N0 /2
s1 s2
2
(z s2 )2 (z s1 )2 s N0 ln 2
2
s1
N0 ln 2 2zs2 + + 2zs1 2 s1 2z b + b + 2z b b s N0 ln 2
2
s2 2
s1 s2 s 1
N0 z s 4 ln 2 2 b
s1
Thus, the optimum threshold is =
N0 4 b
ln 2.
(b) The average probability of error is determined, as follows. P (e) = P (e|s1 )p(s1 ) + P (e|s2 )p(s2 ) 1 2 = f (z|s1 )dz + f (z|s2 )dz. 3 3 From the previous question, this simplies to P (e) = 1 Q 3 + 1 N0 /2 2 + Q 3 +1 N0 /2 .
2
3. (Proakis and Salehi 7.9) A binary digital communication system employs the signals s0 (t) = 0,0 t T s1 (t) = A,0 t T for transmitting the information. This is called on-o signalling. The demodulator cross-correlates the received signal r(t) with s1 (t) (Note: this is dierent to the way the correlator is described in lectures where the received signal is correlated the with basis functions) and samples the output of the correlator at t = T . (a) Because it is an AWGN environment, with equally likely symbols, we can use either the distance or correlation metrics. We choose the correlation metrics. Because the symbols have dierent energies we must include the energy components in the correlation metrics. That is, C(z, sm ) = 2zsm sm
2
Note that this is a one-dimensional system. The energy of s1 is
T
1 =
0
s2 (t)dt = 1
T 0
A2 dt = A2 T.
The energy of s0 is, of course, 0. The basis function, (t), looks like s1 (t) but with a height of 1/ T for an energy of 1. Thus, the projection of s1 onto the basis function is s1 = A T . To nd the decision threshold, we compare the correlation metrics, as follows. Note that the energy and dot product with z of s0 are both 0. C(z, s1 ) 2zs1 s1 2 2zA T A2 T z
s1 s0 s1 s0 s1 s0 s1 s0
C(z, s0 ) 0 0 A T = . 2
(b) For equally likely symbols, the probability of error is given by P (e) = 1 1 P (e|s1 ) + P (e|s0 ) 2 2 = P (e|s0 ) (zs0 )2 1 e 2N0 /2 dz = A T 2N0 /2 2 2 1 z = e 2N0 /2 dz A T 2N0 /2 2 N0 /2dx and when z = A T /2, x = A 1 x2 e 2 dx 2 =Q b 2N0 T /2N0 .
Change of variables, x = z/ So,
N0 /2, so, dz =
P (e) =
A
T /(2N0 )
=Q A
T 2N0
where b = 1 . Recall from lectures that for antipodal signalling, the probability of error is P (e) = Q 3 b N0 .
The probability of error for the same bit energy is a factor of 1/ 2 better for antipodal signalling, as far as the argument of the Q function goes. Now, assume the same pulse en...