Solution_3_2007 - ECE 154B Winter 2007 Homework#3 Due...

ECE 154B Winter 2007
Homework #3 Due: Monday, February 5, 2007
1. Prove that for M a positive even integer,
1
2
+3
2
+…+(M-3)
2
+ (M-1)
2
= M(M
2
-1)/6. (Hint: Use induction)
Proof by induction:
Let M=2. (2-1) 2 =(2)((2) 2 -1)/6=1.
Assume true for M.
Test for M+2: 1 2 +3 2 +…+(M+2-3) 2 + (M+2-1) 2 = (M+2)((M+2) 2 -1)/6
1 2 +3 2 +…+(M-1) 2 + (M+1) 2 = (M+2)(M 2 +4M+4-1)/6
1 2 +3 2 +…+(M-1) 2 + (M+1) 2 = (M)(M 2 -1 )/6 +(2)(M 2 -1)/6+(M+2)(4M+4)/6
(M+1)
2
= (2M
2
-2+4M
2
+12M+8)/6
(M+1)
2
= (6M
2
+12M+6)/6
(M+1)
2
= (M+1)
2
2. The following applies to M= 4, PAM transmission with unequal a priori
probabilities. Assume that one of four signals was sent of the form:
{-3A σ (t), -A σ (t), +A σ (t),and +3A σ (t)}
with a priori probabilities
{1/8, 3/8, 3/8 and 1/8}
respectively. σ (t) has duration T and has unit energy. The received signal is equal
to the transmitted signal plus additive white Gaussian noise with power spectral
density N
0
/2 . Let Z be the output of a correlator-integrator receiver which
multiplies by σ (t) and integrates over the duration of each pulse T.
(a) Evaluate the average symbol energy, E
s
, and the average energy per bit, E
b
. Hint:
You need to take into account the a priori probabilities.
E
0
=
∫
0
T
− 3A   t 
2
dt = 9A
2
∫
0
T
 t 
2
dt = 9A
2
E
1
=
∫
0
T
− A   t 
2
dt = A
2
, E
2
= E
1
, E
3
= E
0
E
s
= 
1
8
 E
0

3
8
 E
1

3
8
 E
2

1
8
 E
3
= 3A
2
E
b
=
E
s
log
2
4
=
3A
2
2
(b) Find the optimal decision regions for Z. That is find the decision regions that lead
to a minimum probability of error.


N
2
= E [
∫
0
T