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.
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N
2
=
E
[
∫
0
T

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- Winter '07
- WOlf
- Conditional Probability, Probability, Probability theory
-
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