This preview shows page 1. Sign up to view the full content.
Unformatted text preview: tober 3, 2012 72 / 93 Lecture Notes 7 Simpliﬁed Calculation Then
(r , s0 ) = x0 + x2
(r , s1 ) = x1 + x3
(r , s2 ) = x0 − x2 (r , s3 ) = x1 − x3 Thus the calculation requires only 6 additions/subtractions. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 73 / 93 Lecture Notes 7 Implementation 0
r0 r1
r2 +
−
+
− EECS 455 (Univ. of Michigan) x0 = r0 +
− x1 = −r0 x2 = r1 + r2
x3 = r1 − r2 Fall 2012 +
− (r , s0 ) = r0 + r1 + r2 (r , s2 ) = r0 − r1 − r2
(r , s1 ) = −r0 + r1 − r2
(r , s3 ) = −r0 − r1 + r2 October 3, 2012 74 / 93 Lecture Notes 7 Performance The performance of the optimum demodulation is usually very
difﬁcult to evaluate exactly.
Usually upper bounds on the error probability are employed.
One bound is called the union bound. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 75 / 93 Lecture Notes 7 Example 3 s0 (t ) = 1ϕ0 (t ) + 0ϕ1 (t )
s1 (t ) = −1ϕ0 (t ) + 0ϕ1 (t )
√
s2 (t ) = 2ϕ0 (t ) + 3ϕ1 (t )
√
s3 (t ) = 0ϕ0 (t ) + 3ϕ1 (t )
√
s4 (t ) = −2ϕ0 (t ) + 3ϕ1 (t )
√
s5 (t ) = 2ϕ0 (t ) − 3ϕ1 (t )
√
s6 (t ) = 0ϕ0 (t ) − 3ϕ1 (t )
√
s7 (t ) = −2ϕ0 (t ) − 3ϕ1 (t ) EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 76 / 93 Lecture Notes 7 Example 3
The energy of the signals are
E0 = 1, E1 = 1, E2 = 7, E3 = 3, E4 = 7, E5 = 7, E6 = 3, E7 = 7. The average energy is E = 36/8 = 4.5.
The energy per bit is Eb = E /3 = 1.5. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 77 / 93 Lecture Notes 7 Example 3
ϕ1 (t )
s3 s4 s1 s7 EECS 455 (Univ. of Michigan) s2 s0 s6 Fall 2012 ϕ0 (t ) s5 October 3, 2012 78 / 93 Lecture Notes 7 Example 3
3 2 φ1(t ) 1 0 −1 −2 −3
−3
EECS 455 (Univ. of Michigan) −2 −1 0
φ0(t )
Fall 2012 1 2 3
October 3, 2012 79 / 93 Lecture Notes 7 Example 3: Decision Region for s0
ϕ1 (t ) R4 s4 R1
R7
s7 EECS 455 (Univ. of Michigan) R3 s3 R2 s2 R0 s1 ϕ0 (t ) s0 R6 s6 Fall 2012 R5 s5 October 3, 2012 80 / 93 Lecture Notes 7 Union Bound To calculate the probability of error (exactly) we need to determine
the probability that a two dimensional Gaussian random vector,
centered at the point s0 is not in the decision region for signal s0 .
This is a two dimensional integration over a subset of the plane of
the Gaussian density.
Let Ri be the region of received signals where it is decided that
signal i is transmitted.
Let Ri ,j be the region where signal i is chosen when compared
only to signal j . EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 81 / 93 Lecture Notes 7 Union Bound
Then
R 1 ∪ R 2 ∪ · · · ∪ R M − 1 = R 1, 0 ∪ R 2, 0 · · · ∪ R M − 1, 0
Pe,0 = P {errors0 transmitted} = P {r ∈ R1 ∪ R2 ∪ R3 ∪ · · · ∪ RM −1 s0 transmitted} = P {r ∈ R1,0 ∪ R2,0 ∪ R3,0 ∪ · · · ∪ RM −1,0 s0 transmitted}
M −1 ≤ i =1 P {r ∈ Ri ,0s0 transmitted} Let P2 (s0 → s1 ) be the pairwise error probability of deciding s1
given s0 was transmitted when the receiver assumes there is only
two possible decisions, either s0 or s1 .
EECS 455 (Uni...
View
Full
Document
This note was uploaded on 02/12/2014 for the course EECS 455 taught by Professor Stark during the Fall '08 term at University of Michigan.
 Fall '08
 Stark

Click to edit the document details