e2 205 error-control coding lecture 18pvkece/pdfs/ecc19/lec18.pdf= q(p e b d p n o=2 = q(r 2e b d n...
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E2 205 Error-Control CodingLecture 18
Himani Kamboj
October 14, 2019
1 More on Error Probability
Setting: Consider an AWGN channel. C is a linear code of block length n.
ci ∈ Fn2 =⇒ si = (−1)ci 0 ≤ i ≤M − 1
Let c0 = 0 (Since code is linear, all zero codeword exists).
Define
ci =[ci1 ci2 . . . cin
]T
Ai =
(−1)ci1
(−1)ci2
...
(−1)cin
Goal: To show that the probability of codeword error is independent un-der (appropriate) MLD of the transmitted codeword in a linear block code.Furthermore, the same holds true for the residual error vector.
e = s− s , s = (−1)c
1
By si = (−1)ci , we mean[si1 si2 . . . sin
]=[(−1)ci1 (−1)ci2 . . . (−1)cin
]Note: Over the AWGN channel, MLD is equivalent to minimum Euclideandistance decoding.
Proof:Let y is the received vector.Define R0 = {y | P (y|c0) ≥ P (y|ci) , i 6= 0} = {y | dE(y, s0) ≤ dE(y, si)}and Ri = AiR0
Declare c = ci if y ∈ Ri, 0 ≤ i ≤M − 1.
Claim: Decoding using decision regions Ri ensures MLD.
Suppose y ∈ Ri =⇒ P (y|ci) ≥ P (y|cj) ∀ i 6= j
then dE(y, si) ≤ dE(y, sj) , i 6= j
Proof:Let y = Ai x , x ∈ R0
dE(y, si) = dE(Ai x, si) = dE(Ai(Ai x), Ai si)(Since Ai is an orthogonal matrix, it is isometric hence preservers norm)dE(y, si) = dE(x,Ai si)Similarly dE(y, sj) = dE(x,Ai sj)Let ci + cj = ckdE(y, sj) = dE(x, sk)Thus dE(x, s0) ≤ dE(x, sk), x ∈ R0 , k 6= 0=⇒ dE(y, si) ≤ dE(y, sj) y ∈ Ri
Next we want to show that the probability of codeword error is independentof the transmitted codeword.
Pr(Correct | c0) =
∫R0
P (y|s0) dy
Pr(Correct | ci) =
∫Ri
P (y|si) dy =
∫Ri
P (y|Ai s0) dy =
∫Ri
P (Ai y|s0) dy
2
Let xi = Ai y =⇒ dx = |Ai|dy
y ∈ Ri =⇒ y ∈ AiRi = R0
Pr(Correct | ci) =
∫R0
P (x0|s0)dx
|Ai|
=
∫R0
P (x0|s0) dx (Since|Ai| = 1|)
Next we will show that the residual error is independent of the transmittedcodeword.
Let ci + cj = ck
=⇒ si sj = sk
Ai Aj = Ak
Pr(cj | ci) =
∫Rj
P (y|ci) dy =
∫Rj
P (y|si) dy =
∫Rj
P (y|Ai s0) dy =
∫Rj
P (Ai y|s0) dy
=
∫Ai, Rj
P (x|s0) dx =
∫Rk
P (x|s0) dx
3
(Since Ai Rj = Ai Aj R0 = Ak R0)
Therefore, residual error = ci + cj
2 Bit Error Probability (BEP) of rate k/n co-
volutional code:
Assume K(N − ν) message bits and Kν flush bits.
Pbe =1
K(N − ν)
K(N−ν)∑n=1
Pbe, i
Set J = K(N − ν) (counting in two different ways)Assuming all zero codeword is transmitted.
Pbe =1
J
∑IP
Pr(P = IP ) IW (IP )
whereP = path associated with message sequenceP = decoded pathIP = incorrect pathDPS = diverged path segmentIW = input weight (hamming weight of input message sequence associatedto IP)
Pbe =1
J
∑IP
Pr(P = IP )∑
DPS in IP
IW (DPS)
=1
J
∑DPS
IW (DPS)∑
IP through DPS
Pr(P = IP )
=1
J
N−ν∑j=1
∑DPSj
IW (DPSj)∑
IP through DPSj
Pr(P = IP )
(DPSj : DPS that diverges at node level j in the trellis)
4
Pr(P = IP ) ≤ Pr( d(DPSj, y) ≤ d(0, y) )
Let d(DPSj, y) ≤ d(0, y) = DPSj prevails
Pbe ≤1
J
N−ν∑j=0
∑DPSj
IW (DPSj)∑
IP through DPSj
Pr(DPSj prevails)
≤ 1
J
N−ν∑j=0
∑all DPS
IW (DPS)∑
IP through DPS
Pr(DPS prevails)
=1
K
∑all DPS
IW (DPS)∑
IP through DPS
Pr(DPS prevails)
5
= Q (
√Eb d√No/2
)
= Q (
√2Eb d
N0
)
= Q (
√2Eb (dfree + (d− dfree))
N0
)
≤ Q (
√2Eb dfree
N0
) exp (−(d− dfree) E0
N0
)
= Q (
√2Eb dfree
N0
) exp (dfree EbN0
) exp (−d EbN0
)
Therefore,
Pbe ≤1
K
∂AEND(L, I,D)
∂IQ (
√2Eb dfree
N0
) exp (dfree EbN0
)
where L = 1, I = 1, D = exp(−Eb/N0), Q(x) =1√2π
∫ ∞x
e−y2/2 dy
and using Q(√x+ y) ≤ Q(
√x) exp(−y/2)
For BSC,
Pbe ≤1
K
∂AEND(L, I,D)
∂I
where L = 1, I = 1, D = 2√ε(1− ε)
6
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