University Problem Set #3: Solutions ECE 456
of Illinois Page 1 of 2 Fall 2002
1. On the AWGN channel, q = (1/2)•exp(–A), and Au(q) = –(ln q + ln 2) for the uncoded system. For the coded
system, q ≈ 3p2 = (3/4)•exp(–2A/3), and thus, Ac(q) = –(3/2)•(ln q + ln 4/3). The coding gain is thus
approximately 2
3 • ln q + ln 2
ln q + ln 4/3 ≈ 2
3 as q → 0.
This is consistent with the result R•(t+1) since R = 1/3 and t = 1 for this code. On the nonselective
Rayleigh fading channel, q = 1/[2(1+A)] for the uncoded system, and so Au(q) = 1/(2q) – 1 ≈ 1/(2q) for small
q. Similarly, for the coded system, q ≈ 3p2 = 3/[2(1+A/3)]2 so that Ac(q) = 3[(1/2)• 3/q – 1] ≈ (1/2) 27/q
Hence, the coding gain is ≈ 1 27/q → 5 as q → 0. This is not consistent with the stated result. Also,
similar results are obtained for other codes as well: the concept of asymptotic coding gain does not apply
very well on fading channels.
2.(a) P(correct decoding) = (1–p)6. Since A(z) = 1 + 4z3 + 3z4, P(error) = (1–p)6A(p/(1–p)) – (1–p)6
= 4p3(1–p)3 + 3p4(1–p)2 = 4p3 – … , and P(decoding failure) = 1 – P(error) – P(correct decoding)
= 6p(1–p)5 + 15p2(1–p)4 + 16p3(1–p)3 + 12p4(1–p)2 + 6p5(1–p) + p6 = 6p – …
(b) Now, P(correct decoding) = (1 – p)6 + 6p(1 – p)5.
P(decoding failure) = 3p2(1 – p)4 + 4p3(1 – p)3 + p6 = 3p2 – ….
P(error) = 1 – P(correct decoding) – P(failure)
= 12p2(1 – p)4 + 16p3(1 – p)3 + 15p4(1 – p)2 + 6p5(1 – p) = 12p 2 – … Note that the probability of
decoding error increased from ≈ 4p3 to ≈ 12p2 while the probability of decoding failure decreased from ≈ 6p
to ≈ 3p2. P(correct decoding) increased by 6p(1–p)6.
(c) P1 = P(first bit is in error)
= Probability that the error pattern is in the fourth, sixth, seventh, or eighth column of the standard array
= [3p2(1–p)4 + 2p3(1–p)3 + 3p4(1–p)2]
+ [p2(1–p)4 + 4p3(1–p)3 + p4(1–p)2 + 2p5(1–p)]
+ [4p3(1–p)3 + p4(1–p)2 + 2p5(1–p) + p6]
+ [3p2(1–p)4 + 2p3(1–p)3 + 3p4(1–p)2]
= 7p2(1–p)4 + 12p3(1–p)3 + 8p4(1–p)2 + 4p5(1–p) + p6 = 7p2 – …
P2 = P(second bit is in error)
= Probability that the error pattern is in third, fifth, seventh, or eighth column of the standard array = P1.
P3 = P(third bit is in error)
= Probability that the error pattern is in the second, fifth, sixth, or eighth column of the standard array
= [3p2(1–p)4 + 2p3(1–p)3 + 3p4(1–p)2]
+ [p2(1–p)4 + 4p3(1–p)3 + p4(1–p)2 + 2p5(1–p)]
+ [3p2(1–p)4 + 2p3(1–p)3 + 3p4(1–p)2]
= 8p2(1–p)4 + 12p3(1–p)3 + 8p4(1–p)2 + 4p5(1–p) = 8p2 – … ≠ P1.
The difference in the bit error probabilities P1, P2, and P3 is quite small. However, note that the dominant
term is slightly larger for P3.
(d) Average bit error probability for a complete decoder = (P1 + P2 + P3)/3 from part (c)
= (22/3)p2(1–p)4 + 12p3(1–p)3 + 8p4(1–p)2 + 4p5(1–p) + (2/3)p6.
(e) Average bit error probability for a single-error-correcting decoder can be obtained by ignoring the last row in
the standard array. We have that the average bit error probability = ^
P1= ^
P2 = ^
P3
= 6p2(1–p)4 + 10p3(1–p)3+ 8p4(1–p)2 + 4p5(1–p).
(f) Q6(x) = (1–p+xp)6A[(x–xp+p)/(1–p+xp)]
= (1–p+xp)6 + 4(1–p+xp)3(x–xp+p)3 + 3(1–p+xp)2(x–xp+p)4.
Q1(x) = all terms of degree no greater than 1 in Q6(x)
= (1–p)6 + 6xp(1–p)5 + 4[(1–p)3•p3 + 3(1–p)2xp•p3 + 3p2(x–xp)•(1–p)3]
+ 3[(1–p)2•p4 + 2(1–p)xp•p4 + 4(x–xp)p3•(1–p)2] giving
Q1(1) = (1–p)6+6p(1–p)5+12p2(1–p)4+16p3(1–p)3+15p4(1–p)2+6p5(1–p). This is the probability of
decoding (correctly or incorrectly). Since P(correct decoding) = (1–p)6 + 6p(1–p)5, the probability of
decoding error matches the answer found in part (b).
3. The set of syndromes is a vector space of dimension n–k. Its image under any linear map cannot be of
larger dimension. Since the n single-error bit patterns are the standard basis for the n dimensional space, the
image of the syndromes under a linear map can include at most n–k of them, and hence at most n–k single
errors can be corrected by a linear decoder.
4.(a) Suppose that P is a square matrix. Then, the code has rate 1/2 and so does its dual. Furthermore, since
PPT = –I, and G = [I|PT], we have GGT = I + PTP = 0 showing that the codewords of the code are