University Problem Set #2: Solutions ECE 456
of Illinois Page 1 of 1 Fall 2002
1.(a) G =
1
1
0 0
1
1 1
1
1 1
1
0 1
1
1 0
0
1 0
1
1 →
1
0
0 0
1
1 1
0
1 1
0
0 1
0
1 0
0
1 0
1
1 →
1
0
0 0
1
0 1
0
1 1
0
0 1
0
1 0
0
1 0
1
0 →
1
0
0 0
1
0 0
0
1 1
0
0 0
0
1 1
0
1 0
1
0 = [I | P]
on subtracting row 1 from row 2, then subtracting row 2 from row 3, and finally subtracting row 3 from
row 1. Hence, H = [–PT | I] =
1
0
1
0
0
0
0
1
0
1
1
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
(c) The minimum weight is no more than 2, since a linear combination of rows of G has weight 2 (where?).
Now, the weight of the linear sum of i rows of the rightmost generator matrix is at least i (why?) and there
are no rows of weight 1. Hence, there are no codewords of weight 1; the minimum weight is exactly 2.
For a linear code, minimum distance = minimum weight. Hence, the minimum distance is also 2.
2.(a) For a, b ∈ {0,1}, it is easy to show that (a + b) mod 2 = a + b – 2ab. Since the weight of a vector is the
sum of the weights of the coordinates, the result follows. An important corollary is that the weight of
x + y is even if and only if both x and y have even weight, or both have odd weight.
(b) Since wt(y) ≥ wt(x∧y), the result follows from (a). Equality will hold if and only if wt(y) = wt(x∧y),
which holds whenever yi = xiyi ∀ i, 1 ≤ i ≤ n, or equivalently, ∀ i, 1 ≤ i ≤ n, yi = 1 ⇒ xi = 1. Some
people say that x must cover y for equality to hold These results apply to binary vectors only.
3.(a) Let C0 and C1 denote the sets of codewords with zeroes and ones respectively in the i-th position. If C1 is
empty, all the codewords have zero in the i-th position. (We note that C0 is never empty because 0 ∈ C0
always). If x ∈ C1, then ∀ y ∈ C0, x + y ∈ C1. This shows that |C0| ≤ |C1|. Similarly, ∀ z ∈ C1,
x + z ∈ C0. Hence, |C1| ≤ |C0|, and therefore |C0| = |C1|.
(b) The argument is exactly the same as in (a) except that now we let Ce and Co denote the sets of codewords
with even and odd weights respectively. The "important corollary" of 2(a) is used to deduce that if x ∈ Co,
then ∀ y ∈ Ce, x + y ∈ Co, and similarly, ∀ z ∈ Co, x + z ∈ Ce. This shows that |Ce| = |Co|.
4.(a) The decoder output is incorrect if two or three of the received channel symbols are in error. The probability
of this is 3p2(1 – p) + p3 = 3p2 – 2p3 = Pe,coded.
(b) From p = (1/2)•exp(–A/3), we get that exp(–A) = (2p)3. Hence,
Pe,uncoded = (1/2)exp(–A) = 4p3 and Pe,coded – Pe,uncoded = 3p2 – 6p3 = 3p2(1 – 2p) > 0 for 0 < p < 1/2.
(c) Now, p = 1/[2(1 + A/3)] ⇒ Pe,uncoded = 1/[2(1+A)] = p/(3 – 4p).
Thus, Pe,coded – Pe,uncoded = 3p2 – 2p3 – p
3 – 4p = p(8p3 – 18p2 + 9p – 1)
3–4p = p(p–1/2)(8p2 – 14p + 2)
3–4p
= p(p–1/2)(p–(7– 33)/8)(p–(7+ 33)/8)
3–4p .
The numerator is negative for 0 < p < (7– 33)/8) ≈ 0.1569… and positive for (7– 33)/8) < p < 1/2. The
denominator is positive for p < 3/4. Hence, Pe,coded < Pe,uncoded if p < (7– 33)/8) ≈ 0.1569…
i.e. if A > (9 + 3 33)/4 ≈ 6.5584… which correspomds to an SNR of 8.1680… dB