University of Illinois
Spring 2011
ECE 361:
Problem Set 6: Solutions
Erasure Channels; Gaussian Channel Capacity
1.
[Single erasure correcting code]
A simple “overall paritycheck code” with generator matrix
G
=
1
0
0
0
1
0
1
0
0
1
0
0
1
0
1
0
0
0
1
1
suffices.
Note that the transmitted codeword is [
u
0
, u
1
, u
2
, u
3
, u
0
⊕
u
1
⊕
u
2
⊕
u
3
] = [
c
0
, c
1
, c
2
, c
3
, c
4
]
where
c
4
is the overall paritycheck bit. If
c
4
is erased, no problems! [
c
0
, c
1
, c
2
, c
3
,
?] = [
u
0
, u
1
, u
2
, u
3
,
?]
and the four data bits are received without error. But if one of the other bits is erased, then it can be
obtained as the XOR sum of the four unerased bits. For example, if the receiver gets [
c
0
, c
1
,
?
, c
3
, c
4
],
then
c
0
⊕
c
1
⊕
c
3
⊕
c
4
=
u
0
⊕
u
1
⊕
u
3
⊕
(
u
0
⊕
u
1
⊕
u
2
⊕
u
3
) =
u
2
, the erased data bit! Of course, the
unerased
data bits are just
c
0
, c
1
, and
c
3
.
2.
[Decoding a ReedSolomon on an erasure channel]
(a) We have to interpolate
ˆ
d
(
z
) of degree 3 through the 4 points (0
,
1)
,
(1
,
3)
,
(2
,
0)
,
(4
,
0). But since
x
i
3
=
x
i
4
= 0, the corresponding terms disappear from the sum defining
ˆ
d
(
z
), and we have
ˆ
d
(
z
) = 1
×
(
z

1)(
z

2)(
z

4)
(0

1)(0

2)(0

4)
+ 3
×
(
z

0)(
z

2)(
z

4)
(1

0)(1

2)(1

4)
= (
z

2)(
z

4)
z

1
6
+
3
z
3
= (
z

2)(
z

4) since
z/
6
≡ 
z
and

1
/
6 = 1 modulo 7
=
z
2

6
z
+ 8
≡
z
2
+
z
+ 1 modulo 7
.
We can crosscheck the arithmetic calculations by computing
ˆ
d
(
i
) for
i
= 0
,
1
,
2
,
4 to verify that
we do get 1
,
3
,
0
,
0 respectively Since
ˆ
d
(
z
) = 1 +
z
+
z
2
+ 0
z
3
must equal
d
(
z
), we have that the
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 University of Illinois, gaussian channel capacity

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