1
ECE 534: Elements of Information Theory
Solutions to Midterm Exam (Spring 2006)
Problem 1
[20 pts.]
A discrete memoryless source has an alphabet of three letters,
x
i
, i
= 1
,
2
,
3
, with probabilities
0
.
4
,
0
.
4
,
and
0
.
2
, respectively.
(a) Find the binary Huffman code for this source and determine the average number of bits needed for
each source letter.
(b) Suppose two letters at a time are encoded into a binary sequence. Find the Huffman code and the
average number of bits needed per source letter.
Solution:
(a) One possible Huffman code is:
C
(1) = 0
,
C
(2) = 10
, and
C
(3) = 11
. The average number of bits
per source letter is
0
.
4 + (0
.
4 + 0
.
2)
×
2 = 1
.
6
.
(b) One possible code construction is:
C
(11) = 000
,
C
(12) = 001
,
C
(13) = 100
,
C
(21) = 010
,
C
(22) =
011
,
C
(23) = 110
,
C
(31) = 111
,
C
(32) = 1010
, and
C
(33) = 1011
. The average number of bits per
source letter is
[3 + (0
.
08 + 0
.
04)]
/
2 = 1
.
56
.
Problem 2
[20 pts.]
A source
X
produces letters from a threesymbol alphabet with the probability assignment
P
X
(0) = 1
/
4
,
P
X
(1) = 1
/
4
, and
P
X
(2) = 1
/
2
. Each source letter
x
is transmitted through two channels simultaneously
with outputs
y
and
z
and the transition probabilities indicated below:
2
1
0
1
0
x
y
P
(
y

x
)
1
1
1
2
1
2
a0
a0
a0
a0
a0
a0
a0
a0
a0
a0
a0
a0
a0
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
2
1
0
1
0
x
z
P
(
z

x
)
1
1
1
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
a8
Calculate
H
(
X
)
,
H
(
Y
)
,
H
(
Z
)
,
H
(
Y, Z
)
,
I
(
X
;
Y
)
,
I
(
X
;
Z
)
,
I
(
X
;
Y

Z
)
, and
I
(
X
;
Y, Z
)
.
Solution:
H
(
X
) =
1
4
log
2
4 +
1
4
log
2
4 +
1
2
log
2
2 = 1
.
5
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The probability distribution of
Y
is
P
Y
(0) =
1
4
+
1
2
1
2
=
1
2
. Therefore,
H
(
Y
) = 1
Similarly,
P
Z
(0) =
P
Z
(1) =
1
2
and
H
(
Z
) = 1
If
Z
= 1
, then
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 Spring '11
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 Information Theory, Px, sum channel, ith channel

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