solutions convolutional codes
DESCRIPTION
notesTRANSCRIPT
Solutions Solutions
Solutions to Chapte 7: Channel Coding(Bernard Sklar)
Note: State diagram, tree diagram and trellis diagram for K=3 are sameonly changes will occur in the output that depends upon the connectionvector.
Manjunatha. P (JNNCE) Coding Techniques July 5, 2013 1 / 11
Solutions Solutions
7.1 Draw the state diagram, tree diagram and trellis diagram for the K=3, rate=1/3 code generated by
g1(X ) = X + X 2, g2(X ) = 1 + X + X 2, g3(X ) = 1 + X
Output
Input U
1
2
3
States represent possible contents of therightmost K-1 register content.
For this example there are only two transitionsfrom each state corresponding to two possibleinput bits.
Solid line denotes for input bit zero, and dashedline denotes for input bit one.
Tuple State00 a10 b01 c11 d
Table: State transition table (rate=1/2, K=3)
Input State at State at Outputbit time ti time ti + 10 00 00 0001 00 10 0110 01 00 1101 01 10 1010 10 01 1111 10 11 1000 11 01 0011 11 11 010
b=10 c=01
d=11
0(000)
a=000(110)1(011)
1(101)
0(111)
1(010)
1(100)
0(001)
Input bit
Output bit
Figure: State diagram for rate=1/3 and K=3
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Solutions Solutions
1
0
000
011
100
000000
111
011a
b011
a
010
111110
001
101c
d100
b
000a
100
100
11000
111
11a
b101c
010
001110
001
101c
d010d
011b
000
111
011
101
100
000000
111
011a
b011
a
010
111110
001
101c
d100
b
111c
010
100
110000
111
011a
b101c
010
001110
001
101c
d010d
100d
110
001
a
b
a
t1 t2 t3 t4 t5
Figure: Tree diagram for rate=1/3 andK=3
Steady State
a=00000
d=11
c=01
b=10
000 000000000
011 011
111
011011 011
010
100
111111
100100100
010010
001 001001
110110110101101101
t1 t5t4t3t2 t6
Figure: Trellis diagram for rate=1/3 and K=3
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Solutions Solutions
7.2 Given a K=3 and rate 1/2 binary convolutional code with the partially completed state diagram shown in Figure 7.1 find thecomplete state diagram, and sketch a diagram for encoder
10 01
11
001(11)
0(10)
1(00)
Input bit
Output bit
Figure: State diagram for rate=1/2, K=3
States represent possible contents of therightmost K-1 register content.
For this example there are only two transitionsfrom each state corresponding to two possibleinput bits.
OutputInput U
1
2
Figure: Encoder diagram
Table: State transition table (rate=1/2, K=3)
Input State at State at Outputbit time ti time ti + 10 00 00 001 00 10 110 01 00 011 01 10 100 10 01 101 10 11 010 11 01 111 11 11 00
10 01
11
0(00)
000(01)1(11)
1(10)
0(10)
1(00)
1(01)
0(11)
Input bit
Output bit
Figure: Modified State diagram
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Solutions
7.3 Draw the state diagram, tree diagram and trellis diagram for the convolutional encoder characterized by the block diagramin Figure p7.2
Output
Input U
1
2
States represent possible contents of therightmost K-1 register content.
For this example there are only two transitionsfrom each state corresponding to two possibleinput bits.
Solid line denotes for input bit zero, and dashedline denotes for input bit one.
Tuple State00 a10 b01 c11 d
Table: State transition table (rate=1/2, K=3)
Input State at State at Outputbit time ti time ti + 10 00 00 001 00 10 100 01 00 111 01 10 010 10 01 111 10 11 010 11 01 001 11 11 10
10 01
11
0(00)
a=000(11)1(10)
1(01)
0(11)
1(10)
1(01)
0(00)
Input bit
Output bit
Figure: State diagram for rate=1/3 and K=3Manjunatha. P (JNNCE) Coding Techniques July 5, 2013 5 / 11
Solutions
1
0
00
10
01
0000
11
10a
b10
a
10
1111
00
01c
d01
b
00a
01
01
1100
11
11a
b01c
10
0011
00
01c
d10d
10b
00
11
10
01
01
0000
11
10a
b10
a
10
1111
00
01c
d01
b
11c
10
01
1100
11
10a
b01c
10
0011
00
01c
d10d
01d
11
00
a
b
a
t1 t2 t3 t4 t5
Figure: Tree diagram for rate=1/3 andK=3
Steady State
a=0000
d=11
c=01
b=10
00 000000
10 10
11
1010 10
10
01
1111
010101
1010
00 0000
111111010101
t1 t5t4t3t2 t6
Figure: Trellis diagram for rate=1/3 and K=3
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Solutions Solutions
7.5 Consider the convolutional encoder shown in Fiugre a)Write the connection vectors and polynomials for this encoder.b)Draw the state diagram, tree diagram and trellis diagram
Solution: g1 = [1 0 1], g2 = [0 1 1]
g1(X ) = 1 + X 2, g2(X ) = X + X 2
Output
Input U
1
2
States represent possible contents of therightmost K-1 register content.
For this example there are only two transitionsfrom each state corresponding to two possibleinput bits.
Tuple State00 a10 b01 c11 d
Table: State transition table (rate=1/2, K=3)
Input State at State at Outputbit time ti time ti + 10 00 00 001 00 10 100 01 00 111 01 10 010 10 01 011 10 11 110 11 01 101 11 11 00
10 01
11
0(00)
000(11)1(10)
1(01)
0(01)
1(00)
1(11)
0(10)
Input bit
Output bit
Figure: State diagram for rate=1/3 and K=3Manjunatha. P (JNNCE) Coding Techniques July 5, 2013 7 / 11
Solutions Solutions
1
0
00
10
11
0000
01
10a
b10
a
00
0111
10
01c
d11
b
00a
11
11
1100
01
11a
b01c
00
1011
10
01c
d00d
10b
00
01
10
01
11
0000
01
10a
b10
a
00
0111
10
01c
d11
b
01c
00
11
1100
01
10a
b01c
00
1011
10
01c
d00d
11d
11
10
a
b
a
t1 t2 t3 t4 t5
Figure: Tree diagram for rate=1/3 andK=3
Steady State
a=0000
d=11
c=01
b=10
00 000000
10 10
01
1010 10
00
11
0101
111111
0000
10 1010
111111010101
t1 t5t4t3t2 t6
Figure: Trellis diagram for rate=1/3 and K=3
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Solutions Solutions
7.6 An encoder diagram is shown in Figure. Find the encoder output for an input sequence 1 0 0 1 0 1 0
Solution: g1 = [1 0 1], g2 = [1 1 1]
g1(X ) = 1 + X 2, g2(X ) = 1 + X + X 2
Output
Input U
1
2
Table: State transition table (rate=1/2, K=3)
Input Register State at State at Outputbit Contents time ti time ti + 1- 000 00 00 001 100 00 10 110 010 10 01 010 001 01 00 111 100 00 10 110 010 10 01 011 101 01 10 000 010 10 01 01
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Solutions Solutions
7.6 Figure shows an encoder for a (3,2) convolutional code. Find the transfer function T (D) and minimum free distance for thiscode. Also, draw the state diagram for the code.
Output
Input U1
2
2
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References
S. Lin and D. J. C. Jr., Error Control Coding, 2nd ed. Pearson / Prentice Hall, 2004.
R. Blahut, Theory and Practice of Error Control Codes, 2nd ed. Addison Wesley, 1984.
J. G. Proakis, Digital communications, 4th ed. Prentice Hall, 2001.
J. G. Proakis and M. Salehi, Communication Systems Engineering, 2nd ed. Prentice Hall,2002.
S. Haykin, Digital communications, 2nd ed. Wiley, 1988.
Manjunatha. P (JNNCE) Coding Techniques July 5, 2013 11 / 11