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QUESTION
The following figure shows the encoder for a 1/2 rate convolutional code.
Determine the encoder output produced by the message sequence 101111111…
1. Assume that the initial state of the encoder is zero.
Flip Flop (shift register) • Modulo 2 adder Yi Output 7 D D2 D3 12
Show transcribed image text Flip Flop (shift register) • Modulo 2 adder Yi Output
7 D D2 D3 12
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The given figure is sy + x D ng D3 34 I X3 >Y 캥고 the status assumed are
arb,c,d,eitigih. State table : Next state x 42, outputs Ч. Ч. 2 O OOO O 0 1 Ooo (a)
Ioo (e) OOO 1 OO input present state X, X, X3 ofa 136 18] Oiided O 1 0 0 OOO (a)
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I 0 Ole) 0 o ol 0 010 OO I (b) 1 1 o 01 (1) 6 O 1 Oo, 0) I 01 (f) O — o 010 © 16 23 %
1 1 loo I10 (9) 0 0 0 1013: 010) 1069) 0 1 I 1 o . o 11032 0 1 011 (d) 1 1 1 1h) 0 0 1
Oil (d) I lich) With O outputs y = x + 24 + 2 + 13 Y ₂ = x+x+ CS Scanned with
CamScanner
The given message sequence is 101111 111...1 Initial state of the encoder is 2010.
So, initid state is a Code Tree : for input message sequeuce 101111111...1 of input
os finput "1" — down woord movement upward movement soutput 427 a 11 12
01 e f 12 10 ů h 01 h lol ► the encoded output is h lol h 11 11 01 11 10 01 01 01
01 ..... 01. CS Scanned with CamScanner
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