The convolutional encoder

Aya2002 said:

I have this question: In the book "Introduction to CDMA Wireless Communications" by: Mosa Ali Abu-Rgheff, page 108 the following example;



Example 2.11
A convolutional encoder that provides the best error performance in satellite communication
systems has the following parameters:

G = [133 171]
R = 1/2
K = 7
Determine the structure of the encoder.
Solution
The two octal number are converted to binary forms as:
133=001 011 011=1 011 011
171=001 111 001=1 111 001
The generator polynomials are:

g1(x) = 1 · (x^0) + 0 · (x^1) + 1 · (x^2) + 1 · (x^3) + 0 · (x^4) + 1 · (x^5) + 1 · (x^6)
g2(x) = 1 · (x^0) + 1 · (x^1) + 1 · (x^2) + 1 · (x^3) + 0 · (x^4) + 0 · (x^5) + 1 · (x^6)

Denote the input as i(x), the 1st digit is computed from i(x). g1(x). The 2nd digit is computed from i(x). g2(x).

Thus for i(x)=101=1+x^2,
1st digit=(1+x^2)(1+x^2 +x^3 +x^5 +x^6)=10 01 10 00 1
2nd digit=(1+x^2)(1+x+x^2 +x^3 +x^6)=11 00 11 10 1

The encoded sequence is 11 01 00 10 11 01 01 00 11

My question is how the encoded sequence is 11 01 00 10 11 01 01 00 11 ?

can i have a detailed procedure for this result?

Many thanks

Montadar

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