Convolution encoder

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i Venky
i Venky 2011 年 10 月 23 日
I got a doubt in this convolution encoder.
When one or two incoming bits per symbol enter a rate-1/2 or rate 2/3 binary convolutional encoder, the result would be two or three coded bits per symbol respectively.
How is that so?
thanks in advance.

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Wayne King
Wayne King 2011 年 10 月 23 日
Not sure what your question is here. Assume you have a encoder with 3 memory registers, 1 input bit, and 3 output bits.
To encode your output bits, you have generator polynomials:
(1,1,1), (1,0,1), and (0,1,1)
so one output bit would be the sum of the bits in the three memory registers, one would be the sum of the last and the first memory register, and one would be the sum of the second to last and last.
This is a rate 1/3 code.
  2 件のコメント
i Venky
i Venky 2011 年 10 月 23 日
What do you mean by the word "symbol" in my question? I am quite confused with that term
Wayne King
Wayne King 2011 年 10 月 23 日
symbols are the meaningful elements in a digital code. If you have a code with 1 bit per symbol, then your alphabet is 0,1
If you have two bits per symbol, then your alphabet is 00 01 10 11

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