How to create generator matrix from polynomial for linear block code?
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Anyone can help me step by step to create generator matrix from polynomial? I have checked in many blogs but didn't understand.( Even i don't know what are mathematics required).
linear-algebra polynomials coding-theory
asked Aug 24 at 5:15
Chandramohan V
61
add a comment |Â
up vote
1
down vote
favorite
Anyone can help me step by step to create generator matrix from polynomial? I have checked in many blogs but didn't understand.( Even i don't know what are mathematics required).
linear-algebra polynomials coding-theory
asked Aug 24 at 5:15
Chandramohan V
61
3
I would look at books rather than blogs.
– Lord Shark the Unknown
Aug 24 at 5:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Anyone can help me step by step to create generator matrix from polynomial? I have checked in many blogs but didn't understand.( Even i don't know what are mathematics required).
linear-algebra polynomials coding-theory
asked Aug 24 at 5:15
Chandramohan V
61
Anyone can help me step by step to create generator matrix from polynomial? I have checked in many blogs but didn't understand.( Even i don't know what are mathematics required).
linear-algebra polynomials coding-theory
asked Aug 24 at 5:15
Chandramohan V
61
asked Aug 24 at 5:15
Chandramohan V
61
asked Aug 24 at 5:15
Chandramohan V
61
asked Aug 24 at 5:15
Chandramohan V
61
61
3
I would look at books rather than blogs.
– Lord Shark the Unknown
Aug 24 at 5:32
add a comment |Â
3
I would look at books rather than blogs.
– Lord Shark the Unknown
Aug 24 at 5:32
3
3
I would look at books rather than blogs.
– Lord Shark the Unknown
Aug 24 at 5:32
I would look at books rather than blogs.
– Lord Shark the Unknown
Aug 24 at 5:32
add a comment |Â
1 Answer
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If you have a finite field $k$, then a polynomial $f(X)$ is a generating polynomial for a linear code of length $n$ over $k$ iff it is a factor of $X^n-1$ over $k$.
Then we can write $f(X)=a_0+a_1X+a_2X^2+cdots+a_dX^d$ where $d$ is the degree
of $f(X)$ and $a_dne0$. This generates a code with dimension $n-d$
and generator matrix
$$G=pmatrixa_0&a_1&a_2&cdots&a_d&0&0&cdots&0\
0&a_0&a_1&cdots&a_d-1&a_d&0&cdots&0\
0&0&a_0&cdots&a_d-2&a_d-1&a_d&cdots&0\
vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots\
0&0&cdots&0&a_0&a_1&a_2&cdots&a_d.$$
There are $n-d$ rows, and each row after the first is obtained by shifting
the previous row cyclically right by one position.
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
That's the way it works in theory. In practice telcomm folks often use shortened cyclic codes. That amounts to leaving out a fixed number of columns from the right end as well as the same number of rows from the bottom. You see, in practice somebody other than the coding theorist decides the rank and the length of the code.
– Jyrki Lahtonen
Aug 24 at 6:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you have a finite field $k$, then a polynomial $f(X)$ is a generating polynomial for a linear code of length $n$ over $k$ iff it is a factor of $X^n-1$ over $k$.
Then we can write $f(X)=a_0+a_1X+a_2X^2+cdots+a_dX^d$ where $d$ is the degree
of $f(X)$ and $a_dne0$. This generates a code with dimension $n-d$
and generator matrix
$$G=pmatrixa_0&a_1&a_2&cdots&a_d&0&0&cdots&0\
0&a_0&a_1&cdots&a_d-1&a_d&0&cdots&0\
0&0&a_0&cdots&a_d-2&a_d-1&a_d&cdots&0\
vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots\
0&0&cdots&0&a_0&a_1&a_2&cdots&a_d.$$
There are $n-d$ rows, and each row after the first is obtained by shifting
the previous row cyclically right by one position.
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
That's the way it works in theory. In practice telcomm folks often use shortened cyclic codes. That amounts to leaving out a fixed number of columns from the right end as well as the same number of rows from the bottom. You see, in practice somebody other than the coding theorist decides the rank and the length of the code.
– Jyrki Lahtonen
Aug 24 at 6:07
add a comment |Â
up vote
0
down vote
If you have a finite field $k$, then a polynomial $f(X)$ is a generating polynomial for a linear code of length $n$ over $k$ iff it is a factor of $X^n-1$ over $k$.
Then we can write $f(X)=a_0+a_1X+a_2X^2+cdots+a_dX^d$ where $d$ is the degree
of $f(X)$ and $a_dne0$. This generates a code with dimension $n-d$
and generator matrix
$$G=pmatrixa_0&a_1&a_2&cdots&a_d&0&0&cdots&0\
0&a_0&a_1&cdots&a_d-1&a_d&0&cdots&0\
0&0&a_0&cdots&a_d-2&a_d-1&a_d&cdots&0\
vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots\
0&0&cdots&0&a_0&a_1&a_2&cdots&a_d.$$
There are $n-d$ rows, and each row after the first is obtained by shifting
the previous row cyclically right by one position.
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
That's the way it works in theory. In practice telcomm folks often use shortened cyclic codes. That amounts to leaving out a fixed number of columns from the right end as well as the same number of rows from the bottom. You see, in practice somebody other than the coding theorist decides the rank and the length of the code.
– Jyrki Lahtonen
Aug 24 at 6:07
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you have a finite field $k$, then a polynomial $f(X)$ is a generating polynomial for a linear code of length $n$ over $k$ iff it is a factor of $X^n-1$ over $k$.
Then we can write $f(X)=a_0+a_1X+a_2X^2+cdots+a_dX^d$ where $d$ is the degree
of $f(X)$ and $a_dne0$. This generates a code with dimension $n-d$
and generator matrix
$$G=pmatrixa_0&a_1&a_2&cdots&a_d&0&0&cdots&0\
0&a_0&a_1&cdots&a_d-1&a_d&0&cdots&0\
0&0&a_0&cdots&a_d-2&a_d-1&a_d&cdots&0\
vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots\
0&0&cdots&0&a_0&a_1&a_2&cdots&a_d.$$
There are $n-d$ rows, and each row after the first is obtained by shifting
the previous row cyclically right by one position.
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
If you have a finite field $k$, then a polynomial $f(X)$ is a generating polynomial for a linear code of length $n$ over $k$ iff it is a factor of $X^n-1$ over $k$.
Then we can write $f(X)=a_0+a_1X+a_2X^2+cdots+a_dX^d$ where $d$ is the degree
of $f(X)$ and $a_dne0$. This generates a code with dimension $n-d$
and generator matrix
$$G=pmatrixa_0&a_1&a_2&cdots&a_d&0&0&cdots&0\
0&a_0&a_1&cdots&a_d-1&a_d&0&cdots&0\
0&0&a_0&cdots&a_d-2&a_d-1&a_d&cdots&0\
vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots&vdots\
0&0&cdots&0&a_0&a_1&a_2&cdots&a_d.$$
There are $n-d$ rows, and each row after the first is obtained by shifting
the previous row cyclically right by one position.
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
answered Aug 24 at 5:31
Lord Shark the Unknown
88.1k955115
88.1k955115
That's the way it works in theory. In practice telcomm folks often use shortened cyclic codes. That amounts to leaving out a fixed number of columns from the right end as well as the same number of rows from the bottom. You see, in practice somebody other than the coding theorist decides the rank and the length of the code.
– Jyrki Lahtonen
Aug 24 at 6:07
add a comment |Â
That's the way it works in theory. In practice telcomm folks often use shortened cyclic codes. That amounts to leaving out a fixed number of columns from the right end as well as the same number of rows from the bottom. You see, in practice somebody other than the coding theorist decides the rank and the length of the code.
– Jyrki Lahtonen
Aug 24 at 6:07
That's the way it works in theory. In practice telcomm folks often use shortened cyclic codes. That amounts to leaving out a fixed number of columns from the right end as well as the same number of rows from the bottom. You see, in practice somebody other than the coding theorist decides the rank and the length of the code.
– Jyrki Lahtonen
Aug 24 at 6:07
That's the way it works in theory. In practice telcomm folks often use shortened cyclic codes. That amounts to leaving out a fixed number of columns from the right end as well as the same number of rows from the bottom. You see, in practice somebody other than the coding theorist decides the rank and the length of the code.
– Jyrki Lahtonen
Aug 24 at 6:07
add a comment |Â
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3
I would look at books rather than blogs.
– Lord Shark the Unknown
Aug 24 at 5:32