Recall that we can view the Reed-Solomon code of dimension k = m + 1 over[a] =0,.|i=0,...,m,j=1,...,q where {a₁,..., ag}GF(q) as having generator matrix GGF (q).===(i) Show that the encoding of (bo, ..., bm) is (ƒ(a₁),..., f(aq)), where f(x) =bo+b1 + +bmxm.

Answered: Show that the encoding of (bo, bo+b1 +…

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Question

Recall that we can view the Reed-Solomon code of dimension k = m + 1 over
[a] =0,.
|i=0,...,m,j=1,...,q where {a₁,..., ag}
GF(q) as having generator matrix G
GF (q).
=
=
=
(i) Show that the encoding of (bo, ..., bm) is (ƒ(a₁),..., f(aq)), where f(x) =
bo+b1 + +bmxm.

Expert Solution

Step 1: step 1

Define the vector $b = (b_{0} , . . ., b_{m} ) .$

Define the generator matrix $G$ as follows:

$(\begin{matrix} a_{1}^{0} & a_{1}^{1} & \dots & a_{1}^{m} \\ a_{2}^{0} & a_{2}^{1} & \dots & a_{2}^{m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{q}^{0} & a_{q}^{1} & \dots & a_{q}^{m} \end{matrix})$

Define the codeword $c = G b .$

We want to show that $c = (f (a_{1} ), . . ., f (a_{q} )) .$

to do this, we can note that the i-th element of $c$ is given by:

$c_{i} = G_{i} b$

where $G_{i} $ is the i-th row of the generator matrix $G$ .

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Show that the encoding of (bo, bo+b1 + +bmxm. ‚bm) is (ƒ(a₁),..., ƒ(aq)), where f(x) = =