Question

Find the minimum distance of the encoding function e : B2→ B

4

given by

e( 0,0) = 0000 , e(10) = 0110 , e(01) = 1011 , e(11) = 1100.​

Answer 1

Answer:

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Answer 2

Answer:

The minimum distance is 3 of the encoding function e : B2→ B4

Step-by-step explanation:

Show that the (2,5) encoding function e:B2→B4 defined by

e(00) = 00000

e(01) = 01110

e(10) = 10101

e(11) = 11011

is a group code.

How many errors will it detect and correct?

Solution: Encoding function e:B2→B4defined as

e(00) = 0000 = x0

e(01) = 1011 = x1

e(10) = 0110 = x2

e(11) = 1100 = x3

∴ Range of encoding function is,

Range (e) = { x0=00000, x1=01110, x2=11011 , x3=11011 }

Encoding function e : B2→B4 is said to be a group code if range of e is subgroup of B2

∴ Prepare composition table for (Bn,(+))

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∵ All the entries of composition table is closed.

∴ Encoding function e:B2→B4 is a group code.

∵ e:B2→B4 is a group codes

∴ mm distance of encoding function e is the min of weight of non-zero code words.

∴ |X_1 | = 3

| X_2 | = 3

| X_3 | = 4

∴ Minimum distance = 3