Question A [10 marks in total]. Louise needs to send to Madhu sequences which consist of lettersT, E, B, L. Each letter is represented by a two-bit message, as follows:T= [0 0], E= [0 1], B= [1 0], L= [1 1].Since Louise and Madhu have to use a one-way, noisy channel, BSC(𝑝) with small 𝑝, Louise encodeseach message using the generator matrix𝐺 = [1 0 0 1 10 1 1 0 1]of a binary linear code 𝐶.A1 [2 marks] Represent the word LET as three two-bit messages, encode each message as describedabove and write down the resulting three codevectors of 𝐶.
Question
Question A [10 marks in total]. Louise needs to send to Madhu sequences which consist of lettersT, E, B, L. Each letter is represented by a two-bit message, as follows:T= [0 0], E= [0 1], B= [1 0], L= [1 1].Since Louise and Madhu have to use a one-way, noisy channel, BSC(𝑝) with small 𝑝, Louise encodeseach message using the generator matrix𝐺 = [1 0 0 1 10 1 1 0 1]of a binary linear code 𝐶.A1 [2 marks] Represent the word LET as three two-bit messages, encode each message as describedabove and write down the resulting three codevectors of 𝐶.
Solution
To answer this question, we first need to convert the word "LET" into the corresponding two-bit messages.
- The letter "L" is represented by [1 1].
- The letter "E" is represented by [0 1].
- The letter "T" is represented by [0 0].
Next, we need to encode each of these messages using the generator matrix G. The generator matrix G is a 2x5 matrix, which means we will multiply each two-bit message by G to get a five-bit codevector.
The generator matrix G is:
1 0 0 1 1
0 1 1 0 1
To encode the message for "L" ([1 1]), we perform the following multiplication:
1*1 + 1*0 = 1
1*0 + 1*1 = 1
1*0 + 1*1 = 1
1*1 + 1*0 = 1
1*1 + 1*1 = 0 (since we are working in binary)
So, the encoded message for "L" is [1 1 1 1 0].
We repeat this process for "E" ([0 1]) and "T" ([0 0]) to get their encoded messages:
For "E":
0*1 + 1*0 = 0
0*0 + 1*1 = 1
0*0 + 1*1 = 1
0*1 + 1*0 = 0
0*1 + 1*1 = 1
So, the encoded message for "E" is [0 1 1 0 1].
For "T":
0*1 + 0*0 = 0
0*0 + 0*1 = 0
0*0 + 0*1 = 0
0*1 + 0*0 = 0
0*1 + 0*1 = 0
So, the encoded message for "T" is [0 0 0 0 0].
Therefore, the three codevectors for "LET" are [1 1 1 1 0], [0 1 1 0 1], and [0 0 0 0 0].
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