Exercise 1.4: Maximum Likelihood Decision

• Return to book

Maximum Likelihood Decoding Model

We consider the digital transmission system according to the graph. Considered are:

• a systematic  $(5, 2)$ block code  $\mathcal{C}$  with the code words

$$\underline{x}_{0} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (0, 0, 0, 0, 0) \hspace{0.05cm},$$ $$\underline{x}_{1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (0, 1, 0, 1, 0) \hspace{0.05cm},$$ $$\underline{x}_{2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (1, 0, 1, 0, 1) \hspace{0.05cm},$$ $$\underline{x}_{3} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (1, 1, 1, 1, 1) \hspace{0.05cm};$$

• a digital (binary) channel model that crosses the vector  $\underline{x} \in {\rm GF} (2^{5})$  over into the vector  $\underline{y} \in {\rm GF} (2^{5})$  (Noah)
• a  Maximum Likelihood Decoder  (short:   ML–Decoder)  with the decision rule

$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm}|\hspace{0.05cm} \underline{y} ) = {\rm arg} \min_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm H}(\underline{y} \hspace{0.05cm}, \hspace{0.1cm}\underline{x}_{\hspace{0.03cm}i}).$$

Here,  $d_{\rm H} (\underline{y}, \ \underline{x_{i}})$  the  Hamming distance  between the received word  $\underline{y}$  and the (possibly) sent codeword  $\underline{x_{i}}$.

Hints:

• This exercise belongs to the chapter  Channel Models and Decision Structures.

Questions

1

Let  $\underline{y} = (1, 0, 0, 1)$. Which decisions satisfy the maximum likelihood criterion?

	$\underline{z} = \underline{x}_{0} = (0, 0, 0, 0)$,
	$\underline{z} = \underline{x}_{1} = (0, 1, 0, 1, 0)$,
	$\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$,
	$\underline{z} = \underline{x}_{3} = (1, 1, 1, 1)$.

2

Let  $\underline{y} = (0, 0, 0, 1, 0)$. Which decisions satisfy the maximum likelihood criterion?

	$\underline{z} = \underline{x}_{0} = (0, 0, 0, 0)$,
	$\underline{z} = \underline{x}_{1} = (0, 1, 0, 1, 0)$,
	$\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$,
	$\underline{z} = \underline{x}_{3} = (1, 1, 1, 1)$.

3

What decision does the ML decoder make for  $\underline{y} = (1, 0, 1, 1, 1)$ when it is told that the last two symbols are uncertain?

	$\underline{z} = \underline{x}_{0} = (0, 0, 0, 0)$,
	$\underline{z} = \underline{x}_{1} = (0, 1, 0, 1, 0)$,
	$\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$,
	$\underline{z} = \underline{x}_{3} = (1, 1, 1, 1)$.

4

What information word  $v = (v_{1}, v_{2})$  does the decision lead to according to the last subtask?

$v_{1} \ = \ $

$v_{2} \ = \ $

Solution

Solution

(1)  Correct answer 3:

• The Hamming distances between the specific received word $\underline{y} = (1, 0, 0, 1)$ and the four possible codewords $\underline{x}_{i}$ are as follows:

$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 3\hspace{0.05cm}.$$

• A decision is made for the sequence with the smallest Hamming distance $d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1$.

(2)  For $\underline{y} = (0, 0, 0, 1, 0)$ the answers 1 and 2 are correct, as the following calculation shows:

$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 4\hspace{0.05cm}.$$

(3)  Correct answer 3:

• According to the Hamming distance, a decision in favor of $x_{2}$ would be just as possible as for $x_{3}$ if the vector $\underline{y} = (1, 0, 1, 1, 1)$ is received:

$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 1\hspace{0.05cm}.$$

• But the received vector $\underline{y}$ is different from $x_{2}$ with respect to the fourth bit and from $x_{3}$ in the second bit.
• Since the fourth bit is more uncertain than the second, it will choose $x_{2}$ .

(4)  Since this is a systematic code, the decision for $\underline{z} = (1, 0, 1, 0, 1)$ is equivalent to the decision

$$v_{1} \ \underline{ = 1}, \ v_{2} \ \underline{= 0}.$$

• It is not certain that $\underline{u} = (1, 0)$ was actually sent.
• But the probability is highest for this given the received vector $\underline{y} = (1, 0, 1, 1)$.