Aufgaben:Exercise 4.09: Decision Regions at Laplace

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Three Laplace decision regions

We consider a transmission system based on the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  The two equally probable transmitted signals are given by the signal points

$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm} \boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}.$$

In the following we normalize the energy parameter to  $E = 1$  for simplification and thus obtain

$$\boldsymbol{ s }_0 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (-1, \hspace{0.1cm}-1) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_0\hspace{0.05cm}, $$

$$ \boldsymbol{ s }_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (+1, \hspace{0.1cm}+1)\hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_1\hspace{0.05cm}.$$

The messages  $m_0$  and  $m_1$  are uniquely assigned to the signals  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$  defined in this way.

Let the noise components  $n_1(t)$  and  $n_2(t)$  be independent of each other and each be Laplace distributed with parameter  $a = 1$:

$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- | \eta_1|} \hspace{0.05cm}, \hspace{0.2cm} p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- | \eta_2|} \hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \boldsymbol{ p }_{\boldsymbol{ n }} (\eta_1, \eta_2) = {1}/{4} \cdot {\rm e}^{- | \eta_1|- | \eta_2|} \hspace{0.05cm}. $$

The properties of such a Laplace noise will be discussed in detail in  "Exercise 4.9Z". 

The received signal  $\boldsymbol{r}$  is composed additively of the transmitted signal  $\boldsymbol{s}$  and the  noise component  $\boldsymbol{n}$:

$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n } \hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2) \hspace{0.05cm},\hspace{0.45cm} \boldsymbol{ s } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} ( s_1, s_2) \hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n } = ( n_1, n_2) \hspace{0.05cm}. $$

The corresponding realizations are denoted as follows:

$$\boldsymbol{ s }\text{:} \hspace{0.4cm} (s_{01},s_{02}){\hspace{0.15cm}\rm and \hspace{0.15cm}} (s_{11},s_{12}) \hspace{0.05cm},\hspace{0.8cm} \boldsymbol{ r }\text{:} \hspace{0.4cm} (\rho_{1},\rho_{2}) \hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n }\text{:} \hspace{0.4cm} (\eta_{1},\eta_{2}) \hspace{0.05cm}.$$

The decision rule of the MAP and ML receivers  $($both are identical due to the same symbol probabilities$)$  are:

⇒   Decide for the symbol  $m_0$, if   $p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_0 ) > p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } (\rho_{1},\rho_{2} |m_1 ) \hspace{0.05cm}.$

⇒   With the further conditions for the  "decision for  $m_0$"  can also be written:

$${1}/{4} \cdot {\rm exp}\left [- | \rho_1 +1|- | \rho_2 +1| \hspace{0.1cm} \right ] > {1}/{4} \cdot {\rm exp}\left [- | \rho_1 -1|- | \rho_2 -1| \hspace{0.1cm} \right ] $$

$$\Rightarrow \hspace{0.3cm} | \rho_1 +1|+ | \rho_2 +1| < | \rho_1 -1|+ | \rho_2 -1|$$

$$\Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = | \rho_1 +1|+ | \rho_2 +1| - | \rho_1 -1|- | \rho_2 -1| < 0 \hspace{0.05cm}.$$

⇒   This function  $L(\rho_1, \rho_2)$  is frequently referred to in the questions.

The graph shows three different decision regions  $(I_0, \ I_1)$.

• For AWGN noise,  only the upper variant  $\rm A$  would be optimal.

• Also for the considered Laplace noise, variant  $\rm A$  leads to the smallest possible error probability, see  "Exercise 4.9Z":

$$p_{\rm min} = {\rm Pr}({\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} {\rm optimal\hspace{0.15cm} receiver}) = {\rm e}^{-2} \approx 13.5\,\%\hspace{0.05cm}.$$

• It is to be examined whether variant  $\rm B$  or variant  $\rm C$  is also optimal, i.e. whether their error probabilities are also as small as possible equal to  $p_{\rm min}$. 

Note:   The exercise belongs to the chapter  "Approximation of the Error Probability".

Questions

1

Which of the decision rules are correct?  Decide for  $m_0$,  if

	$p_{\it r\hspace{0.03cm}|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}|\hspace{0.03cm}m_0) > p_{\it r\hspace{0.03cm}|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}|\hspace{0.03cm}m_1)$,
	$L(\rho_1, \ \rho_2) = |\rho_1+1| \, -|\rho_1 \, –1| + |\rho_2+1| \, -|\rho_2 \, –1| < 0$,
	$L(\rho_1, \ \rho_2) = \rho_1 + \rho_2 ≥ 0$.

2

How can the expression  $|x+1| \ -|x \ -1|$  be transformed?

	For  $x ≥ +1$:    $|x + 1| \, -|x -1| = 2$.
	For  $x ≤ \, -1$:    $|x+1| \,-|x \, -1| = \, -2$.
	For  $-1 ≤ x ≤ +1$:    $|x+1| \, -|x \, -1| = 2x$.

3

What is the decision rule in the range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$?

	Decision for  $m_0$,  if  $\rho_1 + \rho_2 < 0$.
	Decision for  $m_1$,  if  $\rho_1 + \rho_2 < 0$.

4

What is the decision rule in the range  $\rho_1 > +1$?

	Decision for  $m_0$  in the whole range.
	Decision for  $m_1$  in the whole range.
	Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

5

What is the decision rule in the range  $\rho_1 < \, -1$?

	Decision for  $m_0$  in the whole range.
	Decision for  $m_1$  in the whole range.
	Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

6

What is the decision rule in the range  $\rho_2 > +1$?

	Decision for  $m_0$  in the whole range.
	Decision for  $m_1$  in the whole range.
	Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

7

What is the decision rule in the range  $\rho_2 < -1$?

	Decision for  $m_0$  in the whole range.
	Decision for  $m_1$  in the whole range.
	Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

8

Which of the following statements are true?

	Variant  $\rm A$  leads to the minimum error probability.
	Variant  $\rm B$  leads to the minimum error probability.
	Variant  $\rm C$  leads to the minimum error probability.

Solution

Solution

(1)  Solutions 1 and 2  are correct:

• The joint probability densities under conditions  $m_0$  and  $m_1$,  respectively,  are:

$$p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_0 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- | \rho_1 +1|- | \rho_2 +1| \hspace{0.05cm} \right ]\hspace{0.05cm},$$

$$p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_1 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- | \rho_1 -1|- | \rho_2 -1| \hspace{0.05cm} \right ]\hspace{0.05cm}.$$

• For equally probable symbols   ⇒   ${\rm Pr}(m_0) = {\rm Pr}(m_1) = 0.5$,  the MAP decision rule is:   Decide for symbol  $m_0$    ⇔   signal  $s_0$,  if

$$p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_0 ) > p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } (\rho_{1},\rho_{2} |m_1 ) \hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {1}/{4} \cdot {\rm exp}\left [- | \rho_1 +1|- | \rho_2 +1| \hspace{0.05cm} \right ] > {1}/{4} \cdot {\rm exp}\left [- | \rho_1 -1|- | \rho_2 -1|\hspace{0.05cm} \right ] $$

$$\Rightarrow \hspace{0.3cm} | \rho_1 +1|+ | \rho_2 +1| < | \rho_1 -1|+ | \rho_2 -1|\hspace{0.3cm} \Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = | \rho_1 +1|- | \rho_1 -1|+ | \rho_2 +1| - | \rho_2 -1| < 0 \hspace{0.05cm}.$$

(2)  All statements are true:

• For $x ≥ 1$:

$$| x +1|- | x -1| = x +1 -x +1 =2 \hspace{0.05cm}.$$

• Similarly,  for $x ≤ \, –1$,  e.g.  $x = \, –3$:

$$| x +1|- | x -1| = | -3 +1|- | -3 -1| = 2-4 = -2 \hspace{0.05cm}.$$

• On the other hand,  in the middle range $–1 ≤ x ≤ +1$:

$$| x+1|- | x -1| = x +1 -1 +x =2x \hspace{0.05cm}.$$

(3)  Solution 1  is correct:

• The result of subtask  (1)  was:  Decide for the symbol  $m_0$,  if

$$L (\rho_1, \rho_2) = | \rho_1 +1| - | \rho_1 -1|+ | \rho_2 +1| - | \rho_2 -1| < 0 \hspace{0.05cm}.$$

• In the considered  (inner)  range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$  holds with the result of subtask  (2):

$$| \rho_1+1| - | \rho_1 -1| = 2\rho_1 \hspace{0.05cm}, \hspace{0.2cm} | \rho_2+1| - | \rho_2 -1| = 2\rho_2 \hspace{0.05cm}.$$

• If we insert this result above,  we have to decide for  $m_0$  exactly if

$$L (\rho_1, \rho_2) = 2 \cdot ( \rho_1+\rho_2) < 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \rho_1+\rho_2 < 0\hspace{0.05cm}.$$

(4)  Correct is here  solution 2:

• For  $\rho_1 > 1$   ⇒   $|\rho_1+1| \, -|\rho_1 \, -1| = 2$,  while for  $D_2 = |\rho_2+1| \,-|\rho_2 \, -1|$   ⇒   all values between  $-2$  and  $+2$  are possible.

• Thus,  the decision variable is  $L(\rho_1, \rho_2) = 2 + D_2 ≥ 0$.  In this case,  the rule leads to an  $m_1$  decision.

(5)  Solution 1  is correct:

• After similar calculation as in subtask  (3),  one arrives at the result:

$$L (\rho_1, \rho_2) = -2 + D_2 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$

(6)  Solution 2  is correct:  Decision on  $m_1$.

• Similar to subtask  (4),  the following holds here:

$$D_1 = | \rho_1 +1| - | \rho_1 -1| \in \{-2, ... \hspace{0.05cm} , +2 \} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}L (\rho_1, \rho_2) = 2 + D_1 \ge 0 \hspace{0.05cm}.$$

(7)  Solution 1  is correct:  Decision on  $m_0$.

• After similar reasoning as in the last subtask,  we arrive at the result:

$$L (\rho_1, \rho_2) = -2 + D_1 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$

Summary of the results

(8)  The results of subtasks  (3)  to  (7)  are summarized in the graph:

• Subarea  $T_0$:   Decision on  $m_0$  or  $m_1$  according to task  (3).
• Subarea  $T_1$:   Decision on  $m_1$  according to task  (4).
• Subarea  $T_2$:   Decision on  $m_0$  according to task  (5).
• Subarea  $T_3$:   Decision on  $m_1$  according to task  (6).
• Subarea  $T_4$:   Decision on $m_0$  according to task  (7).
• Subarea  $T_5$:   Decision on  $m_0$  according to task  (5), and on  $m_1$  according to task  (6)
  ⇒   For Laplace noise,  it does not matter whether one assigns  $T_5$  to region  $I_0$  or  $I_1$.
• Subarea  $T_6$:   Again,  based on the results of task  (6)  and  (7),  one can assign this region to region  $I_0$  or region  $I_1$.

It can be seen:

1. For subarea  $T_0$,  ... ,  $T_4$  there is a fixed assignment to the decision regions  $I_0$  (red)  and  $I_1$  (blue).
2. In contrast,  the two yellow regions  $T_5$  and  $T_6$  can be assigned to both,  $I_0$  and  $I_1$  without loss of optimality.

Comparing this graph with variants  $\rm A$,  $\rm B$  and  $\rm C$  on the specification page,  we see that  suggestions 1 and 2  are correct:

1. Variants  $\rm A$  and  $\rm B$ are equally good.  Both are optimal.  The error probability in both cases is  $p_{\rm min} = {\rm e}^{\rm -2}$.
2. Variant  $\rm C$  is not optimal;  with respect to the subareas  $T_1$  and  $T_2$  there are mismatches.  The error probability is therefore greater than  $p_{\rm min}$.