Aufgaben:Exercise 4.06: Optimal Decision Boundaries

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Signal space constellation with
$N = 2, \ M = 2$

We consider a binary transmission system  $(M = 2)$  that is defined by the drawn two-dimensional signal space constellation  $(N = 2)$.  The following applies to the two possible transmitted vectors that are directly coupled to the messages  $m_0$  and  $m_1$: 

$$\boldsymbol{ s }_0 \hspace{-0.1cm} \ =\ \hspace{-0.1cm} \sqrt {E} \cdot (1,\hspace{0.1cm} 5) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_0 \hspace{0.05cm},$$

$$ \boldsymbol{ s }_1 \hspace{-0.1cm} \ =\ \hspace{-0.1cm} \sqrt {E} \cdot (4, \hspace{0.1cm}1) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_1 \hspace{0.05cm}.$$

The optimal decision boundary between the regions  $I_0 ⇔ m_0$  and  $I_1 ⇔ m_1$ is sought.  The following assumptions are made:

• It applies to subtasks  (1)  to  (3):

$${\rm Pr}(m_0 ) = {\rm Pr}(m_1 ) = 0.5 \hspace{0.05cm}. $$

• For subtasks  (4)  and  (5)  should apply:

$${\rm Pr}(m_0 ) = 0.817 \hspace{0.05cm},\hspace{0.2cm} {\rm Pr}(m_1 ) = 0.183\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} = 1.5 \hspace{0.05cm}.$$

For AWGN noise with variance  $\sigma_n^2$,  the decision boundary is the solution of the following vectorial equation with respect to the vector  $\boldsymbol{ \rho } =(\rho_1, \rho_2)$:

$$|| \boldsymbol{ s }_1||^2 - || \boldsymbol{ s }_0||^2 + 2 \cdot \sigma_n^2 \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} = 2 \cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm}.$$

In addition,  two received values ​​

$$\boldsymbol{ A }= \sqrt {E} \cdot (1.5, \hspace{0.1cm}2)\hspace{0.05cm},$$

$$\boldsymbol{ B }= \sqrt {E} \cdot (3, \hspace{0.1cm}3.5) $$

are drawn in the graphic.  It must be checked whether these should be assigned to the regions  $I_0$  $($and thus the message $m_0)$  or to  $I_1$  $($message $m_1)$  given the corresponding boundary conditions.

Notes:

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

• For numeric calculations,  the energy  $E = 1$  can be set for simplification.

Questions

1

Where lies the optimal decision limit for equally probable symbols?  At

	$\rho_2 = 3/4 \cdot \rho_1 + 9/8$,
	$\rho_2 = \, –4/3 \cdot \rho_1 + 19/3$,
	$\rho_2 = 3$.

2

To which decision area does the received value  $A = (1.5, \ \, 2)$  belong?

	To decision area  $I_0$,
	to decision area  $I_1$.

3

To which decision area does the received value  $B = (3, \ \, 3.5)$  belong?

	To decision area  $I_0$,
	to decision area  $I_1$.

4

What is the equation of the decision line for  ${\rm Pr}(m_0) = 0.817, \ \sigma_n = 1$?

	$\rho_2 = 3/4 \cdot \rho_1 + 9/8$,
	$\rho_2 = 3/4 \cdot \rho_1 + 3/4$,
	$\rho_2 = 3/4 \cdot \rho_1 + 3/2$,
	$\rho_2 = 3/4 \cdot \rho_1$.

5

Which decisions are made with these new regions  $I_0$  and  $I_1$? 

	The received vector  $A$  is interpreted as message  $m_0$.
	The received vector  $A$  is interpreted as message  $m_1$.
	The received vector  $B$  is interpreted as message  $m_0$.
	The received vector  $B$  is interpreted as message  $m_1$.

Solution

Solution

(1)  With  ${\rm Pr}(m_0) = {\rm Pr}(m_1) = 0.5$,  the equation of the boundary line between the decision areas  $I_0$  and  $I_1$  reads:

$$|| \boldsymbol{ s }_1||^2 - || \boldsymbol{ s }_0||^2 = 2 \cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm}.$$

• With the given vector values,  i.e. the numerical values

$$|| \boldsymbol{ s }_1||^2 = 4^2 + 1^2 = 17\hspace{0.05cm}, \hspace{0.2cm} || \boldsymbol{ s }_0||^2 = 1^2 + 5^2 = 26\hspace{0.05cm}, \hspace{0.2cm} \boldsymbol{ s }_1 - \boldsymbol{ s }_0 = (3,\hspace{0.1cm}-4) \hspace{0.05cm}$$

one obtains the following equation for the decision limits:

$$3 \cdot \rho_1 - 4 \cdot \rho_2 = ({17-26})/{2} = - {9}/{2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\rho_2 = 3/4 \cdot \rho_1 + 9/8 \hspace{0.05cm}.$$

Decision line and decision regions for  $K=0$

• The decision line lies in the middle between  $s_0$  and  $s_1$  and is rotated by  $90^\circ$  compared to the connecting line between the two symbols.  It goes through the point $(2.5, \ \, 3)$.  So the  first solution  is correct.

• Solution 2,  on the other hand,  describes the connecting line itself and  $\rho_2 = 3$  is a horizontal line.

(2)  The decision region  $I_1$  should of course contain the point  $s_1$   ⇒   region below the decision line. 

• Point $A = (1.5, \ \, 2)$  belongs to this decision region,  as shown in the graphic.
• This can be shown mathematically,  since the decision line goes through the point $(1.5, \ \, 2.25)$,  for example,  and thus  $(1.5, \ \, 2)$  lies below the decision line.
• So,  solution 2  is correct.

(3)  The decision line also goes through the point  $(3, \ \, 3.375)$.

• $B = (3, \ \, 3.5)$  lies above and therefore belongs to the decision region  $I_0$  according to  solution 1.

(4)  According to the equation in the information section and the calculations for subtask  (1),  the following now applies:

Decision regions for different  $K$

$$|| \boldsymbol{ s }_1||^2 - || \boldsymbol{ s }_0||^2 + 2 \cdot \sigma_n^2 \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} = 2 \cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm}.$$

• With  $|| \boldsymbol{ s }_1||^2 = 17$,  $|| \boldsymbol{ s }_0||^2 = 26$,  $ \boldsymbol{ s }_1 \, –\boldsymbol{ s }_0 = (3, \ \, –4)$  we obtain:

$$\rho_2 = 3/4 \cdot \rho_1 + 9/8 - K /8 \hspace{0.05cm}.$$

• The following abbreviation was used here:

$$K = 2 \cdot \sigma_n^2 \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} = 2 \cdot 1^2 \cdot 1.5 = 3 \hspace{0.05cm}.$$

• From this it follows:

$$\rho_2 = 3/4 \cdot \rho_1 + 9/8 - 3 /8 = 3/4 \cdot \rho_1 + 3/4 \hspace{0.05cm}.$$

• The decision line is shifted down by  $3/8$  $($black curve,  labeled  "$K = 3$"  in the graphic$)$.  So,  solution 2  is correct.

1. The first equation describes the optimal decision line for equally probable symbols  $(K = 0$,  dashed gray$)$.
2. The third equation is valid for  $K = \, –3$.  This results with  $\sigma_n^2 = 1$  for the symbol probabilities  ${\rm Pr}(m_1) \approx 0.817$  and  ${\rm Pr}(m_0) \approx 0.138$  (green curve).
3. The violet straight line results with  $K = 9$,  i.e. with the same probabilities as for the black curve,  but now with the variance $\sigma_n^2 = 3$.

(5)  The graphic above already shows that both  $A$  and  $B$  now belong to the decision region  $I_0$.  Solutions 1 and 3  are correct.