Aufgaben:Exercise 4.07: Decision Boundaries once again

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PDF with unequal symbol probabilities

We consider a transmission system with

• only one basis function  $(N = 1)$,
• two signals  $(M = 2)$  with  $s_0 = \sqrt{E_s}$  and  $s_1 = -\sqrt{E_s}$ ,
• an AWGN channel with variance  $\sigma_n^2 = N_0/2$.

Since this exercise deals with the general case  ${\rm Pr}(m_0) ≠ {\rm Pr}(m_1)$,  it is not sufficient to consider the conditional density functions  $p_{\it r\hspace{0.05cm}|\hspace{0.05cm}m_i}(\rho\hspace{0.05cm} |\hspace{0.05cm}m_i)$.  Rather, these must still be multiplied by the symbol probabilities   ${\rm Pr}(m_i)$.  For   $i$,  the values  $0$  and  $1$  have to be used here.

If the decision boundary between the two regions  $I_0$  and  $I_1$  is at  $G = 0$, i.e., in the middle between  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$, the error probability is independent of the occurrence probabilities  ${\rm Pr}(m_0)$  and  ${\rm Pr}(m_1)$:

$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {\rm Q} \left ( \frac{d/2}{\sigma_n} \right ) \hspace{0.05cm}.$$

Here  $d$  indicates the distance between the signal points  $s_0$  and  $s_1$  and  $d/2$  accordingly the respective distance of  $s_0$  and  $s_1$  from the decision boundary  $G = 0$. The rms value (root of the variance) of the AWGN noise is  $\sigma_n$.

If, on the other hand, the occurrence probabilities are different   ⇒  ${\rm Pr}(m_0) ≠ {\rm Pr}(m_1)$, a smaller error probability can be obtained by shifting the decision boundary  $G$: 

$$p_{\rm S} = {\rm Pr}(m_1) \cdot {\rm Q} \left ( \frac{d/2}{\sigma_n} \cdot (1 + \gamma) \right ) + {\rm Pr}(m_0) \cdot {\rm Q} \left ( \frac{d/2}{\sigma_n} \cdot (1 - \gamma) \right )\hspace{0.05cm},$$

where the auxiliary quantity  $\gamma$  is defined as follows:

$$\gamma = 2 \cdot \frac{ \sigma_n^2}{d^2} \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_1)}{{\rm Pr}( m_0)} \hspace{0.05cm},\hspace{0.2cm} G_{\rm opt} = \gamma \cdot \sqrt {E_{\rm S}}\hspace{0.05cm}.$$

Notes:

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

• You can find the values of the Q–function with interactive applet  "Complementary Gaussian Error Functions". 

Questions

1

What are the underlying symbol probabilities of the graph, if the blue Gaussian curve is exactly twice as high as the red one?

${\rm Pr}(m_0)\ = \ $

${\rm Pr}(m_1)\ = \ $

2

What is the error probability with noise variance  $\sigma_n^2 = E_{\rm S}/9$  and the given threshold  $G = 0$?

$G = 0 \text{:} \hspace{0.85cm} p_{\rm S} \ = \ $ $\ \%$

3

What is the optimal threshold for the given probabilities?

$G_{\rm opt}\ = \ $ $\ \cdot \sqrt{E_s}$

4

What is the probability of error for optimal threshold  $G = G_{\rm opt}$?

$G = G_{\rm opt} \text{:} \hspace{0.2cm} p_{\rm S}\ = \ $ $\ \%$

5

What error probabilities are obtained with the noise variance  $\sigma_n^2 = E_{\rm S}$?

$G = 0 \text{:} \hspace{0.85cm} p_{\rm S}\ = \ $ $\ \%$

$G = G_{\rm opt} \text{:} \hspace{0.2cm} p_{\rm S}\ = \ $ $\ \%$

6

Which statements are true for the noise variance  $\sigma_n^2 = 4 \cdot E_{\rm S}$?

	With  $G = 0$,  the probability of error is greater than  $30\%$.
	The optimal decision threshold is to the right of  $s_0$.
	With optimal threshold, the error probability is about  $27\%$.
	The estimated value  $m_0$  is possible only with sufficiently large noise.

Solution

Solution

(1)  The heights of the two drawn density functions are proportional to the occurrence probabilities ${\rm Pr}(m_0)$ and ${\rm Pr}(m_1)$. From

$${\rm Pr}(m_1) = 2 \cdot {\rm Pr}(m_0) \hspace{0.05cm},\hspace{0.2cm} {\rm Pr}(m_0) + {\rm Pr}(m_1) = 1$$

it follows directly  ${\rm Pr}(m_0) = 1/3 \ \underline {\approx 0.333}$  and  ${\rm Pr}(m_1) = 2/3 \ \underline {\approx 0.667}$.

(2)  With the decision boundary $G = 0$, independently of the occurrence probabilities:

$$p_{\rm S} = {\rm Q} \left ( \frac{d/2}{\sigma_n} \right ) \hspace{0.05cm}.$$

With $d = 2 \cdot \sqrt{E_{\rm S}}$ and $\sigma_n = \sqrt{E_{\rm S}}/3$, this gives:   $p_{\rm S} = {\rm Q} (3) \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.135 \%} \hspace{0.05cm}.$

(3)  According to the specification, for the "normalized threshold":

$$\gamma = \frac{G_{\rm opt}}{E_{\rm S}^{1/2}} = 2 \cdot \frac{ \sigma_n^2}{d^2} \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_1)}{{\rm Pr}( m_0)} = \frac{ 2 \cdot E_{\rm S}/9}{4 \cdot E_{\rm S}} \cdot {\rm ln} \hspace{0.15cm} \frac{2/3}{1/3} \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.04} \hspace{0.05cm}.$$

• Thus, $G_{\rm opt} = \gamma \cdot \sqrt{E_{\rm S}} = 0.04 \cdot \sqrt{E_{\rm S}}$.
• Thus, the optimal decision boundary is shifted to the right (towards the more improbable symbol $s_0$), because  ${\rm Pr}(m_0) < {\rm Pr}(m_1)$.

(4)  With this optimal decision boundary, the error probability is slightly smaller compared to subtask (2):

$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {2}/{3} \cdot {\rm Q} (3 \cdot 1.04) + {1}/{3} \cdot {\rm Q} (3 \cdot 0.96) = {2}/{3} \cdot 0.090 \cdot 10^{-2} + {1}/{3} \cdot 0.199 \cdot 10^{-2} \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.126 \%} \hspace{0.05cm}.$$

(5)  With the decision boundary in the middle between the symbols ($G = 0$), the result is analogous to subtask (2) with the noise variance now larger:

Density functions with  $\sigma_n^2 = E_{\rm S}$

$$p_{\rm S} = {\rm Q} \left ( \frac{d/2}{\sigma_n} \right ) = {\rm Q} \left ( \frac{\sqrt{E_{\rm S}}}{\sqrt{E_{\rm S}}} \right )= {\rm Q} (1)\hspace{0.1cm} \hspace{0.15cm}\underline {\approx 15.9 \%} \hspace{0.05cm}.$$

The parameter $\gamma$ (normalized best possible displacement of the decision boundary) is

$$\gamma = 2 \cdot \frac{ \sigma_n^2}{d^2} \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_1)}{{\rm Pr}( m_0)} = \frac{ 2 \cdot E_{\rm S}}{4 \cdot E_{\rm S}} \cdot {\rm ln} \hspace{0.15cm} \frac{2/3}{1/3} = \frac{{\rm ln} \hspace{0.15cm} 2}{2} \approx 0.35 $$

$$\Rightarrow \hspace{0.3cm} G_{\rm opt} = 0.35 \cdot \sqrt{E_{\rm S}} \hspace{0.05cm}.$$

The more frequent symbol is now less frequently falsified ⇒ the mean error probability becomes smaller:

$$p_{\rm S} = {2}/{3} \cdot {\rm Q} (1.35) + {1}/{3} \cdot {\rm Q} (0.65) = {2}/{3} \cdot 0.0885 +{1}/{3} \cdot 0.258 \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 14.5 \%} \hspace{0.05cm}.$$

The figure makes it clear that the optimal decision boundary can also be determined graphically as the intersection of the two (weighted) probability density functions:

(6)  All solutions of this rather academic subtask are correct:

Density functions with  $\sigma_n^2 = 4 \cdot E_{\rm S}$

• With threshold $G = 0$, we get $p_{\rm S} = {\rm Q}(0.5) \ \underline {\approx 0.309}$.
• The parameter $\gamma = 1.4$ now has four times the value compared to subtask (5),
  so that the optimal decision boundary is now $G_{\rm opt} = \underline {1.4 \cdot s_0}$.
• Thus, the (undisturbed) signal value $s_0$ does not belong to the decision region $I_0$, but to $I_1$, characterized by $\rho < G_{\rm opt}$.
• Only with a (positive) noise component $I_0 (\rho > G_{\rm opt})$ is possible at all. For the error probability, with $G_{\rm opt} = 1.4 \cdot s_0$:

$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {2}/{3} \cdot {\rm Q} (0.5 \cdot (1 + 1.4)) + {1}/{3} \cdot {\rm Q} (0.5 \cdot (1 - 1.4)) = \hspace{0.15cm}\underline {\approx 27\%} \hspace{0.05cm}.$$

The accompanying graph illustrates the statements made here.