# Exercise 1.3Z: Threshold Optimization

Optimizing the threshold  $E$

In this exercise,  a bipolar binary system with AWGN noise  ("Additive White Gaussian Noise")  is considered,  so that for the bit error probability:

$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {1}/{2} \cdot {\rm erfc} \left( \frac{s_0}{\sqrt{2} \cdot \sigma_d}\right) \hspace{0.05cm}.$$

Here,  the following functions are used:

$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u} \hspace{0.05cm},\hspace{1cm}{\rm erfc} ({\it x}) = \frac{\rm 2}{\sqrt{\rm \pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}}\,d \it u \hspace{0.05cm}.$$
• The above equation holds for the threshold  $E = 0$,  regardless of the symbol probabilities  $p_{\rm L}$  and  $p_{\rm H}$.
• However,  a smaller error probability can be obtained with a different threshold  $E$  if the two occurrence probabilities are different   $(p_{\rm L} ≠ p_{\rm H})$.

The standard deviation of the noise component is always  $σ_{d} = 0.5 \ \rm V$.  The two amplitudes of the detection signal component are fixed at  $±1 \ \rm V$.

The following symbol probabilities are to be investigated:

• $p_{\rm L} = 0.88$   ⇒   $p_{\rm H} = 1- p_{\rm L} =0.12$,
• $p_{\rm L} = 0.31$   ⇒   $p_{\rm H} = 1- p_{\rm L} =0.69$.

The diagram includes this last set of parameters and the threshold  $E = 0.1 \cdot s_{\rm 0}$.  The probability density function of the detection samples  $d$  is shown.

Notes:

• The following applies to the derivative of the Q-function:
$$\frac{{\rm d\hspace{0.05cm}Q} ({\it x})} {{\rm d}\hspace{0.05cm}x} = \frac{\rm 1}{\sqrt{\rm 2\pi}} \cdot \rm e^{\it -x^{\rm 2}/\rm 2} \hspace{0.05cm}.$$

### Questions

1

What is the relationship between the functions  ${\rm Q}(x)$  and  ${\rm erfc}(x)$?

 ${\rm erfc}(x) =2 \cdot {\rm Q}(\sqrt{2} \cdot x)$, ${\rm erfc}(x) =\sqrt{2} \cdot {\rm Q}(x/\sqrt{2})$, ${\rm erfc}(x) \approx {\rm Q}(x)$.

2

What is the bit error probability with  $\underline{p_{\rm L} = 0.88}$  and  $\underline{E = 0}$?

 $p_{\rm B} \ =\$ $\ \%$

3

What is the bit error probability with  $p_{\rm L} = 0.88$  and  $\underline{E = 0.1 \hspace{0.05cm} \rm V}$?

 $p_{\rm B} \ =\$ $\ \%$

4

Determine the optimal threshold  $E_{\rm opt}$  for  $p_{\rm L} = 0.88$   ⇒   $p_{\rm H} = 0.12$.

 $E_{\rm opt} \ =\$ $\ \rm V$

5

Let  $E = E_{\rm opt}$.  What is the minimum bit error probability with  $p_{\rm L} = 0.88$?

 $p_{\rm B,\ min} \ =\$ $\ \%$

6

Determine the optimal threshold  $E_{\rm opt}$  for  $\underline{p_{\rm L} = 0.31}$   ⇒   $p_{\rm H} = 0.69$.

 $E_{\rm opt} \ =\$ $\ \rm V$

7

What is the minimum bit error probability for this case  $(p_{\rm L} = 0.31)$?

 $p_{\rm B,\ min} \ =\$ $\ \%$

### Solution

#### Solution

(1)  From the Q-function,  substitution  $t^{2} = u^{2}/2$  yields:

$$\rm Q (\it x)= \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d \it u = \frac{\rm 1}{\sqrt{\rm \pi}}\int_{\it x/\sqrt{\rm 2}}^{+\infty}\rm e^{\it -t^{\hspace{0.05cm}\rm 2}}\,{d \it t} = {1}/{2} \cdot {\rm erfc} \left( {x}/{\sqrt{2}} \right)\hspace{0.05cm}.$$

From this it follows that the  first solution  is correct:

$${\rm erfc} (x)= 2 \cdot {\rm Q} (\sqrt{2} \cdot x) \hspace{0.05cm}.$$

(2)  Independently of the symbol probabilities,  with the decision threshold  $E = 0$,  we obtain:

$$p_{\rm B} = {\rm Q} \left( {s_0}/{\sigma_d}\right)= {\rm Q} (2) \hspace{0.1cm}\underline {= 2.27 \, \% }\hspace{0.05cm}.$$

(3)  Now the general equation for the bit error probability,  where  $d_{\rm N}$  denotes the noise component of  $d(t)$,  is:

$$p_{\rm B} \ = \ p_{\rm L} \cdot {\rm Pr}( d_{\rm N}> E + s_0)+ p_{\rm H} \cdot {\rm Pr}( d_{\rm N}< E - s_0) \ = \ p_{\rm L} \cdot {\rm Q} \left( \frac{s_0 + E}{\sigma_d}\right)+ p_{\rm H} \cdot {\rm Q} \left( \frac{s_0 - E}{\sigma_d}\right)\hspace{0.05cm}.$$
• Here,  the PDF symmetry is taken into account.  With  $p_{\rm L} = 0.88$   ⇒   $p_{\rm H} = 0.12$  and  $E = 0.1 \hspace{0.05cm} \rm V$  one obtains:
$$p_{\rm B} \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} + 0.1\, {\rm V}}{0.5\, {\rm V}}\right)\hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm}{\rm Q} \left( \frac{1\, {\rm V} - 0.1\, {\rm V}}{0.5\, {\rm V}}\right) \hspace{-0.05cm}=\hspace{-0.05cm} 0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} {\rm Q}(2.2) \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \cdot {\rm Q}(1.8) \hspace{-0.05cm}= \hspace{-0.05cm}0.88 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 1.39\,\% \hspace{-0.05cm}+\hspace{-0.05cm} 0.12 \hspace{-0.03cm}\cdot \hspace{-0.03cm} 3.59\,\%\ \hspace{0.1cm}\underline { = 1.65\,\% } \hspace{0.05cm}.$$
• Threshold shifting to the right by  $E = s_{0}/10$  results in an improvement from  $p_{\rm B} = 2.27 \ \%$  to  $p_{\rm B} = 1.65 \ \%$.

(4)  This optimization task is solved by setting the derivative to zero,  taking into account the note on the information section:

$$\frac{{\rm d\hspace{0.05cm}} p_{\rm B}(E)} {{\rm d}\hspace{0.05cm}E} = - \frac{\rm p_{\rm L}}{\sqrt{\rm 2\pi}\cdot {\sigma_d}} \cdot {\rm exp}\left(- \frac{(s_0 + E)^2}{{\rm 2}\cdot {\sigma_d}^2}\right)+\frac{\rm p_{\rm H}}{\sqrt{\rm 2\pi}\cdot {\sigma_d}} \cdot {\rm exp}\left(- \frac{(s_0 - E)^2}{{\rm 2}\cdot {\sigma_d}^2}\right) = 0$$
$$\Rightarrow \hspace{0.3cm} \frac{p_{\rm L}} {p_{\rm H}} = - \frac{ {\rm exp} \left(-\frac{(s_0 - E)^2}{{\rm 2}\cdot {\sigma_d}^2}\right)}{{\rm exp} \left(-\frac{(s_0 + E)^2}{{\rm 2}\cdot {\sigma_d}^2}\right)} = {\rm exp} \left( \frac{2 \cdot E \cdot s_0 }{ {\sigma_d}^2}\right)\hspace{0.05cm}.$$
• Thus,  we obtain for the optimal threshold value in general:
$$E_{\rm opt} = \frac{\sigma_d^2} {2 \cdot s_0} \cdot {\rm ln}\hspace{0.05cm}\frac{p_{\rm L}} {p_{\rm H}} \hspace{0.05cm}.$$
• With $σ_{d} = 0.5 \ \rm V$,  $s_{\rm 0} = 1 \ \rm V$, $p_{\rm L} = 0.88$  and  $p_{\rm H} = 0.12$,  the following optimum is obtained:
$$E_{\rm opt} = \frac{(0.5\, {\rm V})^2} {2 \cdot 1\, {\rm V}} \cdot {\rm ln}\hspace{0.05cm}\frac{0.88} {0.12} \hspace{0.1cm}\underline { \approx 0.25\, {\rm V}}\hspace{0.05cm}.$$

(5)  The minimum error probability for the optimal threshold  $E_{\rm opt} = 0.25 \hspace{0.05cm} \rm V$  is thus:

$$p_{\rm B, \hspace{0.05cm}min} = 0.88 \cdot {\rm Q}(2.5) + 0.12 \cdot {\rm Q}(1.5) = 0.88 \cdot 0.62\,\% + 0.12 \cdot 6.68\,\%\ \hspace{0.1cm}\underline {= 1.35\,\% }\hspace{0.05cm}.$$
• Compared to  $E = 0$,  the error probability is now about  $40 \%$  smaller.

(6)  Using the result from (4),  the optimal threshold value is now:

$$E_{\rm opt} = \frac{(0.5\, {\rm V})^2} {2 \cdot 1\, {\rm V}} \cdot {\rm ln}\hspace{0.05cm}\frac{0.31} {0.69}\hspace{0.1cm}\underline { \approx -0.1\, {\rm V}}\hspace{0.05cm}.$$

$$p_{\rm B, \hspace{0.05cm}min} = 0.31 \cdot {\rm Q}(1.8) + 0.69 \cdot {\rm Q}(2.2) = 0.31 \cdot 3.59\,\% + 0.69 \cdot 1.39\,\%\ \hspace{0.1cm}\underline { = 2.07\,\% } \hspace{0.05cm}.$$
• Due to the less severe asymmetry,  the achievable improvement of  $9 \%$  is lower than calculated in subtask  (5).