Aufgaben:Exercise 3.7: Bit Error Rate (BER)

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To illustrate the bit error rate

We consider a binary transmission system with.

• the source symbol sequence  $\langle q_\nu \rangle $  and
• the sink symbol sequence  $\langle v_\nu \rangle $.

If sink symbol  $v_\nu$  and source symbol  $q_\nu$  do not match, there is a bit error   ⇒   $e_\nu = 1$.

Otherwise  $e_\nu = 0$ holds.

$\rm (A)$  The most important evaluation criterion of such a digital system is

the  Bit Error Probability .

• With the expected value  ${\rm E}\big[\text{ ...} \big]$  this is defined as follows:

$$\it p_{\rm B} = \rm E\big[\rm Pr(\it v_{\nu} \ne q_{\nu} \rm )\big]=\rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\lim_{{\it N}\to\infty}\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N}\rm Pr(\it e_{\nu}=\rm 1). $$

• The right part of this equation describes a time averaging; this must always be applied, for example, to time-varying channels.
• If the error probability is the same for all symbols (which is assumed here), the above equation can be simplified:

$$\it p_{\rm B} = \rm E\big[\rm Pr(\it e_{\nu}=\rm 1)\big]=\rm E\big[\it e_{\nu} \rm \big].$$

The bit error probability is an a priori parameter, so it allows a prediction for the expected result.

$\rm (B)$  On the other hand, for the metrological determination of the transmission quality or for the system simulation, it is necessary to rely on

the comparable A-posteriori parameter  Bit Error Rate  must be ignored.

$$h_{\rm B}=\frac{n_{\rm B}}{N}=\frac{\rm 1}{\it N}\cdot\sum\limits_{\it \nu=\rm 1}^{\it N} e_{\nu}.$$

• $h_{\rm B}$  is a  relative frequency.  $n_{\rm B}$  indicates the number of bit errors occurred when a total of  $N$  symbols (bits) transmitted.

• In the limiting case  $N \to \infty$  the relative frequency  $h_{\rm B}$  coincides with the probability  $p_{\rm B}$ .
• Here now the question shall be clarified, which statistical uncertainty has to be expected with finite  $N$ .

Hints:

• The exercise belongs to the chapter  Gaussian random variables.

• Read the exercises as far as possible in general.
• Use the parameter values  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$ for control input.
• The following are some values of the so-called Q-function:

$$\rm Q(\rm 1.00)=\rm 0.159,\hspace{0.5cm}\rm Q(\rm 1.65)=\rm 0.050,\hspace{0.5cm}\rm Q(\rm 1.96)=\rm 0.025,\hspace{0.5cm}\rm Q(\rm 2.59)=\rm 0.005.$$

Questions

1

Which of the following statements are true?

	For  $n_{\rm B}$  all values  $(0$, ... , $N)$  are equally likely.
	The random variable  $n_{\rm B}$  is binomially distributed.
	With  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$  we get  ${\rm E}\big[n_{\rm B}\big] = 100$.

2

How large; is the rms of the random variable  $n_{\rm B}$  for  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$?

$\sigma_{n{\rm B}} \ = \ $

3

What values can the bit error rate  $h_{\rm B}$  take? 
Show that the linear mean  $m_{h{\rm B}}$  of this random variable is equal to the actual bit error probability  $p_{\rm B}$  What is its dispersion?

$\sigma_{h{\rm B}} \ = \ $

4

Under certain conditions, a binomially distributed random variable can be approximated by a Gaussian distribution with equal mean  $(m_{h{\rm B}})$  and equal dispersion  $(\sigma_{h{\rm B}})$  Which statement is true?

	${\rm Pr}(\hspace{0.05cm}|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}| \le \varepsilon)=1- 2\cdot \rm Q({\varepsilon}/{\sigma_{\it h}{\rm B}}).$
	${\rm Pr}(\hspace{0.05cm}|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}| \le \varepsilon)=1- \rm Q({\varepsilon}/{2\cdot \sigma_{\it h}{\rm B}}).$

5

For abbreviation, we use the confidence level  $p_\varepsilon = {\rm Pr}(\hspace{0.05cm}|\hspace{0.05cm}h_{\rm B} - p_{\rm B}\hspace{0.05cm}| \le \varepsilon)$.   Which  $p_\varepsilon$  results with  $\varepsilon = 10^{-4}$,  $p_{\rm B} = 10^{-3}$  and  $N = 10^{5}$ ?

$p_\varepsilon \ = \ $

6

Let the argument of the Q-function be  $\alpha$.  What is the minimum   $\alpha$  that must be chosen for the confidence level  $p_\varepsilon = 95\%$  to be ?

$\alpha_{\rm min} \ = \ $

7

It still holds  $p_{\rm B} = 10^{-3}$  and  $p_\varepsilon = 95\%$.   Over how many symbols  $(N_\text{min})$  must be averaged at least,
so that the determined bit error rate in the range between  $0. 9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$  lies   $(\varepsilon = 10^{-4}, \ \text{10% of nominal value)}$ ?

$N_\text{min} \ = \ $

Solution

Solution

(1)  The last two statements are true:

• Relative to the random variable  $n_{\rm B}$  there is the classical case of a binomial distribution.
• The sum over  $N$  binary random variables is formed.
• The possible values of  $n_{\rm B}$  thus lie between  $0$  and  $N$.
• The linear mean gives   $m_{n{\rm B}}=p_{\rm B}\cdot N=\rm 10^{-3}\cdot 10^{5}=\rm 100.$

(2)  Für the rms of the binomial distribution holds with good approximation:

$$\sigma_{n{\rm B}}=\sqrt{N\cdot p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}} \hspace{0.15cm}\underline{\approx 10}.$$

(3)  Possible values of  $h_{\rm B}$  are all integer multiples of  $1/N$.  These all lie between  $0$  and  $1$.

• For the mean value, one obtains:

$$m_{h{\rm B}}=m_{n{\rm B}}/N=p_{\rm B} = 10^{-3}.$$

• The scatter results in

$$\sigma_{h{\rm B}}=\frac{\sigma_{n{\rm B}}{N}=\sqrt{\frac{ p_{\rm B}\cdot (\rm 1- \it p_{\rm B}{\rm )}}{N}}\hspace{0.15cm}\underline{\approx \rm 0.0001}.$$

(4)  Correct is the first proposition. It holds:

$${\rm Pr}(h_{\rm B} > p_{\rm B} + \varepsilon)=\rm Q({\it\varepsilon}/{\it\sigma_{h{\rm B}}),$$

$$\rm Pr(\it h_{\rm B} < p_{\rm B} - \itvarepsilon {\rm )}=\rm Q(\it{\varepsilon}/{\sigma_{h{\rm B}}{\rm )}$$

$$\Rightarrow \hspace{0.5cm}\rm Pr(\it |h_{\rm B} - p_{\rm B}| \le \varepsilon \rm )=\rm 1-\rm 2\cdot \rm Q({\it \varepsilon}/{\it \sigma_{h{\rm B}}).$$

(5)  One obtains with the numerical values  $\varepsilon = \sigma_{h{\rm B}} = 10^{-4}$:

$$p_{\varepsilon}=\rm 1-\rm 2\cdot \rm Q(\frac{\rm 10^{\rm -4}}{\rm 10^{\rm -4}} {\rm )}=\rm 1-\rm 2\cdot\rm Q(\rm 1)\hspace{0.15cm}\underline{\approx\rm 0.684}.$$

In words:

• If one determines the bit error rate by simulation over  $10^5$  symbols,
• with a confidence level of  $\underline{68.4\%}$  one obtains a value between  $0.9 \cdot 10^{-3}$  and  $1.1 \cdot 10^{-3}$,
• if  $p_{\rm B} = 10^{-3}$  is.

(6)  From the relation  $p_{\varepsilon}=\rm 1-\rm 2\cdot {\rm Q}(\alpha) = 0.95$  it follows directly:

$$\alpha_{\rm min}=\rm Q^{\rm -1}\Big(\frac{\rm 1-\it p_{\varepsilon}}{\rm 2}\Big)=\rm Q^{\rm -1}(\rm 0.025)\hspace{0.15cm}\underline{=\rm 1.96}\hspace{0.15cm}{\approx\rm 2}.$$

(7)  It must  $\alpha = \varepsilon/\sigma_{h{\rm B}}$  With the result of the subtask  (2)  then follows:

$$\frac{\varepsilon}{\sqrt{p_{\rm B}\cdot(\rm 1-\it p_{\rm B})/N}}\ge {\rm 2} \hspace{0.5cm}\Rightarrow\hspace{0.5cm} N\ge \frac{\rm 4\cdot \it p_{\rm B}\cdot(\rm 1-\it p_{\rm B})}{\varepsilon^{\rm 2}}\approx \frac{\rm 4\cdot 10^{-3}}{10^{-8}}\hspace{0.15cm}\underline{=\rm 400\hspace{0.08cm}000}.$$