Exercise 3.7: Bit Error Rate (BER)

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 the 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 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.

$\rm (B)$ For the metrological determination of the transmission quality or for a system simulation, it is necessary to rely on the Bit Error Rate $\rm (BER)$:

The bit error rate is an "a posteriori parameter" derived from a performed statistical experiment as a relative frequency.

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

$n_{\rm B}$ indicates the number of bit errors occurred when a total of $N$ binary symbols ("bits") were 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.
Solve this exercise 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

	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 ${\rm E}\big[n_{\rm B}\big] = 100$.

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

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

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

$p_\varepsilon \ = \ $

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

$N_\text{min} \ = \ $

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 standard deviation 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 standard deviation 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:

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

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

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

(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 $p_{\rm B} = 10^{-3}$.

(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:

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