Aufgaben:Exercise 3.7Z: Error Performance

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Every operator of ISDN systems must comply with certain minimum requirements regarding the bit error rate  $\rm (BER)$,  which are specified for example in the  CCITT Recommendation G.821  under the name  "Error Performance".

On the right you can see an excerpt from this recommendation:

• This states,  among other things,  that – averaged over a sufficiently long time – at least  $99.8\%$  of all one-second intervals must have a bit error rate less than  $10^{-3}$  (one per thousand).
• For a bit rate of  $\text{64 kbit/s}$  this corresponds to the condition that in one second  $($and thus for  $N = 64\hspace{0.08cm}000$  transmitted symbols$)$  no more than  $64$  bit errors may occur:

$$\rm Pr(\it f \le \rm 64) \ge \rm 0.998.$$

Hints:

• The exercise belongs to the chapter  Gaussian distributed random variables.
• Always assume bit error probability  $p = 10^{-3}$  for the first three subtasks.
• In addition,  throughout the task,  let $N = 64\hspace{0.08cm}000$  hold.
• Under certain conditions – which are all fulfilled here – the binomial distribution can be approximated by a Gaussian distribution with equal mean and equal rms value.  Use this approximation for the subtask (4).

Questions

1

Which of the following statements are true regarding the random variable $f$ ?

	The random variable $f$  is binomially distributed.
	$f$  can be approximated by a Poisson distribution.

2

What is the mean value of the random variable  $f$?

$m_f \ = \ $

3

How large is the rms value?  Use appropriate approximations.

$\sigma_f \ = \ $

4

Calculate the probability that no more than  $64$  bit errors occur.  Use the Gaussian approximation.

${\rm Pr}(f ≤ 64) \ = \ $ $ \ \rm \%$

5

What is the maximum bit error probability  $p_\text{B, max}$  that the condition  "64  (or more)  bit errors only in at most 0.2% of the one-second intervals"  can be met? 
It holds  ${\rm Q}(2.9) \approx 0.002$.

$p_\text{B, max}\ = \ $ $ \ \rm \%$

Solution

Solution

(1)  Both statements are correct:

• The random vairable $f$  defined here is the classical case of a binomially distributed random variable:  Sum over  $N$  binary values  $(0$ or $1)$.
• Because the product  $N \cdot p = 64$  and thus is much larger than  $1$ ,
• the binomial distribution can be approximated with good approximation by a Poisson distribution with rate  ${\it \lambda} = 64$  .

(2)  The mean is obtained as  $m_f = N \cdot p \hspace{0.15cm}\underline{= 64}$  regardless of whether one assumes the binomial– or the Poisson distribution.

(3)  For the rms one obtains  

$$\it \sigma_f=\rm\sqrt{\rm 64000\cdot 10^{-3}\cdot 0.999}\hspace{0.15cm}\underline{\approx\sqrt{64}=8}.$$

• The error by applying Poissonl distribution instead of binomial distribution here is smaller than  $0.05\%$.

(4)  For a Gaussian random variable $f$  with mean  $m_f {= 64}$  the probability  ${\rm Pr}(f \le 64) \hspace{0.15cm}\underline{\approx 50\%}$.   Note:

• For a continuous random size, the probability would be exactly $50\%$.
• Since $f$  can only take integer values, it is slightly larger here.

(5)  With  $\lambda = N \cdot p$  the corresponding condition is:

$$\rm Q\big (\frac{\rm 64-\it \lambda}{\sqrt{\it \lambda}} \big )\le \rm 0.002\hspace{0.5cm}\rm or.\hspace{0.5cm}\frac{\rm 64-\it \lambda}{\sqrt{\it \lambda}}>\rm 2.9.$$

• The maximum value of  $\lambda$  can be determined according to the following equation:

$$ \lambda+\rm 2.9\cdot\sqrt{\it\lambda}-\rm 64 = \rm 0.$$

• The solution of this quadratic equation is thus:

$$\sqrt{\it \lambda}=\frac{\rm -2.9\pm\rm\sqrt{\rm 8.41+256}}{\rm 2}=\rm 6.68 \hspace{0.5cm}\rightarrow \hspace{0.5cm} \lambda = 44.6 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} {\it p}_\text{B, max}= \frac{44.6}{64000} \hspace{0.15cm}\underline{\approx 0.069\%}.$$

• The second solution is negative and need not be considered further.