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 (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.
• 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?  Use appropriate approximations.

$\sigma_f \ = \ $

4

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