Aufgaben:Exercise 5.7Z: McCullough Model once more

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EDD and ECF of GE model and equivalent MC model

As in  "Exercise 5.6",  "Exercise 5.6Z"  and  "Exercise 5.7",  we consider the burst error channel model according to Gilbert and Elliott (GE model) with the parameters

$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,

\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm} p(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.1, \hspace{0.2cm} p(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

From these four probabilities, the corresponding characteristics of the channel model according to McCullough (MC model) can be determined in such a way that both models have exactly the same statistical properties, namely

• exactly the same error distance distribution (EDD)  $V_a(k)$,
• exactly the same error correlation function (ECF)  $\varphi_e(k)$.

The probabilities of the MC model were determined in  "Exercise 5.7"  as follows $($labels according to the graph for Exercise 5.7, all with  $q$  instead of  $p)$:

$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.0061,

\hspace{0.2cm}q_{\rm B} = 0.1949,\hspace{0.2cm}

q(\rm

G\hspace{0.05cm}|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5528, \hspace{0.2cm} q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.3724\hspace{0.05cm}.$$

The upper graph shows the functions  $V_a(k)$  and  $\varphi_e(k)$  simulatively determined from  $N = 10^6$  sequence elements for the GE and MC models. There are still slight discrepancies here. In the limiting case for  $N → ∞$,  on the other hand, error correlation function and error distance distribution of both models agree exactly.

In this exercise, important descriptive variables of the GE model such as

• state probabilities,
• mean error probabilities, and
• correlation duration

should be determined directly from the $q$ parameters of the MC model.

Notes:

• The exercise belongs to the chapter  "Burst Error Channels".

• From the above exercises, the following results can be further used:

(a) The state probabilities of the GE model are

$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}{p(\rmG\hspace{0.05cm}|\hspace{0.05cm} B) + p(\rmB\hspace{0.05cm}|\hspace{0.05cm} G)}\hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G}\hspace{0.05cm}.$$

(b) The mean error probability of the GE model is

$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}= \varphi_{e}(k = 0 )\hspace{0.05cm}.$$

(c) The correlation duration of the GE model is calculated as

$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm}B ) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )}-1\hspace{0.05cm}.$$

Questions

1 Calculate the probabilities  $\alpha_{\rm G}$  and  $\alpha_{\rm B}$ that the MC model is in the state "Good" and the state "Bad".

$\alpha_{\rm G} \hspace{0.05cm} = \ $

$\alpha_{\rm B} \ = \ $

2 Determine the mean error distance of the MC model.

${\rm E}\big[a\big] \ = \ $

3 What is the error correlation function value for  $k = 0$?

$\varphi_e(k = 0) \ = \ $

4 Give the error correlation duration  $D_{\rm K}$  as a function of the MC parameters  $q_{\rm G},  q_{\rm B},  q(\rm G\hspace{0.05cm}|\hspace{0.05cm}B)$  and  $q(\rm B\hspace{0.05cm}|\hspace{0.05cm}G)$. 
Which result is correct?

	$D_{\rm K} = \big [q({\rm B\hspace{0.05cm}|\hspace{0.05cm}G}) + q({\rm G\hspace{0.05cm}|\hspace{0.05cm}B})\big]^{-1} \ -1$,
	$D_{\rm K} = \big [q_{\rm G} \cdot q({\rm G|B}) + q_{\rm B} \cdot q({\rm G|B}) \big]^{-1} \ -1$.

Solution

Solution

(1)  For the state probabilities of the GE model was determined in Exercise 5.6Z:

$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}{p(\rmG\hspace{0.05cm}|\hspace{0.05cm} B) + p(\rmB\hspace{0.05cm}|\hspace{0.05cm} G)} = 0.909\hspace{0.05cm},\hspace{0.5cm} w_{\rm B} = 1 - w_{\rm G}= 0.091\hspace{0.05cm}.$$

• In contrast, for the MC model we obtain:

$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{q(\rm

G\hspace{0.05cm}|\hspace{0.05cm} B)}{q(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) + q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}= \frac{0.5528}{0.5528 + 0.3724}\hspace{0.15cm}\underline {= 0.5975}\hspace{0.05cm},\hspace{0.5cm} \alpha_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 - \alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$

• In subtask (3) of Exercise 5.7, these values have already been determined once, but from the parameters of the equivalent Gilbert-Elliott model.

(2)  The mean error distance in the channel state "GOOD" is equal to the reciprocal of the associated error probability $q_{\rm G}$.

• Accordingly, the mean error distance in the state "BAD" is $1/q_{\rm B}$.
• By weighting with the two state probabilities $\alpha_{\rm G}$ and $\alpha_{\rm B}$, the mean error distance of the MC model as a whole is given by

$${\rm E}[a] =\frac{\alpha_{\rm G}}{q_{\rm G}} + \frac{\alpha_{\rmB}}{q_{\rm B}}=\frac{0.5975}{0.0061} + \frac{0.4025}{0.1949} =97.95 + 2.06\hspace{0.15cm}\underline { = 100.1}\hspace{0.05cm}.$$

• Of course, this value should be exactly the same as for the corresponding GE model.
• The small deviation of $0.1$ is due to rounding errors.

(3)  Again, the relation $\varphi_e(k = 0) = p_{\rm M}$ holds.

• However, the mean error probability is equal to the reciprocal of the mean error distance ${\rm E}[a]$.
• It follows that $\varphi_e(k = 0) \ \underline {= 0.01}$.

(4)  In the GE model, the correlation duration is given as follows ($S$ stands for sum):

$$D_{\rm K} = {1}/{S}-1 \hspace{0.05cm},\hspace{0.2cm}S = {\rmPr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B ) + {\rm Pr}(\rmB\hspace{0.05cm}|\hspace{0.05cm} G )\hspace{0.05cm}.$$

• Further, using the data for Exercise 5.7:

$$q({\rm B\hspace{0.05cm}|\hspace{0.05cm} G }) = \frac{\alpha_{\rm

B} \cdot S}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \cdot q_{\rm G}} \hspace{0.05cm}, \hspace{0.2cm}q({\rm G\hspace{0.05cm}|\hspace{0.05cm} B })= \frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )$$

$$\Rightarrow \hspace{0.3cm} S = q_{\rm G} \cdot q({\rm

B\hspace{0.05cm}|\hspace{0.05cm} G }) + q_{\rm B} \cdot \frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) = q_{\rm G} \cdot q({\rm B\hspace{0.05cm}|\hspace{0.05cm} G })+ q_{\rm B} \cdot q({\rm G\hspace{0.05cm}|\hspace{0.05cm} B }) \hspace{0.05cm}.$$

$$\Rightarrow \hspace{0.3cm}D_{\rm K} =\frac{1}{q_{\rm G} \cdotq({\rm B\hspace{0.05cm}|\hspace{0.05cm} G })+ q_{\rm B} \cdotq({\rm G\hspace{0.05cm}|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$

• So, the correct solution is solution 2. With the given parameter values, we obtain, for example:

$$D_{\rm K} =\frac{1}{0.0061 \cdot 0.3724 + 0.1949 \cdot0.5528}-1=\frac{1}{0.11}-1 {\approx 8.09}\hspace{0.05cm}.$$

• The result is exactly the same value as in subtask (3) of Exercise 5.6.