Aufgaben:Exercise 3.14: Error Probability Bounds

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Bhattacharyya and Viterbi bound with the BSC model  $($incomplete table$)$.

For the frequently used convolutional code with

• the code rate  $R = 1/2$,

• the memory  $m = 2$,  and

• the transfer function matrix

$${\boldsymbol{\rm G}}(D) = \big ( 1 + D + D^2\hspace{0.05cm},\hspace{0.1cm} 1 + D^2 \hspace{0.05cm}\big ), $$

the  "extended path weighting enumerator function" is:

$$T_{\rm enh}(X, U) = \frac{UX^5}{1- 2 \hspace{0.05cm}U \hspace{-0.05cm}X} \hspace{0.05cm}.$$

With the series expansion   $1/(1 \, –x) = 1 + x + x^2 + \text{...} \ $   can also be written for this purpose:

$$T_{\rm enh}(X, U) = U X^5 \cdot \left [ 1 + (2 \hspace{0.05cm}U \hspace{-0.05cm}X) + (2 \hspace{0.05cm}U\hspace{-0.05cm}X)^2 + (2 \hspace{0.05cm}U\hspace{-0.05cm}X)^3 +\text{...} \hspace{0.10cm} \right ] \hspace{0.05cm}.$$

The  "simple path weighting enumerator function"  $T(X)$  results from setting the second variable  $U = 1$ .

Using these two functions,  error probability bounds can be specified:

• The  "burst error probability"  is limited by the  Bhattacharyya bound:

$${\rm Pr(burst\:error)} \le {\rm Pr(Bhattacharyya)} = T(X = \beta) \hspace{0.05cm}.$$

• In contrast,  the  "bit error probability"  is always less than  $($or equal to$)$  the  Viterbi bound:

\[{\rm Pr(bit\:error)} \le {\rm Pr(Viterbi)} = \left [ \frac {\rm d}{ {\rm d}U}\hspace{0.2cm}T_{\rm enh}(X, U) \right ]_{\substack{X=\beta \\ U=1} } \hspace{0.05cm}.\]

Hints:

• The exercise belongs to the chapter  "Distance Characteristics and Error Probability Barriers".

• The Bhattacharyya parameter for BSC is:   $\beta = 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)}$.

• In the table,  for some values of the BSC parameter  $\varepsilon$  are given:

•   the Bhattacharyya parameter  $\beta$,
•   the Bhattacharyya bound  ${\rm Pr}(\rm Bhattacharyya)$,  and
•   the Viterbi bound  $\rm Pr(Viterbi)$.

• Throughout this exercise,  you are to compute the corresponding quantities for  $\varepsilon = 10^{-2}$  and  $\varepsilon = 10^{-4}$.

• You can find the complete table in the sample solution.

Questions

1

What Bhattacharyya parameter results for the BSC model?

$\varepsilon = 10^{–2} \text{:} \hspace{0.4cm} \beta \ = \ $

$\varepsilon = 10^{–4} \text{:} \hspace{0.4cm} \beta \ = \ $

2

What is the Bhattacharyya bound?

$\varepsilon = 10^{-2} \text{:} \hspace{0.4cm} {\rm Pr(Bhattacharyya)} \ = \ $ $\ \cdot 10^{–4}$

$\varepsilon = 10^{-4} \text{:} \hspace{0.4cm} {\rm Pr(Bhattacharyya)} \ = \ $ $\ \cdot 10^{–9}$

3

What is the Viterbi bound?

$\varepsilon = 10^{-2} \text{:} \hspace{0.4cm} {\rm Pr(Viterbi)} \ = \ $ $\ \cdot 10^{–4}$

$\varepsilon = 10^{-4} \text{:} \hspace{0.4cm} {\rm Pr(Viterbi)} \ = \ $ $\ \cdot 10^{–9}$

4

For which values  $\varepsilon < \varepsilon_0$  are both bounds not applicable?

$\varepsilon_0 \ = \ $

Solution

Solution

(1)  The Bhattacharyya parameter results for the BSC model with  $\varepsilon = 0.01$  to

$$\beta = 2 \cdot \sqrt{\varepsilon \cdot (1- \varepsilon)} = 2 \cdot \sqrt{0.01 \cdot 0.99} \hspace{0.2cm}\underline {\approx 0.199} \hspace{0.05cm}.$$

• For even smaller falsification probabilities  $\varepsilon$  can be written approximately:

$$\beta \approx 2 \cdot \sqrt{\varepsilon } \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \varepsilon = 10^{-4}\hspace{-0.1cm}: \hspace{0.2cm} \beta \hspace{0.2cm}\underline {\approx 0.02} \hspace{0.05cm}.$$

(2)  It holds  ${\rm Pr(burst\:error)} ≤ {\rm Pr(Bhattacharyya)}$  with  ${\rm Pr(Bhattacharyya)} = T(X = \beta)$.

• For the considered convolutional code of rate 1/2,  memory $m = 2$  and  $\mathbf{G}(D) = (1 + D + D^2, \ 1 + D^2)$,  the path weighting enumerator function is:

$$T(X) = \frac{X^5 }{1- 2X} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr(Bhattacharyya)} = T(X = \beta) = \frac{\beta^5 }{1- 2\beta}$$

$$\Rightarrow \hspace{0.3cm}\varepsilon = 10^{-2}\hspace{-0.1cm}: \hspace{0.1cm} {\rm Pr(Bhattacharyya)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac{0.199^5 }{1- 2\cdot 0.199} \hspace{0.2cm}\underline {\approx 5.18 \cdot 10^{-4}}\hspace{0.05cm},$$

$$\hspace{0.85cm} \varepsilon = 10^{-4}\hspace{-0.1cm}: \hspace{0.1cm} {\rm Pr(Bhattacharyya)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac{0.02^5 }{1- 2\cdot 0.02} \hspace{0.38cm}\underline {\approx 3.33 \cdot 10^{-9}}\hspace{0.05cm}.$$

(3)  To calculate the Viterbi bound,  we assume the  "extended path weighting enumerator function":

$$T_{\rm enh}(X, U) = \frac{U X^5}{1- 2UX} \hspace{0.05cm}.$$

• The derivative of this function with respect to the input parameter  $U$ gives:

$$\frac {\rm d}{{\rm d}U}\hspace{0.1cm}T_{\rm enh}(X, U) = \frac{(1- 2UX) \cdot X^5 - U X^5 \cdot (-2X)}{(1- 2UX)^2} = \frac{ X^5}{(1- 2UX)^2} \hspace{0.05cm}.$$

• This equation provides the Viterbi bound for  $U = 1$  and  $X = \beta$:

$$\frac {\rm d}{{\rm d}U}\hspace{0.1cm}T_{\rm enh}(X, U) = \frac{(1- 2UX) \cdot X^5 - U X^5 \cdot (-2X)}{(1- 2UX)^2} = \frac{U X^5}{(1- 2UX)^2} \hspace{0.05cm}.$$

$$\Rightarrow \hspace{0.3cm}\varepsilon = 10^{-2}\hspace{-0.1cm}: \hspace{0.1cm} {\rm Pr(Viterbi)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac{0.199^5 }{(1- 2\cdot 0.199)^2} = \hspace{0.2cm}\underline {\approx 8.61 \cdot 10^{-4}}\hspace{0.05cm},$$

$$\hspace{0.85cm} \varepsilon = 10^{-4}\hspace{-0.1cm}: \hspace{0.1cm} {\rm Pr(Viterbi)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac{0.02^5 }{(1- 2\cdot 0.02)^2} = \hspace{0.2cm}\underline {\approx 3.47 \cdot 10^{-9}}\hspace{0.05cm}.$$

• We check the result using the following approximation:

$$T_{\rm enh}(X, U) = U X^5 + 2\hspace{0.05cm}U^2 X^6 + 4\hspace{0.05cm}U^3 X^7 + 8\hspace{0.05cm}U^4 X^8 + \text{...} $$

$$\Rightarrow \hspace{0.3cm}\frac {\rm d}{{\rm d}U}\hspace{0.1cm}T_{\rm enh}(X, U) = X^5 + 4\hspace{0.05cm}U X^6 + 12\hspace{0.05cm}U^2 X^7 + 32\hspace{0.05cm}U^3 X^8 + \text{...} $$

• Setting  $U = 1$  and  $X = \beta$  we get again the Viterbi bound:

$${\rm Pr(Viterbi)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \beta^5 + 4\hspace{0.05cm}\beta^6 + 12\hspace{0.05cm}\beta^7 + 32\hspace{0.05cm}\beta^8 +\text{...} = \beta^5 \cdot (1+ 4\hspace{0.05cm}\beta + 12\hspace{0.05cm}\beta^2 + 32\hspace{0.05cm}\beta^3 + ... )\hspace{0.05cm}. $$

• For  $\varepsilon = 10^{–2} \ \Rightarrow \ \beta = 0.199$  is obtained by truncating the infinite sum after the  $\beta^3$  term:

$${\rm Pr(Viterbi)} \approx 3.12 \cdot 10^{-4} \cdot (1 + 0.796 + 0.475 + 0.252) = 7.87 \cdot 10^{-4} \hspace{0.05cm}.$$

• The termination error  $($related to  $8.61 \cdot 10^{–4})$  is here about  $8.6\%$.  For $\varepsilon = 10^{–4} \ \Rightarrow \ \beta = 0.02$ the termination error is even smaller:

$${\rm Pr(Viterbi)} \approx 3.20 \cdot 10^{-9} \cdot (1 + 0.086 + 0.0048 + 0.0003) = 3.47 \cdot 10^{-9} \hspace{0.05cm}.$$

Bhattacharyya and Viterbi bound in the BSC model  $($full table$)$.

(4)  For  $\beta = 0.5$  both bounds result in the value  "infinite".

• For even larger  $\beta$  values,  the Bhattacharyya bound becomes negative and the result for the Viterbi bound is then also not applicable.  It follows:

$$\beta_0 = 2 \cdot \sqrt{\varepsilon_0 \cdot (1- \varepsilon_0)} = 0.5$$

$$\Rightarrow \hspace{0.3cm} {\varepsilon_0 \cdot (1- \varepsilon_0)} = 0.25^2 = 0.0625$$

$$\Rightarrow \hspace{0.3cm} \varepsilon_0^2 - \varepsilon_0 + 0.0625 = 0$$

$$\Rightarrow \hspace{0.3cm} \varepsilon_0 = 0.5 \cdot (1 - \sqrt{0.75}) \hspace{0.15cm} \underline {\approx 0.067}\hspace{0.05cm}.$$