Aufgaben:Exercise 2.15: Block Error Probability with AWGN

• Return to book

Incomplete table of results

Using the example of  $\rm RSC \, (7, \, 3, \, 5)_8$  with parameters

• $n = 7$  $($number of code symbols$)$,

• $k =3$  $($number of information symbols$)$,

• $t = 2$  $($correction capability$)$.

the calculation of the block error probability in  "Bounded Distance Decoding"  $\rm (BDD)$  shall be shown.  The corresponding equation is:

$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) = \sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

⇒   The calculation is performed for the  "AWGN channel"   characterized by the parameter  $E_{\rm B}/N_0$:

• The quotient  $E_{\rm B}/{N_0}$  can be expressed by the relation

$$\varepsilon = {\rm Q} \big (\sqrt{{2 \cdot R \cdot E_{\rm B}}/{N_0}} \big ) $$

into the  "BSC model"  where  $R$  denotes the code rate  $($here:  $R = 3/7)$  and  ${\rm Q}(x)$  indicates the  "complementary Gaussian error integral".

• But since in the considered code the symbols come from  $\rm GF(2^3)$,  the BSC model with parameter  $\varepsilon$  must also still be adapted to the task.

• For the falsification probability of the  "m– BSC"  model  applies: 

$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m \hspace{0.05cm}.$$

Here it is to be set  $m = 3$  $($three bits per code symbol$)$.

⇒   For some  $E_{\rm B}/N_0$ values the results are entered in the table above.  The two rows with yellow background are briefly explained here:

• For  $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$  we get  $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$  and  $\varepsilon_{\rm S} \approx 0.2$.   The block error probability here can most easily be calculated using the complement:

$${\rm Pr(block\:error)} = 1 - \left [ {7 \choose 0} \cdot 0.8^7 + {7 \choose 1} \cdot 0.2 \cdot 0.8^6 + {7 \choose 2} \cdot 0.2^2 \cdot 0.8^5\right ] \approx 0.148 \hspace{0.05cm}.$$

• For  $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$  one gets  $\varepsilon \approx 1.2 \cdot 10^{-4}$  and  $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$.  With this very small falsification probability,  the  $f = 3$  term dominates,  and we obtain:

$${\rm Pr(block\:error)} \approx {7 \choose 3} \cdot (3.5 \cdot 10^{-4})^3 \cdot (1- 3.5 \cdot 10^{-4})^4 \approx 1.63 \cdot 10^{-9} \hspace{0.05cm}.$$

⇒   You are to calculate the block error probabilities for the rows highlighted in red   $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,  $10 \ \rm dB)$.

• The rows with blue background show some results of  "Exercise 2.15Z".  There  ${\rm Pr}(\underline{v} ≠ \underline{u})$  is calculated for  $\varepsilon_{\rm S} = 10\%,  \ 1\%$  $0.1\%$.
• In subtasks (4) and (5) you are to establish the relationship between the quantity  $\varepsilon_{\rm S}$  and the AWGN parameter  $E_{\rm B}/N_0$  thus completing the above table.

Hints:

• The exercise belongs to the chapter  "Error Probability and Application Areas".

• We refer you here to the two interactive applets 

• "Complementary Gaussian error functions",  and 

• "Binomial and Poisson Distribution".

Questions

1

What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 5 \ \rm dB}$?

${\rm Pr(block\:error)} \ = \ $ $\ \cdot 10^{-2}$

2

What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 8 \ \rm dB}$?

${\rm Pr(block\:error)} \ = \ $ $\ \cdot 10^{-4}$

3

What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0}\hspace{0.15cm}\underline{ = 10 \ \rm dB}$?

${\rm Pr(block\:error)} \ = \ $ $\ \cdot 10^{-6}$

4

How is  $\varepsilon_{\rm S} = 0.1$  related to  $10 \cdot \lg {E_{\rm B}/N_0}$ ?   Note:  Use the given applet to calculate  ${\rm Q}(x)$.

$\varepsilon_{\rm S} = 10^{-1} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ $\ \rm dB$

5

Find also the  $E_{\rm B}/N_0$ values  $($in  $\rm dB)$  for  $\varepsilon_{\rm S} = 0.01$  and  $\varepsilon_{\rm S} = 0.001$. Complete the table.

$\varepsilon_{\rm S} = 10^{-2} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ $\ \rm dB$

$\varepsilon_{\rm S} = 10^{-3} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ $\ \rm dB$

Solution

Solution

(1)  From the table on the information page, the BSC parameter $\varepsilon = 0.0505$ can be read.

• This gives $\varepsilon_{\rm S}$ for the symbol corruption probability with $m = 3$:

$$1 - \varepsilon_{\rm S} = (1 - 0.0505)^3 \approx 0.856 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \varepsilon_{\rm S} \approx 0.144 \hspace{0.05cm}.$$

• The fastest way to calculate the block error probability here is to use the formula

$${\rm Pr(Block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - {\rm Pr}(f=0) - {\rm Pr}(f=1) - {\rm Pr}(f=2) = 1 - 1 \cdot 0.856^7 - 7 \cdot 0.144^1 \cdot 0.856^6 - 21 \cdot 0.144^2 \cdot 0.856^5$$

$$\Rightarrow \hspace{0.3cm} {\rm Pr(Block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{v} \ne \underline{u}) =1 - 0.3368 - 0.3965 - 0.2001 \hspace{0.15cm} \underline{=0.0666} \hspace{0.05cm}.$$

(2)  Following the same calculation procedure as in subtask (1), the following is obtained with $\varepsilon_{\rm S} \approx 0.03 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.97$:

$${\rm Pr(Block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \hspace{-0.05cm}-\hspace{-0.05cm} 1 \cdot 0.97^7 \hspace{-0.05cm}-\hspace{-0.05cm} 7 \cdot 0.03^1 \cdot 0.97^6 \hspace{-0.05cm}-\hspace{-0.05cm} 21 \cdot 0.03^2 \cdot 0.97^5 =1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.8080 \hspace{-0.05cm}-\hspace{-0.05cm} 0.1749\hspace{-0.05cm}-\hspace{-0.05cm} 0.0162= 1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.9991 = 9 \cdot 10^{-4} \hspace{0.05cm}.$$

• You can see that here the difference between two numbers of almost the same size must be formed, so that the result could be affected by an error.
• Therefore we still calculate the following quantities:

$${\rm Pr}(f=3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 3} \cdot \varepsilon_{\rm S}^3 \cdot (1 - \varepsilon_{\rm S})^4 = 35 \cdot 0.03^3 \cdot 0.97^4 = 8.366 \cdot 10^{-4}\hspace{0.05cm},$$

$${\rm Pr}(f=4) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 4} \cdot \varepsilon_{\rm S}^4 \cdot (1 - \varepsilon_{\rm S})^3 = 35 \cdot 0.03^4 \cdot 0.97^3 = 0.259 \cdot 10^{-4}\hspace{0.05cm},$$

$${\rm Pr}(f=5) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 5} \cdot \varepsilon_{\rm S}^5 \cdot (1 - \varepsilon_{\rm S})^2 = 21 \cdot 0.03^5 \cdot 0.97^2 = 0.005 \cdot 10^{-4}$$

$$\Rightarrow \hspace{0.3cm} {\rm Pr(Blockfehler)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) + {\rm Pr}(f=4) + {\rm Pr}(f=5) \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$

• The terms for $f = 6$ and $f = 7$ can be omitted here. They do not provide a relevant contribution.

(3)  Here $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$ is already given in the table.

• The (by far) dominant term in the calculation of the block error probability is ${\rm Pr}(f = 3)$:.

$${\rm Pr(Blockfehler)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) = {7 \choose 3} \cdot 0.005^3 \cdot 0.995^4 \hspace{0.15cm} \underline{\approx 4.3 \cdot 10^{-6}} \hspace{0.05cm}.$$

(4)  For the BSC parameter $\varepsilon$ holds with $\varepsilon_{\rm S} = 0.1$:

$$\varepsilon = 1 -(1 - \varepsilon_{\rm S})^{1/3} = 1 - 0.9^{1/3} \approx 0.0345 \hspace{0.05cm}.$$

• The relation between $\varepsilon$ and $E_{\rm B}/N_0$ is:

$$\varepsilon = {\rm Q}(x)\hspace{0.05cm}, \hspace{0.5cm} x = \sqrt{2 \cdot R \cdot E_{\rm B}/N_0}\hspace{0.05cm}.$$

• The inverse $x = {\rm Q}^{-1}(0.0345)$ is obtained with the applet "Complementary Gaussian Error Functions" to $x = 1.82$.
• This further gives:

$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{1.82^2}{2R \cdot 3/7} \approx 3.864 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0) \hspace{0.15cm} \underline{\approx 5.87 \,\, {\rm dB}} \hspace{0.05cm}. $$

(5)  After the same calculation one obtains

• für $\varepsilon_{\rm S} = 10^{-2} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-2} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 2.71$

$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{2.71^2}{2R \cdot 3/7} \approx 8.568 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0) \hspace{0.15cm} \underline{\approx 9.32 \,\, {\rm dB}} \hspace{0.05cm}, $$

• for $\varepsilon_{\rm S} = 10^{-3} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-3} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 3.4$:

$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{3.4^2}{2R \cdot 3/7} \approx 13.487 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0) \hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$

Results for $rm RSC \, (7, \, 3, \, 5)_8$ decoding

The graph shows the course of the block error probability as a function of $10 \cdot \lg {E_{\rm B}/N_0}$ as well as the completely filled result table.

One can see the clearly less favorable (asymptotic) behavior of this short (green) code $\rm RSC \, (7, \, 5, \, 3)_8$ compared to the (red) comparison code $\rm RSC \, (255, \, 223, \, 33)_8$:

• For abscissa values smaller than $10 \ \rm dB$ the result is even worse than without coding.
• Therefore it should be pointed out again that this $\rm RSC \, (7, \, 3, \, 5)_8$ has little practical meaning.
• It was chosen for this exercise only to be able to demonstrate with reasonable effort the calculation of the block error probability for "Bounded Distance Decoding" (BDD).