Aufgaben:Exercise 1.16: Block Error Probability Bounds for AWGN

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Function  ${\rm Q}(x)$  and approximations;
it holds:  ${\rm Q_u}(x)\le{\rm Q}(x)\le{\rm Q_o}(x)$

We assume the following constellation:

• A linear block code with code rate  $R = k/n$  and distance spectrum  $\{W_i\}, \ i = 1, \ \text{...} \ , n$,

• an AWGN channel characterized by  $E_{\rm B}/N_{0}$   ⇒   convertible to noise power  $\sigma^2$,

• a receiver based on  "soft decision"  as well as the  "maximum likelihood criterion".

Under the assumption valid for the entire exercise that always the zero-word  $\underline{x}_{1} = (0, 0, \text{... } \ , 0)$  is sent, the  "pairwise error probability"  with a different code word  $\underline{x}_{l} (l = 2,\ \text{...} \ , 2^k)$:

$$ {\rm Pr}[\hspace{0.05cm}\underline{x}_{\hspace{0.02cm}1} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}l}\hspace{0.05cm}] = {\rm Q}\left ( \sqrt{w_{\rm H}(\underline{x}_{\hspace{0.02cm}l})/\sigma^2} \right ) \hspace{0.05cm}.$$

The derivation of this relation can be found in  [Liv10].  Used in this equation are:

• the  "complementary Gaussian error function"  ${\rm Q}(x)$,

• the  "Hamming weight"  $w_{\rm H}(\underline{x}_{l})$  of the code word  $\underline{x}_{l}$,

• the  "AWGN noise power"  $\sigma^2 = (2 \cdot R \cdot E_{\rm B}/N_{0})^{-1}.$

This allows various bounds to be specified for the block error probability:

• the so called  "Union Bound"  $\rm (UB)$:

$$p_1 = \sum_{l = 2}^{2^k}\hspace{0.05cm}{\rm Pr}[\hspace{0.05cm}\underline{x}_{\hspace{0.02cm}1} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}l}\hspace{0.05cm}] = \sum_{l \hspace{0.05cm}= \hspace{0.05cm}2}^{2^k}\hspace{0.05cm}{\rm Q}\left ( \sqrt{w_{\rm H}(\underline{x}_{\hspace{0.02cm}l})/\sigma^2} \right ) \hspace{0.05cm},$$

• the so called  "Truncated Union Bound"  $\rm (TUB)$:

$$p_2 = W_{d_{\rm min}} \cdot {\rm Q}\left ( \sqrt{d_{\rm min}/\sigma^2} \right ) \hspace{0.05cm},$$

• the  "Bhattacharyya Bound":

$$p_3 = W(\beta) - 1\hspace{0.05cm},\hspace{0.2cm} {\rm with}\hspace{0.15cm} \beta = {\rm e}^{ - 1/(2\sigma^2) } \hspace{0.05cm}.$$

In this case,  replace the distance spectrum  $\{W_i\}$  with the weight enumerator function:

$$\left \{ \hspace{0.05cm} W_i \hspace{0.05cm} \right \} \hspace{0.3cm} \Leftrightarrow \hspace{0.3cm} W(X) = \sum_{i=0 }^{n} W_i \cdot X^{i} = W_0 + W_1 \cdot X + W_2 \cdot X^{2} + ... \hspace{0.05cm} + W_n \cdot X^{n}\hspace{0.05cm}.$$

In the transition from the  "Union Bound"  $p_{1}$  to the more imprecise bound  $p_{3}$  among others

• the function  ${\rm Q}(x)$  is replaced by the  "Chernoff-Rubin bound"  ${\rm Q}_{\rm CR}(x)$.

• Both functions are shown in the above graph  (red and green curve, resp.).

In the  "Exercise 1.16Z"  the relationship between these functions is evaluated numerically and referenced to the bounds  ${\rm Q}_{\rm o}(x)$ and ${\rm Q}_{\rm u}(x)$  which are also drawn in the above graph.

Hints:

• This exercise belongs to the chapter  "Bounds for block error probability".

• The above cited reference  "[Liv10]"  refers to the lecture manuscript "Liva, G.:  Channel Coding.  Chair of Communications Engineering, TU Munich and DLR Oberpfaffenhofen, 2010."

• Further we refer to the interactive HTML5/JavaScript applet  "Complementary Gaussian error functions".

Questions

1

Which equation applies to the  "Union Bound"?

	$p_{1} = \sum_{l\hspace{0.05cm}=\hspace{0.05cm}2}^{2^k} W_{l} · {\rm Q}\big[(l/\sigma^2)^{0.5}\big],$
	$p_{1} = \sum_{i\hspace{0.05cm}=\hspace{0.05cm}1}^{n} W_{i} · {\rm Q}\big[(i/\sigma^2)^{0.5}\big].$

2

Specify the Union Bound for the  $(8, 4, 4)$  code and various  $\sigma$.

$\sigma = 1.0 \text{:} \hspace{0.4cm} p_{1} \ = \ $ $\ \%$

$\sigma = 0.5 \text{:} \hspace{0.4cm} p_{1} \ = \ $ $\ \%$

3

Given the same boundary conditions, what does the  "Truncated Union Bound"  provide?

$\sigma = 1.0 \text{:} \hspace{0.4cm} p_{2} \ = \ $ $\ \%$

$\sigma = 0.5 \text{:} \hspace{0.4cm} p_{2} \ = \ $ $\ \%$

4

Which statement is always true  (for all constellations)?

	The block error probability is never greater than  $p_{1}$.
	The block error probability is never greater than  $p_{2}$.

5

How do you get from  $p_{1}$  to the  "Bhattacharyya Bound"  $p_{3}$? 

	Replace the error function  ${\rm Q}(x)$  with the function  ${\rm Q}_{\rm CR}(x)$.
	Set the Bhattacharyya parameter  $\beta = 1/\sigma$.
	Instead of  $\{W_i\}$  uses the weight enumerator function  $W(X)$.

6

Specify the Bhattacharyya Bound for  $\sigma = 1$  and  $\sigma = 0.5$ .

$\sigma = 1.0 \text{:} \hspace{0.4cm} p_{3} \ = \ $ $\ \%$

$\sigma = 0.5 \text{:} \hspace{0.4cm} p_{3} \ = \ $ $\ \%$

Solution

Solution

(1)  The correct solution is suggestion 2:

• The distance spectrum  $\{W_i\}$  is defined for  $i = 0, \ \text{...} \ , \ n$:

1. $W_{1}$  indicates how often the Hamming weight  $w_{\rm H}(\underline{x}_{i}) = 1$  occurs.
2. $W_{n}$  indicates how often the Hamming weight  $w_{\rm H}(\underline{x}_{i}) = n$  occurs.

• With that,  the  "Union Bound"  is:

$$p_1 = {\rm Pr(Union \hspace{0.15cm}Bound)}= \sum_{i = 1}^{n}\hspace{0.05cm}W_i \cdot {\rm Q}\left ( \sqrt{i/\sigma^2} \right ) \hspace{0.05cm}.$$

(2)  The distance spectrum of the  $(8, 4, 4)$  code was given as  $W_{0} = 1 , \ W_{4} = 14, \ W_{8} = 1$. 

• Thus,  one obtains for $\sigma = 1$:

$$p_1 = W_4 \cdot {\rm Q}\left ( 2 \right ) + W_8 \cdot {\rm Q}\left ( 2 \cdot \sqrt{2} \right ) = 14 \cdot 2.28 \cdot 10^{-2}+ 1 \cdot 0.23 \cdot 10^{-2} \hspace{0.15cm}\underline{\approx 32.15\%}\hspace{0.05cm},$$

• For  $\sigma = 0.5$:

$$p_1 = 14 \cdot {\rm Q}\left ( 4 \right ) + {\rm Q}\left ( 4 \cdot \sqrt{2} \right ) = 14 \cdot 3.17 \cdot 10^{-5}+ 1.1 \cdot 10^{-8} \hspace{0.15cm}\underline{\approx 0.0444 \%}\hspace{0.05cm}.$$

(3)  With the minimum distance  $d_{\rm min} = 4$  we get:

$$\sigma = 1.0\text{:} \hspace{0.4cm} p_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} W_4 \cdot {\rm Q}\left ( 2 \right ) \hspace{0.15cm}\underline{= 31.92\%}\hspace{0.05cm},$$

$$\sigma = 0.5\text{:} \hspace{0.4cm} p_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm}W_4 \cdot {\rm Q}\left ( 4 \right ) \approx p_1 \hspace{0.15cm}\underline{ = 0.0444 \%}\hspace{0.05cm}.$$

(4)  The correct solution is  suggestion 1:

• The  "Union Bound"  - denoted here by  $p_{1}$ - is an upper bound on the block error probability in all cases.

• For the bound  $p_{2}$  ("Truncated Union Bound")  this is not always true.

• For example,  in the  $(7, 4, 3)$  Hamming code   ⇒   $W_{3} = W_{4} = 7, \ W_{7} = 1$  is obtained with standard deviation  $\sigma = 1$:

$$p_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} 7 \cdot {\rm Q}\left ( \sqrt{3} \right ) = 7 \cdot 4.18 \cdot 10^{-2} \approx 0.293\hspace{0.05cm},$$

$$p_1 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} p_2 + 7 \cdot {\rm Q}\left ( \sqrt{4} \right )+ 1 \cdot {\rm Q}\left ( \sqrt{7} \right ) \approx 0.455 \hspace{0.05cm}.$$

• The actual block error probability is likely to be between  $p_{2} = 29.3\%$  and  $p_{1} = 45.5\%$  (but this has not been verified).
  That is,   $p_{2}$ is not an upper bound.

(5)  Correct are  suggested solutions 1 and 3,  as the following calculation for the  $(8, 4, 4)$  code shows:

• It holds  ${\rm Q}(x) ≤ {\rm Q_{CR}}(x) = {\rm e}^{-x^2/2}$.  Thus,  for the Union Bound

$$p_1 = W_4 \cdot {\rm Q}\left ( \sqrt{4/\sigma^2} \right ) +W_8 \cdot {\rm Q}\left ( \sqrt{8/\sigma^2} \right )$$

another upper bound can be specified:

$$p_1 \le W_4 \cdot {\rm e}^{ - {4}/(2 \sigma^2) } +W_8 \cdot {\rm e}^{ - {8}/(2 \sigma^2) } \hspace{0.05cm}.$$

• With  $\beta = {\rm e}^{-1/(2\sigma^2)}$  can be written for this also  (so the given  $\beta = 1/\sigma$  is wrong):

$$p_1 \le W_4 \cdot \beta^4 + W_8 \cdot \beta^8 \hspace{0.05cm}.$$

• The weight function of the  $(8, 4, 4)$  code is:

$$W(X) = 1 + W_4 \cdot X^4 + W_8 \cdot X^8 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} W(\beta) - 1 = W_4 \cdot \beta^4 + W_8 \cdot \beta^8\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_3 = W(\beta) - 1 \ge p_1\hspace{0.05cm}.$$

(6)  With  $\sigma = 1$,  the Bhattacharyya parameter is  $\beta = {\rm e}^{-0.5} = 0.6065$,  and thus one obtains for the Bhattacharyya Bound:

$$p_3 = 14 \cdot \beta^4 + \beta^8 = 14 \cdot 0.135 + 0.018= 1.913 \hspace{0.15cm}\underline{= 191.3%}\hspace{0.05cm}.$$

• Considering that  $p_{3}$  is a bound for a probability,  $p_{3} = 1.913$  is only a trivial bound.

• For  $\sigma = 0.5$,  on the other hand,  $\beta = {\rm e}^{-2} \approx 0.135.$  Then holds:

$$p_3 = 14 \cdot \beta^4 + \beta^8 = 14 \cdot 3.35 \cdot 10^{-4} + 1.1 \cdot 10^{-7} \hspace{0.15cm}\underline{= 0.47 \%}\hspace{0.05cm}.$$

A comparison with subtask  (2)  shows that in the present example the Bhattacharyya Bound  $p_{3}$  is above the  "Union Bound"  $p_{1}$  by a factor 

$$(0.47 - 10^{-2})/(0.044 - 10^{-2}) > 10.$$

• The reason for this large deviation is the Chernoff-Rubin bound,  which is well above the  ${\rm Q}$ function.

• In  "Exercise 1.16Z",  the deviation between  ${\rm Q}_{\rm CR}$  and  ${\rm Q}(x)$  is also calculated quantitatively:

$${{\rm Q_{CR}}( x )}/{{\rm Q}( x )} \approx 2.5 \cdot x \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {{\rm Q_{CR}}( x = 4 )}/{{\rm Q}( x = 4)} \approx 10 \hspace{0.05cm}.$$