Aufgaben:Exercise 4.2: Channel Log Likelihood Ratio at AWGN

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Conditional Gaussian functions

We consider two channels  $\rm A$  and  $\rm B$,  each with

• binary bipolar input  $x ∈ \{+1, \, -1\}$,  and

• continuous-valued output  $y ∈ {\rm \mathcal{R}}$  (real number).

The graph shows for both channels

• as blue curve the probability density functions  $f_{y\hspace{0.05cm}|\hspace{0.05cm}x=+1}$,

• as red curve the probability density functions  $f_{y\hspace{0.05cm}|\hspace{0.05cm}x=-1}$.

In the   "theory section"  the channel  $($German:  "Kanal"   ⇒   subscript:  "K"$)$  log likelihood ratio was derived for this AWGN constellation as follows:

$$L_{\rm K}(y) = L(y\hspace{0.05cm}|\hspace{0.05cm}x) = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y \hspace{0.05cm}|\hspace{0.05cm}x=+1) }{{\rm Pr}(y \hspace{0.05cm}|\hspace{0.05cm}x = -1)}\hspace{0.05cm}.$$

Evaluating this equation analytically,  we obtain with the proportionality constant  $K_{\rm L} = 2/\sigma^2$:

$$L_{\rm K}(y) =K_{\rm L} \cdot y\hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  "Soft–in Soft–out Decoder".

• Reference is made in particular to the sections 

• "Reliability Information – Log Likelihood Ratio", 

• "AWGN–Channel at Binary Input".

Questions

1 What are the characteristics of the channels shown in the diagram?

	They describe the binary transmission under Gaussian noise.
	The bit error probability without coding is  ${\rm Q}(1/\sigma)$.
	The channel log likelihood ratio is given as  $L_{\rm K}(y) = K_{\rm L} \cdot y$.

2 Which constant  $K_{\rm L}$  characterizes the channel  $\rm A$?

$K_{\rm L} \ = \ $

3 For channel  $\rm A$  what information do the received values  $y_1 = 1, \ y_2 = 0.5$,  $y_3 = \, -1.5$  provide about the transmitted binary symbols  $x_1, \ x_2$  and  $x_3$?

	$y_1 = 1.0$  states that probably  $x_1 = +1$  was sent.
	$y_2 = 0.5$  states that probably  $x_2 = +1$  was sent.
	$y_3 = \, -1.5$  states that probably  $x_3 = \, -1$  was sent.
	The decision  "$y_1 → x_1$"  is safer than  "$y_2 → x_2$".
	The decision  "$y_1 → x_1$"  is safer than  "$y_3 → x_3$".

4 Which  $K_{\rm L}$  identifies the channel  $\rm B$?

$K_{\rm L} \ = \ $

5 What information does channel  $\rm B$  provide about the received values  $y_1 = 1, \ y_2 = 0.5$,  $y_3 = -1.5$  about the transmitted binary symbols  $x_1, \ x_2$  and  $x_3$?

	For  $x_1, \ x_2, \ x_3$  is decided the same as for channel  $\rm A$.
	The estimate  "$x_2 = +1$"  is four times more certain than for channel  $\rm A$.
	The estimate  "$x_3 = \, -1$"  at channel  $\rm A$  is more reliable than the estimate  "$x_2 = +1$"  at channel  $\rm B$.

Solution

Solution

(1)  All proposed solutions  are correct:

• The transfer equation is always  $y = x + n$,  with  $x ∈ \{+1, \, -1\}$. 

• The variable  $n$  is a Gaussian random variable with standard deviation  $\sigma$   ⇒   variance  $\sigma^2$   ⇒   "AWGN Channel".

• The  "AWGN bit error probability"  is calculated using the standard deviation   $\sigma$   to   ${\rm Q}(1/\sigma)$   where  ${\rm Q}(x)$  denotes the  "complementary Gaussian error function".

• For each AWGN channel,  according to the  "theory section",  the channel log likelihood ratio always results in  $L_{\rm K}(y) = L(y|x) = K_{\rm L} \cdot y$.

• The constant  $K_{\rm L}$  is different for the two channels.

(2)  For the AWGN channel   ⇒   $L_{\rm K}(y) = K_{\rm L} \cdot y$   with constant   $K_{\rm L} = 2/\sigma^2$.

• The standard deviation  $\sigma$  can be read from the graph on the data page as the distance of the inflection points within the Gaussian curves from their respective midpoints.  For channel A   ⇒   $\sigma = 1$  results.

• The same result is obtained by evaluating the Gaussian function

$$\frac{f_{\rm G}( y = \sigma)}{f_{\rm G}( y = 0)} = {\rm e} ^{ - y^2/(2\sigma^2) } \Bigg |_{\hspace{0.05cm} y \hspace{0.05cm} = \hspace{0.05cm} \sigma} = {\rm e} ^{ -0.5} \approx 0.6065\hspace{0.05cm}.$$

• This means:  At the abscissa value  $y = \sigma$  the mean-free Gaussian function  $f_{\rm G}(y)$  has decayed to  $60.65\%$  of its maximum value. 

• Thus,  for the constant at  channel A:   $K_{\rm L} = 2/\sigma^2 \ \underline{= 2}$.

(3)  Correct are the  solutions 1 to 4:

• We first give the respective log likelihood ratios of  Channel A:

$$L_{\rm K}(y_1 = +1.0) = +2\hspace{0.05cm},\hspace{0.3cm}L_{\rm K}(y_2 = +0.5) = +1\hspace{0.05cm},\hspace{0.3cm}L_{\rm K}(y_3 = -1.5) = -3\hspace{0.05cm}. $$

• This results in the following consequences:

1. The decision for the  $($most probable$)$  code bit  $x_i$  is based on the sign of  $L_{\rm K}(y_i)$:
       $x_1 = +1, \ x_2 = +1, \ x_3 = \, -1$   ⇒   the  proposed solutions 1, 2 and 3  are correct.
2. The decision  "$x_1 = +1$"  is more reliable than the decision  "$x_2 = +1$"   ⇒   Proposition 4  is also correct.
3. However,  the decision  "$x_1 = +1$"  is less reliable than the decision  "$x_3 = \, –1$"  because  $|L_{\rm K}(y_1)<|L_{\rm K}(y_3)|$   ⇒   proposed solution 5 is incorrect.

• This can also be interpreted as follows:  The quotient between the red and the blue PDF value at  $y_3 = \, -1.5$  is larger than the quotient between the blue and the red PDF value at  $y_1 = +1$.

(4)  Following the same considerations as in subtask  (2),  the standard deviation of  channel B  is given by:  

$$\sigma = 1/2 \ \Rightarrow \ K_{\rm L} = 2/\sigma^2 \ \underline{= 8}.$$

(5)  For  channel B,  the following applies:  

$$L_{\rm K}(y_1 = +1.0) = +8, \ L_{\rm K}(y_2 = +0.5) = +4, \ L_{\rm K}(y_3 = \, -1.5) = \, -12.$$

• It is obvious that  the first two proposed solutions  are true,  but not the third,  because

$$|L_{\rm K}(y_3 = -1.5, {\rm channel\hspace{0.15cm} A)}| = 3\hspace{0.2cm} <\hspace{0.2cm}|L_{\rm K}(y_2 = 0.5, {\rm channel\hspace{0.15cm} B)}| = 4\hspace{0.05cm} . $$