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

In  "theory section"  the channel LLR 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 pages  "Reliability Information – Log Likelihood Ratio"  and  [[Channel_Coding/Channel_Models_and_Decision_Structures#AWGN_channel_at_Binary_Input|"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 interference.
	The bit error probability without coding is  ${\rm Q}(1/\sigma)$.
	The channel–LLR is given as  $L_{\rm K}(y) = K_{\rm L} \cdot y$  representable.

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  $x_3$, respectively?

	$y_1 = 1.0$  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 more certain 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  $x_1, \ x_2$  respectively.  $x_3$?

	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 $x ∈ \{+1, \, -1\}$; $n$ gives a Gaussian random variable with variance $\sigma$   ⇒   variance $\sigma^2$   ⇒   "AWGN Channel".
• The "AWGN Bit Error Probability" is calculated using the dispersion $\sigma$ 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–LLR 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, however.

(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)  The correct solutions are 1 to 4:

• We first give the respective LLRs 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) codebit $x_i$ is made 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 $|L_{\rm K}(y_1)|$ is smaller than $|L_{\rm K}(y_3)|$   ⇒   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 scattering 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:   $L_{\rm K}(y_1 = +1.0) = +8, \ L_{\rm K}(y_2 = +0.5) = +4$ und $L_{\rm K}(y_3 = \, -1.5) = \, -12$.

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

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