Exercise 4.08Z: Error Probability with Three Symbols

Decision regions with $M = 3$

The diagram shows exactly the same signal space constellation as in "Exercise 4.8":

the $M = 3$ possible transmitted signals, viz.

$$\boldsymbol{ s }_0 = (-1, \hspace{0.1cm}1)\hspace{0.05cm}, \hspace{0.2cm} \boldsymbol{ s }_1 = (1, \hspace{0.1cm}2)\hspace{0.05cm}, \hspace{0.2cm} \boldsymbol{ s }_2 = (2, \hspace{0.1cm}-1)\hspace{0.05cm}.$$

the $M = 3$ decision boundaries

$$G_{01}\text{:} \hspace{0.4cm} y \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1.5 - 2 \cdot x\hspace{0.05cm},$$

$$G_{02}\text{:} \hspace{0.4cm} y \hspace{-0.1cm} \ = \ \hspace{-0.1cm} -0.75 +1.5 \cdot x\hspace{0.05cm},$$

$$G_{12}\text{:} \hspace{0.4cm} y \hspace{-0.1cm} \ = \ \hspace{-0.1cm} x/3\hspace{0.05cm}.$$

The two axes of the two-dimensional signal space are simplistically denoted here as $x$ and $y$; actually, $\varphi_1(t)/\sqrt {E}$ and $\varphi_2(t)/\sqrt {E}$ should be written for these, respectively.

These decision boundaries are optimal under the two conditions:

equal probability symbol probabilities,

circularly–symmetric PDF of the noise (e.g. AWGN).

In contrast, in this exercise we consider a two–dimensional uniform distribution for the noise PDF:

$$\boldsymbol{ p }_{\boldsymbol{ n }} (x,\hspace{0.15cm} y) = \left\{ \begin{array}{c} K\\ 0 \end{array} \right.\quad \begin{array}{*{1}c}{\rm for} \hspace{0.15cm}|x| <A, \hspace{0.15cm} |y| <A \hspace{0.05cm}, \\ {\rm else} \hspace{0.05cm}.\\ \end{array}$$

Such an amplitude-limited noise is admittedly without any practical meaning.

However, it allows an error probability calculation without extensive integrals, from which the principle of the procedure can be seen.

Notes:

The exercise belongs to the chapter "Approximation of the Error Probability".

To simplify the notation, the following is used:

$$x = {\varphi_1(t)}/{\sqrt{E}}\hspace{0.05cm}, \hspace{0.2cm} y = {\varphi_2(t)}/{\sqrt{E}}\hspace{0.05cm}.$$

Questions

$\boldsymbol{K} \ = \ $

$p_{\rm S} \ = \ $

$\ \%$

	All messages $m_i$ are falsified in the same way.
	Conditional error probability ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_0) = 1/64$.
	Conditional error probability ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_1) = 0$.
	Conditional error probability ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_2) = 0$.

$p_{\rm S} \ = \ $

$\ \%$

$p_{\rm S} \ = \ $

$\ \%$

	Yes.
	No.

Solution

Noise regions with $A = 0.75$

(1) The volume of the two-dimensional PDF must give $p_n(x, y) =1$, that is:

$$2A \cdot 2A \cdot K = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \frac{1}{4A^2}\hspace{0.05cm}.$$

With $A = 0.75$ ⇒ $2A = 3/2$, we get $K = 4/9 \ \underline {=0.444}$.

(2) In the accompanying graph, the noise component $\boldsymbol{n}$ is plotted by the squares of edge length $1.5$ around the signal space points $\boldsymbol{s}_i$.

It can be seen that no decision boundary is exceeded by noise components.

It follows: The symbol error probability is $p_{\rm S}\ \underline { \equiv 0}$ under the conditions given here.

Noise regions with $A = 1$

(3) Statements 2 and 4 are correct, as can be seen from the second graph:

The message $m_2$ cannot be falsified because the square around $\boldsymbol{s}_2$ lies entirely in the lower right quadrant and thus in the decision region $I_2$.

Likewise, $m_2$ was sent with certainty if the received value lies in decision region $I_2$.
The reason: None of the squares around $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$ extends into the region $I_2$.

$m_0$ can only be falsified to $m_1$. The (conditional) falsification probability is equal to the ratio of the areas of the small yellow triangle $($area $1/16)$ and the square $($area $4)$:

$${\rm Pr}({ \cal E}\hspace{0.05cm}|\hspace{0.05cm} m_0 ) = \frac{1/2 \cdot 1/2 \cdot 1/4}{4}= {1}/{64} \hspace{0.05cm}.$$

For symmetry reasons, equally:

$${\rm Pr}({ \cal E}\hspace{0.05cm}|\hspace{0.05cm} m_1 ) = {\rm Pr}({ \cal E}\hspace{0.05cm}|\hspace{0.05cm} m_0 )={1}/{64} \hspace{0.05cm}. $$

(4) For equal probability symbols, we obtain for the (average) error probability:

$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \big [{\rm Pr}({ \cal E}\hspace{0.05cm}|\hspace{0.05cm} m_0 ) + {\rm Pr}({ \cal E}\hspace{0.05cm}|\hspace{0.05cm} m_1 )+{\rm Pr}({ \cal E}\hspace{0.05cm}|\hspace{0.05cm} m_2 )\big ]$$

$$ \Rightarrow \hspace{0.3cm} p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{3} \cdot \left [{1}/{64} + {1}/{64} + 0 )\right ]= \frac{2}{3 \cdot 64} = {1}/{96}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 1.04 \%} \hspace{0.05cm}.$$

(5) Now we obtain a smaller average error probability, viz.

$$p_{\rm S} = {\rm Pr}({ \cal E} ) = {1}/{4} \cdot {1}/{64} + {1}/{4} \cdot {1}/{64}+ {1}/{2} \cdot0 = {1}/{128}\hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.78 \% } \hspace{0.05cm}. $$

(6) Correct is YES:

For example, $I_1$: first quadrant, $I_0$: second quadrant, $I_2 \text{:} \ y < 0$ would give zero error probability.

This means that the given bounds are optimal only in the case of circularly symmetric PDF of the noise, for example, the AWGN model.

	All messages $m_i$ are falsified in the same way.
	Conditional error probability ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_0) = 1/64$.
	Conditional error probability ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_1) = 0$.
	Conditional error probability ${\rm Pr({ \cal E}} \hspace{0.05cm} \| \hspace{0.05cm} {\it m}_2) = 0$.