Exercise 2.5: Ternary Signal Transmission

Probability density function $\rm (PDF)$ of a noisy ternary signal

A ternary transmission system $(M = 3)$ with the possible amplitude values $-s_0$, $0$ and $+s_0$ is considered.

During transmission, additive Gaussian noise with rms value $\sigma_d$ is added to the signal.

The recovery of the three-level digital signal at the receiver is done with the help of two decision thresholds at $E_{–}$ and $E_{+}$.

First, the occurrence probabilities of the three input symbols are assumed to be equally probable:

$$p_{\rm -} = {\rm Pr}(-s_0) = {1}/{ 3}, \hspace{0.15cm} p_{\rm 0} = {\rm Pr}(0) = {1}/{ 3}, \hspace{0.15cm} p_{\rm +} = {\rm Pr}(+s_0) ={1}/{ 3}\hspace{0.05cm}.$$

For the time being, the decision thresholds are centered at $E_{–} = \, –s_0/2$ and $E_{+} = +s_0/2$.

From subtask (3) on, the symbol probabilities are $p_{–} = p_+ = 1/4$ and $p_0 = 1/2$, as shown in the diagram.

For this constellation, the symbol error probability $p_{\rm S}$ is to be minimized by varying the decision thresholds $E_{–}$ and $E_+$.

Notes:

The exercise refers to the chapter "Redundancy-Free Coding".

For the symbol error probability $p_{\rm S}$ of a $M$–level transmission system

with equally probable input symbols
and threshold values exactly in the middle between two adjacent amplitude levels holds:

$$p_{\rm S} = \frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot \sigma_d}}\right) \hspace{0.05cm}.$$

You can numerically determine the error probability values according to our applet "Complementary Gaussian Error Functions".

To check your results, use our (German language) SWF applet "Symbol error probability of digital communications systems".

Questions

$p_0 = 1/3, \ \sigma_d = 0.25 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $

$\ \%$

$p_0 = 1/3, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $

$\ \%$

$p_0 = 1/2, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $

$\ \%$

$p_0 = 1/2, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} E_{\rm +, \ opt} \ = \ $

${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $

$\ \%$

$p_0 = 0.2, \ \sigma_d = 0.5 \text{:} \hspace{0.4cm} E_{\rm +, \ opt} \ = \ $

${\rm optimal \ thresholds} \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $

$\ \%$

Solution

(1) According to the given equation, with $M = 3$ and $\sigma_d/s_0 = 0.25$:

$$p_{\rm S} = \frac{ 2 \cdot (M-1)}{M} \cdot {\rm Q} \left( {\frac{s_0}{(M-1) \cdot \sigma_d}}\right)= {4}/{ 3}\cdot {\rm Q}(2) ={4}/{ 3}\cdot 0.0228\hspace{0.15cm}\underline {\approx 3 \,\%} \hspace{0.05cm}.$$

(2) When the noise rms value is doubled, the error probability increases significantly:

$$p_{\rm S} = {4}/{ 3}\cdot {\rm Q}(1)= {4}/{ 3}\cdot 0.1587 \hspace{0.15cm}\underline {\approx 21.2 \,\%} \hspace{0.05cm}.$$

(3) The two outer symbols are each falsified with probability $p = {\rm Q}(s_0/(2 \cdot \sigma_d)) = 0.1587$.

The falsification probability of the symbol $0$ is twice as large (it is limited by two thresholds).
Considering the individual symbol probabilities, we obtain:

$$p_{\rm S} = {1}/{ 4}\cdot p + {1}/{ 2}\cdot 2p +{1}/{ 4}\cdot p = 1.5 \cdot p = 1.5 \cdot 0.1587 \hspace{0.15cm}\underline {\approx 23.8 \,\%} \hspace{0.05cm}.$$

(4) Since the symbol $0$ occurs more frequently and can also be biased in both directions, the thresholds should be shifted outward.

The optimal decision threshold $E_{\rm +, \ opt}$ is obtained from the intersection of the two Gaussian functions shown in the graph. It must hold:

Optimal thresholds for subtask (4)

$$\frac{ 1/2}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ E_{\rm +}^2}{2 \cdot \sigma_d^2}\right] = \frac{ 1/4}{ \sqrt{2\pi} \cdot \sigma_d} \cdot {\rm exp} \left[ - \frac{ (s_0 -E_{\rm +})^2}{2 \cdot \sigma_d^2}\right]$$

$$\Rightarrow \hspace{0.3cm} {\rm exp} \left[ \frac{ (s_0 -E_{\rm +})^2 - E_{\rm +}^2}{2 \cdot \sigma_d^2}\right]= {1}/{ 2} \Rightarrow \hspace{0.3cm} {\rm exp} \left[ \frac{ 1 -2 \cdot E_{\rm +}/s_0}{2 \cdot \sigma_d^2/s_0^2}\right]= {1}/{ 2}$$

$$\Rightarrow \hspace{0.3cm}\frac{ E_{\rm +}}{s_0}= \frac{1} { 2}+ \frac{\sigma_d^2} {s_0^2} \cdot {\rm ln}(2)\hspace{0.15cm}\underline {=0.673}\hspace{0.15cm}\approx {2}/ {3} \hspace{0.05cm}.$$

(5) Using the approximate result from subtask (4), we obtain:

$$p_{\rm S} \ = \ { 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{ \sigma_d}}\right)+ 2 \cdot { 1}/{2} \cdot {\rm Q} \left( {\frac{2s_0/3}{ \sigma_d}}\right) +{ 1}/{4} \cdot {\rm Q} \left( {\frac{s_0/3}{ \sigma_d}}\right)$$

Optimal thresholds for subtask (6)

$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \ = { 1}/{2} \cdot {\rm Q} \left( 2/3 \right)+ {\rm Q} \left( 4/3 \right)= { 1}/{2} \cdot 0.251 + 0.092 \hspace{0.15cm}\underline {\approx 21.7 \,\%} \hspace{0.05cm}.$$

(6) After a similar calculation as in subtask (4) we get

$E_+ = 1 \, –0.0673 \ \underline{= 0.327} \approx 1/3$.
$E_{–} = \, –E_+$ is still valid.

(7) Similar to the solution for subtask (5), we now obtain:

$$p_{\rm S} \ = \ 0.4 \cdot {\rm Q} \left( 4/3 \right)+ 2 \cdot 0.2 \cdot{\rm Q} \left( 2/3 \right)+0.4 \cdot {\rm Q} \left( 4/3 \right)$$

$$\Rightarrow \hspace{0.3cm}p_{\rm S} \ = \ 0.4 \cdot (0.092 + 0.251 + 0.092) \hspace{0.15cm}\underline {\approx 17.4 \,\%} \hspace{0.05cm}.$$

$\text{Discussion of the result:}$

Accordingly, there is a smaller symbol error probability $(17.4 \ \%$ versus $21.2 \ \%)$ than with equal probability amplitude coefficients.

However, redundancy-free coding is no longer present, even if the amplitude coefficients are statistically independent of each other.

While for equally probable ternary symbols the entropy is $H = {\rm log}_2(3) = 1.585 \ {\rm bit/ternary \ symbol}$.

The equivalent bit rate (the information flow) $R_{\rm B} = H/T$, with probabilities $p_0 = 0.2$ and $p_{–} = p_+ = 0.4$:

$$H \ = \ 0.2 \cdot {\rm log_2} (5) + 2 \cdot 0.4 \cdot {\rm log_2} (2.5)= 0.2 \cdot 2.322 + 0.8 \cdot 1.322 \hspace{0.15cm}\underline {\approx 1.522\,\, {\rm bit/ternary \ symbol}} \hspace{0.05cm}.$$

Thus, the equivalent bit rate here is $\approx 4 \ \%$ smaller than the maximum possible equivalent bit rate for $M = 3$.