Aufgaben:Exercise 4.17: Non-Coherent On-Off Keying

• Return to book

Rayleigh and Rice distribution

The figure shows the two density functions resulting from a non-coherent demodulation of On–Off–Keying  (OOK). It is assumed that the two OOK signal space points are located at  $\boldsymbol{s}_0 = C$  $($message  $m_0)$  and at $\boldsymbol{s}_1 = 0$  $($message  $m_1)$. 

The symbol error probability of this system is described by the following equation:

$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}({\cal{E}}) = {1}/{ 2} \cdot \int_{0}^{G} p_{y\hspace{0.05cm}|\hspace{0.05cm}m} (\eta\hspace{0.05cm} | \hspace{0.05cm}m_0) \,{\rm d} \eta

+{1}/{ 2} \cdot \int_{G}^{\infty} p_{y\hspace{0.05cm}|\hspace{0.05cm}m} (\eta\hspace{0.05cm} |\hspace{0.05cm} m_1) \,{\rm d} \eta  
\hspace{0.05cm}.$$

With the standard deviation  $\sigma_n = 1$, which is assumed in the following, the resulting Rayleigh distribution for  $m = m_1$  (blue curve) is:

$$p_{y\hspace{0.05cm}|\hspace{0.05cm}m} (\eta\hspace{0.05cm} \hspace{0.05cm}| m_1) = \eta \cdot {\rm e }^{-\eta^2/2}

\hspace{0.05cm}.$$

The Rice distribution (red curve) can be approximated in the present case $($because of  $C\gg \sigma_n)$  by a Gaussian curve:

$$p_{y\hspace{0.05cm}|\hspace{0.05cm}m} (\eta\hspace{0.05cm} |\hspace{0.05cm} m_0) = \frac{1}{\sqrt{2\pi}} \cdot {\rm e }^{-(\eta-C)^2/2}

\hspace{0.05cm}.$$

The optimal decision boundary  $G_{\rm opt}$  is obtained from the intersection of the red and blue curves.

• From the two sketches it can be seen that  $G_{\rm opt}$  depends on  $C$. 
• For the upper graph  $C = 4$, for the lower  $C = 6$.
• All quantities are normalized and  $\sigma_n = 1$  is always assumed.

Notes:

• The exercise belongs to the topic of the chapter  "Carrier Frequency Systems with Non-Coherent Demodulation".
• For the complementary Gaussian error integral, you can use the following approximations:

$${\rm Q }(1.5) \approx 0.0668\hspace{0.05cm}, \hspace{0.5cm}{\rm Q }(2.5) \approx 0.0062\hspace{0.05cm}, \hspace{0.5cm}

{\rm Q }(2.65) \approx 0.0040 
\hspace{0.05cm}.$$

• You can check your results with the interactive applet  "Non-coherent On–Off–Keying". 

Questions

1 What is the relationship between the mean symbol energy  $E_{\rm S}$  and the constant  $C$  of the Rice distribution?

	$E_{\rm S} = C$,
	$E_{\rm S} = C^2$,
	$E_{\rm S} = C^2/2$.

2 What is the governing equation for the optimal decision boundary  $G_{\rm opt}$?

	$G = C/2$,
	$G \, –1/C \cdot {\rm ln} \, (G) = C/2 + 1/(2C) \cdot {\rm ln} \, (2\pi)$,
	$G \, –1/C \cdot {\rm ln} \, (G)$.

3 Determine the optimal decision boundary for  $C = 4$.

$G_{\rm opt} \ = \ $

4 What is the symbol error probability for  $C = 4$  and  $G = 2.5 \approx G_{\rm opt}$?

$p_{\rm S} \ = \ $ $\ \% $

5 Determine the optimal decision threshold for  $C = 6$.

$G_{\rm opt} \ = \ $

6 What is the symbol error probability with  $C = 6$  and  $G = 3.5\approx G_{\rm opt}$?

$p_{\rm S} \ = \ $ $\ \% $

Solution

Solution

(1)  Solution 3 is correct:

• The energy is equal to the value $\boldsymbol{s}_0 = C$ in the signal space constellation squared, divided by $2$.
• The factor $1/2$ takes into account that the message $m_1$ does not contribute any energy ($\boldsymbol{s}_1 = 0$).

(2)  Solution 2 is correct here:

• The optimal decision boundary $G$ lies at the intersection of the two curves shown.
• The factor $1/2$ considers the equally probable messages $m_0$ and $m_1$. Thus, the following determination equation is obtained:

$${G}/{2} \cdot {\rm exp } \left [ - {G^2 }/{2 }\right ] = \frac{1}{2 \cdot \sqrt{2\pi}} \cdot

{\rm exp } \left [ - \frac{G^2 - 2 C \cdot G + C^2}{2 }\right ]$$

$$\Rightarrow \hspace{0.3cm} \sqrt{2\pi} \cdot G = {\rm exp } \left [ C \cdot G - C^2/2 \right ]

\hspace{0.3cm}\Rightarrow \hspace{0.3cm} C \cdot G - {\rm ln }\hspace{0.15cm} (\sqrt{2\pi} \cdot G) - C^2/2 = 0$$

$$\Rightarrow \hspace{0.3cm} G - {1}/{C} \cdot {\rm ln }\hspace{0.15cm} ( G) = C/2 + {1}/({2C}) \cdot {\rm ln }\hspace{0.15cm} (\sqrt{2\pi}) = C/2 + {1}/({2C}) \cdot {\rm ln }\hspace{0.15cm} ({2\pi})\hspace{0.05cm}.$$

(3)  With $C = 4$, the governing equation given in (2) is

$$f(G) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} G - {1}/{C} \cdot {\rm ln }\hspace{0.15cm} ( G) - C/2 - {1}/({2C}) \cdot {\rm ln }\hspace{0.15cm} ({2\pi})= G - 0.25 \cdot {\rm ln }\hspace{0.15cm} ( G) - 2 - {\rm ln }\hspace{0.15cm} ({2\pi})/8

\approx G - 0.25 \cdot {\rm ln }\hspace{0.15cm} ( G) - 2.23 = 0
 \hspace{0.05cm}.$$

• This equation can only be solved numerically:

$$G = 2.0\text{:}\hspace{0.15cm}f(G) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} -0.403 \hspace{0.05cm}, \hspace{0.2cm}G = 3.0\text{:}\hspace{0.15cm}f(G) = 0.495

\hspace{0.05cm}, \hspace{0.2cm}G = 2.5\text{:}\hspace{0.15cm}f(G) = 0.041\hspace{0.05cm},$$

$$ G = 2.4\text{:}\hspace{0.15cm}f(G) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} -0.049 \hspace{0.05cm}, \hspace{0.2cm}G = 2.46\text{:}\hspace{0.15cm}f(G) \approx 0

 \hspace{0.05cm}.$$

• Thus, the optimal decider boundary is $G_{\rm opt} \underline {= 2.46 \approx 2.5}$.

(4)  The error probability is composed of two parts:

$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {1}/{ 2} \cdot {\rm Pr}({\cal{E}}\hspace{0.05cm}| \hspace{0.05cm} m = m_1)+{1}/{ 2}\cdot {\rm Pr}({\cal{E}}\hspace{0.05cm}| \hspace{0.05cm} m = m_0)\hspace{0.05cm}.$$

• The first part (falsification from $m_1$ nach $m_0$) results from the crossing of the limit $G$ by the Rayleigh distribution

$${\rm Pr}({\cal{E}} \hspace{0.05cm}| \hspace{0.05cm} m = m_1) = \int_{G}^{\infty} p_{y\hspace{0.05cm}| \hspace{0.05cm}m} (\eta \hspace{0.05cm}| \hspace{0.05cm} m_1) \,{\rm d} \eta = {\rm e }^{-G^2/2}= {\rm e }^{-3.125}\approx 0.044

\hspace{0.05cm}.$$

• The second part (falsification from $m_0$ to $m_1$) results from the Rice distribution, which is approximated here by the Gaussian distribution:

$${\rm Pr}({\cal{E}}| m = m_0) = \int_{0}^{G} p_{y\hspace{0.05cm}| \hspace{0.05cm}m} (\eta \hspace{0.05cm}| \hspace{0.05cm} m_0) \,{\rm d} \eta =

  \frac{1}{\sqrt{2\pi}} \cdot \int_{0}^{G} {\rm e }^{-(\eta-C)^2/2} \,{\rm d} \eta 
\hspace{0.05cm}.$$

• This part can be given by the complementary Gaussian error integral ${\rm Q}(x)$:

$${\rm Pr}({\cal{E}}\hspace{0.05cm}| \hspace{0.05cm} m = m_0) = {\rm Pr}(y < G-C) = {\rm Pr}(y > C-G) = {\rm Q }(\frac{C-G}{\sigma_n})= {\rm Q }(\frac{4-2.5}{1})= {\rm Q }(1.5) \approx 0.0688

\hspace{0.05cm}.  $$

• This gives a total of:

$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {1}/{ 2} \cdot 0.0440 +{1}/{ 2} \cdot 0.0668 \approx \underline{5.54\, \%}\hspace{0.05cm}.$$

Note:   A system simulation has shown that a slightly smaller error probability results if the actual Rice distribution is used instead of the Gaussian approximation. Then with $G = 2.5$:

$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {1}/{ 2} \cdot 0.0440 + {1}/{ 2} \cdot 0.0484 \approx \underline{4.62\, \%}\hspace{0.05cm}.$$

Thus, the Gaussian approximation provides an upper bound on the true error probability.

(5)  With  $C = 6$,  the governing equation given in (3) is

$$f(G)= G - {1}/{C} \cdot {\rm ln }\hspace{0.15cm} ( G) - C/2 - \frac{1}{2C} \cdot {\rm ln }\hspace{0.15cm} ({2\pi}) \approx G - {\rm ln }\hspace{0.15cm} ( G)/6 - 3.153 = 0

 \hspace{0.05cm},$$

$$G = 3.0\hspace{-0.1cm}:\hspace{0.15cm}f(G) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} -0.336 \hspace{0.05cm}, \hspace{0.2cm}G = 3.50\hspace{-0.1cm}:\hspace{0.15cm}f(G) = 0.138

\hspace{0.05cm},$$

$$ G = 3.3\hspace{-0.1cm}:\hspace{0.15cm}f(G) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} -0.052 \hspace{0.05cm}, \hspace{0.2cm}G = 3.35\hspace{-0.1cm}:\hspace{0.15cm}f(G) \approx 0

\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{G_{\rm opt} \approx 3.35}\hspace{0.05cm}.$$

(6)  Analogous to subtask (4), we obtain with $G = 3.5$:

$$p_{\rm S} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}({\cal{E}}) = {1}/{ 2} \cdot {\rm e }^{-G^2/2} +{1}/{ 2} \cdot {\rm Q }(C-G)= {1}/{ 2} \cdot {\rm e }^{-6.125} + {1}/{ 2} \cdot {\rm Q }(2.5)=

{1}/{ 2} \cdot 2.2 \cdot 10^{-3} + {1}/{ 2} \cdot 6.2 \cdot 10^{-3} \underline{= 0.42 \,\%}
\hspace{0.05cm}.$$

• For  $C = 6$,  the optimal decision boundary ($G_{\rm opt} = 3.35$) results in an error probability that is about a factor of $10$ smaller than with $C = 4$:

$$p_{\rm S} = {1}/{ 2} \cdot {\rm e }^{-5.61} + {1}/{ 2} \cdot {\rm Q }(2.65)=

{1}/{ 2} \cdot 3.6 \cdot 10^{-3} +{1}/{ 2} \cdot 4 \cdot 10^{-3}= {0.38 \,\%}

\hspace{0.05cm}.$$

• The actual error probability using the Rice distribution (no Gaussian approximation) gives a slightly smaller value:   $0.33\%$.