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

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Rayleigh and Rice PDF

The figure shows the two density functions resulting from a non-coherent demodulation of  "On–Off–Keying"  $\rm (OOK)$.  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  (red)  Rice distribution 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 graph  $C = 6$.

• 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 theHTML5/JavaScript applet  "Coherent and 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?

	$E_{\rm S} = C$,
	$E_{\rm S} = C^2$,
	$E_{\rm S} = C^2\hspace{-0.1cm}/2$.

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 threshold for  $C = 4$.

$G_{\rm opt} \ = \ $

4 What is the symbol error probability with  $C = 4$  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 subtask  (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 decision threshold 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  $m_1$  to  $m_0)$  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 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 subtask  (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:   $0.33\%$.