Exercise 4.17: Non-Coherent On-Off Keying

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

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

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

$G_{\rm opt} \ = \$

$p_{\rm S} \ = \$

$\ \%$

$G_{\rm opt} \ = \$

$p_{\rm S} \ = \$

$\ \%$

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: $0.33\%$ .