Exercise 4.11: On-Off Keying and Binary Phase Shift Keying

Two signal space constellations for OOK and BPSK

The graphic shows signal space constellations for carrier-modulated modulation methods:

"On–off keying" $\rm (OOK)$, also known as "Amplitude Shift Keying" $\rm (ASK)$ in some books,

"Binary Phase Shift Keying" $\rm (BPSK)$.

For the error probability calculation we start from the AWGN channel. In this case the error probability is
$($related to symbols or to bits alike$)$:

$$p_{\rm S} = p_{\rm B} = {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right ) \hspace{0.05cm}.$$

Here

$d$ denotes the distance between the signal space points, and

$\sigma_n^2 = N_0/2$ the variance of the AWGN noise.

In the questions from (3) onwards, reference is also made to the mean symbol energy $E_{\rm S}$.

Notes:

The exercise belongs to the chapter "Carrier Frequency Systems with Coherent Demodulation".

Reference is also made to the chapter "Linear Digital Modulation – Coherent Demodulation" and the chapter "Linear Digital Modulation" in the book "Modulation Methods".

For the complementary Gaussian error function, use the following approximation:

$${\rm Q}(x) \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2} \hspace{0.05cm}.$$

Questions

$b \hspace{0.35cm} = \ $

$M \ = \ $

	The representation in the (actual) band-pass range,
	the representation in the (equivalent) low-pass range.

$E_{\rm S}/N_0 = 9 \text{:} \hspace{2.3cm} p_{\rm S} \ = \ $

$\ \cdot 10^{\rm –4}$

$10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ {\rm dB} \text{:} \hspace{0.2cm} p_{\rm S} \ = \ $

$\ \cdot 10^{\rm –4}$

$E_{\rm S}/N_0 = 9 \text{:} \hspace{2.3cm} p_{\rm S} \ = \ $

$\ \cdot 10^{\rm –8}$

$10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ {\rm dB} \text{:} \hspace{0.2cm} p_{\rm S} \ = \ $

$\ \cdot 10^{\rm –8}$

Solution

(1) Both "'On–off Keying" $\rm (OOK)$ and "Binary Phase Shift Keying" $\rm (BPSK)$ are binary modulation methods:

$$\underline{b = 1 }\hspace{0.05cm},\hspace{0.5cm} \underline{M = 2} \hspace{0.05cm}.$$

(2) Solution 2 is correct, recognizable by the imaginary basis function $\varphi_2(t) = {\rm j} \cdot \varphi_1(t)$.

If described in the band-pass range, the basis functions would be real: "cosine" and "(minus) sine".

(3) The given equation is for "On–off Keying" with

$d = \sqrt {E}$,

$E_{\rm S} = E/2$ (assuming equally probable symbols $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$),

$\sigma_n^2 = N_0/2$:

$$p_{\rm S} \hspace{-0.1cm} = \hspace{-0.1cm} {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right )= {\rm Q} \left ( \frac{ \sqrt{E}/2}{ \sqrt{N_0/2}}\right ) = {\rm Q} \left ( \sqrt{ \frac{ E/2}{ N_0} }\right ) = {\rm Q} \left ( \sqrt{ { E_{\rm S}}/{ N_0} }\right ) \hspace{0.05cm}.$$

For $E_{\rm S}/N_0 = 9 = 3^2$ this results in:

$$p_{\rm S} = {\rm Q} (3) \approx \frac{1}{\sqrt{2\pi} \cdot 3} \cdot {\rm e}^{-9/2} = \underline{14.8 \cdot 10^{-4}} \hspace{0.05cm}.$$

Accordingly, for $10 \cdot {\rm lg} \, (E_{\rm S}/N_0) = 12 \ \rm dB$ ⇒ $E_{\rm S}/N_0 = 15.85$:

$$p_{\rm S} = {\rm Q} (\sqrt{15.85}) \approx \frac{1}{\sqrt{2\pi\cdot 15.85} } \cdot {\rm e}^{-15.85/2} = \underline{0.362 \cdot 10^{-4}} \hspace{0.05cm}.$$

(4) In contrast to subtask (3), "Binary Phase Shift Keying" $\rm (BPSK)$ applies

$d = 2 \cdot \sqrt {E}$,

$E_{\rm S} = E$,

both even independent of the occurrence probabilities for $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$.

It follows:

$$p_{\rm S} = {\rm Q} \left ( \frac{ \sqrt{E_{\rm S}}}{ \sqrt{N_0/2}}\right ) = {\rm Q} \left ( \sqrt{ { 2E_{\rm S}}/{ N_0} }\right ) \hspace{0.05cm}.$$

With $E_{\rm S}/N_0 = 9$, this results in the numerical value:

$$p_{\rm S} = {\rm Q} (\sqrt{18}) \approx \frac{1}{\sqrt{2\pi\cdot 18} } \cdot {\rm e}^{-18/2} = \underline{117 \cdot 10^{-8}} \hspace{0.05cm}.$$

And with $10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ \rm dB$ ⇒ $2E_{\rm S}/N_0 = 31.7$:

$$p_{\rm S} = {\rm Q} (\sqrt{31.7}) \approx \frac{1}{\sqrt{2\pi\cdot 31.7} } \cdot {\rm e}^{-31.7/2} = \underline{0.926 \cdot 10^{-8}}\hspace{0.05cm}.$$