Difference between revisions of "Aufgaben:Exercise 4.11: On-Off Keying and Binary Phase Shift Keying"
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[[File:P_ID2060__Dig_A_4_11.png|right|frame|Two signal space constellations for OOK and BPSK]] | [[File:P_ID2060__Dig_A_4_11.png|right|frame|Two signal space constellations for OOK and BPSK]] | ||
The graphic shows signal space constellations for carrier-modulated modulation methods: | 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 | + | |
+ | For the error probability calculation we start from the AWGN channel. In this case the error probability is <br>$($related to symbols or to bits alike$)$: | ||
:$$p_{\rm S} = p_{\rm B} = {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right ) | :$$p_{\rm S} = p_{\rm B} = {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right ) | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
Here | Here | ||
− | * $d$ denotes the distance between the signal space points, and | + | * $d$ denotes the distance between the signal space points, and |
+ | |||
* $\sigma_n^2 = N_0/2$ the variance of the AWGN noise. | * $\sigma_n^2 = N_0/2$ the variance of the AWGN noise. | ||
− | In the questions from '''(3)''' onwards, reference is also made to the mean symbol energy $E_{\rm S}$. | + | In the questions from '''(3)''' onwards, reference is also made to the mean symbol energy $E_{\rm S}$. |
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+ | Notes: | ||
+ | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]]. | ||
− | + | *Reference is also made to the chapter [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation|"Linear Digital Modulation – Coherent Demodulation"]] and the chapter [[Modulation_Methods/Linear_Digital_Modulation|"Linear Digital Modulation"]] in the book "Modulation Methods". | |
− | |||
− | *Reference is also made to the chapter [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation|"Linear Digital Modulation – Coherent Demodulation"]] and the chapter [[Modulation_Methods/Linear_Digital_Modulation|"Linear Digital Modulation"]] in the book "Modulation Methods". | ||
− | *For the complementary Gaussian error function, use the following approximation: | + | *For the complementary Gaussian error function, use the following approximation: |
:$${\rm Q}(x) \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2} | :$${\rm Q}(x) \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
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===Questions=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | {How many bits $(b)$ does each symbol represent? What is the level number $M$? | + | {How many bits $(b)$ does each symbol represent? What is the level number $M$? |
|type="{}"} | |type="{}"} | ||
$b \hspace{0.35cm} = \ $ { 1 3% } | $b \hspace{0.35cm} = \ $ { 1 3% } | ||
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+ the representation in the (equivalent) low-pass range. | + the representation in the (equivalent) low-pass range. | ||
− | {What error probability results for | + | {What error probability results for "On–Off Keying" depending on $E_{\rm S}/N_0$? |
|type="{}"} | |type="{}"} | ||
$E_{\rm S}/N_0 = 9 \text{:} \hspace{2.3cm} p_{\rm S} \ = \ $ { 14.8 3% } $\ \cdot 10^{\rm –4}$ | $E_{\rm S}/N_0 = 9 \text{:} \hspace{2.3cm} p_{\rm S} \ = \ $ { 14.8 3% } $\ \cdot 10^{\rm –4}$ | ||
$10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ {\rm dB} \text{:} \hspace{0.2cm} p_{\rm S} \ = \ $ { 0.362 3% } $\ \cdot 10^{\rm –4}$ | $10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ {\rm dB} \text{:} \hspace{0.2cm} p_{\rm S} \ = \ $ { 0.362 3% } $\ \cdot 10^{\rm –4}$ | ||
− | {What is the error probability for | + | {What is the error probability for "Binary Phase Shift Keying" depending on $E_{\rm S}/N_0$? |
|type="{}"} | |type="{}"} | ||
$E_{\rm S}/N_0 = 9 \text{:} \hspace{2.3cm} p_{\rm S} \ = \ $ { 117 3% } $\ \cdot 10^{\rm –8}$ | $E_{\rm S}/N_0 = 9 \text{:} \hspace{2.3cm} p_{\rm S} \ = \ $ { 117 3% } $\ \cdot 10^{\rm –8}$ |
Revision as of 14:17, 20 August 2022
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
Solution
(1) Both On–Off Keying (OOK) and Binary Phase Shift Keying (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 cosine and (minus)sine real.
(3) The given equation is for On–Off Keying (OOK) with
- $d = \sqrt {E}$,
- $E_{\rm S} = E/2$ (assuming equally probable symbols $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$) and
- $\sigma_n^2 = N_0/2$:
- $$p_{\rm S} \hspace{-0.1cm} = \hspace{-0.1cm} {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right )= {\rm Q} \left ( \frac{ \sqrt{E}/2}{ \sqrt{N_0/2}}\right ) = {\rm Q} \left ( \sqrt{ \frac{ E/2}{ N_0} }\right ) = {\rm Q} \left ( \sqrt{ { E_{\rm S}}/{ N_0} }\right ) \hspace{0.05cm}.$$
- 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 (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 $10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ \rm dB$ ⇒ $2E_{\rm S}/N_0 = 31.7$:
- $$p_{\rm S} = {\rm Q} (\sqrt{31.7}) \approx \frac{1}{\sqrt{2\pi\cdot 31.7} } \cdot {\rm e}^{-31.7/2} = \underline{0.926 \cdot 10^{-8}}\hspace{0.05cm}.$$