Difference between revisions of "Aufgaben:Exercise 4.11: On-Off Keying and Binary Phase Shift Keying"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}} |
− | [[File:P_ID2060__Dig_A_4_11.png|right|frame| | + | [[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: | |
− | * | + | * "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 <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 | |
− | * $d$ | + | * $d$ 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 $E_{\rm S}$. | ||
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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". | ||
− | * | + | *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=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {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% } | ||
$M \ = \ $ { 2 3% } | $M \ = \ $ { 2 3% } | ||
− | { | + | {Which representation do the signal space constellations show? |
|type="()"} | |type="()"} | ||
− | - | + | - The representation in the (actual) band-pass range, |
− | + | + | + the representation in the (equivalent) low-pass range. |
− | { | + | {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 "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% } $\ \cdot 10^{\rm –8}$ | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Both "'On–Off Keying" $\rm (OOK)$ and "Binary Phase Shift Keying" $\rm (BPSK)$ are binary modulation methods: |
:$$\underline{b = 1 }\hspace{0.05cm},\hspace{0.5cm} \underline{M = 2} \hspace{0.05cm}.$$ | :$$\underline{b = 1 }\hspace{0.05cm},\hspace{0.5cm} \underline{M = 2} \hspace{0.05cm}.$$ | ||
− | '''(2)''' | + | '''(2)''' <u>Solution 2</u> 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)''' | + | '''(3)''' The given equation is for "On–Off Keying" with |
− | *$d = \sqrt {E}$, | + | *$d = \sqrt {E}$, |
− | *$E_{\rm S} = E/2$ ( | + | |
+ | *$E_{\rm S} = E/2$ (assuming equally probable symbols $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$), | ||
+ | |||
*$\sigma_n^2 = N_0/2$: | *$\sigma_n^2 = N_0/2$: | ||
:$$p_{\rm S} \hspace{-0.1cm} = \hspace{-0.1cm} {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right )= {\rm Q} \left ( \frac{ \sqrt{E}/2}{ \sqrt{N_0/2}}\right ) = {\rm Q} \left ( \sqrt{ \frac{ E/2}{ N_0} }\right ) = {\rm Q} \left ( \sqrt{ { E_{\rm S}}/{ N_0} }\right ) | :$$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}.$$ | \hspace{0.05cm}.$$ | ||
− | * | + | *For $E_{\rm S}/N_0 = 9 = 3^2$ this results in: |
:$$p_{\rm S} = {\rm Q} (3) \approx \frac{1}{\sqrt{2\pi} \cdot 3} \cdot {\rm e}^{-9/2} = \underline{14.8 \cdot 10^{-4}} | :$$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}.$$ | \hspace{0.05cm}.$$ | ||
− | * | + | *Accordingly, for $10 \cdot {\rm lg} \, (E_{\rm S}/N_0) = 12 \ \rm dB$ ⇒ $E_{\rm S}/N_0 = 15.85$: |
:$$p_{\rm S} = {\rm Q} (\sqrt{15.85}) \approx \frac{1}{\sqrt{2\pi\cdot 15.85} } \cdot {\rm e}^{-15.85/2} = \underline{0.362 \cdot 10^{-4}} | :$$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}.$$ | \hspace{0.05cm}.$$ | ||
− | '''(4)''' | + | '''(4)''' In contrast to subtask '''(3)''', "Binary Phase Shift Keying" $\rm (BPSK)$ applies |
− | *$d = 2 \cdot \sqrt {E}$, | + | *$d = 2 \cdot \sqrt {E}$, |
+ | |||
*$E_{\rm S} = E$, | *$E_{\rm S} = E$, | ||
− | + | both even independent of the occurrence probabilities for $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$. | |
− | * | + | |
+ | *It follows: | ||
:$$p_{\rm S} = {\rm Q} \left ( \frac{ \sqrt{E_{\rm S}}}{ \sqrt{N_0/2}}\right ) = {\rm Q} \left ( \sqrt{ { 2E_{\rm S}}/{ N_0} }\right ) | :$$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}.$$ | \hspace{0.05cm}.$$ | ||
− | * | + | *With $E_{\rm S}/N_0 = 9$, this results in the numerical value: |
:$$p_{\rm S} = {\rm Q} (\sqrt{18}) \approx \frac{1}{\sqrt{2\pi\cdot 18} } \cdot {\rm e}^{-18/2} = \underline{117 \cdot 10^{-8}} \hspace{0.05cm}.$$ | :$$p_{\rm S} = {\rm Q} (\sqrt{18}) \approx \frac{1}{\sqrt{2\pi\cdot 18} } \cdot {\rm e}^{-18/2} = \underline{117 \cdot 10^{-8}} \hspace{0.05cm}.$$ | ||
− | * | + | *And with $10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ \rm dB$ ⇒ $2E_{\rm S}/N_0 = 31.7$: |
:$$p_{\rm S} = {\rm Q} (\sqrt{31.7}) \approx \frac{1}{\sqrt{2\pi\cdot 31.7} } \cdot {\rm e}^{-31.7/2} = \underline{0.926 \cdot 10^{-8}}\hspace{0.05cm}.$$ | :$$p_{\rm S} = {\rm Q} (\sqrt{31.7}) \approx \frac{1}{\sqrt{2\pi\cdot 31.7} } \cdot {\rm e}^{-31.7/2} = \underline{0.926 \cdot 10^{-8}}\hspace{0.05cm}.$$ | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category:Digital Signal Transmission: Exercises|^4.4 | + | [[Category:Digital Signal Transmission: Exercises|^4.4 Coherent Demodulation^]] |
Latest revision as of 20:48, 1 September 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" $\rm (OOK)$ 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 $10 \cdot {\rm lg} \, E_{\rm S}/N_0 = 12 \ \rm dB$ ⇒ $2E_{\rm S}/N_0 = 31.7$:
- $$p_{\rm S} = {\rm Q} (\sqrt{31.7}) \approx \frac{1}{\sqrt{2\pi\cdot 31.7} } \cdot {\rm e}^{-31.7/2} = \underline{0.926 \cdot 10^{-8}}\hspace{0.05cm}.$$