Difference between revisions of "Aufgaben:Exercise 4.16: Comparison between Binary PSK and Binary FSK"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Modulationsverfahren/Nichtlineare_digitale_Modulation |
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− | [[File: | + | [[File:EN_Mod_A_4_15_neu.png|right|frame|Bit error probability curves: <br>Binary PSK and binary FSK]] |
− | + | The graph shows the bit error probability for binary [[Modulation_Methods/Nonlinear_Digital_Modulation#FSK_.E2.80.93_Frequency_Shift_Keying| FSK modulation]] $\rm (BFSK)$ in | |
− | + | *[[Modulation_Methods/Nonlinear_Digital_Modulation#Coherent_demodulation_of_FSK|coherent demodulation]] , as well as in | |
+ | *[[Modulation_Methods/Nonlinear_Digital_Modulation#Error_probability_of_orthogonal_FSK|non-coherent demodulation]] | ||
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− | + | in comparison with [[Modulation_Methods/Lineare_digitale_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying|binary phase modulation]] $\rm (BPSK)$. | |
+ | Orthogonality is always assumed. For coherent demodulation, the modulation index can be a multiple of $h = 0.5$ , so that the middle plot can also apply to [[Modulation_Methods/Nonlinear_Digital_Modulation#MSK_.E2.80.93_Minimum_Shift_Keying|Minimum Shift Keying]] $\rm (MSK)$ . In contrast, for non-coherent demodulation of BFSK, the modulation index must be a multiple of $h = 1$ . | ||
− | === | + | This system comparison is once again based on the [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|AWGN channel]] , characterized by the relationship $E_{\rm B}/N_0$. The equations for the bit error probabilities are as follows |
+ | * for ''Binary Phase Shift Keying'' $\rm (BPSK)$: | ||
+ | :$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$ | ||
+ | * for ''Binary Frequency Shift Keying'' $\rm (BFSK)$ with ''coherent'' demodulation: | ||
+ | :$$p_{\rm B} = {\rm Q}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/(2 N_0 )} \hspace{0.1cm}\right ),$$ | ||
+ | * for ''Binary Frequency Shift Keying'' $\rm (BFSK)$ with ''incoherent'' demodulation: | ||
+ | :$$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$ | ||
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+ | It was shown in [[Aufgaben:Exercise_4.8:_Different_Error_Probabilities|Exercise 4.8]] , that for BPSK, the log ratio $10 · \lg \ E_{\rm B}/N_0$ must be at least $9.6 \ \rm dB$ so that the bit error probability does not exceed $p_{\rm B} = 10^{–5}$ . | ||
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+ | ''Hints:'' | ||
+ | *This exercise belongs to the chapter [[Modulation_Methods/Nonlinear_Digital_Modulation|Nonlinear_Digital_Modulation]]. | ||
+ | *However, reference is also made to the page [[Modulation_Methods/Lineare_digitale_Modulation|Linear Digital Modulation]]. | ||
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+ | *Use the approximation $\lg(2) ≈ 0.3$. | ||
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+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What $E_{\rm B}/N_0$ (in dB) is required in for MSK and coherent demodulation to satisfy $p_{\rm B} \le 10^{–5}$ ? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $10 · \lg \ E_{\rm B}/N_0 \ = \ $ { 12.6 3% } $\ \rm dB$ |
− | { | + | {Which of the following statements is correct: The same result is obtained for |
|type="[]"} | |type="[]"} | ||
− | - | + | - FSK with modulation index $h = 0.7$, |
− | + | + | + FSK with modulation index $h = 1$? |
− | { | + | {What $E_{\rm B}/N_0$ (in dB) is required for BFSK with $h = 1$ and incoheren demodulation to satisfy $p_{\rm B} \le 10^{–5}$ ? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $10 · \lg \ E_{\rm B}/N_0 \ = \ $ { 13.4 3% } $\ \rm dB$ |
− | { | + | {What is the bit error probability $p_{\rm B}$ results from incoherent BFSK demodulation when $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$? |
|type="{}"} | |type="{}"} | ||
− | $ | + | $p_{\rm B} \ = \ $ { 1.12 3% } $\ \cdot 10^{-4}$ |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''1 | + | '''(1)''' A comparison of the first two equations on the exercise page makes it clear that for MSK with coherent demodulation, the AWGN ratio $E_{\rm B}/N_0$ must be doubled to achieve the same error probability as for BPSK. |
− | $$10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}} /{N_{\rm 0}}= 9.6\,\,{\rm dB} + 3\,\,{\rm dB} = \underline{12.6\,\,{\rm dB}}\hspace{0.05cm}.$$ | + | |
+ | *In other words: the coherent BFSK curve is $10 · \lg (2) ≈ 3 \ \rm dB$ to the right of the the BPSK curve. | ||
+ | *Thus, to guarantee $p_{\rm B} \le 10^{–5}$ it must hold that: | ||
+ | :$$10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}} /{N_{\rm 0}}= 9.6\,\,{\rm dB} + 3\,\,{\rm dB} = \underline{12.6\,\,{\rm dB}}\hspace{0.05cm}.$$ | ||
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+ | |||
+ | '''(2)''' <u>Answer 2</u> is correct: | ||
+ | *The equation given does not just hold for MSK $($this is a FSK with $h = 0.5)$, but also for every form of orthogonal FSK. | ||
+ | *This is the case whenever the modulation index $h$ is an integer multiple of $0.5$ , such as when $h = 1$. | ||
+ | *When $h = 0.7$ , there is no orthogonal FSK. | ||
+ | |||
− | ''' | + | '''(3)''' From the inverse function of the equation given we get: |
+ | :$$\frac{E_{\rm B}} {2 \cdot N_{\rm 0}}= {\rm ln}\hspace{0.05cm}\frac{1}{2 p_{\rm B}}= {\rm ln}(50000)\approx 10.82 \hspace{0.3cm} | ||
+ | \Rightarrow \hspace{0.3cm}{E_{\rm B}} /{N_{\rm 0}}= 21.64 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}}/ {N_{\rm 0}}\approx \underline{13.4\,\,{\rm dB}}\hspace{0.05cm}.$$ | ||
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− | '''4 | + | '''(4)''' From $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$ it follows that: |
− | $${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.25cm}\Rightarrow \hspace{0.25cm} ({E_{\rm B}} /{N_{\rm 0}})/2 \approx 8.4 \hspace{0.25cm} \Rightarrow \hspace{0.25cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4} \approx \underline{1.12 \cdot 10^{-4}}\hspace{0.05cm}.$$ | + | :$${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.25cm}\Rightarrow \hspace{0.25cm} ({E_{\rm B}} /{N_{\rm 0}})/2 \approx 8.4 \hspace{0.25cm} \Rightarrow \hspace{0.25cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4} \approx \underline{1.12 \cdot 10^{-4}}\hspace{0.05cm}.$$ |
− | + | This means that for the same $E_{\rm B}/N_0$, the error probability of incoherent demodulation is increased by a factor of $11$ compared to that of coherent demodulation (see answer to question '''1''') . | |
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Modulation Methods: Exercises|^4.4 Non-linear Digital Modulation^]] |
Latest revision as of 11:52, 12 April 2022
The graph shows the bit error probability for binary FSK modulation $\rm (BFSK)$ in
- coherent demodulation , as well as in
- non-coherent demodulation
in comparison with binary phase modulation $\rm (BPSK)$.
Orthogonality is always assumed. For coherent demodulation, the modulation index can be a multiple of $h = 0.5$ , so that the middle plot can also apply to Minimum Shift Keying $\rm (MSK)$ . In contrast, for non-coherent demodulation of BFSK, the modulation index must be a multiple of $h = 1$ .
This system comparison is once again based on the AWGN channel , characterized by the relationship $E_{\rm B}/N_0$. The equations for the bit error probabilities are as follows
- for Binary Phase Shift Keying $\rm (BPSK)$:
- $$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$
- for Binary Frequency Shift Keying $\rm (BFSK)$ with coherent demodulation:
- $$p_{\rm B} = {\rm Q}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/(2 N_0 )} \hspace{0.1cm}\right ),$$
- for Binary Frequency Shift Keying $\rm (BFSK)$ with incoherent demodulation:
- $$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$
It was shown in Exercise 4.8 , that for BPSK, the log ratio $10 · \lg \ E_{\rm B}/N_0$ must be at least $9.6 \ \rm dB$ so that the bit error probability does not exceed $p_{\rm B} = 10^{–5}$ .
Hints:
- This exercise belongs to the chapter Nonlinear_Digital_Modulation.
- However, reference is also made to the page Linear Digital Modulation.
- Use the approximation $\lg(2) ≈ 0.3$.
Questions
Solution
- In other words: the coherent BFSK curve is $10 · \lg (2) ≈ 3 \ \rm dB$ to the right of the the BPSK curve.
- Thus, to guarantee $p_{\rm B} \le 10^{–5}$ it must hold that:
- $$10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}} /{N_{\rm 0}}= 9.6\,\,{\rm dB} + 3\,\,{\rm dB} = \underline{12.6\,\,{\rm dB}}\hspace{0.05cm}.$$
(2) Answer 2 is correct:
- The equation given does not just hold for MSK $($this is a FSK with $h = 0.5)$, but also for every form of orthogonal FSK.
- This is the case whenever the modulation index $h$ is an integer multiple of $0.5$ , such as when $h = 1$.
- When $h = 0.7$ , there is no orthogonal FSK.
(3) From the inverse function of the equation given we get:
- $$\frac{E_{\rm B}} {2 \cdot N_{\rm 0}}= {\rm ln}\hspace{0.05cm}\frac{1}{2 p_{\rm B}}= {\rm ln}(50000)\approx 10.82 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{E_{\rm B}} /{N_{\rm 0}}= 21.64 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.05cm}{E_{\rm B}}/ {N_{\rm 0}}\approx \underline{13.4\,\,{\rm dB}}\hspace{0.05cm}.$$
(4) From $10 · \lg \ E_{\rm B}/N_0 = 12.6 \ \rm dB$ it follows that:
- $${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.25cm}\Rightarrow \hspace{0.25cm} ({E_{\rm B}} /{N_{\rm 0}})/2 \approx 8.4 \hspace{0.25cm} \Rightarrow \hspace{0.25cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4} \approx \underline{1.12 \cdot 10^{-4}}\hspace{0.05cm}.$$
This means that for the same $E_{\rm B}/N_0$, the error probability of incoherent demodulation is increased by a factor of $11$ compared to that of coherent demodulation (see answer to question 1) .