Difference between revisions of "Aufgaben:Exercise 4.18Z: BER of Coherent and Non-Coherent FSK"
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}} | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}} | ||
− | [[File:EN_Dig_Z_4_18.png|right|frame|Bit error probabilities of | + | [[File:EN_Dig_Z_4_18.png|right|frame|Bit error probabilities of BPSK and BFSK]] |
− | The diagram shows the bit error probability for | + | The diagram shows the bit error probability for [[Modulation_Methods/Non-Linear_Digital_Modulation#FSK_.E2.80.93_Frequency_Shift_Keying|"binary FSK modulation"]] $\rm (BFSK)$ with |
− | * coherent demodulation or | + | * coherent demodulation, or |
+ | |||
* incoherent demodulation | * incoherent demodulation | ||
− | in comparison with binary phase modulation (BPSK). Orthogonality is always assumed. For coherent demodulation, the modulation index $h$ can be a multiple of $0.5$, so that the | + | in comparison with binary phase modulation $\rm (BPSK)$. Orthogonality is always assumed. |
+ | *For coherent demodulation, the modulation index $h$ can be a multiple of $0.5$, so that the purple curve is also valid for "Minimum Shift Keying" $\rm (MSK)$. | ||
+ | |||
+ | *On the other hand, for non-coherent demodulation of a BFSK, the modulation index $h$ must be a multiple of $1$. | ||
+ | |||
− | This system comparison is based on the AWGN channel, characterized by the ratio $E_{\rm B}/N_0$. The equations for the bit error probabilities are as follows for | + | This system comparison is based on the AWGN channel, characterized by the ratio $E_{\rm B}/N_0$. The equations for the bit error probabilities are as follows for |
− | * | + | * BFSK with <U>coherent</u> demodulation: |
:$$p_{\rm B} = {\rm Q } \left ( \sqrt {{E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$ | :$$p_{\rm B} = {\rm Q } \left ( \sqrt {{E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$ | ||
− | * < | + | * BFSK with <U>non-coherent</u> demodulation: |
:$$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$ | :$$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$ | ||
− | * | + | * BPSK, only <U>coherent</u> demodulation possible: |
:$$p_{\rm B} = {\rm Q } \left ( \sqrt {{2 \cdot E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$ | :$$p_{\rm B} = {\rm Q } \left ( \sqrt {{2 \cdot E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$ | ||
− | |||
− | For binary modulation methods, $p_{\rm B}$ can also be replaced by $p_{\rm S}$ and $E_{\rm B}$ by $E_{\rm S}$. Then we speak of the symbol error probability $p_{\rm S}$ and the symbol energy $E_{\rm S}$. | + | <u>Remember:</u> |
+ | #For BPSK, the log ratio $10 \cdot {\rm lg} \, (E_{\rm B}/N_0)$ must be at least $9.6 \, \rm dB$ so that the bit error probability does not exceed the value $p_{\rm B} = 10^{\rm -5}$. | ||
+ | #For binary modulation methods, $p_{\rm B}$ can also be replaced by $p_{\rm S}$ and $E_{\rm B}$ by $E_{\rm S}$. | ||
+ | #Then we speak of the symbol error probability $p_{\rm S}$ and the symbol energy $E_{\rm S}$. | ||
+ | <u>Notes:</u> | ||
+ | * The exercise belongs to the chapter [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation| "Carrier Frequency Systems with Non-Coherent Demodulation"]]. | ||
+ | * However, is also made to the chapter [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "Carrier Frequency Systems with Coherent Demodulation"]]. | ||
− | + | * Further information can be found in the book [[Modulation_Methods|"Modulation Methods"]]. | |
− | |||
− | |||
− | |||
− | * Further information can be found in the book [[Modulation_Methods|"Modulation Methods"]]. | ||
* Use the approximation ${\rm lg}(2) \approx 0.3$. | * Use the approximation ${\rm lg}(2) \approx 0.3$. | ||
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===Questions=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | {For | + | {For BFSK and <u>coherent demodulation</u>, which $E_{\rm B}/N_0$ is required to satisfy the requirement $p_{\rm B} ≤ 10^{\rm -5}$? |
|type="{}"} | |type="{}"} | ||
$10 \cdot {\rm lg} \, E_{\rm B}/N_0 \ = \ $ { 12.6 3% } $\ \rm dB$ | $10 \cdot {\rm lg} \, E_{\rm B}/N_0 \ = \ $ { 12.6 3% } $\ \rm dB$ | ||
− | {Are the following statements correct: The same result as in '''(1)''' is obtained for | + | {Are the following statements correct: The same result as in '''(1)''' is obtained for |
|type="[]"} | |type="[]"} | ||
- the coherent FSK with modulation index $\eta = 0.7$, | - the coherent FSK with modulation index $\eta = 0.7$, | ||
+ the coherent FSK with modulation index $\eta = 1$. | + the coherent FSK with modulation index $\eta = 1$. | ||
− | {For | + | {For BFSK with modulation index $h = 1$ and <u>non-coherent demodulation</u>, which $E_{\rm B}/N_0$ is required for $p_{\rm B} ≤ 10^{\rm -5}$ to be satisfied? |
|type="{}"} | |type="{}"} | ||
$10 \cdot {\rm lg} \, E_{\rm B}/N_0 \ = \ ${ 13.4 3% } $\ \rm dB$ | $10 \cdot {\rm lg} \, E_{\rm B}/N_0 \ = \ ${ 13.4 3% } $\ \rm dB$ | ||
− | {What is the error probability with $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 12.6 \ \rm dB$ for | + | {What is the error probability with $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 12.6 \ \rm dB$ for BFSK and <u>non-coherent demodulation</u>? |
|type="{}"} | |type="{}"} | ||
$p_{\rm B} \ = \ ${ 0.012 5% } $\ \%$ | $p_{\rm B} \ = \ ${ 0.012 5% } $\ \%$ |
Revision as of 16:48, 30 August 2022
The diagram shows the bit error probability for "binary FSK modulation" $\rm (BFSK)$ with
- coherent demodulation, or
- incoherent demodulation
in comparison with binary phase modulation $\rm (BPSK)$. Orthogonality is always assumed.
- For coherent demodulation, the modulation index $h$ can be a multiple of $0.5$, so that the purple curve is also valid for "Minimum Shift Keying" $\rm (MSK)$.
- On the other hand, for non-coherent demodulation of a BFSK, the modulation index $h$ must be a multiple of $1$.
This system comparison is based on the AWGN channel, characterized by the ratio $E_{\rm B}/N_0$. The equations for the bit error probabilities are as follows for
- BFSK with coherent demodulation:
- $$p_{\rm B} = {\rm Q } \left ( \sqrt {{E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$
- BFSK with non-coherent demodulation:
- $$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$
- BPSK, only coherent demodulation possible:
- $$p_{\rm B} = {\rm Q } \left ( \sqrt {{2 \cdot E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$
Remember:
- For BPSK, the log ratio $10 \cdot {\rm lg} \, (E_{\rm B}/N_0)$ must be at least $9.6 \, \rm dB$ so that the bit error probability does not exceed the value $p_{\rm B} = 10^{\rm -5}$.
- For binary modulation methods, $p_{\rm B}$ can also be replaced by $p_{\rm S}$ and $E_{\rm B}$ by $E_{\rm S}$.
- Then we speak of the symbol error probability $p_{\rm S}$ and the symbol energy $E_{\rm S}$.
Notes:
- The exercise belongs to the chapter "Carrier Frequency Systems with Non-Coherent Demodulation".
- However, is also made to the chapter "Carrier Frequency Systems with Coherent Demodulation".
- Further information can be found in the book "Modulation Methods".
- Use the approximation ${\rm lg}(2) \approx 0.3$.
Questions
Solution
(1) A comparison of the equations in the information section makes it clear that for binary FSK with coherent demodulation, the AWGN ratio $E_{\rm B}/N_0$ must be doubled to achieve the same error probability as for BPSK.
- In other words: The coherent BFSK curve is $10 \cdot {\rm lg} \, (2) \approx 3 \ \rm dB$ to the right of the BPSK curve. To guarantee $p_{\rm B} ≤ 10^{\rm –5}$, it must hold:
- $$10 \cdot {\rm lg}\hspace{0.05cm} {E_{\rm B}}/ {N_{\rm 0}}\approx 9.6\,\,{\rm dB} + 3\,\,{\rm dB}\hspace{0.15cm} \underline{=12.6\,\,{\rm dB}}\hspace{0.05cm}.$$
(2) Solution 2 is correct:
- The given equation is valid not only for the MSK (this is an FSK with $h = 0.5$), but for any form of orthogonal FSK.
- Such a FSK exists if the modulation index $h$ is an integer multiple of $0.5$, for example for $h = 1$.
- With $h = 0.7$ there is no orthogonal FSK.
- It can be shown that for $h = 0.7$ there is even a smaller error probability than with orthogonal FSK.
- With $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 12.6 \ \rm dB$ one even achieves $p_{\rm B} \approx 10^{\rm –6}$, here, i.e. an improvement by one power of ten.
(3) From the inverse function of the given equation one obtains:
- $$\frac{E_{\rm B}} {2 \cdot N_{\rm 0}}= {\rm ln}\hspace{0.05cm}\frac{1}{2 p_{\rm B}}= {\rm ln}(50000)\approx 10.82\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {E_{\rm B}}/ {N_{\rm 0}}= 21.64 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.09cm} {E_{\rm B}}/ {N_{\rm 0}}\hspace{0.15cm} \underline{\approx 13.4\,\,{\rm dB}}\hspace{0.05cm}.$$
(4) From $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 12.6 \ \rm dB$ follows:
- $${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{E_{\rm B}} {2 \cdot N_{\rm 0}}\approx 8.4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4}\hspace{0.15cm} \underline{ \approx 0.012 \%}\hspace{0.05cm}.$$
This means: For the same $E_{\rm B}/N_0$, the error probability for non-coherent demodulation is increased by a factor of about 11 compared to coherent demodulation according to subtask (1).