Difference between revisions of "Aufgaben:Exercise 4.18Z: BER of Coherent and Non-Coherent FSK"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}} |
− | [[File: | + | [[File:EN_Dig_Z_4_18.png|right|frame|Bit error probabilities of BPSK and BFSK]] |
− | + | 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 |
− | |||
+ | * 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 <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: |
:$$p_{\rm B} = {\rm Q } \left ( \sqrt {{2 \cdot E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$ | :$$p_{\rm B} = {\rm Q } \left ( \sqrt {{2 \cdot E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$ | ||
− | |||
− | |||
− | |||
+ | <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, reference 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"]]. | |
− | * | + | * Use the approximation ${\rm lg}(2) \approx 0.3$. |
− | |||
− | |||
− | * | ||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {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 |
|type="[]"} | |type="[]"} | ||
− | - | + | - the coherent FSK with modulation index $\eta = 0.7$, |
− | + | + | + the coherent FSK with modulation index $\eta = 1$. |
− | { | + | {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 BFSK and <u>non-coherent demodulation</u>? |
|type="{}"} | |type="{}"} | ||
$p_{\rm B} \ = \ ${ 0.012 5% } $\ \%$ | $p_{\rm B} \ = \ ${ 0.012 5% } $\ \%$ | ||
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' A comparison of the equations in the information section makes it clear that for BFSK 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 lies $10 \cdot {\rm lg} \, (2) \approx 3 \ \rm dB$ to the right of the BPSK curve. | ||
− | * | + | *To guarantee $p_{\rm B} ≤ 10^{\rm –5}$, it must hold: |
:$$10 \cdot {\rm lg}\hspace{0.05cm} {E_{\rm B}}/ {N_{\rm 0}}\approx | :$$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}.$$ | 9.6\,\,{\rm dB} + 3\,\,{\rm dB}\hspace{0.15cm} \underline{=12.6\,\,{\rm dB}}\hspace{0.05cm}.$$ | ||
− | '''(2)''' | + | '''(2)''' <u>Solution 2</u> is correct: |
− | * | + | *The given equation is valid not only for the MSK $($this is a BFSK with $h = 0.5)$, but for any form of orthogonal BFSK. |
− | * | + | |
− | * | + | *Such a BFSK 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)''' | + | '''(3)''' From the inverse function of the given equation, one obtains: |
:$$\frac{E_{\rm B}} {2 \cdot N_{\rm 0}}= {\rm ln}\hspace{0.05cm}\frac{1}{2 p_{\rm B}}= {\rm | :$$\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} | ln}(50000)\approx 10.82\hspace{0.3cm} | ||
Line 86: | Line 98: | ||
− | '''(4)''' | + | '''(4)''' From $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 12.6 \ \rm dB$ follows: |
:$${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | :$${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} | \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}.$$ | + | \Rightarrow \hspace{0.3cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4}\hspace{0.15cm} \underline{ \approx 0.012 \%= 12 \cdot 10^{\rm -5}}\hspace{0.05cm}.$$ |
− | + | This means: For the same $E_{\rm B}/N_0$, the error probability for non-coherent demodulation is increased by a factor of about $12$ compared to coherent demodulation according to subtask '''(1)'''. | |
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Digital Signal Transmission: Exercises|^4.5 Non-Coherent Demodulation^]] |
Latest revision as of 05:39, 11 September 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, reference 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
- In other words: The coherent BFSK curve lies $10 \cdot {\rm lg} \, (2) \approx 3 \ \rm dB$ 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 a BFSK with $h = 0.5)$, but for any form of orthogonal BFSK.
- Such a BFSK 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:
- $${E_{\rm B}} /{N_{\rm 0}}= 10^{1.26} \approx 16.8 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{E_{\rm B}} {2 \cdot N_{\rm 0}}\approx 8.4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4}\hspace{0.15cm} \underline{ \approx 0.012 \%= 12 \cdot 10^{\rm -5}}\hspace{0.05cm}.$$
This means: For the same $E_{\rm B}/N_0$, the error probability for non-coherent demodulation is increased by a factor of about $12$ compared to coherent demodulation according to subtask (1).