Difference between revisions of "Aufgaben:Exercise 4.18Z: BER of Coherent and Non-Coherent FSK"
Line 53: | Line 53: | ||
$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 FSK and non-coherent demodulation? |
|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 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 | + | *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: |
:$$10 \cdot {\rm lg}\hspace{0.05cm} {E_{\rm B}}/ {N_{\rm 0}}\approx | :$$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 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)''' | + | '''(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 86: | ||
− | '''(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}.$$ | \Rightarrow \hspace{0.3cm} p_{\rm B} = {1}/{2} \cdot {\rm e}^{- 8.4}\hspace{0.15cm} \underline{ \approx 0.012 \%}\hspace{0.05cm}.$$ | ||
− | + | 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). | |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 19:37, 16 August 2022
The diagram shows the bit error probability for a binary "FSK modulation" (BFSK) with
- coherent demodulation or
- 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 average curve is also valid for Minimum Shift Keying (MSK). On the other hand, for non-coherent demodulation of an FSK, 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
- Binary Frequency Shift Keying (BFSK) with coherent demodulation:
- $$p_{\rm B} = {\rm Q } \left ( \sqrt {{E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$
- Binary Frequency Shift Keying (BFSK) with incoherent demodulation:
- $$p_{\rm B} = {1}/{2} \cdot {\rm e}^{- E_{\rm B}/{(2N_0) }}\hspace{0.05cm}.$$
- Binary Phase Shift Keying (BPSK), only coherent demodulation possible:
- $$p_{\rm B} = {\rm Q } \left ( \sqrt {{2 \cdot E_{\rm B}}/{N_0} }\right ) \hspace{0.05cm}.$$
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 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).