Difference between revisions of "Aufgaben:Exercise 1.2Z: Bit Error Measurement"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission |
}} | }} | ||
− | [[File:EN_Dig_Z_1_2.png|right|frame| | + | [[File:EN_Dig_Z_1_2.png|right|frame|Simulated bit error frequencies]] |
− | + | The bit error probability | |
:$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$ | :$$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$ | ||
− | + | of a binary system was simulatively determined by a measurement of the bit error rate (BER) | |
− | :$$h_{\rm B} = {n_{\rm B}}/{N}$$ | + | :$$h_{\rm B} = {n_{\rm B}}/{N}$$. |
− | + | Often, $h_{\rm B}$ is also called bit error frequency. | |
− | In | + | In above equations mean: |
− | *$E_{\rm B}$: | + | *$E_{\rm B}$: energy per bit, |
− | *$N_0$: AWGN | + | *$N_0$: AWGN noise power density, |
− | *$n_{\rm B}$: | + | *$n_{\rm B}$: number of bit errors occurred, |
− | *$N$: | + | *$N$: number of simulated bits of a test series. |
− | + | The table shows the results of some test series with $N = 6.4 \cdot 10^4 $, $N = 1. 28 \cdot 10^5$ and $N = 1.6 \cdot 10^6$. The last column named $N \to \infty $ gives the bit error probability $p_{\rm B}$. | |
− | + | The following properties are referred to in the exercise questionnaire: | |
− | * | + | *The bit error frequency $h_{\rm B}$ is, to a first approximation, a Gaussian distributed random variable with mean $m_h = p_{\rm B}$ and variance $\sigma_h^2 \approx p_{\rm B}$. |
− | * | + | *The relative deviation of the bit error frequency from the probability is |
:$$\varepsilon_{\rm rel}= \frac {h_{\rm B}-p_{\rm B}}{p_{\rm B}}\hspace{0.05cm}.$$ | :$$\varepsilon_{\rm rel}= \frac {h_{\rm B}-p_{\rm B}}{p_{\rm B}}\hspace{0.05cm}.$$ | ||
− | * | + | *As a rough rule of thumb on the required accuracy, the number of measured bit errors should be $n_{\rm B} \ge 100$. |
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− | '' | + | ''Note:'' |
− | * | + | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|Error Probability for Baseband Transmission]]. |
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which of the following statements are true? |
|type="[]"} | |type="[]"} | ||
− | - | + | - The accuracy of the BER measurement is independent of $N$. |
− | + | + | + The larger $N$ is, the more accurate the BER measurement is on average. |
− | - | + | - The larger $N$ is, the more accurate each individual BER measurement is. |
− | { | + | {Give the standard deviation $\sigma_h$ for different $N$. Let $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$. |
|type="{}"} | |type="{}"} | ||
$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 1.1 3% } $\ \cdot 10^{ -3 }\ $ | $N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 1.1 3% } $\ \cdot 10^{ -3 }\ $ | ||
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− | { | + | {What is the respective relative deviation for $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$? |
|type="{}"} | |type="{}"} | ||
$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} ε_{\rm rel} \ = \ $ { -0.927--0.873 } $\ \% $ | $N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} ε_{\rm rel} \ = \ $ { -0.927--0.873 } $\ \% $ | ||
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− | { | + | {Give the standard deviation $\sigma_h$ for different $N$ Let $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$. |
|type="{}"} | |type="{}"} | ||
$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 2.3 3% } $\ \cdot 10^{ -5 }\ $ | $N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} σ_h \ = \ $ { 2.3 3% } $\ \cdot 10^{ -5 }\ $ | ||
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− | { | + | {What is the respective relative deviation for $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$? |
|type="{}"} | |type="{}"} | ||
$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} ε_{\rm rel} \ = \ $ { 86 3% } $\ \% $ | $N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm} ε_{\rm rel} \ = \ $ { 86 3% } $\ \% $ | ||
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− | { | + | {Up to what (logarithmic) $E_{\rm B}/N_0$ value is $N = 1.6 \cdot 10^6$ sufficient due to the condition $n_{\rm B} \ge 100$? |
|type="{}"} | |type="{}"} | ||
$\text{Maximum} \ \big [10 \cdot \lg \ E_{\rm B}/N_0 \big] \ = \ $ { 8 3% } $\ \rm dB $ | $\text{Maximum} \ \big [10 \cdot \lg \ E_{\rm B}/N_0 \big] \ = \ $ { 8 3% } $\ \rm dB $ | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Only the <u>second solution</u> is correct: |
− | * | + | *Of course, the accuracy of the BER measurement is influenced by the parameter $N$ to a large extent. On statistical average, the BER measurement naturally becomes better when $N$ is increased. |
− | * | + | *However, there is no deterministic relationship between the number of simulated bits and the accuracy of the BER measurement, as shown, for example, by the results for $10 \cdot \lg \ E_{\rm B}/N_0 = 6 \ \rm dB$: |
− | * | + | *For $N = 6.4 \cdot 10^4\ (n_{\rm B} = 0.258 \cdot 10^{-2})$, the deviation from the true value $(0.239 \cdot 10^{-2})$ is smaller than for $N = 1.28 \cdot 10^5\ (n_{\rm B} = 0.272 \cdot 10^{-2})$. |
− | '''(2)''' | + | '''(2)''' At $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$, i.e. $E_{\rm B} = N_0$, the following values are obtained: |
:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{64000}}\hspace{0.1cm}\underline {\approx 1.1 | :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{64000}}\hspace{0.1cm}\underline {\approx 1.1 | ||
\cdot10^{-3}}\hspace{0.05cm},$$ | \cdot10^{-3}}\hspace{0.05cm},$$ | ||
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− | '''(3)''' | + | '''(3)''' For this, $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$ yields the following values: |
:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}} | :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}} | ||
= \frac{0.0779-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.9\% }\hspace{0.05cm}$$ | = \frac{0.0779-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.9\% }\hspace{0.05cm}$$ | ||
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− | '''(4)''' | + | '''(4)''' Due to the smaller error probability, the values are now smaller than in subtask (2): |
:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{6.4 \cdot 10^{4}}}\hspace{0.1cm}\underline {\approx | :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{6.4 \cdot 10^{4}}}\hspace{0.1cm}\underline {\approx | ||
2.3 \cdot 10^{-5}}\hspace{0.05cm},$$ | 2.3 \cdot 10^{-5}}\hspace{0.05cm},$$ | ||
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− | '''(5)''' | + | '''(5)''' Despite the much smaller standard deviation $\sigma_h$, the smaller error probability results in larger relative deviations for $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ than for $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$: |
:$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.625 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline { \approx 86\% } \hspace{0.05cm},$$ | :$$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.625 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline { \approx 86\% } \hspace{0.05cm},$$ | ||
:$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.325 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline {\approx -3.3\%}\hspace{0.05cm}.$$ | :$$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.325 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline {\approx -3.3\%}\hspace{0.05cm}.$$ | ||
− | '''(6)''' | + | '''(6)''' The number of measured bit errors should be $n_{\rm B} \ge 100$. Therefore, approximately (rounding errors should be taken into account): |
:$$n_{\rm B} = {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | :$$n_{\rm B} = {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} | ||
p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$ | p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$ | ||
− | * | + | *It further follows that in the simulation for $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$ still a sufficient number of bit errors occurred $(n_{\rm B} =315)$, while for $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ on average only $n_{\rm B} =52$ errors are to be expected. |
− | * | + | *For this dB value, about twice the number of bits would have to be simulated. |
{{ML-Fuß}} | {{ML-Fuß}} |
Revision as of 16:51, 2 March 2022
The bit error probability
- $$p_{\rm B} = {1}/{2} \cdot{\rm erfc} \left( \sqrt{{E_{\rm B}}/{N_0}}\right)$$
of a binary system was simulatively determined by a measurement of the bit error rate (BER)
- $$h_{\rm B} = {n_{\rm B}}/{N}$$.
Often, $h_{\rm B}$ is also called bit error frequency.
In above equations mean:
- $E_{\rm B}$: energy per bit,
- $N_0$: AWGN noise power density,
- $n_{\rm B}$: number of bit errors occurred,
- $N$: number of simulated bits of a test series.
The table shows the results of some test series with $N = 6.4 \cdot 10^4 $, $N = 1. 28 \cdot 10^5$ and $N = 1.6 \cdot 10^6$. The last column named $N \to \infty $ gives the bit error probability $p_{\rm B}$.
The following properties are referred to in the exercise questionnaire:
- The bit error frequency $h_{\rm B}$ is, to a first approximation, a Gaussian distributed random variable with mean $m_h = p_{\rm B}$ and variance $\sigma_h^2 \approx p_{\rm B}$.
- The relative deviation of the bit error frequency from the probability is
- $$\varepsilon_{\rm rel}= \frac {h_{\rm B}-p_{\rm B}}{p_{\rm B}}\hspace{0.05cm}.$$
- As a rough rule of thumb on the required accuracy, the number of measured bit errors should be $n_{\rm B} \ge 100$.
Note:
- The exercise belongs to the chapter Error Probability for Baseband Transmission.
Questions
Solution
- Of course, the accuracy of the BER measurement is influenced by the parameter $N$ to a large extent. On statistical average, the BER measurement naturally becomes better when $N$ is increased.
- However, there is no deterministic relationship between the number of simulated bits and the accuracy of the BER measurement, as shown, for example, by the results for $10 \cdot \lg \ E_{\rm B}/N_0 = 6 \ \rm dB$:
- For $N = 6.4 \cdot 10^4\ (n_{\rm B} = 0.258 \cdot 10^{-2})$, the deviation from the true value $(0.239 \cdot 10^{-2})$ is smaller than for $N = 1.28 \cdot 10^5\ (n_{\rm B} = 0.272 \cdot 10^{-2})$.
(2) At $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$, i.e. $E_{\rm B} = N_0$, the following values are obtained:
- $$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{64000}}\hspace{0.1cm}\underline {\approx 1.1 \cdot10^{-3}}\hspace{0.05cm},$$
- $$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.0786}{1600000}}\hspace{0.1cm}\underline {\approx 0.22 \cdot10^{-3}}\hspace{0.05cm}.$$
(3) For this, $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$ yields the following values:
- $$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}} = \frac{0.0779-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.9\% }\hspace{0.05cm}$$
- $$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm} \varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.0782-0.0786}{0.0786}\hspace{0.1cm}\underline {\approx -0.5\% } \hspace{0.05cm}.$$
(4) Due to the smaller error probability, the values are now smaller than in subtask (2):
- $$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{6.4 \cdot 10^{4}}}\hspace{0.1cm}\underline {\approx 2.3 \cdot 10^{-5}}\hspace{0.05cm},$$
- $$ N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\sigma_h = \sqrt{{p}/{N}}= \sqrt{\frac{0.336 \cdot 10^{-4}}{1.6 \cdot 10^{6}}}\hspace{0.1cm}\underline {\approx 0.46 \cdot10^{-5}}\hspace{0.05cm}.$$
(5) Despite the much smaller standard deviation $\sigma_h$, the smaller error probability results in larger relative deviations for $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ than for $10 \cdot \lg \ E_{\rm B}/N_0 = 0 \ \rm dB$:
- $$N = 6.4 \cdot 10^4\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.625 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline { \approx 86\% } \hspace{0.05cm},$$
- $$N = 1.6 \cdot 10^6\text{:}\hspace{0.4cm}\varepsilon_{\rm rel}= \frac{h_{\rm B}- p_{\rm B}}{h_{\rm B}}= \frac{0.325 \cdot 10^{-4} - 0.336 \cdot 10^{-4}}{0.336 \cdot 10^{-4}}\hspace{0.1cm}\underline {\approx -3.3\%}\hspace{0.05cm}.$$
(6) The number of measured bit errors should be $n_{\rm B} \ge 100$. Therefore, approximately (rounding errors should be taken into account):
- $$n_{\rm B} = {p_{\rm B}}\cdot {N} > 100 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm B} > \frac{100}{1.6 \cdot 10^6} = 0.625 \cdot 10^{-4}\hspace{0.05cm}.$$
- It further follows that in the simulation for $10 \cdot \lg \ E_{\rm B}/N_0\hspace{0.05cm}\underline{ = 8 \ \rm dB}$ still a sufficient number of bit errors occurred $(n_{\rm B} =315)$, while for $10 \cdot \lg \ E_{\rm B}/N_0 = 9 \ \rm dB$ on average only $n_{\rm B} =52$ errors are to be expected.
- For this dB value, about twice the number of bits would have to be simulated.