Difference between revisions of "Aufgaben:Exercise 5.3Z: Analysis of the BSC Model"
From LNTwww
| (One intermediate revision by the same user not shown) | |||
| Line 2: | Line 2: | ||
{{quiz-Header|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}} | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}} | ||
| − | [[File:P_ID1832__Dig_Z_5_3.png|right|frame| | + | [[File:P_ID1832__Dig_Z_5_3.png|right|frame|The given error sequence]] |
We consider two different BSC models with the following parameters: | We consider two different BSC models with the following parameters: | ||
* Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$, | * Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$, | ||
| Line 8: | Line 8: | ||
| − | The graph shows an error sequence of length $N = 1000$, but it is not known from which of the two models this sequence originates. | + | The graph shows an error sequence of length $N = 1000$, but it is not known from which of the two models this sequence originates. |
The two models are to be analyzed on the basis of | The two models are to be analyzed on the basis of | ||
| − | * the error distance probabilities | + | * the '''error distance probabilities''' |
:$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$ | :$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$ | ||
| − | * the error distance distribution | + | * the '''error distance distribution''' $\rm (EDD)$ |
:$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$ | :$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$ | ||
| − | * the error correlation function | + | * the '''error correlation function''' $\rm (ECF)$ |
:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm | :$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm | ||
E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \ | E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \ | ||
| Line 25: | Line 25: | ||
| + | Notes: | ||
| + | * The exercise belongs to the chapter [[Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)| "Binary Symmetric Channel"]]. | ||
| − | + | *By counting, we would see that the error sequence of length $N = 1000$ contains exactly $22$ "ones". | |
| − | |||
| − | |||
| − | |||
| − | * | ||
| − | |||
| Line 49: | Line 46: | ||
|type="()"} | |type="()"} | ||
- Model $M_1$, | - Model $M_1$, | ||
| − | + | + | + model $M_2$. |
{What is the mean error distance of model $M_1$? | {What is the mean error distance of model $M_1$? | ||
| Line 70: | Line 67: | ||
===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' In the BSC model, the mean error probability $p_{\rm M}$ is always equal to the characteristic probability $p$. | + | '''(1)''' In the BSC model, the mean error probability $p_{\rm M}$ is always equal to the characteristic probability $p$. |
| + | |||
*For the error correlation function and the error distance distribution are valid | *For the error correlation function and the error distance distribution are valid | ||
:$$\varphi_{e}(k) = | :$$\varphi_{e}(k) = | ||
| Line 76: | Line 74: | ||
p^2 \end{array} \right.\quad | p^2 \end{array} \right.\quad | ||
\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, | \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, | ||
| − | \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array} | + | \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}$$ |
| − | + | :$$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$ | |
| − | *$p$ can be determined from all the given characteristics, except $V_a(k = 1)$. This EDD value is independent of $p$ equal to $(1–p)^0 = 1$. | + | *$p$ can be determined from all the given characteristics, except $V_a(k = 1)$. This EDD value is independent of $p$ equal to $(1–p)^0 = 1$. |
| − | *Therefore, the <u>solutions 1, 2, 4 and 5</u> are correct. | + | |
| + | *Therefore, the <u>solutions 1, 2, 4 and 5</u> are correct. | ||
| − | '''(2)''' The relative error frequency of the given sequence is equal to $h_{\rm F} = 22/1000 \approx 0.022$. | + | '''(2)''' The relative error frequency of the given sequence is equal to $h_{\rm F} = 22/1000 \approx 0.022$. |
| − | *It is quite obvious that the error sequence was generated by the model $M_2$ ⇒ $p_{\rm M} = 0.02$. | + | *It is quite obvious that the error sequence was generated by the model $M_2$ ⇒ $p_{\rm M} = 0.02$. |
| − | *Because of the short sequence, $h_{\rm F}$ does not match $p_{\rm M}$ exactly, but at least approximates ⇒ <u>solution 2</u>. | + | |
| + | *Because of the short sequence, $h_{\rm F}$ does not match $p_{\rm M}$ exactly, but at least approximates ⇒ <u>solution 2</u>. | ||
| − | '''(3)''' The mean error distance – that is, the expected value of the random variable $a$ – is equal to the inverse of the mean error probability ⇒ ${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}$ | + | '''(3)''' The mean error distance – that is, the expected value of the random variable $a$ – is equal to the inverse of the mean error probability ⇒ |
| + | :$${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}.$$ | ||
| − | '''(4)''' According to the equation ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$ we obtain: | + | '''(4)''' According to the equation ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$ we obtain: |
:$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {= | :$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {= | ||
0.1}\hspace{0.05cm},$$ | 0.1}\hspace{0.05cm},$$ | ||
| Line 112: | Line 113: | ||
\hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10} | \hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10} | ||
\hspace{0.15cm}\underline {=0.3487}.$$ | \hspace{0.15cm}\underline {=0.3487}.$$ | ||
| − | To check in comparison with subtask (4): | + | *To check in comparison with subtask '''(4)''': |
:$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k = | :$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k = | ||
11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$ | 11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$ | ||
Latest revision as of 15:52, 5 September 2022
The given error sequence
We consider two different BSC models with the following parameters:
- Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$,
- Model $M_2 \text{:} \hspace{0.4cm} p = 0.02$.
The graph shows an error sequence of length $N = 1000$, but it is not known from which of the two models this sequence originates.
The two models are to be analyzed on the basis of
- the error distance probabilities
- $${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$
- the error distance distribution $\rm (EDD)$
- $$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$
- the error correlation function $\rm (ECF)$
- $$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \ \left\{ \begin{array}{c} p \\ p^2 \end{array} \right.\quad \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$
Notes:
- The exercise belongs to the chapter "Binary Symmetric Channel".
- By counting, we would see that the error sequence of length $N = 1000$ contains exactly $22$ "ones".
Questions
Solution
(1) In the BSC model, the mean error probability $p_{\rm M}$ is always equal to the characteristic probability $p$.
- For the error correlation function and the error distance distribution are valid
- $$\varphi_{e}(k) = \left\{ \begin{array}{c} p \\ p^2 \end{array} \right.\quad \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}$$
- $$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$
- $p$ can be determined from all the given characteristics, except $V_a(k = 1)$. This EDD value is independent of $p$ equal to $(1–p)^0 = 1$.
- Therefore, the solutions 1, 2, 4 and 5 are correct.
(2) The relative error frequency of the given sequence is equal to $h_{\rm F} = 22/1000 \approx 0.022$.
- It is quite obvious that the error sequence was generated by the model $M_2$ ⇒ $p_{\rm M} = 0.02$.
- Because of the short sequence, $h_{\rm F}$ does not match $p_{\rm M}$ exactly, but at least approximates ⇒ solution 2.
(3) The mean error distance – that is, the expected value of the random variable $a$ – is equal to the inverse of the mean error probability ⇒
- $${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}.$$
(4) According to the equation ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$ we obtain:
- $${\rm Pr}(a = 1) \hspace{0.15cm}\underline {= 0.1}\hspace{0.05cm},$$
- $${\rm Pr}(a = 2) = 0.9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.09}\hspace{0.05cm},$$
- $${\rm Pr}(a = {\rm E}[a]) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(a = 10)= 0.9^9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.0387}\hspace{0.05cm}.$$
(5) From the relation $V_a(k) = (1–p)^{k–1}$ we obtain
- $$V_a(k = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^1 \hspace{0.15cm}\underline {= 0.9 } \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}(a = 1) = V_a(k = 1) - V_a(k = 2) = 0.1\hspace{0.05cm},$$
- $$V_a(k = 10)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^9 \hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10} \hspace{0.15cm}\underline {=0.3487}.$$
- To check in comparison with subtask (4):
- $${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k = 11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$
