Difference between revisions of "Aufgaben:Exercise 4.8Z: What does the AWGN Channel Capacity Curve say?"
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− | [[File:EN_Inf_Z_4_8.png|right|frame| | + | [[File:EN_Inf_Z_4_8.png|right|frame|Channel capacity as a function of $10 \cdot \lg (E_{\rm B}/{N_0})$ ]] |
− | We consider the channel capacity of the AWGN channel as in [[Aufgaben:Aufgabe_4.8:_Numerische_Auswertung_der_AWGN-Kanalkapazität|Exercise 4.8]] | + | We consider the channel capacity of the AWGN channel as in [[Aufgaben:Aufgabe_4.8:_Numerische_Auswertung_der_AWGN-Kanalkapazität|Exercise 4.8]]: |
− | :$$C_{\rm | + | :$$C_{\rm Gaussian}( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) . $$ |
− | * The curve is shown on the right with logarithmic abscissa between $-2 \ \rm dB$ and $+6 \ \rm dB$ dargestellt. | + | * The curve is shown on the right with logarithmic abscissa between $-2 \ \rm dB$ and $+6 \ \rm dB$ dargestellt. |
− | * The addition of "Gaussian" indicates that a Gaussian distribution was assumed for this curve at the AWGN input. | + | * The addition of "Gaussian" indicates that a Gaussian distribution was assumed for this curve at the AWGN input. |
− | Three system variants are indicated by dots in the | + | Three system variants are indicated by dots in the graph: |
* System $X$: with $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$ and $R = 1$, | * System $X$: with $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$ and $R = 1$, | ||
* System $Y$: with $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ and $R = 2$, | * System $Y$: with $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ and $R = 2$, | ||
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In the questions for this exercise, we still use the following terms: | In the questions for this exercise, we still use the following terms: | ||
− | * Digital system: | + | * Digital system: Symbol set size $M_X = |X|$ beliebig, |
− | * Binary system: | + | * Binary system: Symbol set size $M_X = 2$, |
− | * Quaternary system: | + | * Quaternary system: Symbol set size $M_X = 4$. |
− | |||
− | |||
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Hints: | Hints: | ||
*The exercise belongs to the chapter [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang|AWGN channel capacity with discrete value input]]. | *The exercise belongs to the chapter [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang|AWGN channel capacity with discrete value input]]. | ||
− | *Reference is made in particular to the page [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang#The_channel_capacity_.7F.27.22.60UNIQ-MathJax128-QINU.60.22.27.7F_as_a_function_of_.7F.27.22.60UNIQ-MathJax129-QINU.60.22.27.7F|The channel capacity $C$ as a function of $E_{\rm B}/{N_0}$]]. | + | *Reference is made in particular to the page [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang#The_channel_capacity_.7F.27.22.60UNIQ-MathJax128-QINU.60.22.27.7F_as_a_function_of_.7F.27.22.60UNIQ-MathJax129-QINU.60.22.27.7F|The channel capacity $C$ as a function of $E_{\rm B}/{N_0}$]]. |
− | *Since the results are to be given in "bit" | + | *Since the results are to be given in "bit" ⇒ "log" ⇒ "log<sub>2</sub>" is used in the equations. |
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|type="[]"} | |type="[]"} | ||
- For $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ a digital system with rate $R = 2$ and error probability zero can be imagined. | - For $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ a digital system with rate $R = 2$ and error probability zero can be imagined. | ||
− | + For $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ $R = 0.5$ would be sufficient. | + | + For $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ ⇒ $R = 0.5$ would be sufficient. |
− | - For rate $R = 2$ | + | - For rate $R = 2$ ⇒ $10 \cdot \lg (E_{\rm B}/{N_0}) = 5 \ \rm dB$ would be sufficient. |
{What statement does point $Z$ provide for binary transmission? | {What statement does point $Z$ provide for binary transmission? | ||
|type="[]"} | |type="[]"} | ||
+ A binary system does not meet the requirements in any case. | + A binary system does not meet the requirements in any case. | ||
− | - The curve $C_\text{ | + | - The curve $C_\text{Gaussian}(E_{\rm B}/{N_0})$ is not sufficient for this evaluation. |
{Which statement does the point $Z$ provide for the quaternary transmission? | {Which statement does the point $Z$ provide for the quaternary transmission? | ||
|type="[]"} | |type="[]"} | ||
- A quaternary system does not meet the requirements in any case. | - A quaternary system does not meet the requirements in any case. | ||
− | + The curve $C_\text{ | + | + The curve $C_\text{Gaussian}(E_{\rm B}/{N_0})$ is not sufficient for this evaluation. |
Revision as of 13:02, 4 November 2021
We consider the channel capacity of the AWGN channel as in Exercise 4.8:
- $$C_{\rm Gaussian}( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) . $$
- The curve is shown on the right with logarithmic abscissa between $-2 \ \rm dB$ and $+6 \ \rm dB$ dargestellt.
- The addition of "Gaussian" indicates that a Gaussian distribution was assumed for this curve at the AWGN input.
Three system variants are indicated by dots in the graph:
- System $X$: with $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$ and $R = 1$,
- System $Y$: with $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ and $R = 2$,
- System $Z$: with $10 \cdot \lg (E_{\rm B}/{N_0}) = 6 \ \rm dB$ and $R = 1.5$.
In the questions for this exercise, we still use the following terms:
- Digital system: Symbol set size $M_X = |X|$ beliebig,
- Binary system: Symbol set size $M_X = 2$,
- Quaternary system: Symbol set size $M_X = 4$.
Hints:
- The exercise belongs to the chapter AWGN channel capacity with discrete value input.
- Reference is made in particular to the page The channel capacity $C$ as a function of $E_{\rm B}/{N_0}$.
- Since the results are to be given in "bit" ⇒ "log" ⇒ "log2" is used in the equations.
Questions
Solution
(1) Proposed solutions 1 and 3 are correct:
- Since the point $X$ lies to the right of the channel capacity curve $C_\text{Gauß}(E_{\rm B}/{N_0})$ , there is (at least) one message system of rate $R = 1$that provides quasi–error–free transmission with $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$ .
- Despite the code rate $R = 1$ , this system includes a channel coding with an infinitely long code, but unfortunately this code is unknown.
- However, a binary system of rate $R = 1$ does not allow channel coding.
(2) Only the proposed solution 2 is correct. Here the following statements are valid:
- The required $E_{\rm B}/{N_0}$ for the rate $R = 2$ results in
- $$(E_{\rm B}/{N_0})_{\rm min} = \frac{2^{2R} - 1} { 2 \cdot R} = \frac{2^4 - 1} { 4 } = 3.75 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10\cdot {\rm lg} \hspace{0.1cm}(E_{\rm B}/{N_0})_{\rm min} = 15.74\,{\rm dB} \hspace{0.05cm}. $$
- The maximum code rate $R_{\rm max}$ for $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ ⇒ $E_{\rm B}/{N_0} = 1$ is calculated as follows:
- $$C = R = \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 2^{2R} - 1 \stackrel{!}{=} 2 R \hspace{0.3cm}\Rightarrow \hspace{0.3cm} R_{\rm max} = 0.5 \hspace{0.05cm}. $$
- Both calculations show that the point $Y$ with characteristics $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ and $R = 1$ does not satisfy the channel coding theorem.
(3) With a binary system, the rate $R = 1.5$ can never be realized ⇒ proposed solution 1.
(4) Correct is the proposed solution 2:
- The point $Z$ ies to the right of the boundary curve and for the code rate of a quaternary system $R \le 2$ holds.
- So the rate $R =1.5$ would be quite realizable with $M_X = 4$ .
- The proposed solution 1 is wrong. On the other hand, the second solution suggestion is correct:
- The given curve $C_\text{Gauß}(E_{\rm B}/{N_0})$ always assumes a Gaussian distributed input.
- For a binary system, a different boundary curve results, namely $C_\text{BPSK} ≤ 1 \ \rm bit/channel use$. $C_\text{Gauß}$ and $C_\text{BPSK}$ are significantly different.
- For the quaternary system $(M_X = 4)$ one would have to calculate and analyze the curve $C_{M=4}$ . Again, $C_{M=4} ≤ C_\text{Gauß}$ .
- For small $E_{\rm B}/{N_0}$ , $C_{M=4} \approx C_\text{Gauß}$, holds, after which the curve diverges significantly and ends in a horizontal at $C_{M=4} = 2 \ \rm bit/channel use$.
The point $Z$ ⇒ $10 \cdot \lg (E_{\rm B}/{N_0}) = 6 \ \rm dB, \ \ R = 1.5$ lies below $C_{M=4}$.
- Such a quaternary system would thus be feasible, as will be shown in Exercise 4.10 .
- But only from knowledge of $C_\text{Gauß}$ the question cannot be answered (proposed solution 2).