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 capacitance of the AWGN channel as in [[Aufgaben:Aufgabe_4.8:_Numerische_Auswertung_der_AWGN-Kanalkapazität|Exercise 4.8]] : | |
:$$C_{\rm Gauß}( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) . $$ | :$$C_{\rm Gauß}( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) . $$ | ||
− | * | + | * The curve is shown on the right with logarithmic abscissa between $-2 \ \rm dB$ und $+6 \ \rm dB$ 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 above 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 $R = 1.5$. | ||
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+ | In the questions for this exercise, we still use the following terms: | ||
+ | * Digital system: symbol range $M_X = |X|$ beliebig, | ||
+ | * Binary system: symbol range $M_X = 2$, | ||
+ | * Quaternary system: symbol range $M_X = 4$. | ||
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+ | Hints: | ||
+ | *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 $E_{\rm B}/{N_0}$]]. | ||
+ | *Since the results are to be given in "bit", "log" ⇒ "log<sub>2</sub>" is used in the equations. | ||
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− | + | ===Questions=== | |
− | === | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What statement does the point $X$ provide for digital signal transmission? |
|type="[]"} | |type="[]"} | ||
− | + | + | + For $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$ a digital system with rate $R = 1$ and error probability zero can be imagined. |
− | - | + | - Such a system does not require channel coding. |
− | + | + | + Such a system uses an infinitely long code. |
− | - | + | - A binary system can also meet the requirements. |
− | { | + | {What statement does the point $Y$ provide for digital signal transmission? |
|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 $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ $R = 0.5$ would be sufficient. |
− | - | + | - For rate $R = 2$ würde $10 \cdot \lg (E_{\rm B}/{N_0}) = 5 \ \rm dB$ would be sufficient. |
− | { | + | {What statement does point $Z$ provide for binary transmission? |
|type="[]"} | |type="[]"} | ||
− | + | + | + A binary system does not meet the requirements in any case. |
− | - | + | - The curve $C_\text{Gauß}(E_{\rm B}/{N_0})$ is not sufficient for this evaluation. |
− | { | + | {Which statement does the point $Z$ provide for the quaternary transmission? |
|type="[]"} | |type="[]"} | ||
− | - | + | - A quaternary system does not meet the requirements in any case. |
− | + | + | + The curve $C_\text{Gauß}(E_{\rm B}/{N_0})$ is not sufficient for this evaluation. |
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' <u>Proposed solutions 1 and 3</u> 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)''' | + | '''(2)''' Only the <u>proposed solution 2</u> is correct. Here the following statements are valid: |
− | * | + | * The required $E_{\rm B}/{N_0}$ for the rate $R = 2$ results in |
:$$(E_{\rm B}/{N_0})_{\rm min} = \frac{2^{2R} - 1} { 2 \cdot R} | :$$(E_{\rm B}/{N_0})_{\rm min} = \frac{2^{2R} - 1} { 2 \cdot R} | ||
= \frac{2^4 - 1} { 4 } = 3.75 | = \frac{2^4 - 1} { 4 } = 3.75 | ||
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10\cdot {\rm lg} \hspace{0.1cm}(E_{\rm B}/{N_0})_{\rm min} = 15.74\,{\rm dB} | 10\cdot {\rm lg} \hspace{0.1cm}(E_{\rm B}/{N_0})_{\rm min} = 15.74\,{\rm dB} | ||
\hspace{0.05cm}. $$ | \hspace{0.05cm}. $$ | ||
− | * | + | * 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: |
:$$C = R = \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) | :$$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} 2^{2R} - 1 \stackrel{!}{=} 2 R | ||
\hspace{0.3cm}\Rightarrow \hspace{0.3cm} R_{\rm max} = 0.5 \hspace{0.05cm}. $$ | \hspace{0.3cm}\Rightarrow \hspace{0.3cm} R_{\rm max} = 0.5 \hspace{0.05cm}. $$ | ||
− | * | + | *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)''' | + | '''(3)''' With a binary system, the rate $R = 1.5$ can never be realized ⇒ <u>proposed solution 1</u>. |
− | '''(4)''' | + | '''(4)''' Correct is the <u>proposed solution 2</u>: |
− | * | + | *The point $Z$ ies to the right of the limit 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 [[Aufgaben:Aufgabe_4.Zehn:_QPSK–Kanalkapazität|Exercise 4.10]] . |
− | * | + | *But only from knowledge of $C_\text{Gauß}$ the question cannot be answered (<u>proposed solution 2</u>). |
{{ML-Fuß}} | {{ML-Fuß}} |
Revision as of 14:45, 12 October 2021
We consider the channel capacitance of the AWGN channel as in Exercise 4.8 :
- $$C_{\rm Gauß}( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) . $$
- The curve is shown on the right with logarithmic abscissa between $-2 \ \rm dB$ und $+6 \ \rm dB$ 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 above 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 range $M_X = |X|$ beliebig,
- Binary system: symbol range $M_X = 2$,
- Quaternary system: symbol range $M_X = 4$.
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 limit 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).