View source for Aufgaben:Exercise 4.8Z: What does the AWGN Channel Capacity Curve say?


{{quiz-Header|Buchseite=Information_Theory/AWGN_Channel_Capacity_for_Discrete_Input
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[[File:EN_Inf_Z_4_8.png|right|frame|Channel capacity as a function of&nbsp; $10 \cdot \lg (E_{\rm B}/{N_0})$ ]]
We consider the channel capacity of the AWGN channel as in&nbsp; [[Aufgaben:Aufgabe_4.8:_Numerische_Auswertung_der_AWGN-Kanalkapazität|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&nbsp; $-2 \ \rm dB$&nbsp; and&nbsp; $+6 \ \rm dB$&nbsp; dargestellt.   
* The addition of&nbsp; "Gaussian"&nbsp; 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 $X$: &nbsp;&nbsp; with&nbsp; $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$&nbsp; and&nbsp; $R = 1$,
* System $Y$: &nbsp;&nbsp; with&nbsp; $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$&nbsp; and&nbsp; $R = 2$,
* System $Z$: &nbsp;&nbsp; with&nbsp; $10 \cdot \lg (E_{\rm B}/{N_0}) = 6 \ \rm dB$&nbsp; and&nbsp; $R = 1.5$.  


In the questions for this exercise, we still use the following terms:
* Digital system:&nbsp;&nbsp; Symbol set size&nbsp; $M_X = |X|$&nbsp; beliebig,
* Binary system:&nbsp;&nbsp; Symbol set size&nbsp; $M_X = 2$,
* Quaternary system:&nbsp;&nbsp; Symbol set size&nbsp; $M_X = 4$.   





Hints:
*The exercise belongs to the chapter&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang|AWGN channel capacity with discrete value input]].
*Reference is made in particular to the page&nbsp; [[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&nbsp; $C$&nbsp; as a function of $E_{\rm B}/{N_0}$]]. 
*Since the results are to be given in&nbsp; "bit" &nbsp; &rArr; &nbsp; "log"&nbsp; &nbsp; &#8658; &nbsp; "log<sub>2</sub>"&nbsp; is used in the equations.


===Questions===

<quiz display=simple>
{What statement does the point &nbsp;$X$&nbsp; provide for digital signal transmission?
|type="[]"}
+ For &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$&nbsp; a digital system with rate &nbsp;$R = 1$&nbsp; 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 &nbsp;$Y$&nbsp; provide for digital signal transmission?
|type="[]"}
- For &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$&nbsp; a digital system with rate &nbsp;$R = 2$&nbsp;and error probability zero can be imagined.
+ For &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$ &nbsp; &rArr; &nbsp; $R = 0.5$&nbsp; would be sufficient.
- For rate &nbsp;$R = 2$ &nbsp; &rArr; &nbsp; $10 \cdot \lg (E_{\rm B}/{N_0}) = 5 \ \rm dB$&nbsp; would be sufficient.

{What statement does point &nbsp;$Z$&nbsp; provide for binary transmission?
|type="[]"}
+ A binary system does not meet the requirements in any case.
- The curve &nbsp;$C_\text{Gaussian}(E_{\rm B}/{N_0})$&nbsp; is not sufficient for this evaluation.

{Which statement does the point &nbsp;$Z$&nbsp; provide for the quaternary transmission?
|type="[]"}
- A quaternary system does not meet the requirements in any case.
+ The curve &nbsp;$C_\text{Gaussian}(E_{\rm B}/{N_0})$&nbsp; is not sufficient for this evaluation.





</quiz>

===Solution===
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'''(1)'''&nbsp; <u>Proposed solutions 1 and 3</u>&nbsp; are correct:
*Since the point &nbsp;$X$&nbsp; lies to the right of the channel capacity curve &nbsp;$C_\text{Gaussian}(E_{\rm B}/{N_0})$,&nbsp; there is (at least) one transmission system of rate &nbsp;$R = 1$&nbsp; that provides&nbsp; "quasi&ndash;error&ndash;free"&nbsp; transmission with&nbsp; $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$. 
*Despite the code rate &nbsp;$R = 1$,&nbsp; this system includes a channel coding with an infinitely long code,&nbsp; but unfortunately this code is unknown.
*However,&nbsp; a binary system of rate &nbsp;$R = 1$&nbsp; does not allow channel coding.




'''(2)'''&nbsp; Only the&nbsp; <u>proposed solution 2</u>&nbsp; is correct.&nbsp; Here the following statements are valid:
* The required &nbsp;$E_{\rm B}/{N_0}$&nbsp; for the rate &nbsp;$R = 2$&nbsp; 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 &nbsp;$R_{\rm max}$&nbsp; for &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$&nbsp;  &nbsp; &#8658; &nbsp; $E_{\rm B}/{N_0} = 1$&nbsp; 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  &nbsp;$Y$&nbsp;with characteristics &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$&nbsp; and &nbsp;$R = 1$&nbsp; does not satisfy the channel coding theorem.




'''(3)'''&nbsp; With a binary system,&nbsp; the rate &nbsp;$R = 1.5$&nbsp; can never be realized &nbsp; &#8658; &nbsp; <u>proposed solution 1</u>.




'''(4)'''&nbsp; Correct is the&nbsp; <u>proposed solution 2</u>:
*The point &nbsp;$Z$&nbsp; lies to the right of the boundary curve and for the code rate of a quaternary system holds&nbsp; $R \le 2$.&nbsp;
*So the code rate &nbsp;$R =1.5$&nbsp; would be quite realizable with&nbsp; $M_X = 4$. 
*The proposed solution 1 is wrong.&nbsp; On the other hand,holds the second solution suggestion is correct:

* The given curve &nbsp;$C_\text{Gaussian}(E_{\rm B}/{N_0})$&nbsp; always assumes a Gaussian distributed input.
* For a binary system,&nbsp; a different boundary curve results,&nbsp; namely &nbsp;$C_\text{BPSK} &#8804; 1 \ \rm  bit/channel \ use$.&nbsp; $C_\text{Gaussian}$&nbsp; and &nbsp;$C_\text{BPSK}$&nbsp; are significantly different.
* For the quaternary system &nbsp;$(M_X = 4)$&nbsp; one would have to calculate and analyze the curve &nbsp;$C_{M=4}$&nbsp;.&nbsp; Again, &nbsp;$C_{M=4} &#8804; C_\text{Gaussian}$&nbsp;. 
*For small &nbsp;$E_{\rm B}/{N_0}$:&nbsp; $C_{M=4} \approx C_\text{Gaussian}$&nbsp; holds,&nbsp; after which the curve diverges significantly and ends in a horizontal at &nbsp;$C_{M=4}  = 2 \ \rm bit/channel \ use$.


The point &nbsp;$Z$&nbsp;   &nbsp; &#8658; &nbsp; $10 \cdot \lg (E_{\rm B}/{N_0}) = 6 \ \rm dB, \ \ R = 1.5$&nbsp; lies below &nbsp;$C_{M=4}$. 
*Such a quaternary system would thus be feasible,&nbsp; as will be shown in&nbsp; [[Aufgaben:Aufgabe_4.Zehn:_QPSK–Kanalkapazität|Exercise 4.10]]&nbsp;.
*But only from knowledge of &nbsp;$C_\text{Gaussian}$&nbsp; the question cannot be answered&nbsp; (<u>proposed solution 2</u>).

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[[Category:Information Theory: Exercises|^4.3 AWGN and Value-Discrete Input^]]