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Difference between revisions of "Aufgaben:Exercise 3.14: Channel Coding Theorem"

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Hints:
 
Hints:
 
*The exercise belongs to the chapter  [[Information_Theory/Anwendung_auf_die_Digitalsignalübertragung|Application to Digital Signal Transmission]].
 
*The exercise belongs to the chapter  [[Information_Theory/Anwendung_auf_die_Digitalsignalübertragung|Application to Digital Signal Transmission]].
*Reference is made in particular to the page     [[Information_Theory/Anwendung_auf_die_Digitalsignalübertragung#Definition_and_meaning_of_channel_capacity|Definition and meaning of channel capacity]].
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*Reference is made in particular to the section     [[Information_Theory/Anwendung_auf_die_Digitalsignalübertragung#Definition_and_meaning_of_channel_capacity|Definition and meaning of channel capacity]].
 
   
 
   
  
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'''(3)''' &nbsp; <u>Proposed solution 1</u>&nbsp; is correct:
 
'''(3)''' &nbsp; <u>Proposed solution 1</u>&nbsp; is correct:
*From the graph on the information page,&nbsp; it can be read for the capacities of the two channels:
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*From the graph in the information section,&nbsp; it can be read for the capacities of the two channels:
 
:CBSC=0.1887 bit/use,CrmEUC=0.3219 bit/use.
 
:CBSC=0.1887 bit/use,CrmEUC=0.3219 bit/use.
 
*According to the channel coding theorem,&nbsp; a special channel coding can be found at&nbsp; RC&nbsp; with which the error probability can be made zero.
 
*According to the channel coding theorem,&nbsp; a special channel coding can be found at&nbsp; RC&nbsp; with which the error probability can be made zero.

Latest revision as of 14:50, 17 November 2022

Information-theoretical quantities of the  BSC   and  EUCmodels

Shannon's  channel coding theorem states that an error-free transmission can be made over a  "discrete memoryless channel"  (DMC)  with the code rate  R  as long as  R  is not greater than the channel capacity

C=max

The  "Channel Coding Theorem"  is to be evaluated numerically in this task, whereby two typical channel models are to be considered:

  • The  \rm BSC  model  ("Binary Symmetric Channel")  with error probability ε = 0.25  and channel capacity  C = 1 - H_{\rm bin}(ε),
  • the  \rm EUC  model  (from  "Extremely Unsymmetric Channel";  this designation originates from us and is not common) according to  Exercise 3.11Z.


The graphs show the numerical values of the information-theoretical quantities for the two models  \rm BSC  and  \rm EUC:

  • The source entropy  H(X),
  • the equivocation  H(X|Y),
  • the mutual information  I(X; Y),
  • the irrelevance  H(Y|X), 
  • the sink entropy  H(Y).


The parameter in these tables is  p_0 = {\rm Pr}(X = 0)  in the range between  p_0 = 0.3  to  p_0 = 0.7
For the second source symbol probability,   p_1 = {\rm Pr}(X = 1) =1 - p_0.




Hints:



Questions

1

Which statements are true for uncoded transmission   ⇒  \underline{R = 1} assuming  p_0 = p_1 = 0.5 ?

BSC results in a smaller bit error probability.
EUC results in a smaller bit error probability.
Both models lead to the same bit error probability.

2

With  \underline{R = 1}  the result can be (formally) improved by other values of  p_0  or  $p_1 ?

With both channels.
With the BSC model.
With the EUC model.
Not for any model.

3

Over which channel can error-free transmission be achieved with the  \underline{R = 0.16} ?

With both channels.
With the BSC model.
With the EUC model.
With none of the models.

4

Over which channel can error-free transmission be achieved with the  \underline{R = 0.32} ?

Over both channels.
With the BSC model.
With the EUC model.
With none of the models.

5

Over which channel can error-free transmission be achieved with the rate  \underline{R = 0.48} ?

With both channels.
With the BSC model.
With the EUC model.
With none of the models.


Solution

(1)   Proposed solution 3  is correct:

  • The BSC error probability is with p_0 = p_1 = 0.5  for uncoded transmission  ⇒   R = 1:  
p_{\rm B} = 0.5 \cdot 0.25 + 0.5 \cdot 0.25=0.25 \hspace{0.05cm}.
  • Correspondingly, with the same boundary conditions for the EUC model::
p_{\rm B} = 0.5 \cdot 0 + 0.5 \cdot 0.5=0.25 \hspace{0.05cm}.


(2)   Proposed solution 3  is correct:

  • In the BSC model with error probability  ε = 0.25,  for uncoded transmission   ⇒   R = 1  the bit error probability is  p_{\rm B} = 0.25,  regardless of  p_0  and  p_1
  • In contrast, with the EUC model, for example, a smaller bit error probability is obtained with  p_0 = 0.6  and  p_1 = 0.4 :
p_{\rm B} = 0.6 \cdot 0 + 0.4 \cdot 0.5=0.2 \hspace{0.05cm}.
  • Note, however, that now the source entropy is no longer  H(X) = 1\ \rm (bit)  but only  H(X) = H_{bin} (0.6) = 0.971 \ \rm (bit).
  • In the limiting case  p_0 = 1,  only zeros are transmitted and  H(X) = 0.  For the bit error probability, however, the following then actually applies:
p_{\rm B} = 1 \cdot 0 + 0 \cdot 0.5=0 \hspace{0.05cm}.
So no information is transmitted, but it is transmitted with the bit error probability "zero".


(3)   Proposed solution 1  is correct:

  • From the graph in the information section,  it can be read for the capacities of the two channels:
C_{\rm BSC} = 0.1887 \ \rm {bit/use}, \hspace{0.5cm}C_{rm EUC} = 0.3219 \ \rm {bit/use}.
  • According to the channel coding theorem,  a special channel coding can be found at  R ≤ C  with which the error probability can be made zero.
  • For both channels this condition is true with rate  R = 0.16 .


(4)   Proposed solution 3  is correct:

  • With the EUC model, the necessary condition   R ≤ C  for an error-free transmission is fulfilled with  R = 0.32  and  C = 0.3219
  • However, the prerequisite for this is the probability function   P_X(X) = (0.6,\ 0.4).
  • In contrast, for equally probable symbols   ⇒   P_X(X) = (0.5,\ 0.5)  the mutual information  I(X; Y) = 0.3113  would result,
    i.e. a smaller value than for the channel capacity  C, and  I(X; Y) < R  also applies.
  • It can be seen that the EUC model offers more potential for the application of channel coding than the BSC model.  Here, for example, it can be exploited in the code that a transmitted  "0"  is always transmitted without errors.


(5)   The comments on subtasks  (3)  and  (4)  show that the  proposed solution 4  applies.