Digital Communication Systems 452 - 3 - Discrete...1 Digital Communication Systems ECS 452 Asst....
Transcript of Digital Communication Systems 452 - 3 - Discrete...1 Digital Communication Systems ECS 452 Asst....
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1
Digital Communication SystemsECS 452
Asst. Prof. Dr. Prapun [email protected]
3 Discrete Memoryless Channel (DMC)
Office Hours: Check Google Calendar on the course website.Dr.Prapun’s Office:6th floor of Sirindhralai building, BKD
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Digital communication system
System considered back in CH2
System Under Consideration
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Digital communication system
System considered in Section 3.1
System Under Consideration
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Noisy Channel
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EquivalentChannel
X: channel input
Y: channel output
10
1111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000111111…
10
1101101111011110111110111111101010011110111101111111100111111111111001111111011111001000000000101101…
[BSC_demo_image.m]
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MATLAB
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x_raw = imresize(imread('SIIT_25LOGO.png'),[150 150]); x_raw = [rgb2gray(x_raw) > 128]; % convert to 0 and 1n = prod(size(x_raw)); x = reshape(x_raw,1,n); % convert to row vector
figurex_img = reshape(x,size(x_raw));imshow(x_img*255);
[BSC_demo_image.m]
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Another Noisy Channel
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EquivalentChannel
X: channel input
Y: channel output
[DMC_demo_2.m]
TTTTTTTTTTTTTTTHTHTTTTTHTTTHTTTTTTTHTTTTHTTHTTTTTTTTTTTTTTTTHTTTHTTTHTHTTTTTTTTHHTTTTTTTTTHTTHTHTTTT
HEHHEHTTTHEETETHETHHHETTHEEHEEHEHTHHEHEETTHTHEETETETEEEEHHTHHETEHEETHTEEHEEHEHEEHHEHTTTEEHTTTHEHEEHH
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MATLAB
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%% Generating the channel input xx = randsrc(1,n,[S_X;p_X]); % channel input
%% Applying the effect of the channel to create the channel output yy = DMC_Channel_sim(x,S_X,S_Y,Q); % channel output
function y = DMC_Channel_sim(x,S_X,S_Y,Q)%% Applying the effect of the channel to create the channel output yy = zeros(size(x)); % preallocationfor k = 1:length(x)
% Look at the channel input one by one. Choose the corresponding row% from the Q matrix to generate the channel output.y(k) = randsrc(1,1,[S_Y;Q(find(S_X == x(k)),:)]);
end
[DMC_Analysis_demo.m]
[DMC_Channel_sim.m]
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Ex 3.3: BSC
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>> BSC_demoans =1 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1ans =1 1 1 1 1 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1
Elapsed time is 0.134992 seconds.
%% Simulation parameters% The number of symbols to be transmittedn = 20; % Channel Input S_X = [0 1]; S_Y = [0 1];p_X = [0.3 0.7];% Channel Characteristicsp = 0.1; Q = [1-p p; p 1-p];
[BSC_demo.m]
p_X =0.3000 0.7000
p_X_sim =0.1500 0.8500
q =0.3400 0.6600
q_sim =0.1500 0.8500
Q =0.9000 0.10000.1000 0.9000
Q_sim =0.6667 0.33330.0588 0.9412
PE_sim =0.1000
PE_theretical =0.1000
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Rel. freq. from the simulation
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%% Statistical Analysis% The probability values for the channel inputsp_X % Theoretical probabilityp_X_sim = hist(x,S_X)/n % Relative frequencies from the simulation% The probability values for the channel outputsq = p_X*Q % Theoretical probabilityq_sim = hist(y,S_Y)/n % Relative frequencies from the simulation% The channel transition probabilities from the simulationQ_sim = [];for k = 1:length(S_X)
I = find(x==S_X(k)); LI = length(I);rel_freq_Xk = LI/n; yc = y(I);cond_rel_freq = hist(yc,S_Y)/LI; Q_sim = [Q_sim; cond_rel_freq];
endQ % Theoretical probabilityQ_sim % Relative frequencies from the simulation
[DMC_Analysis_demo.m]
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Ex 3.3: BSC
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>> BSC_demoans =1 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1ans =1 1 1 1 1 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1
Elapsed time is 0.134992 seconds.
%% Simulation parameters% The number of symbols to be transmittedn = 20; % Channel Input S_X = [0 1]; S_Y = [0 1];p_X = [0.3 0.7];% Channel Characteristicsp = 0.1; Q = [1-p p; p 1-p];
[BSC_demo.m]
p_X =0.3000 0.7000
p_X_sim =0.1500 0.8500
q =0.3400 0.6600
q_sim =0.1500 0.8500
Q =0.9000 0.10000.1000 0.9000
Q_sim =0.6667 0.33330.0588 0.9412
PE_sim =0.1000
PE_theretical =0.1000
Because there are only 20 samples, we can’t expect the relative freq. from the simulation to match the specified or calculated probabilities.
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Ex 3.3: BSC
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%% Simulation parameters% The number of symbols to be transmittedn = 1e4; % Channel Input S_X = [0 1]; S_Y = [0 1];p_X = [0.3 0.7];% Channel Characteristicsp = 0.1; Q = [1-p p; p 1-p];
[BSC_demo.m]
>> BSC_demo
p_X =0.3000 0.7000
p_X_sim =0.3037 0.6963
q =0.3400 0.6600
q_sim =0.3407 0.6593
Elapsed time is 0.922728 seconds.
Q =0.9000 0.10000.1000 0.9000
Q_sim =0.9078 0.09220.0934 0.9066
PE_sim =0.0930
PE_theretical =0.1000
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Ex 3.9: DMC
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p_X =0.2000 0.8000
p_X_sim =0.2000 0.8000
q =0.3400 0.3600 0.3000
q_sim =0.4000 0.3500 0.2500
Q =0.5000 0.2000 0.30000.3000 0.4000 0.3000
Q_sim =0.7500 0 0.25000.3125 0.4375 0.2500
>> sym(Q_sim)ans =[ 3/4, 0, 1/4][ 5/16, 7/16, 1/4]PE_sim =
0.7500
>> DMC_demoans =1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 1ans =1 3 2 2 1 2 1 2 2 3 1 1 1 3 1 3 2 3 1 2
x:
y:
0
1
1
3
0.5
0.3
X Y
%% Simulation parameters% The number of symbols to be transmittedn = 20; % General DMC% Ex. 3.6 amd 3.12 in lecture note% Channel Input S_X = [0 1]; S_Y = [1 2 3];p_X = [0.2 0.8];% Channel CharacteristicsQ = [0.5 0.2 0.3; 0.3 0.4 0.3];
[DMC_demo.m]
2𝒑 𝟎 𝟎. 𝟐
𝒑 𝟏 𝟎. 𝟖
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Ex 3.9 & 3.13: DMC
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>> p_X_sim * Q_simans =
0.4000 0.3500 0.2500
p_X =0.2000 0.8000
p_X_sim =0.2000 0.8000
q =0.3400 0.3600 0.3000
q_sim =0.4000 0.3500 0.2500
Q =0.5000 0.2000 0.30000.3000 0.4000 0.3000
Q_sim =0.7500 0 0.25000.3125 0.4375 0.2500
>> sym(Q_sim)ans =[ 3/4, 0, 1/4][ 5/16, 7/16, 1/4]PE_sim =
0.7500
q pQ
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Review: Evaluation of Probability from the Joint PMF Matrix
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Consider two random variables X and Y.
Suppose their joint pmf matrix is
Find
0.1 0.1 0 0 0 0.1 0 0 0.1 00 0.1 0.2 0 00 0 0 0 0.3
2 3 4 5 6xy
1346
3 4 5 6 75 6 7 8 96 7 8 9 108 9 10 11 12
2 3 4 5 6xy
1346
Step 1: Find the pairs (x,y) that satisfy the condition“x+y < 7”
One way to do this is to first construct the matrix of x+y.
,X YP
x y
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Review: Evaluation of Probability from the Joint PMF Matrix
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Consider two random variables X and Y.
Suppose their joint pmf matrix is
Find
0.1 0.1 0 0 0 0.1 0 0 0.1 00 0.1 0.2 0 00 0 0 0 0.3
2 3 4 5 6xy
1346
3 4 5 6 75 6 7 8 96 7 8 9 108 9 10 11 12
2 3 4 5 6xy
1346
Step 2: Add the corresponding probabilities from the joint pmf (matrix)
,X YP
x y𝑃 𝑋 𝑌 7 0.1 0.1 0.1
0.3
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Review: Evaluation of Probability from the Joint PMF Matrix
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Consider two random variables X and Y.
Suppose their joint pmf matrix is
Find
0.1 0.1 0 0 0 0.1 0 0 0.1 00 0.1 0.2 0 00 0 0 0 0.3
2 3 4 5 6xy
1346
,X YP
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Review: Sum of two discrete RVs
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Consider two random variables X and Y.
Suppose their joint pmf matrix is
Find
0.1 0.1 0 0 00.1 0 0 0.1 00 0.1 0.2 0 00 0 0 0 0.3
2 3 4 5 6xy
1346
3 4 5 6 75 6 7 8 96 7 8 9 108 9 10 11 12
2 3 4 5 6xy
1346
x y𝑃 𝑋 𝑌 7 0.1
,X YP
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Summary: DMC
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|
,
Channel Input Alphabet
𝑃 𝑋 𝑥
𝑃 𝑌 𝑦
𝑃 𝑌 𝑦|𝑋 𝑥
𝑃 𝑋 𝑥, 𝑌 𝑦
matrix
matrixrow vector
row vector “AND”
Put together All values at corresponding positions in the matrix.
y
Q
1
i
m
p x
p x
p x
,
1
,X Y i ji
m
x
x
x
p x y
y 1 j ny y y
jq y q pQ
P
1
i
m
x
x
x
1
j
n
y
y
y
j iQ y x
1
j ii
m
x
yx
x
Q x
1 j ny y y
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Matrix Multiplication
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diag
diag
0 0 00 0 0
1 1 1
0 0 0 5 4 6 6 5 4 6
5 4 6 66
60 0 0 6 1 6 3 6 6 30
1 6 31 1 1 1 18 11 19 15
1 2 1
10 0 00 0
0 0 1 2 1 5 2 50 0 0 6 4 6 1 6 4
56 4 6 1
6
0a b c d
a b c d
a a a a ab b b b b
c c c c cd d d d d
ab
cd
a b c d
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Ex: DMC
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>> p = [0.2 0.8];>> Q = [0.5 0.2 0.3; 0.3 0.4 0.3];>> p*Qans =
0.3400 0.3600 0.3000>> P = (diag(p))*QP =
0.1000 0.0400 0.06000.2400 0.3200 0.2400
>> sum(P)ans =
0.3400 0.3600 0.3000
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for Naïve Decoder
23 [DMC_Analysis_demo.m]
%% Naive Decoderx_hat = y;
%% Error ProbabilityPE_sim = 1-sum(x==x_hat)/n % Error probability from the simulation
% Calculation of the theoretical error probabilityPC = 0;for k = 1:length(S_X)
t = S_X(k);i = find(S_Y == t);if length(i) == 1
PC = PC+ p_X(k)*Q(k,i);end
endPE_theretical = 1-PC
Formula derived in 3.24 of lecture notes
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[Ex. 3.25]: Naïve Decoder and BAC
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p_X =0.5000 0.5000
p_X_sim =0.7000 0.3000
q =0.5500 0.4500
q_sim =0.6500 0.3500
Q =0.7000 0.30000.4000 0.6000
Q_sim =0.7143 0.28570.5000 0.5000
PE_sim =0.3500
PE_theretical =0.3500
>> BAC_demoans =0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 1 0 1 0 0ans =0 0 1 1 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 0
x:
y:
0
1
0
1
0.3
0.7
0.4
0.6
X Y
%% Simulation parameters% The number of symbols to be transmittedn = 20; % Binary Assymmetric Channel (BAC)% Ex 3.8 in lecture note (11.3 in [Z&T, 2010])% Channel Input S_X = [0 1]; S_Y = [0 1];p_X = [0.5 0.5];% Channel CharacteristicsQ = [0.7 0.3; 0.4 0.6];
[BAC_demo.m]
𝒑 𝟎 𝟎. 𝟓
𝒑 𝟏 𝟎. 𝟓
720
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25
p_X =0.5000 0.5000
p_X_sim =0.5043 0.4957
q =0.5500 0.4500
q_sim =0.5532 0.4468
Q =0.7000 0.30000.4000 0.6000
Q_sim =0.7109 0.28910.3928 0.6072
PE_sim =0.3405
PE_theretical =0.3500
0
1
0
1
0.3
0.7
0.4
0.6
X Y
%% Simulation parameters% The number of symbols to be transmittedn = 1e4; % Binary Assymmetric Channel (BAC)% Ex 3.8 in lecture note (11.3 in [Z&T, 2010])% Channel Input S_X = [0 1]; S_Y = [0 1];p_X = [0.5 0.5];% Channel CharacteristicsQ = [0.7 0.3; 0.4 0.6];
𝒑 𝟎 𝟎. 𝟓
𝒑 𝟏 𝟎. 𝟓
10 20 30 40 50 60 70 80 90 100
10
20
30
40
50
60
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80
90
10010 20 30 40 50 60 70 80 90 100
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50
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[BAC_demo.m]
[Ex. 3.25]: Naïve Decoder and BAC
[Ex. 3.16]:
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26
p_X_sim =0.2326 0.7674
q =0.4698 0.5302
q_sim =0.4672 0.5328
Q =0.7000 0.30000.4000 0.6000
Q_sim =0.6928 0.30720.3989 0.6011
PE_sim =0.3776
PE_theretical =0.3767
0
1
0
1
0.3
0.7
0.4
0.6
X Y
S_X = [0 1]; S_Y = [0 1];p_X = [0.5 0.5];% Channel CharacteristicsQ = [0.7 0.3; 0.4 0.6];x_raw = imresize(imread('SIIT_25LOGO.png'),[150 150]); x_raw = [rgb2gray(x_raw) > 128]; % convert to 0 and 1x = reshape(x_raw,1,n);
[BAC_demo_image.m]
Naïve Decoder and BAC
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[Ex. 3.26]: Naïve Decoder and DMC
27
p_X =0.2000 0.8000
p_X_sim =0.2000 0.8000
q =0.3400 0.3600 0.3000
q_sim =0.4000 0.3500 0.2500
Q =0.5000 0.2000 0.30000.3000 0.4000 0.3000
Q_sim =0.7500 0 0.25000.3125 0.4375 0.2500
PE_sim =0.7500
PE_theretical =0.7600
>> DMC_demoans =1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 1ans =1 3 2 2 1 2 1 2 2 3 1 1 1 3 1 3 2 3 1 2
x:
y:
0
1
1
3
0.5
0.3
X Y
%% Simulation parameters% The number of symbols to be transmittedn = 20; % General DMC% Ex. 3.16 in lecture note% Channel Input S_X = [0 1]; S_Y = [1 2 3];p_X = [0.2 0.8];% Channel CharacteristicsQ = [0.5 0.2 0.3; 0.3 0.4 0.3];
2𝒑 𝟎 𝟎. 𝟐
𝒑 𝟏 𝟎. 𝟖
20 420
[Same samples as in Ex. 3.6]
[DMC_demo.m]
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28
p_X =0.2000 0.8000
p_X_sim =0.2011 0.7989
q =0.3400 0.3600 0.3000
q_sim =0.3387 0.3607 0.3006
Q =0.5000 0.2000 0.30000.3000 0.4000 0.3000
Q_sim =0.4943 0.1914 0.31430.2995 0.4033 0.2972
PE_sim =0.7607
PE_theretical =0.7600
0
1
1
3
0.5
0.3
X Y
%% Simulation parameters% The number of symbols to be transmittedn = 1e4; % General DMC% Ex. 3.16 in lecture note% Channel Input S_X = [0 1]; S_Y = [1 2 3];p_X = [0.2 0.8];% Channel CharacteristicsQ = [0.5 0.2 0.3; 0.3 0.4 0.3];
2𝒑 𝟎 𝟎. 𝟐
𝒑 𝟏 𝟎. 𝟖
10 20 30 40 50 60 70 80 90 100
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90
10010 20 30 40 50 60 70 80 90 100
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100
[DMC_demo.m]
[Ex. 3.26]: Naïve Decoder and DMC
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Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
DigitalDemodulator
Transmitter
Remove redundancy
Add systematic redundancy
EquivalentChannel
X: channel input
Y: channel outputSource Decoder
Channel Decoder(Detector)
Receiver
Decodedvalue
ˆ ˆX x Y
System Model for Section 3.2-3.3
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Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
DigitalDemodulator
Transmitter
Remove redundancy
Add systematic redundancy
X: channel input
Y: channel outputSource Decoder
Channel Decoder(Detector)
Decodedvalue
[Ex. 3.12, 3.18, 3.26]
0
1
1
3
0.5
0.3
2
10111110110111100101…
21133122121231131311…
Receiver ˆ ˆX x Y
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Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
DigitalDemodulator
Transmitter
Remove redundancy
Add systematic redundancy
X: channel input
Y: channel outputSource Decoder
Decodedvalue
[Ex. 3.26]: Naïve Decoder
0
1
1
3
0.5
0.3
2
10111110110111100101…
21133122121231131311…
ˆ12
1
3 32
x yy
ˆ01
1
3 02
x yy
21133122121231131311…
Naïve Decoder
ˆ 0.76P P X X
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Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
DigitalDemodulator
Transmitter
Remove redundancy
Add systematic redundancy
X: channel input
Y: channel outputSource Decoder
Decodedvalue
[Ex. 3.27]: DIY Decoder
0
1
1
3
0.5
0.3
2
10111110110111100101…
21133122121231131311…
ˆ01
1
3 02
x yy
ˆ01
1
3 02
x yy
10000011010100000000…
ˆ 0.52P P X X
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[Ex. 3.27]: DIY Decoder
33
>> DMC_decoder_DIY_demoans =1 0 1 1 1 1 1 0 1 1 0 1 1 1 1 0 0 1 0 1ans =2 1 1 3 3 1 2 2 1 2 1 2 3 1 1 3 1 3 1 1ans =1 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 0 0 0 0PE_sim =
0.5500PE_theretical =
0.5200Elapsed time is 0.081161 seconds.
%% Simulation parameters% The number of symbols to be transmittedn = 20; % General DMC% Ex. 3.16 in lecture note% Channel Input S_X = [0 1]; S_Y = [1 2 3];p_X = [0.2 0.8];% Channel CharacteristicsQ = [0.5 0.2 0.3; 0.3 0.4 0.3];
%% DIY DecoderDecoder_Table = [0 1 0]; % The decoded values corresponding to the received Y
[DMC_decoder_DIY_demo.m]
X
Y
𝑋
ˆ01
1
3 02
x yy
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[Ex. 3.27]: DIY Decoder
34
%% DIY DecoderDecoder_Table = [0 1 0]; % The decoded values corresponding to the received Y
% Decode according to the decoder tablex_hat = y; % preallocationfor k = 1:length(S_Y)
I = (y==S_Y(k));x_hat(I) = Decoder_Table(k);
end
PE_sim = 1-sum(x==x_hat)/n % Error probability from the simulation
% Calculation of the theoretical error probabilityPC = 0;for k = 1:length(S_X)
I = (Decoder_Table == S_X(k));q = Q(k,:); PC = PC+ p_X(k)*sum(q(I));
endPE_theretical = 1-PC
[DMC_decoder_DIY_demo.m]
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Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
DigitalDemodulator
Transmitter
Remove redundancy
Add systematic redundancy
X: channel input
Y: channel outputSource Decoder
[Ex. 3.27]: DIY Decoder
0
1
1
3
0.5
0.3
2
ˆ01
1
3 02
x yy
ˆ 0.52P P X X
10 20 30 40 50 60 70 80 90 100
10
20
30
40
50
60
70
80
90
100
10 20 30 40 50 60 70 80 90 100
10
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30
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70
80
90
100
10 20 30 40 50 60 70 80 90 100
10
20
30
40
50
60
70
80
90
100
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[Ex. 3.27]: DIY Decoder
36
>> DMC_decoder_DIY_demoPE_sim =
0.5213PE_theretical =
0.5200Elapsed time is 2.154024 seconds.
%% Simulation parameters% The number of symbols to be transmittedn = 1e4; % General DMC% Ex. 3.16 in lecture note% Channel Input S_X = [0 1]; S_Y = [1 2 3];p_X = [0.2 0.8];% Channel CharacteristicsQ = [0.5 0.2 0.3; 0.3 0.4 0.3];
%% DIY DecoderDecoder_Table = [0 1 0]; % The decoded values corresponding to the received Y
10 20 30 40 50 60 70 80 90 100
10
20
30
40
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60
70
80
90
10010 20 30 40 50 60 70 80 90 100
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10010 20 30 40 50 60 70 80 90 100
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[DMC_decoder_DIY_demo.m]
ˆ 0.52P P X X
𝑋 𝑋𝑌
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Recipe for finding of any decoder
37
Use the 𝐏 matrix. If unavailable, can be found by scaling each row of the 𝐐 matrix by its
corresponding 𝑝 𝑥 . Write 𝑥 𝑦 values on top of the 𝑦 values for the 𝐏 matrix. For column 𝑦 in the 𝐏 matrix, circle the element whose corresponding
𝑥 value is the same as 𝑥 𝑦 . 𝑃 the sum of the circled probabilities. 𝑃 1 𝑃 .
0.2
0.8
00 01
0.10 0.04 0.060.24 0.
.5 0.21
0.30.3 0.4 0.3 32 0.24
Q P
1 2 3 1 2 3x y xy1 1 0 x̂ y
0
1
1
3
0.5
0.3
X Y2
ˆ11
1
3 02
x yy
[Ex. 3.28]
[3.29]
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38
Digital Communication SystemsECS 452
Asst. Prof. Dr. Prapun [email protected]
3.3 Optimal Decoder
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Searching for the Optimal Detector
39
>> DMC_decoder_ALL_demoans =
0 0 0 0.80000 0 1 0.62000 1 0 0.52000 1 1 0.34001 0 0 0.66001 0 1 0.48001 1 0 0.38001 1 1 0.2000
Min_PE =0.2000
Optimal_Detector =1 1 1
Elapsed time is 0.003351 seconds.
𝑥 1 𝑥 2 𝑥 3 𝑃 ℰ
Ex. 3.27
Ex. 3.28
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Review: ECS315 (2018)
40
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Guessing Game 1
41
There are 15 cards. Each have a number on it. Here are the 15 cards:
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 One card is randomly selected from the 15 cards.
You need to guess the number on the card.
Have to pay 1 Baht for incorrect guess.
The game is to be repeated n = 10,000 times.
What should be your guess value?
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42
close all; clear all;
n = 5; % number of time to play this game
D = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5];X = D(randi(length(D),1,n));
if n <= 10X
end
g = 1cost = sum(X ~= g)
if n > 1averageCostPerGame = cost/nend
>> GuessingGame_4_1_1X =
3 5 1 2 5g =
1cost =
4averageCostPerGame =
0.8000
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43
close all; clear all;
n = 5; % number of time to play this game
D = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5];X = D(randi(length(D),1,n));
if n <= 10X
end
g = 3.3cost = sum(X ~= g)
if n > 1averageCostPerGame = cost/nend
>> GuessingGame_4_1_1X =
5 3 2 4 1g =
3.3000cost =
5averageCostPerGame =
1
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44
close all; clear all;
n = 1e4; % number of time to play this game
D = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5];X = D(randi(length(D),1,n));
if n <= 10X
end
g = ?cost = sum(X ~= g)
if n > 1averageCostPerGame = cost/nend
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Guessing Game 1
451 1.5 2 2.5 3 3.5 4 4.5 5
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
Guess value
AV
erag
e C
ost P
er G
ame
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1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 60.65
0.7
0.75
0.8
0.85
0.9
0.95
1
Guess value
AV
erag
e C
ost P
er G
ame
Guessing Game 1
46
Optimal Guess: The most-likely value
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Summary: MAP vs. ML Decoders
47
Decoder is derived from the 𝐏 matrix Select (by circling) the maximum value
in each column (for each value of 𝑦) in the 𝐏 matrix.
The corresponding 𝑥 value is the value of 𝑥 𝑦 .
MAP ,
optimal
ˆ argmax ,
ˆ
argmax
argmax
X Yx
x
x
x y p x y
x y
P X x Y y
Q y p xx
MLˆ argmaxx
x y Q y x
Decoder is derived from the 𝐐 matrix Select (by circling) the maximum value
in each column (for each value of 𝑦) in the 𝐐 matrix.
The corresponding 𝑥 value is the value of 𝑥 𝑦 .
Once the decoder (the decoding table) is derived 𝑃 and 𝑃 are calculated by adding the corresponding probabilities in the 𝐏 matrix.
Optimal at least when 𝑝 𝑥 is uniform (the channel inputs are equally likely)
a posteriori probability
likelihood funtion
Can be derived without knowing the channel input probabilities.
prior probability
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Example: MAP Decoder [Ex. 3.36]
48
0.2
0.8
00 01
0.10 0.04 0.060.24 0.
.5 0.21
0.30.3 0.4 0.3 32 0.24
Q P
1 2 3 1 2 3x y xy1 1 1 MAPx̂ y
0
1
1
3
0.5
0.3
X Y2
( ) 0.24 0.32 0.( ) 1 21 24 0.P P
MAPˆ
13 1
1 12
x yy
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Example: ML Decoder [Ex. 3.47]
49
( ) 0.10 0.( 32 0.06) 1 1 0.52PP
( ) 0.10 0.( 32 0.24) 1 1 0.34PP
0.2
0.8
00 01
0.10 0.04 0.060.24 0.
.5 0.21
0.30.3 0.4 0.3 32 0.24
Q P
1 2 3 1 2 3x y xy0 1 0 MLx̂ y
0
1
1
3
0.5
0.3
X Y2
0.2
0.8
00 01
0.10 0.04 0.060.24 0.
.5 0.21
0.30.3 0.4 0.3 32 0.24
Q P
1 2 3 1 2 3x y xy
0
1
1
3
0.5
0.3
X Y2
MLx̂ y0 1 1
Sol 1:
Sol 2:
ML
1ˆ01
3 12
yy x
ML
1ˆ01
3 02
yy x[Same as Ex. 3.27]
[Agree with Ex. 3.33]
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MAP Decoder
50
%% MAP DecoderP = diag(p_X)*Q; % Weight the channel transition probability by the
% corresponding prior probability.[V I] = max(P); % For I, the default MATLAB behavior is that when there are
% multiple max, the index of the first one is returned.Decoder_Table = S_X(I) % The decoded values corresponding to the received Y
%% Decode according to the decoder tablex_hat = y; % preallocationfor k = 1:length(S_Y)
I = (y==S_Y(k));x_hat(I) = Decoder_Table(k);
end
PE_sim = 1-sum(x==x_hat)/n % Error probability from the simulation
%% Calculation of the theoretical error probabilityPC = 0;for k = 1:length(S_X)
I = (Decoder_Table == S_X(k));Q_row = Q(k,:); PC = PC+ p_X(k)*sum(Q_row(I));
endPE_theretical = 1-PC
[DMC_decoder_MAP_demo.m]
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ML Decoder
51
%% ML Decoder[V I] = max(Q); % For I, the default MATLAB behavior is that when there are
% multiple max, the index of the first one is returned.Decoder_Table = S_X(I) % The decoded values corresponding to the received Y
%% Decode according to the decoder tablex_hat = y; % preallocationfor k = 1:length(S_Y)
I = (y==S_Y(k));x_hat(I) = Decoder_Table(k);
end
PE_sim = 1-sum(x==x_hat)/n % Error probability from the simulation
%% Calculation of the theoretical error probabilityPC = 0;for k = 1:length(S_X)
I = (Decoder_Table == S_X(k));Q_row = Q(k,:); PC = PC+ p_X(k)*sum(Q_row(I));
endPE_theretical = 1-PC
[DMC_decoder_ML_demo.m]
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Optimal Decoder for BSC
57
close all; clear all;tic
%% Simulation parameters% Channel Input S_X = [0 1]; S_Y = [0 1];p0 = 0.8; p1 = 1-p0; p_X = [p0 p1];% Channel Characteristicsp = 0.1; Q = [1-p p; p 1-p];
%% All possible "reasonable" decoders% X_hat = Y; X_hat = 1-Y; X_hat = 1; X_hat = 0Decoder_Table_ALL = [0 1; 1 0; 1 1; 0 0]
%% Calculate the error probability for each of the decoder PE_ALL = [];for k = 1:size(Decoder_Table_ALL,1)
Decoder_Table = Decoder_Table_ALL(k,:);PC = 0;for k = 1:length(S_X)
I = (Decoder_Table == S_X(k));Q_row = Q(k,:);PC = PC + p_X(k)*sum(Q_row(I));
endPE_theretical = 1-PC;PE_ALL = [PE_ALL; PE_theretical];
end
%% Display the results[Decoder_Table_ALL PE_ALL]
%% Find the optimal detectors[V I] = min(PE_ALL);Optimal_Detector = Decoder_Table_ALL(I,:)Min_PE = V
toc
>> BSC_decoder_ALL_demoDecoder_Table_ALL =
0 11 01 10 0
ans =0 1 0.10001 0 0.90001 1 0.80000 0 0.2000
Optimal_Detector =0 1
Min_PE =0.1000
Elapsed time is 0.008709 seconds.
[BSC_decoder_ALL_demo.m]
𝑥 0 𝑥 1 𝑃 ℰ
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Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
DigitalDemodulator
Transmitter
Remove redundancy
Add systematic redundancy
EquivalentChannel
X: channel input
Y: channel outputSource Decoder
Channel Decoder(Detector)
Receiver
Decodedvalue
System Model for Section 3.2-3.3
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Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
DigitalDemodulator
Transmitter
Remove redundancy
Add systematic redundancy
EquivalentChannel
Source Decoder
Channel Decoder(Detector)
Receiver
Decodedvalue
System Model for Section 3.4
X: channel input
Y: channel output
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Some results from Section 3.3-3.4
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Information Source
Destination
Message
Recovered Message
Source Encoder
Channel Encoder
Transmitter
Remove redundancy
Add systematic redundancy
EquivalentChannel
X: channel input
Y: channel outputSource Decoder
Channel Decoder(Detector)
Receiver
Decodedvalue
System Model for Section 3.5
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Channel Encoder
[3.62] Block Encoding
k bits k bits k bits n bits n bits n bits
X
𝒏, 𝒌 code
Code rate = 𝒏𝒌
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Channel Encoder
X
[3.62] Block Encoding
k bits n bits
Codebook
𝒔 𝐬𝐬
𝐬
𝐱
𝐱
𝐱
𝐱
𝑀 2 possibilities
Choose 𝑀 2 from 2 possibilities to be used as codewords.
[Figure 13]
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64
Digital Communication SystemsECS 452
Asst. Prof. Dr. Prapun [email protected]
3.5 Repetition Code in Communications Over BSC
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Hypercube
65 https://www.youtube.com/watch?v=Q_B5GpsbSQw
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Hypercube
66 https://www.youtube.com/watch?v=Q_B5GpsbSQw
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n-bit space
67
n = 1
n = 2
n = 3
n = 4
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Channel Encoder and Decoder
68
Noise & In
terferen
ce
Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
Source Decoder
Channel Decoder
DigitalDemodulator
Transmitter
Receiver
Add systematic redundancy
X: channel input
Y: channel output
0
1
0
1
p
1-p
p
1-p
S
𝐒
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Better Equivalent Channel
69
Noise & In
terferen
ce
Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
Source Decoder
Channel Decoder
DigitalDemodulator
Transmitter
Receiver
Remove redundancy
Add systematic redundancy Better
EquivalentChannel
S
𝐒
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Example: Repetition Code
70
Original Equivalent Channel:
BSC with crossover probability p = 0.01
New (and Better) Equivalent Channel:
Use repetition code with n = 5 at the transmitter Use majority vote at the receiver New BSC with
0
1
0
1
p
1-p
p
1-p
0
1
0
1
p
1-p
p
1-p
Repetition Code with n = 5
Majority Vote
0
1
0
1
𝑝 1053 𝑝 1 𝑝 5
4 𝑝 1 𝑝 55 𝑝 1 𝑝
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Example: ASCII Encoder and BSC
71
Noise & In
terferen
ce
Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
Source Decoder
Channel Decoder
DigitalDemodulator
Transmitter
Receiver
Add systematic redundancy
X: channel input
Y: channel output
“LOVE”
“1001100100111110101101000101”
0
1
0
1
p
1-p
p
1-p
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The ASCII Coded Character Set
72[The ARRL Handbook for Radio Communications 2013]
0 16 32 48 64 80 96 112
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Example: ASCII Encoder
73
Character Codeword
⋮E 1000101
⋮L 1001100
⋮O 1001111
⋮V 1010110
⋮
SourceEncoder
Information Source
“LOVE”“1001100100111110101101000101”
>> M = 'LOVE';>> X = dec2bin(M,7);>> X = reshape(X',1,numel(X))X =1001100100111110101101000101
MATLAB:
c(“L”) c(“O”) c(“V”) c(“E”)
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System considered
74
Noise & In
terferen
ce
Information Source
Destination
Channel
ReceivedSignal
TransmittedSignal
Message
Recovered Message
Source Encoder
Channel Encoder
DigitalModulator
Source Decoder
Channel Decoder
DigitalDemodulator
Transmitter
Receiver
Add systematic redundancy
X: channel input
Y: channel output
0
1
0
1
p
1-p
p
1-p
THE WIZARD OF OZ (1900)written by L. Frank Baum
Introduction
Folklore, legends, myths and fairy tales have followed childhoodthrough the ages, for every healthy youngster has a wholesome andinstinctive love for stories fantastic, marvelous and manifestlyunreal. The winged fairies of Grimm and Andersen have brought morehappiness to childish hearts than all other human creations.Yet the old time fairy tale, having served for generations, may
101010010010001000101010000010101111001001101101010000011010010100010001000001001111100011001000001001111101101001000000100000010000001010000110001011100101100000110000010100100010100001001111011111100101101001111010011101001100101110111001000001100010111100101000001001100010111001000001000110111001011000011101110110101101000001000010110000111101011101101000101000010100001010100100111011101110100111001011011111100100111010111000111110100110100111011111101110000101000010100100000010000010001101101111110110011010111101100110111111100101100101010110001000001101100110010111001111100101110111011001001110011010110001000001101101111100111101001101000111001101000001100001110111011001000100000110011011000011101001111001011110010100000111010011000011101100110010111100110100000110100011000011110110110010101000001100110110111111011001101100110111111101111100101110010001000001100011110100011010011101100110010011010001101111110111111001000001010111010011010001110010110111111101011100111110100001000001110100110100011001010100000110000111001111100101111001101011000100000110011011011111110010010000011001011110110110010111100101111001010000011010001100101110000111011001110100110100011110010100000111100111011111110101110111011001111110011111010011001011110010010000011010001100001111001101000001100001010000011101111101000110111111011001100101111001111011111101101110010101000001100001110111011001000001010110100111011101110011111010011010011101110110001111101001101001111011011001010100000110110011011111110110110010101000001100110110111111100100100000111001111101001101111111001011010011100101111001101000001100110110000111011101110100110000111100111110100110100111000110101100010000011011011100001111001011101101100101110110011011111110101111001101000001100001110111011001000100000110110111000011101110110100111001101100101111001111101001101100111100100010101110101110111011100101100101110000111011000101110010000010101001101000110010101000001110111110100111011101100111110010111001000100000110011011000011101001111001011010011100101111001101000001101111110011001000001000111111001011010011101101110110101000001100001110111011001000100000100000111011101100100110010111100101110011110010111011100100000110100011000011110110110010101000001100010111001011011111110101110011111010001110100010000011011011101111111001011001010001010110100011000011110000111000011010011101110110010111100111110011010000011101001101111010000011000111101000110100111011001100100110100111100111101000010000011010001100101110000111100101110100111001101000001110100110100011000011101110010000011000011101100110110001000001101111111010011010001100101111001001000001101000111010111011011100001110111001000001100011111001011001011100001111010011010011101111110111011100110101110000101001000000100000101100111001011110100010000011101001101000110010101000001101111110110011001000100000111010011010011101101110010101000001100110110000111010011110010111100101000001110100110000111011001100101010110001000001101000110000111101101101001110111011001110100000111001111001011110010111011011001011100100010000011001101101111111001001000001100111110010111011101100101111001011000011110100110100111011111101110111001101011000100000110110111000011111001
𝑝 0.01
[ErrorProbabilityoverBSC.m]
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Results
75
101010010010001000101010000010101111001001101101010000011010010100010001000001011111100011001000001001111101101001000000100000010000001010000110001011100101100000110000011100100110100001001111011111100101101001111010011101001100101110111001000001100010111100101000001001100010111001000001000110111001011000011101110110101101100001000010110000111101011101101000101000010100001010100100111011101110100111001011011111100100111010111000111110100110100111011111101110000101000010100100000011000010001101101111110110011010111101100110111111100101100101010111000000001101100110010111001111100101110111011001001010011000110001000001101101111100111101001101000111001101000001100001110111011001000100000110011011000011101001110001011110010100000111010011000011101100110011111100110100000110100011000011110110110010101000001100110110110111011001101100110111111101111100101110010001000001100011110100011010011101100110010011010001101111110111111001000001010111010011010001110010110111111101011100111110100001000001110100110100011001010100000110000111001111100101111001101011000000000110011011011111110010010010011001011110110110010111100101111001010000011011101100101110000111011001110100110100011110010100000111100111011111110101110111011001111110011111010011001011110010010000011010001100001111001101000001100001010000011101111101000110111111011001100101111001111011111101101110010101000001100001110111011001000001010110100111011001110011111000011110011101110110001111101001101001111001011001010100000110110011011111110110110010101000001100110110111111100100100000111001111101001101111111001011010011100101111000101000001100110110000101011101110100110000111100111110100110100111000110101100010000011011011100001111001011101101100101110110011011111110101011001101000001100101110111011001000100000110110111000011101110110100111001101100101111001111101001101100111100100010101110101110111011100101100101110000111011000101110010000010101001101000110010101000001110111110100111011101100111110010111001000100000110011011000011101001111001011010011100101111001101000001101111110011001000001000111111001011010011101101110110101000001100001110111011001000101010100000111011101100100110010111100101110011110010111011100100000110100011000011110110100010101000001100010111001011011111110101110011111010001110100010000011011011101111111001011001010001010110100011000011110000111000011010011101110110010111100111110011000000011101001101111010000011000111101000110100111010001100100110100111100111101000010000011010001100101110000111100101110100111001101000001110100110100011000011101110010000011000011101100110110001000001101111110010011010001110101111001001000001101000111010111011011100001110111001000001100011111001011001011100001111010011010011101111110111011100110101111000101001000000100000101100111001011110100010000011101001101000110010101100001101111110110011001000100010111010011010011101101110110101000001100110110000111010010110010111100101000001110100110000111011001100101010110001000001101000110000111101101101001110111011001110100000111000111001011110010111011011001011100100010000011001101101111111001001000001100111110010111011101100101111001011000011110100110100111011111101110111000101011000100000110110111000011111001
101010010010001000101010000010101111001001101101010000011010010100010001000001001111100011001000001001111101101001000000100000010000001010000110001011100101100000110000010100100010100001001111011111100101101001111010011101001100101110111001000001100010111100101000001001100010111001000001000110111001011000011101110110101101000001000010110000111101011101101000101000010100001010100100111011101110100111001011011111100100111010111000111110100110100111011111101110000101000010100100000010000010001101101111110110011010111101100110111111100101100101010110001000001101100110010111001111100101110111011001001110011010110001000001101101111100111101001101000111001101000001100001110111011001000100000110011011000011101001111001011110010100000111010011000011101100110010111100110100000110100011000011110110110010101000001100110110111111011001101100110111111101111100101110010001000001100011110100011010011101100110010011010001101111110111111001000001010111010011010001110010110111111101011100111110100001000001110100110100011001010100000110000111001111100101111001101011000100000110011011011111110010010000011001011110110110010111100101111001010000011010001100101110000111011001110100110100011110010100000111100111011111110101110111011001111110011111010011001011110010010000011010001100001111001101000001100001010000011101111101000110111111011001100101111001111011111101101110010101000001100001110111011001000001010110100111011101110011111010011010011101110110001111101001101001111011011001010100000110110011011111110110110010101000001100110110111111100100100000111001111101001101111111001011010011100101111001101000001100110110000111011101110100110000111100111110100110100111000110101100010000011011011100001111001011101101100101110110011011111110101111001101000001100001110111011001000100000110110111000011101110110100111001101100101111001111101001101100111100100010101110101110111011100101100101110000111011000101110010000010101001101000110010101000001110111110100111011101100111110010111001000100000110011011000011101001111001011010011100101111001101000001101111110011001000001000111111001011010011101101110110101000001100001110111011001000100000100000111011101100100110010111100101110011110010111011100100000110100011000011110110110010101000001100010111001011011111110101110011111010001110100010000011011011101111111001011001010001010110100011000011110000111000011010011101110110010111100111110011010000011101001101111010000011000111101000110100111011001100100110100111100111101000010000011010001100101110000111100101110100111001101000001110100110100011000011101110010000011000011101100110110001000001101111111010011010001100101111001001000001101000111010111011011100001110111001000001100011111001011001011100001111010011010011101111110111011100110101110000101001000000100000101100111001011110100010000011101001101000110010101000001101111110110011001000100000111010011010011101101110010101000001100110110000111010011110010111100101000001110100110000111011001100101010110001000001101000110000111101101101001110111011001110100000111001111001011110010111011011001011100100010000011001101101111111001001000001100111110010111011101100101111001011000011110100110100111011111101110111001101011000100000110110111000011111001
THE WIZARD OF OZ (1900)written by L. Frank Baum
Introduction
Folklore, legends, myths and fairy tales have followed childhoodthrough the ages, for every healthy youngster has a wholesome andinstinctive love for stories fantastic, marvelous and manifestlyunreal. The winged fairies of Grimm and Andersen have brought morehappiness to childish hearts than all other human creations.Yet the old time fairy tale, having served for generations, may
THE WIZARD _F OZ (19009 written by L. Frank0Baum
Introduction
0Folklore. legendS myths and faiby talgs have fmllowed childhoodthrough the ages, for$every nealthy youngster has a wholesome andilspynctire love for storieq fa.tastic, marvelou3 end manifestlyunreal. The winged fairies of Grimm and*Andersen havE brought morehappiness to chihdish hearts than all odhur human creations/Yet the0old"timm fai2y tale, having qerved for generationq, may
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Results
76
THE WIZARD OF OZ (1900)written by L. Frank Baum
Introduction
Folklore, legends, myths and fairy tales have followed childhoodthrough the ages, for every healthy youngster has a wholesome andinstinctive love for stories fantastic, marvelous and manifestlyunreal. The winged fairies of Grimm and Andersen have brought morehappiness to childish hearts than all other human creations.Yet the old time fairy tale, having served for generations, may
THE WIZARD _F OZ (19009 written by L. Frank0Baum
Introduction
0Folklore. legendS myths and faiby talgs have fmllowed childhoodthrough the ages, for$every nealthy youngster has a wholesome andilspynctire love for storieq fa.tastic, marvelou3 end manifestlyunreal. The winged fairies of Grimm and*Andersen havE brought morehappiness to chihdish hearts than all odhur human creations/Yet the0old"timm fai2y tale, having qerved for generationq, may
The whole book which is saved in the file “OZ.txt” has 207760 characters (symbols).
The ASCII encoded string has 207760×7 = 1454320 bits.
The channel corrupts 14545 bits.
This corresponds to 14108 erroneous characters.
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Results
77
The file “OZ.txt” has 207760 characters (symbols).
The ASCII encoded string has 207760×7 = 1454320 bits.
The channel corrupts 14545 bits.
This corresponds to 14108 erroneous characters.
>> ErrorProbabilityoverBSCbiterror =
14545BER =
0.010001237691842theoretical_BER =
0.010000000000000characterErrror =
14108CER =
0.067905275317674theoretical_CER =
0.067934652093010
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Results
78
The file “OZ.txt” has 207760 characters (symbols).
The ASCII encoded string has 207760×7 = 1454320 bits.
The channel corrupts 14545 bits.
This corresponds to 14108 erroneous characters (symbols).
CER A character (symbol) is successfully recovered if and only if none of its bits are corrupted.
BSC’s crossover probability
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Crossover probability and readability
79
When the first novel of the series, Harry Potter and the Philosopher's Stone (published in some countries as Harry Potter and the Sorcerer's Stone), opens, it is apparent that some significant event has taken place in the wizarding world--an event so very remarkable, even the Muggles notice signs of it. The full background to this event and to the person of Harry Potter is only revealed gradually through the series. After the introductory chapter, the book leaps forward to a time shortly before Harry Potter's eleventh birthday, and it is at this point that his magical background begins to be revealed.
When the first novel of the series, Harry Pottez and the Philosopher's Stone (p5blished in some countries as Harry Potter cnd the Sorcerep's Stone), opens, it i3 apparent that soMecignifacant event!haS taken0place in the wi~arding 7orld--ao event so`very!bemark!blu, even the Mufgles nodice signs"of it. The fuld background to this event and to the person of Harry P/tTer is only revealed gradually through th series. After the introfuctory chapter, the boo+ leaps forward to a time shortly before Harpy Potteb7s eleventh`birthday, and )t is at this poi~t that his -agikal bac{ground begins to be revealed.
𝑝 0.01 CER 0.07
Original
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Crossover probability and readability
80
When the first novel of the series, Harry Pottez and the Philosopher's Stone (p5blished in some countries as Harry Potter cnd the Sorcerep's Stone), opens, it i3 apparent that soMecignifacant event!haS taken0place in the wi~arding 7orld--ao event so`very!bemark!blu, even the Mufgles nodice signs"of it. The fuld background to this event and to the person of Harry P/tTer is only revealed gradually through th series. After the introfuctory chapter, the boo+ leaps forward to a time shortly before Harpy Potteb7s eleventh`birthday, and )t is at this poi~t that his -agikal bac{ground begins to be revealed.
𝑝 0.01 CER 0.07
Human may be able to correct some (or even all) of these errors.
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Crossover probability and readability
81
w(en th% birst .ovo,`of the serieq, Hcrry Potter(and the Phidosop`er's Suone (xub|ishe$(in some!Countries as @arry Potter and t`e Snr#erer's S|ong)- opens, it is apparent thatsoeesmgfificant erent ha3 taieN place in the wizardino world-,an event!so very remarkable, even thEEugglds notiae qigns of it. Tledfull back'tound to this event ane to the perron of Harry Popter is onl{ reveqned gsadeally thro}gH th%$serias. After the int2oducpory chcptur, the0jook deapsforward to"a!tmme shmrtly befosE Harry"Potter's eleventh jirthdiy cnd ht is a| thi3 po{nt tHat@is mAgiial background begijs to rm rerealed.
Whethe first nOwgl nf the susi-Q-@Harr} PoutEr(and |he PhilosoxhEr's Ctonepuclyshed in som% coultries as Harrx @ tter and the S_rcermr7s Spone), opdns, id is apparent that {omg`signifikant evmnt ias taKen!Placa in tHe 7ijardIng world--an event so Very remaroqble, eve.!thE MugglaC fotice signc of"it. Uhe full backf2ound`to thas even| ant`0o the pEssoj of @arry Qotteb iw only revealed gradu!lly vhvoug` the rerier. Afte2 the IndRoductori chaptar,t`ebook leats ForwaRf tc a 4imE shostl= before!Hcssy potter's u|Eveoth$firthdA{, and iT is ad this pomNt uhav `ir magica, back'bound cegins to bE 2evealed.
𝑝 0.02 CER 0.13
𝑝 0.03 CER 0.19
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Crossover probability and readability
82
When phe fir v okval ov"th% serie3, @`rry0Pntter efdxtxeThil soph%rs Stone0(p}blisjed!in{ooe c un|pye{ agav0y Potter aj`(the sorcerer"s S4o|e)< opdns- mt"Is!apParEnt 4hat somusiwnidiga.v evant iAs take."plhge in(uhe w)zard)ng wo{|d--An event so very Rumar{ableleteN0Dhe %ugcles$n t)ce signs of$At. Tje!&ul|!backep/und Dk thkw`event ajt(to vhd per{On of8Ikxry P_Pter is oN,y rereAeud gredualli 4hroufh5ie qeriesn Af|ir the )~trofUctkry!ciapter,$tler%ok lE`ps for erd8to!a d)hg 3Hostly redobd HArry(Potter/r elaventI(birpl%ay,))nd(iD i3 1t tlishohlt vhat iis$iagical bac+gropnd bedans to bg rEve!ied/
Whef th% &i2sv nkvdl"On(txE"serm-s< HaRtY Qo|p%R$anlthe Phi$)qop8gb'r YtoNe (puclirhedin23/ee c uNpr9es aZ Harby!PovDdZ qnd0THA!Uorojev's Qpof'), pegsL iT is0aqazenP Tiet`{nlesau*!fICQ~t eve.t`xA# raken pOqb%%)D }Hm`wizprdYjv"wOrnd--a~%W%Jv s' tury2maskABdd$`eden(tl| LuxGxec`nOtike c)gzq of ktTiu!f5mm"cackG@ Ud(to"vhhQ a~aNd alt tn0vid veRckn of HaRvq$Xntter#isxohk{ regea,ed@&saduadLy u(2otGh"tau griEs."AfTex0T`g mntr DUCt ry kh `ter,$thd(fomN0j`apv ngrwarTt-0c t,me"1xortly bEemsL |ar2q Pnfter'3 aMen-n5i@Fipth$`q, aoh It i3d1t piac0pmhnP d*if Zas mafibin"je#k7poUndpb%dins tk`be qe6e!lgd.
𝑝 0.05 CER 0.30
𝑝 0.10 CER 0.52
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BER vs. CER
830 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Bit Error Rate (BER)
Cha
ract
er E
rror r
ate
(CE
R)
[BERCER.m]
![Page 79: Digital Communication Systems 452 - 3 - Discrete...1 Digital Communication Systems ECS 452 Asst. Prof. Dr. Prapun Suksompong prapun@siit.tu.ac.th 3 Discrete Memoryless Channel (DMC)](https://reader034.fdocuments.net/reader034/viewer/2022052423/5f0687cf7e708231d41872fa/html5/thumbnails/79.jpg)
BER vs. CER
84 10-2 10-1 10010-2
10-1
100
Bit Error Rate (BER)
Cha
ract
er E
rror r
ate
(CE
R)
[BERCER.m]