CODE error correction

Question

milad babaei 2011-9-25

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https://ww2.mathworks.cn/matlabcentral/answers/16676-code-error-correction

评论： Walter Roberson 2024-11-10

HI all,

would somebody please help me find out what should i do to plot and correct this script?

clear
clc;
B=1000; L=2000;Df=200; Sgama=.8;
nsamples=3000;
rmem = zeros(3,nsamples);
qult = zeros(1,nsamples);
for K=1:nsamples
  while true
  C = normrnd(620,147.64);
  if C<=450 || C >= 800 || ismember(C,rmem(1,1:K-1)); continue; end
   gama = normrnd(1.96,0.02);
  if gama <= 1.92 || gama >= 1.98 || ismember(gama,rmem(2,1:K-1)); continue; end
   fi = normrnd(3.76,1.1034);
  if fi <= 2.7 || fi >= 16.3 || ismember(fi,rmem(3,1:K-1)); continue; end
  rmem(:,K) = [C; gama; fi];
   break
  end
   q=107.25+(100*fi);
   dq=1+(.4*tan(fi*pi/180)*(1-sin(fi*pi/180)^2));
   Sq=1+(.5*sin(fi*pi/180));
   Nq = tan((pi/4)+(pi*fi/360)) * tan((pi/4)+(pi*fi/360)) * exp(pi*tan(fi*pi/180));
   Nc = (Nq-1)*cot(fi*pi/180);
   Sc=1+(.5* Nq / Nc);
    Ngama = 2*(Nq+1)*tan(fi*pi/180);
    qult(K)=(1.08*C*Nc*Sc)+(q*Nq*Sq*dq)+(429*Ngama);
  end
surf( C(1,1:nsamples), fi(1,1:nsamples), qult );
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

ERROR :

??? Index exceeds matrix dimensions.

Error in ==> graphQ at 27 surf( C(1,1:nsamples), fi(1,1:nsamples), qult );

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dimuthu chathuranga 2021-4-8

eroor occur communication somebody

help me

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Answer 1

UJJWAL 2011-9-25

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Hi ,

I don't see any reason as to why thic code should not give an error. C is not a vector. It is a single number(or let us say C is a vector of dimension 1 by 1. In the surf function you are trying to access C(1,1:nsamples) which obviously goes beyond the length of C . So an error must occour.

As far as correction is concerned, it is difficult unless you tell the nature of this code and what it is that you want to do. I defined a vector C1 and stored the individual values of C in it. I did the same for fi . However in that case surf would not work as it will require a 2D matrix in the form of qult to work. I believe that you want to plot qult for corresponding values of C and fi. However for that u need to use Plot3. I used it but I got something weird. So unless the nature of the code is not made clear it is extremely difficult to comment on the nature of the solution.

Happy to Help

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Walter Roberson 2011-9-25

Good point about surf() being the wrong call.

Walter Roberson 2011-9-25

UUJWAL is correct: with that flow of code, you will need to store each accepted C and fi value. Just before your q= assignment would seem to be an appropriate location.

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Answer 2

Vikash Anand 2021-12-13

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https://ww2.mathworks.cn/matlabcentral/answers/16676-code-error-correction#answer_853790

在 MATLAB Online 中打开

clc;
clear all;
close all;
n=7;k=4;
num_bit=10000;
genpoly=cyclpoly(n,k,'max');
SNRdB=0:10;
SNR=10.^(SNRdB/10);
for i=1:length(SNR)
 msg = randi(num_bit,k,[0,1]);
 code=encode(msg,n,k,'cyclic/binary',genpoly);
 [row, column]=size(code);
 codevec=reshape(code.',1,row*column);
 noise=awgn(codevec,SNRdB(i));
 y=codevec+noise;
 error=0;
 for j=1:length(y)
 if (y(j)>1&&codevec(j)==0)||(y(j)<0&&codevec(j)==1) 
 error=error+1;
 end
 end
 error=error/num_bit; 
m(i)=error;
end
y(y>0)=1;
 y(y<0)=0;
 decode_y=decode(y,n,k,'cyclic/binary',genpoly);
 decmsg=reshape(decode_y,num_bit,k);
 semilogy(SNRdB,m,'r','linewidth',2),grid on;
title(' Bit Error Rate verses SNR for Cyclic Code');
xlabel(' SNR(dB)');
ylabel('BER');

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Walter Roberson 2024-11-10

@Vikash Anand

I do not understand how this is an answer to the question that was originally asked?

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Answer 3

RAJESH 2024-9-25

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https://ww2.mathworks.cn/matlabcentral/answers/16676-code-error-correction#answer_1522465

编辑：Walter Roberson 2024-9-25

在 MATLAB Online 中打开

clc; clear all; close all;
% Input data points
x = [0 3 4 5 6];    % x-coordinates
y = [3 6 1 7 5];    % y-coordinates
% Number of intervals
N = length(x);
% Spline parameters (can be tuned for smoothness)
alpha = [0.5 0.8 0.7 0.6 0.9];  % Scale control parameters
r = 3 * ones(1, N);              % Shape parameters
intervals = [0 5; 3 5; 0 4; 3 4; 5 6];  % Define intervals
% Derivative values (arbitrary initialization for now)
d = [5.5 -3.5 0.5 2.0 -6.0];
d1 = [3.5 -1.5 5.5 2.5 1.5];
% Prepare X1 and Y1 matrices for the initial interpolation data
X1 = zeros(4, 4);  % Initialize X1 with zeros
x1 = [x(1) x(2) x(3) x(4)];
x2 = [x(2) x(3) x(4)];
x3 = [x(1) x(2) x(3)];
x4 = [x(2) x(3)];
t = [length(x1) length(x2) length(x3) length(x4)];
X1(1, 1:t(1)) = x1;
X1(2, 1:t(2)) = x2;
X1(3, 1:t(3)) = x3;
X1(4, 1:t(4)) = x4;
Y1 = zeros(4, 4);  % Initialize Y1 with zeros
y1 = [y(1) y(2) y(3) y(4)];
y2 = [y(2) y(3) y(4)];
y3 = [y(1) y(2) y(3)];
y4 = [y(2) y(3)];
t = [length(y1) length(y2) length(y3) length(y4)];
Y1(1, 1:t(1)) = y1;
Y1(2, 1:t(2)) = y2;
Y1(3, 1:t(3)) = y3;
Y1(4, 1:t(4)) = y4;
% Loop to compute parameters
a = zeros(1, 4);  % Initialize arrays for a(i) and b(i)
b = zeros(1, 4);
for i = 1:4
    % Compute a(i) and b(i)
    a(i) = (x(i+1) - x(i)) / (X1(i, t(i)) - X1(i, 1));
    b(i) = (X1(i, t(i)) * x(i) - X1(i, 1) * x(i+1)) / (X1(i, t(i)) - X1(i, 1));
end
% Main loop for calculating terms and interpolation
iter = 1;
L = []; L1 = [];
for i = 1:4
    rho = (x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1));  % Calculate rho
    
    % Calculate terms
    term1 = (y(i) - alpha(i) * Y1(i, 1)) * (1 - (x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1))).^3;
    term2 = (y(i+1) - alpha(i) * Y1(i, t(i))) * ((x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1))).^3;
    
    % Prevent accessing out of bounds
    if i < N
        term3 = (r(i) * (y(i) - alpha(i) * Y1(i, 1)) + (x(i+1) - x(i)) * d(i) - alpha(i) * d1(i, 1) * (X1(i, t(i)) - X1(i, 1))) * ...
            (1 - (x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1))).^2 * (x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1));
        term4 = (r(i) * (y(i+1) - alpha(i) * Y1(i, t(i))) - ((x(i+1) - x(i)) * d(i+1)) + alpha(i) * d1(i, t(i)) * (x1(i, t(i)) - x1(i, 1))) * ...
            (1 - (x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1))) * ((x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1))).^2;
    else
        term3 = 0;  % Handle boundary condition
        term4 = 0;  % Handle boundary condition
    end
    
    % Calculate numerator and denominator
    numerator = term1 + term2 + term3 + term4;
    denominator = 1 + (r(i) - 3) * (1 - (x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1))) * ((x(i) - X1(i, 1)) / (X1(i, t(i)) - X1(i, 1)));
    
    % Calculate q(i)
    q(i) = numerator / denominator;
end
Index in position 1 exceeds array bounds. Index must not exceed 1.
% Plot the result
plot(x, y, '.k', 'markersize', 20);  % Original data points
hold on;
plot(X1(1,:), Y1(1,:), 'b-');  % Interpolated fractal curve
xlabel('x');
ylabel('y');
title('Recurrent Rational Fractal Cubic Spline');
grid on;

Index in position 1 exceeds array bounds. Index must not exceed 1.

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EN-NEJJAR 2024-11-10

编辑：Walter Roberson 2024-11-10

在 MATLAB Online 中打开

%************************************************************** 
% PARTIE 2 : Simulation du vent par la distribution de Weibull
%**************************************************************
% On a relevé 50 mesures de vitesses du vent pendant une journée
clear
clc
% Définir les données de vitesse
vitesse = [6.2 16.1 16.3 19.0 12.2 8.1 8.8 5.9 7.3 8.2... 
    16.1 12.8 9.8 11.3 5.1 10.8 6.7 1.2 8.3 2.3 ... 
    4.3 2.9 14.8 4.6 3.1 13.6 14.5 5.2 5.7 6.5 ... 
    5.3 6.4 3.5 11.4 9.3 12.4 18.3 15.9 4.0 10.4 ... 
    8.7 3.0 12.1 3.9 6.5 3.4 8.5 0.9 9.9 7.9]; 
% Tracer l'histogramme des fréquences avec un pas de vitesse de 2m/s
binWidth = 2; 
binCtrs = 1:binWidth:20; 
% Utilisation de hist pour afficher l'histogramme
[counts, edges] = hist(vitesse, binCtrs); 
whos counts edges
  Name        Size            Bytes  Class     Attributes

  counts      1x10               80  double              
  edges       1x10               80  double              
% Tracer l'histogramme
bar(edges(1:end-1), counts / sum(counts), 'hist'); 
Error using barV6 (line 11)
X must be same length as Y.

Error in bar (line 90)
            h = barV6(parent,args{:});
xlabel('vitesse'); 
ylabel('Frequency %');
ylim([0 0.1]); 
hold on;
% Traçage automatique de la distribution de Weibull
% Estimation des paramètres de la distribution Weibull
paramEsts = wblfit(vitesse); 
% Créer une grille de valeurs pour le tracé de la courbe de Weibull
xgrid = linspace(0, 20, 100); 
% Calcul de la densité de probabilité de la distribution de Weibull
pdfEst = wblpdf(xgrid, paramEsts(1), paramEsts(2)); 
% Tracer la distribution de Weibull
line(xgrid, pdfEst, 'Color', 'r');
% Modifier l'apparence des barres de l'histogramme
h = get(gca, 'Children'); 
set(h, 'FaceColor', [0.9 0.9 0.9]);
% Titre et labels
xlabel('vitesse');
ylabel('Fréquence');
ylim([0 0.1]);
% Fin du tracé
hold off;

Walter Roberson 2024-11-10

You have edges(1:end-1) plotted against counts, but edges(1:end-1) is one shorter than counts is because edges and counts are the same size.

Walter Roberson 2024-11-10

在 MATLAB Online 中打开

d1 = [3.5 -1.5 5.5 2.5 1.5];

d1 is 1 x 5

        term3 = (r(i) * (y(i) - alpha(i) * Y1(i, 1)) + (x(i+1) - x(i)) * d(i) - alpha(i) * d1(i, 1) * (X1(i, t(i)) - X1(i, 1))) * ...

You access d1(i,1). When i becomes 2 that would be d1(2,1). However, there is no d1(2,1), only d1(1,2)

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CODE error correction

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