I am currently working on a project in MATLAB that compares two different numerical methods (direct and iterative), each for two different matrix sizes, to the actual analytic solution of the system. The first .m file included below provides an implementation of a tridiagonal matrix solver, where the main function plots a solution for two different sizes of tridiagonal matrices: a 2500x2500 matrix and 50x50 matrix.
function main
n=50;
for i=1:2
di=2*ones(n,1);
up=-ones(n,1);
lo=-ones(n,1);
b=ones(n,1)/(n*n);
[subl,du,supu]=tridiag_factor(lo,di,up);
usol = tridiag_solve(subl,du,supu,b);
%figure(i)
plot(usol);
hold on;
title(['tridiag ' num2str(n) ''])
n=2500;
end
function [subl,du,supu]=tridiag_factor(subd,d,supd)
n=length(d);
for i=1:n-1
subd(i)=subd(i)/d(i);
d(i+1)=d(i+1)-subd(i)*supd(i);
end
subl=subd; supu=supd; du=d;
function x = tridiag_solve(subl,du,supu,b)
% forward solve
n=length(b); z=b;
z(1)=b(1);
for i=2:n
z(i)=b(i)-subl(i)*z(i-1);
end
% back solve
x(n)=z(n);
for i=n-1:-1:1
x(i)=(z(i)-supu(i)*x(i+1))/du(i);
end
I have included a hold on statement after the first usol is plotted so I can capture the second usol on the same figure during the second iteration of the for loop. The next piece of code is a Jacobi method that iteratively solves a system for a tridiagonal matrix. In my case, I have constructed two tridiagonal matrices of different sizes which I will provide after this piece of code:
function [xf,r] = jacobi(A,b,x,tol,K)
%
% The function jacobi applies Jacobi's method to solve A*x = b.
% On entry:
% A coefficient matrix of the linear system;
% b right-hand side vector of the linear system;
% x initial approximation;
% eps accuracy requirement: sx=top when norm(dx) < eps;
% K maximal number of steps allowed.
% On return:
% x approximation for the solution of A*x = b;
% r residual vector last used to update x,
% if success, then norm(r) < eps.[
%
n = size(A,1);
fprintf('Running the method of Jacobi...\n');
for k = 1:K
% r = b - A*x;
for i=1:n
sum=0;
for j = 1:n
if j~=i
sum=sum+A(i,j)*x(j);
end
end
x(i)=(b(i)-sum)/(A(i,i));
end;
k=k+1;
r = b - A*x;
% fprintf(' norm(r) = %.4e\n', norm(r));
if (norm(r) < tol)
fprintf('Succeeded in %d steps\n', k);
return;
end;
end
fprintf('Failed to reached accuracy requirement in %d steps.\n', K);
xf=x;
%r(j) = r(j)/A(j,j);
%x(j) = x(j) + r(j);
Now, lastly, here is my code for the two tridiagonal matrices (and other related information for each system corresponding to the appropriate matrices) I wish to use in my two jacobi function calls:
% First tridiagonal matrix
n=50;
A_50=full(gallery('tridiag', n, -1, 2, -1));
% Second tridiagonal matrix
n=2500;
A_2500=full(gallery('tridiag', n, -1, 2, -1));
K=10000;
tol=1e-6;
b_A50=ones(length(A_50), 1);
for i=1:length(A_50)
b_A50(i,1)=0.0004;
end
x_A50=ones(length(A_50),1);
b_A2500=ones(length(A_2500), 1);
for i=1:length(A_2500)
b_A2500(i,1)= 1.6e-7;
end
x_A25000=ones(length(A_2500),1);
As stated in the question header, I need to plot the vectors produced from jacobi(A_50,b_A50,x_A50,tol,K), jacobi(A_2500,b_A2500,x_A25000,tol,K) and a mathematical function y=#(x) -0.5*(x-1)*x all on the same figure along with the two usol (for n=50 and n=2500) produced by the main function shown in the first piece of code, but I have had no luck due to the different lengths of vectors. I understand this is probably an easy fix, but of course my novice MATLAB skills are not sufficing.
Note: I understand x_A50 and x_A2500 are trivial choices for x, but I figured while I ask for help I better keep it simple for now and not create any more issues in my code.
In MATLAB traces sent to same plot must have same length.
I have allocated just one variable containing all traces, for the respective tridiagonals and resulting traces out of your jacobi function.
I have shortened from 2500 to 250, and the reason is that with 2500, compared to 50, the tridiag traces, and the short jacobi are so flat compared to the last trace than one has to recur to dB scale to find them on same graph window as you asked.
1st generate all data, and then plot all data.
So here you have the script plotting all traces in same graph :
clear all;close all;clc
n=[50 250];
us_ol=zeros(numel(n),max(n));
% generate data
for k=[1:1:numel(n)]
di=2*ones(n(k),1);
up=-ones(n(k),1);
lo=-ones(n(k),1);
b=ones(n(k),1)/(n(k)*n(k));
[subl,du,supu]=tridiag_factor(lo,di,up);
us_ol0 = tridiag_solve(subl,du,supu,b);
us_ol(k,[1:numel(us_ol0)])=us_ol0;
end
n_us_ol=[1:1:size(us_ol,2)];
str1=['tridiag ' num2str(n(1))];
str2=['tridiag ' num2str(n(2))];
legend(str1,str2);
grid on
% the jacobi traces
nstp=1e3;
tol=1e-6;
A1=zeros(max(n),max(n),numel(n));
for k=1:1:numel(n)
A0=full(gallery('tridiag', n(k), -1, 2, -1));
A1([1:1:size(A0,1)],[1:1:size(A0,2)],k)=A0;
end
b_A1=ones(max(n),max(n),numel(n));
for k=1:1:numel(n)
for i=1:n(k)
b_A1(i,1,k)=0.0004;
end
end
n_A1=[1:1:size(A1,1)];
jkb=zeros(numel(n),max(n));
for k=1:1:numel(n)
A0=A1([1:n(k)],[1:n(k)],k);
b_A0=b_A1([1:n(k)],[1:n(k)],k);
n0=[1:1:n(k)];
jkb0=jacobi(A0,b_A0,n0',tol,nstp)
jkb(k,[1:numel(jkb0)])=jkb0';
end
% plot data
figure(1)
ax1=gca
plot(ax1,n_us_ol,us_ol(1,:),n_us_ol,us_ol(2,:));
hold(ax1,'on')
plot(ax1,n_A1,jkb(1,:),n_A1,jkb(2,:))
grid on
legend('3/1','3/2','jkb1','jkb2')
title('3diags and jakobians graph')
As mentioned above, one has to zoom to find some of the traces
One way to combine really small traces with large traces is to use Y log scale
figure(2)
ax2=gca
plot(ax2,n_us_ol,10*log10(us_ol(1,:)),n_us_ol,10*log10(us_ol(2,:)));
hold(ax2,'on')
plot(ax2,n_A1,10*log10(jkb(1,:)),n_A1,10*log10(jkb(2,:)))
grid on
legend('3/1[dB]','3/2[dB]','jkb1[dB]','jkb2[dB]')
title('3diags and jakobians graph in dB')
I have a problem with my code. My objective is to obtain the Fourier coefficients with the passage matrices. Then use the fft and check that my two results are the same. But I get different results and I don't understand why?
clear all;
N=[50];
for k=1:length(N)
Dx=1/(N(k)-1);
x=linspace(0,1-Dx,N(k));
for j=1:N(k)
f(j,k)=100.*exp(-20*x(j))*(x(j)-(x(j)).^2);
end
for j=1:N(k)
for m=1:N(k)
Mphsp(j,m)=exp((2*pi*i*(m-1)*(j-1))/N(k));
Mspph(j,m)=(1/N(k)).*exp(-(2*pi*i*(m-1)*(j-1))/N(k));
end
end
Idd=Mphsp*Mspph;
coeff(1:N(k),k)=Mspph*f(1:N(k),k);
coeff2(1:N(k),k)=fft(f(1:N(k),k));
verf(1:N(k),k)=coeff2(1:N(k),k)-coeff(1:N(k),k);
end
If anyone has any ideas? Please.
Your for loop is wrong. length(N) = 1, because N is a 1x1 matrix.
clear all;
N=[50];
for k=1:length(N)
I suppose you want to do something like
N = zeros(50, 1);
for k = 1:50
N(k) = k;
end
for k=1:length(N)
...
I have a problem when calculate discrete Fourier transform in MATLAB, apparently get the right result but when plot the amplitude of the frequencies obtained you can see values very close to zero which should be exactly zero. I use my own implementation:
function [y] = Discrete_Fourier_Transform(x)
N=length(x);
y=zeros(1,N);
for k = 1:N
for n = 1:N
y(k) = y(k) + x(n)*exp( -1j*2*pi*(n-1)*(k-1)/N );
end;
end;
end
I know it's better to use fft of MATLAB, but I need to use my own implementation as it is for college.
The code I used to generate the square wave:
x = [ones(1,8), -ones(1,8)];
for i=1:63
x = [x, ones(1,8), -ones(1,8)];
end
MATLAB version: R2013a(8.1.0.604) 64 bits
I have tried everything that has happened to me but I do not have much experience using MATLAB and I have not found information relevant to this issue in forums. I hope someone can help me.
Thanks in advance.
This will be a numerical problem. The values are in the range of 1e-15, while the DFT of your signal has values in the range of 1e+02. Most likely this won't lead to any errors when doing further processing. You can calculate the total squared error between your DFT and the MATLAB fft function by
y = fft(x);
yh = Discrete_Fourier_Transform(x);
sum(abs(yh - y).^2)
ans =
3.1327e-20
which is basically zero. I would therefore conclude: your DFT function works just fine.
Just one small remark: You can easily vectorize the DFT.
n = 0:1:N-1;
k = 0:1:N-1;
y = exp(-1j*2*pi/N * n'*k) * x(:);
With n'*k you create a matrix with all combinations of n and k. You then take the exp(...) of each of those matrix elements. With x(:) you make sure x is a column vector, so you can do the matrix multiplication (...)*x which automatically sums over all k's. Actually, I just notice, this is exactly the well-known matrix form of the DFT.
Generate a plot showing the graphs of
y=(2*a+1)*exp(-x)-(a+1)*exp(2*x)
in the range x ∈ <-2, 4> for all integer values of a between -3 and 3
I know how to make typical plot for 2 values and set a range on the axes, but how to draw the graph dependent on the parameter a?
To elaborate on Ben Voigt's comment: A more advanced technique would be to replace the for-loop with a call to bsxfun to generate a matrix of evaluations of M(i,j) = f(x(i),a(j)) and call plot with this matrix. Matlab will then use the columns of the matrix and plot each column with individual colors.
%%// Create a function handle of your function
f = #(x,a) (2*a+1)*exp(-x)-(a+1)*exp(2*x);
%%// Plot the data
x = linspace(-2, 4);
as = -3:3;
plot(x, bsxfun(f,x(:),as));
%%// Add a legend
legendTexts = arrayfun(#(a) sprintf('a == %d', a), as, 'uni', 0);
legend(legendTexts, 'Location', 'best');
You could also create the evaluation matrix using ndgrid, which explicitly returns all combinations of the values of x and as. Here you have to pay closer attention on properly vectorizing the code. (We were lucky that the bsxfun approach worked without having to change the original f.)
f = #(x,a) (2*a+1).*exp(-x)-(a+1).*exp(2*x); %// Note the added dots.
[X,As] = ndgrid(x,as);
plot(x, f(X,As))
However for starters, you should get familiar with loops.
You can do it using a simple for loop as follows. You basically loop through each value of a and plot the corresponding y function.
clear
clc
close all
x = -2:4;
%// Define a
a = -3:3;
%// Counter for legend
p = 1;
LegendText = cell(1,numel(a));
figure;
hold on %// Important to keep all the lines on the same plot.
for k = a
CurrColor = rand(1,3);
y= (2*k+1).*exp(-x)-(k+1).*exp(2.*x);
plot(x,y,'Color',CurrColor);
%// Text for legend
LegendText{p} = sprintf('a equals %d',k);
p = p+1;
end
legend(LegendText,'Location','best')
Which gives something like this:
You can customize the graph as you like. Hope that helps get you started!
Consider the following code, which makes up a random data set and fits polynomials of order 1 to 5 against the data:
x=1:100;
data=exp(-rand(1).*x);
for ii = 1:1:5;
polyfit(x,data,ii);
end
I am wondering what is the most elegant way to store the polyfit coefficients during each loop? I thought
fitCoef(ii,:) = polyfit(x,data,ii)
would work but i am getting a Subscripted assignment dimension mismatch. error at the ii = 2 iteration. I understand why, because it is trying to put a vector of 3 values into a vector who's length is only 2... but am not sure how to properly resolve this problem.
My ideal output is a matrix which displays the coefficients on each row so e.g.
p(1) p(2)
p(1) p(2) p(3)
p(1) p(2) p(3) p(4)
But I don't know how to deal with the black spaces?
Thanks
You could skip the for loop entirely as well:
A = cellfun(#(degree) polyfit(x, data, degree), num2cell(1:5), 'UniformOutput', false);
You could initialise a matrix of zeros before beginning the loop, using fitCoeffs=zeros(5,6). This wont affect polyval.
You could also save the results into a cell array, fitCoeffs{ii}=polyfit(x,data,ii), then to use each set of coefficients do, for example, polyval(fitCoeffs{3},X).
I've illustrated both options below:
x=1:100;
data=exp(-rand(1).*x);
N=5;
P=zeros(N,N+1);
P2=cell(1,N);
for ii = 1:N;
P(ii,1:ii+1)=polyfit(x,data,ii);
P2{ii}=polyfit(x,data,ii);
end
P
P2
polyval(P(3,:),1)
polyval(P(3,1:4),1) % Note it doesn't matter if you leave the zeros in
polyval(P2{3},1)