I'm trying to compute the Fourier coefficients for a waveform using MATLAB. The coefficients can be computed using the following formulas:
T is chosen to be 1 which gives omega = 2pi.
However I'm having issues performing the integrals. The functions are are triangle wave (Which can be generated using sawtooth(t,0.5) if I'm not mistaking) as well as a square wave.
I've tried with the following code (For the triangle wave):
function [ a0,am,bm ] = test( numTerms )
b_m = zeros(1,numTerms);
w=2*pi;
for i = 1:numTerms
f1 = #(t) sawtooth(t,0.5).*cos(i*w*t);
f2 = #(t) sawtooth(t,0.5).*sin(i*w*t);
am(i) = 2*quad(f1,0,1);
bm(i) = 2*quad(f2,0,1);
end
end
However it's not getting anywhere near the values I need. The b_m coefficients are given for a
triangle wave and are supposed to be 1/m^2 and -1/m^2 when m is odd alternating beginning with the positive term.
The major issue for me is that I don't quite understand how integrals work in MATLAB and I'm not sure whether or not the approach I've chosen works.
Edit:
To clairify, this is the form that I'm looking to write the function on when the coefficients have been determined:
Here's an attempt using fft:
function [ a0,am,bm ] = test( numTerms )
T=2*pi;
w=1;
t = [0:0.1:2];
f = fft(sawtooth(t,0.5));
am = real(f);
bm = imag(f);
func = num2str(f(1));
for i = 1:numTerms
func = strcat(func,'+',num2str(am(i)),'*cos(',num2str(i*w),'*t)','+',num2str(bm(i)),'*sin(',num2str(i*w),'*t)');
end
y = inline(func);
plot(t,y(t));
end
Looks to me that your problem is what sawtooth returns the mathworks documentation says that:
sawtooth(t,width) generates a modified triangle wave where width, a scalar parameter between 0 and 1, determines the point between 0 and 2π at which the maximum occurs. The function increases from -1 to 1 on the interval 0 to 2πwidth, then decreases linearly from 1 to -1 on the interval 2πwidth to 2π. Thus a parameter of 0.5 specifies a standard triangle wave, symmetric about time instant π with peak-to-peak amplitude of 1. sawtooth(t,1) is equivalent to sawtooth(t).
So I'm guessing that's part of your problem.
After you responded I looked into it some more. Looks to me like it's the quad function; not very accurate! I recast the problem like this:
function [ a0,am,bm ] = sotest( t, numTerms )
bm = zeros(1,numTerms);
am = zeros(1,numTerms);
% 2L = 1
L = 0.5;
for ii = 1:numTerms
am(ii) = (1/L)*quadl(#(x) aCos(x,ii,L),0,2*L);
bm(ii) = (1/L)*quadl(#(x) aSin(x,ii,L),0,2*L);
end
ii = 0;
a0 = (1/L)*trapz( t, t.*cos((ii*pi*t)/L) );
% now let's test it
y = ones(size(t))*(a0/2);
for ii=1:numTerms
y = y + am(ii)*cos(ii*2*pi*t);
y = y + bm(ii)*sin(ii*2*pi*t);
end
figure; plot( t, y);
end
function a = aCos(t,n,L)
a = t.*cos((n*pi*t)/L);
end
function b = aSin(t,n,L)
b = t.*sin((n*pi*t)/L);
end
And then I called it like:
[ a0,am,bm ] = sotest( t, 100 );
and I got:
Sweetness!!!
All I really changed was from quad to quadl. I figured that out by using trapz which worked great until the time vector I was using didn't have enough resolution, which led me to believe it was a numerical issue rather than something fundamental. Hope this helps!
To troubleshoot your code I would plot the functions you are using and investigate, how the quad function samples them. You might be undersampling them, so make sure your minimum step size is smaller than the period of the function by at least factor 10.
I would suggest using the FFTs that are built-in to Matlab. Not only is the FFT the most efficient method to compute a spectrum (it is n*log(n) dependent on the length n of the array, whereas the integral in n^2 dependent), it will also give you automatically the frequency points that are supported by your (equally spaced) time data. If you compute the integral yourself (might be needed if datapoints are not equally spaced), you might calculate frequency data that are not resolved (closer spacing than 1/over the spacing in time, i.e. beyond the 'Fourier limit').
Related
I need to solve this differential equation using Runge-Kytta 4(5) on Scilab:
The initial conditions are above. The interval and the h-step are:
I don't need to implement Runge-Kutta. I just need to solve this and plot the result on the plane:
I tried to follow these instructions on the official "Scilab Help":
https://x-engineer.org/graduate-engineering/programming-languages/scilab/solve-second-order-ordinary-differential-equation-ode-scilab/
The suggested code is:
// Import the diagram and set the ending time
loadScicos();
loadXcosLibs();
importXcosDiagram("SCI/modules/xcos/examples/solvers/ODE_Example.zcos");
scs_m.props.tf = 5000;
// Select the solver Runge-Kutta and set the precision
scs_m.props.tol(6) = 6;
scs_m.props.tol(7) = 10^-2;
// Start the timer, launch the simulation and display time
tic();
try xcos_simulate(scs_m, 4); catch disp(lasterror()); end
t = toc();
disp(t, "Time for Runge-Kutta:");
However, it is not clear for me how I can change this for the specific differential equation that I showed above. I have a very basic knowledge of Scilab.
The final plot should be something like the picture bellow, an ellipse:
Just to provide some mathematical context, this is the differential equation that describes the pendulum problem.
Could someone help me, please?
=========
UPDATE
Based on #luizpauloml comments, I am updating this post.
I need to convert the second-order ODE into a system of first-order ODEs and then I need to write a function to represent such system.
So, I know how to do this on pen and paper. Hence, using z as a variable:
OK, but how do I write a normal script?
The Xcos is quite disposable. I only kept it because I was trying to mimic the example on the official Scilab page.
To solve this, you need to use ode(), which can employ many methods, Runge-Kutta included. First, you need to define a function to represent the system of ODEs, and Step 1 in the link you provided shows you what to do:
function z = f(t,y)
//f(t,z) represents the sysmte of ODEs:
// -the first argument should always be the independe variable
// -the second argument should always be the dependent variables
// -it may have more than two arguments
// -y is a vector 2x1: y(1) = theta, y(2) = theta'
// -z is a vector 2x1: z(1) = z , z(2) = z'
z(1) = y(2) //first equation: z = theta'
z(2) = 10*sin(y(1)) //second equation: z' = 10*sin(theta)
endfunction
Notice that even if t (the independent variable) does not explicitly appear in your system of ODEs, it still needs to be an argument of f(). Now you just use ode(), setting the flag 'rk' or 'rkf' to use either one of the available Runge-Kutta methods:
ts = linspace(0,3,200);
theta0 = %pi/4;
dtheta0 = 0;
y0 = [theta0; dtheta0];
t0 = 0;
thetas = ode('rk',y0, t0, ts, f); //the output have the same order
//as the argument `y` of f()
scf(1); clf();
plot2d(thetas(2,:),thetas(1,:),-5);
xtitle('Phase portrait', 'theta''(t)','theta(t)');
xgrid();
The output:
I have a scilab function that looks something like this (very simplified code just to get the concept of how it works):
function [A, S, Q]=myfunc(a)
A = a^2;
S = a+a+a;
if S > A then
Q = "Bigger";
else
Q = "Lower";
end
endfunction
And I get the expected result if I run:
--> [A,S,Q]=myfunc(2)
Q =
Bigger
S =
6.
A =
4.
But if I put matrices into the function I expect to get equivalent matrices back as an answer with a result but instead I got this:
--> [A,S,Q]=myfunc([2 4 6 8])
Q =
Lower
S =
6. 12. 18. 24.
A =
4. 16. 36. 64.
Why isn't Q returning matrices of values like S and A? And how do I achieve that it will return "Bigger. Lower. Lower. Lower." as an answer? That is, I want to perform the operation on each element of the matrix.
Because in your program you wrote Q = "Bigger" and Q = "Lower". That means that Q will only have one value. If you want to store the comparisons for every value in A and S, you have to make Scilab do that.
You can achieve such behavior by using loops. This is how you can do it by using two for loops:
function [A, S, Q]=myfunc(a)
A = a^2;
S = a+a+a;
//Get the size of input a
[nrows, ncols] = size(a)
//Traverse all rows of the input
for i = 1 : nrows
//Traverse all columns of the input
for j = 1 : ncols
//Compare each element
if S(i,j) > A(i,j) then
//Store each result
Q(i,j) = "Bigger"
else
Q(i,j) = "Lower"
end
end
end
endfunction
Beware of A = a^2. It can break your function. It has different behaviors if input a is a vector (1-by-n or n-by-1 matrix), rectangle matrix (m-by-n matrix, m ≠ n ), or square matrix (n-by-n matrix):
Vector: it works like .^, i.e. it raises each element individually (see Scilab help).
Rectangle: it won't work because it has to follow the rule of matrix multiplication.
Square: it works and follows the rule of matrix multiplication.
I will add that in Scilab, the fewer the number of loop, the better : so #luispauloml answer may rewrite to
function [A, S, Q]=myfunc(a)
A = a.^2; // used element wise power, see luispauloml advice
S = a+a+a;
Q(S > A) = "Bigger"
Q(S <= A) = "Lower"
Q = matrix(Q,size(a,1),size(a,2)) // a-like shape
endfunction
I have the following code in Octave for implementing the composite trapezoid rule and for some reason the function only stalls whenever I execute it in Octave on f = #(x) x^2, a = 0, b = 4, TOL = 10^-6. Whenever I call trapezoid(f, a, b, TOL), nothing happens and I have to exit the Terminal in order to do anything else in Octave. Here is the code:
% INPUTS
%
% f : a function
% a : starting point
% b : endpoint
% TOL : tolerance
function root = trapezoid(f, a, b, TOL)
disp('test');
max_iterations = 10000;
disp(max_iterations);
count = 1;
disp(count);
initial = (b-a)*(f(b) + f(a))/2;
while count < max_iterations
disp(initial);
trap_0 = initial;
trap_1 = 0;
trap_1_midpoints = a:(0.5^count):b;
for i = 1:(length(trap_1_midpoints)-1)
trap_1 = trap_1 + (trap_1_midpoints(i+1) - trap_1_midpoints(i))*(f(trap_1_midpoints(i+1) + f(trap_1_midpoints(i))))/2;
endfor
if abs(trap_0 - trap_1) < TOL
root = trap_1;
return;
endif
intial = trap_1;
count = count + 1;
disp(count);
endwhile
disp(['Process ended after ' num2str(max_iterations), ' iterations.']);
I have tried your function in Matlab.
Your code is not stalling. It is rather that the size of trap_1_midpoints increases exponentionaly. With that the computation time of trap_1 increases also exponentionaly. This is what you experience as stalling.
I also found a possible bug in your code. I guess the line after the if clause should be initial = trap_1. Check the missing 'i'.
With that, your code still takes forever, but if you increase the tolerance (e.g. to a value of 1) your code return.
You could try to vectorize the for loop for speed up.
Edit: I think inside your for loop, a ) is missing after f(trap_1_midpoints(i+1).
After count=52 or so, the arithmetic sequence trap_1_midpoints is no longer representable in any meaningful fashion in floating point numbers. After count=1075 or similar, the step size is no longer representable as a positive floating point double number. That all is to say, the bound max_iterations = 10000 is ludicrous. As explained below, all computations after count=20 are meaningless.
The theoretical error for stepsize h is O(T·h^2). There is a numerical error accumulation in the summation of O(T/h) numbers that is of that size, i.e., O(mu/h) with mu=1ulp=2^(-52). Which in total means that the lowest error of the numerical integration can be expected around h=mu^(1/3), for double numbers thus h=1e-5 or in the algorithm count=17. This may vary with interval length and how smooth or wavy the function is.
One can expect the behavior that the error divides by four while halving the step size only for step sizes above this boundary 1e-5. This also means that abs(trap_0 - trap_1) is a reliable measure for the error of trap_0 (and abs(trap_0 - trap_1)/3 for trap_1) only inside this range of step sizes.
The error bound TOL=1e-6 should be met for about h=1e-3, which corresponds to count=10. If the recursion does not stop for count = 14 (which should give an error smaller than 1e-8) then the method is not accurately implemented.
Within my daily work, I have got to maximize a particular function making use of fminsearch; the code is:
clc
clear all
close all
f = #(x,c,k) -(x(2)/c)^3*(((exp(-(x(1)/c)^k)-exp(-(x(2)/c)^k))/((x(2)/c)^k-(x(1)/c)^k))-exp(-(x(3)/c)^k))^2;
c = 10.1;
k = 2.3;
X = fminsearch(#(x) f(x,c,k),[4,10,20]);
It works fine, as I expect, but not the issue is coming up: I need to bound x within certain limits, as:
4 < x(1) < 5
10 < x(2) < 15
20 < x(3) < 30
To achieve the proper results, I should use the optimization toolbox, that I unfortunately cannot hand.
Is there any way to get the same analysis by making use of only fminsearch?
Well, not using fminsearch directly, but if you are willing to download fminsearchbnd from the file exchange, then yes. fminsearchbnd does a bound constrained minimization of a general objective function, as an overlay on fminsearch. It calls fminsearch for you, applying bounds to the problem.
Essentially the idea is to transform your problem for you, in a way that your objective function sees as if it is solving a constrained problem. It is totally transparent. You call fminsearchbnd with a function, a starting point in the parameter space, and a set of lower and upper bounds.
For example, minimizing the rosenbrock function returns a minimum at [1,1] by fminsearch. But if we apply purely lower bounds on the problem of 2 for each variable, then fminsearchbnd finds the bound constrained solution at [2,4].
rosen = #(x) (1-x(1)).^2 + 105*(x(2)-x(1).^2).^2;
fminsearch(rosen,[3 3]) % unconstrained
ans =
1.0000 1.0000
fminsearchbnd(rosen,[3 3],[2 2],[]) % constrained
ans =
2.0000 4.0000
If you have no constraints on a variable, then supply -inf or inf as the corresponding bound.
fminsearchbnd(rosen,[3 3],[-inf 2],[])
ans =
1.4137 2
Andrey has the right idea, and the smoother way of providing a penalty isn't hard: just add the distance to the equation.
To keep using the anonymous function:
f = #(x,c,k, Xmin, Xmax) -(x(2)/c)^3*(((exp(-(x(1)/c)^k)-exp(-(x(2)/c)^k))/((x(2)/c)^k-(x(1)/c)^k))-exp(-(x(3)/c)^k))^2 ...
+ (x< Xmin)*(Xmin' - x' + 10000) + (x>Xmax)*(x' - Xmax' + 10000) ;
The most naive way to bound x, would be giving a huge penalty for any x that is not in the range.
For example:
function res = f(x,c,k)
if x(1)>5 || x(1)<4
penalty = 1000000000000;
else
penalty = 0;
end
res = penalty - (x(2)/c)^3*(((exp(-(x(1)/c)^k)-exp(-(x(2)/c)^k))/((x(2)/c)^k-(x(1)/c)^k))-exp(-(x(3)/c)^k))^2;
end
You can improve this approach, by giving the penalty in a smoother way.
I am learning MatLab on my own, and I have this assignment in my book which I don't quite understand. Basically I am writing a function that will calculate sine through the use of Taylor series. My code is as follows so far:
function y = sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
The book is now asking me to include an optional output err which will calculate the difference between sin(x) and y. The book hints that I may use nargout to accomplish this, but there are no examples in the book on how to use this, and reading the MatLab help on the subject did not make my any wiser.
If anyone can please help me understand this, I would really appreciate it!
The call to nargout checks for the number of output arguments a function is called with. Depending on the size of nargout you can assign entries to the output argument varargout. For your code this would look like:
function [y varargout]= sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
if nargout ==2
varargout{1} = sin(x)'-y;
end
Compare the output of
[y] = sine_series(rand(1,10),3)
and
[y err] = sine_series(rand(1,10),3)
to see the difference.