The code below currently plots the fourier series for a square wave for N terms. Is there any way I could change the range from [0;1] to [-1;1]?
% Assignment of variables
syms t
% Function variables
N = 5;
T0 = 1;
w0 = 2*pi/T0;
Imin = 0;
Imax = 0.5;
% Function
ft = 1;
% First term calculation
a0 = (1/T0)*int(ft, t, Imin, Imax);
y = a0;
% Calculation of n terms
for n = 1:N
an = (2/T0)*int(ft*cos(n*w0*t), t, Imin, Imax);
bn = (2/T0)*int(ft*sin(n*w0*t), t, Imin, Imax);
y = y + an*cos(n*w0*t) + bn*sin(n*w0*t);
end
fplot(y, [-4,4], "Black")
grid on
If you are talking about the figure scale, then ylim([-1 1])
1.- The following does what you asked for:
clear all;clc;close all
syms t
assume(t>0 & t<1)
% Function variables
N = 5;
T0 = 1;
w0 = 2*pi/T0;
Imin = 0;
Imax = 1;
% Function
h1=heaviside(t-.5)
h2=heaviside(t+.5)
ht=-2*((h1-h2)+.5)
% First term calculation
a0 = (1/T0)*int(ht, t, Imin, Imax);
y = a0;
% Calculation of n terms
for n = 1:N
an = (2/T0)*int(ht*cos(n*w0*t), t, Imin, Imax);
bn = (2/T0)*int(ht*sin(n*w0*t), t, Imin, Imax);
y = y + an*cos(n*w0*t) + bn*sin(n*w0*t);
end
fplot(y, [-4,4], "Black")
grid on
2.- You allocate a specific group of code lines headed with % Function to precisely define the function.
Yet you actually define the function with Imin and Imax.
It's good practice to constrain the function definition within the lines you intend for such purpose, not to scatter the function all over the place.
Related
I am trying to plot a 10% duty cycle square wave using matlab, but for some reason the amplitude of the series changes in a unpredictable way for different values of N. I was expecting the amplitude to be 1 (i.e. from [-1;1]. I am not sure what to change?
% Assignment of variables
syms t
[enter image description here][1]
% Function variables
N = 5;
T0 = 1;
w0 = 2*pi/T0;
Imin1 = 0;
Imax1 = 0.1;
% If square wave with mean at 0
bool = 1;
Imin2 = 0.1;
Imax2 = 1;
% Function
ft = 1;
% First term calculation
a0 = (1/T0)*int(ft, t, Imin1, Imax1) + (bool)*(1/T0)*int((-ft), t, Imin1, Imax1);
y = a0;
% Calculation of n terms
for n = 1:N
an = (2/T0)*int(ft*cos(n*w0*t), t, Imin1, Imax1) + (bool)*(2/T0)*int((-ft)*cos(n*w0*t), t, Imin2, Imax2);
bn = (2/T0)*int(ft*sin(n*w0*t), t, Imin1, Imax1) + (bool)*(2/T0)*int((-ft)*sin(n*w0*t), t, Imin2, Imax2);
y = y + an*cos(n*w0*t) + bn*sin(n*w0*t);
end
fplot(y, [0,4], "Black")
I have written the below MATLAB code. I want to know how can I optimize it without using for loop.
Any help will be very appreciated.
MATLAB code:
%Some parameters:
s = 50;
k = 50;
r = 0.1;
v = 0.2;
t = 2;
n=10000;
% Calculate CT by calling EurCall function
CT = EurCall(s, k, r, v, t, n);
%Function EurCall to be called
function C = EurCall(s, k, r, v, t, n)
X = zeros(n,1);
hh = zeros(n,1);
for ii = 1 : n
X(ii) = normrnd(0, 1);
SS = s*exp((r - v^2/2)*t + v*X(ii)*sqrt(t));
hh(ii) = exp(-r*t)*max(SS - k, 0);
end %end for loop
C = (1/n) * sum(hh);
end %end function
Vectorized Approach:
Here is a vectorized approach that I think replicates the same functionality as the original script. Instead of looping this example declares X as a vector of size n by 1. By using element-wise multiplication .* we can effectively calculate the remaining vectors SS and hh without need to loop through the indices. In this case SS and hh will also be vectors of size n by 1. I do agree with comment above that MATLAB's for-loops are no longer inherently slow.
%Some parameters:
s = 50;
k = 50;
r = 0.1;
v = 0.2;
t = 2;
n=10000;
% Calculate CT by calling EurCall function
[CT] = EurCall(s, k, r, v, t, n);
%Function EurCall to be called
function [C] = EurCall(s, k, r, v, t, n)
X = zeros(n,1);
hh = zeros(n,1);
mu = 0; sigma = 1;
%Creating a vector of normal random numbers of size (n by 1)%
X = normrnd(mu,sigma,[n 1]);
SS = s*exp((r - v^2/2)*t + v.*X.*sqrt(t));
hh = exp(-r*t)*max(SS - k, 0);
C = (1/n) * sum(hh);
end %end function
Ran using MATLAB R2019b
Find the error as a function of n, where the error is defined as the difference between two the voltage from the Fourier series (vF (t)) and the value from the ideal function (v(t)), normalized to the maximum magnitude (Vm ):
I am given this prompt where Vm = 1 V. Below this line is the code which I have written.
I am trying to write a function to solve this question: Plot the error versus time for n=3,n=5,n=10, and n=50. (10points). What does it look like I am doing incorrectly?
clc;
close all;
clear all;
% define the signal parameters
Vm = 1;
T = 1;
w0 = 2*pi/T;
% define the symbolic variables
syms n t;
% define the signal
v1 = Vm*sin(4*pi*t/T);
v2 = 2*Vm*sin(4*pi*t/T);
% evaluate the fourier series integral
an1 = 2/T*int(v1*cos(n*w0*t),0,T/2) + 2/T*int(v2*cos(n*w0*t),T/2,T);
bn1 = 2/T*int(v1*sin(n*w0*t),0,T/2) + 2/T*int(v2*sin(n*w0*t),T/2,T);
a0 = 1/T*int(v1,0,T/2) + 1/T*int(v2,T/2,T);
% obtain C by substituting n in c[n]
nmax = 100;
n = 1:nmax;
a = subs(an1);
b = subs(bn1);
% define the time vector
ts = 1e-2; % ts is sampling the
t = 0:ts:3*T-ts;
% directly plot the signal x(t)
t1 = 0:ts:T-ts;
v1 = Vm*sin(4*pi*t1/T).*(t1<=T/2);
v2 = 2*Vm*sin(4*pi*t1/T).*(t1>T/2).*(t1<T);
v = v1+v2;
x = repmat(v,1,3);
% Now fourier series reconstruction
N = [3];
for p = 1:length(N)
for i = 1:length(t)
for k = N(p)
x(k,i) = a(k)*cos(k*w0*t(i)) + b(k)*sin(k*w0*t(i));
end
% y(k,i) = a0+sum(x(:,i)); % Add DC term
end
end
z = a0 + sum(x);
figure(1);
plot(t,z);
%Percent error
function [per_error] = percent_error(measured, actual)
per_error = abs(( (measured - actual) ./ 1) * 100);
end
The purpose of the forum is helping with specific technical questions, not doing your homework.
I have 2 features which I expand to contain all possible combinations of the two features under order 6. When I do MATLAB's fminunc, it returns a weight vector where all elements are 0.
The dataset is here
clear all;
clc;
data = load("P2-data1.txt");
m = length(data);
para = 0; % regularization parameter
%% Augment Feature
y = data(:,3);
new_data = newfeature(data(:,1), data(:,2), 3);
[~, n] = size(new_data);
betas1 = zeros(n,1); % initial weights
options = optimset('GradObj', 'on', 'MaxIter', 400);
[beta_new, cost] = fminunc(#(t)(regucostfunction(t, new_data, y, para)), betas1, options);
fprintf('Cost at theta found by fminunc: %f\n', cost);
fprintf('theta: \n');
fprintf(' %f \n', beta_new); % get all 0 here
% Compute accuracy on our training set
p_new = predict(beta_new, new_data);
fprintf('Train Accuracy after feature augmentation: %f\n', mean(double(p_new == y)) * 100);
fprintf('\n');
%% the functions are defined below
function g = sigmoid(z) % running properly
g = zeros(size(z));
g=ones(size(z))./(ones(size(z))+exp(-z));
end
function [J,grad] = regucostfunction(theta,x,y,para) % CalculateCost(x1,betas1,y);
m = length(y); % number of training examples
J = 0;
grad = zeros(size(theta));
hyp = sigmoid(x*theta);
err = (hyp - y)';
grad = (1/m)*(err)*x;
sum = 0;
for k = 2:length(theta)
sum = sum+theta(k)^2;
end
J = (1/m)*((-y' * log(hyp) - (1 - y)' * log(1 - hyp)) + para*(sum) );
end
function p = predict(theta, X)
m = size(X, 1); % Number of training examples
p = zeros(m, 1);
index = find(sigmoid(theta'*X') >= 0.5);
p(index,1) = 1;
end
function out = newfeature(X1, X2, degree)
out = ones(size(X1(:,1)));
for i = 1:degree
for j = 0:i
out(:, end+1) = (X1.^(i-j)).*(X2.^j);
end
end
end
data contains 2 columns of rows followed by a third column of 0/1 values.
The functions used are: newfeature returns the expanded features and regucostfunction computes the cost. When I did the same approach with the default features, it worked and I think the problem here has to do with some coding issue.
Introduction
I am using Matlab to simulate some dynamic systems through numerically solving systems of Second Order Ordinary Differential Equations using ODE45. I found a great tutorial from Mathworks (link for tutorial at end) on how to do this.
In the tutorial the system of equations is explicit in x and y as shown below:
x''=-D(y) * x' * sqrt(x'^2 + y'^2)
y''=-D(y) * y' * sqrt(x'^2 + y'^2) + g(y)
Both equations above have form y'' = f(x, x', y, y')
Question
However, I am coming across systems of equations where the variables can not be solved for explicitly as shown in the example. For example one of the systems has the following set of 3 second order ordinary differential equations:
y double prime equation
y'' - .5*L*(x''*sin(x) + x'^2*cos(x) + (k/m)*y - g = 0
x double prime equation
.33*L^2*x'' - .5*L*y''sin(x) - .33*L^2*C*cos(x) + .5*g*L*sin(x) = 0
A single prime is first derivative
A double prime is second derivative
L, g, m, k, and C are given parameters.
How can Matlab be used to numerically solve a set of second order ordinary differential equations where second order can not be explicitly solved for?
Thanks!
Your second system has the form
a11*x'' + a12*y'' = f1(x,y,x',y')
a21*x'' + a22*y'' = f2(x,y,x',y')
which you can solve as a linear system
[x'', y''] = A\f
or in this case explicitly using Cramer's rule
x'' = ( a22*f1 - a12*f2 ) / (a11*a22 - a12*a21)
y'' accordingly.
I would strongly recommend leaving the intermediate variables in the code to reduce chances for typing errors and avoid multiple computation of the same expressions.
Code could look like this (untested)
function dz = odefunc(t,z)
x=z(1); dx=z(2); y=z(3); dy=z(4);
A = [ [-.5*L*sin(x), 1] ; [.33*L^2, -0.5*L*sin(x)] ]
b = [ [dx^2*cos(x) + (k/m)*y-g]; [-.33*L^2*C*cos(x) + .5*g*L*sin(x)] ]
d2 = A\b
dz = [ dx, d2(1), dy, d2(2) ]
end
Yes your method is correct!
I post the following code below:
%Rotating Pendulum Sym Main
clc
clear all;
%Define parameters
global M K L g C;
M = 1;
K = 25.6;
L = 1;
C = 1;
g = 9.8;
% define initial values for theta, thetad, del, deld
e_0 = 1;
ed_0 = 0;
theta_0 = 0;
thetad_0 = .5;
initialValues = [e_0, ed_0, theta_0, thetad_0];
% Set a timespan
t_initial = 0;
t_final = 36;
dt = .01;
N = (t_final - t_initial)/dt;
timeSpan = linspace(t_final, t_initial, N);
% Run ode45 to get z (theta, thetad, del, deld)
[t, z] = ode45(#RotSpngHndl, timeSpan, initialValues);
%initialize variables
e = zeros(N,1);
ed = zeros(N,1);
theta = zeros(N,1);
thetad = zeros(N,1);
T = zeros(N,1);
V = zeros(N,1);
x = zeros(N,1);
y = zeros(N,1);
for i = 1:N
e(i) = z(i, 1);
ed(i) = z(i, 2);
theta(i) = z(i, 3);
thetad(i) = z(i, 4);
T(i) = .5*M*(ed(i)^2 + (1/3)*L^2*C*sin(theta(i)) + (1/3)*L^2*thetad(i)^2 - L*ed(i)*thetad(i)*sin(theta(i)));
V(i) = -M*g*(e(i) + .5*L*cos(theta(i)));
E(i) = T(i) + V(i);
end
figure(1)
plot(t, T,'r');
hold on;
plot(t, V,'b');
plot(t,E,'y');
title('Energy');
xlabel('time(sec)');
legend('Kinetic Energy', 'Potential Energy', 'Total Energy');
Here is function handle file for ode45:
function dz = RotSpngHndl(~, z)
% Define Global Parameters
global M K L g C
A = [1, -.5*L*sin(z(3));
-.5*L*sin(z(3)), (1/3)*L^2];
b = [.5*L*z(4)^2*cos(z(3)) - (K/M)*z(1) + g;
(1/3)*L^2*C*cos(z(3)) + .5*g*L*sin(z(3))];
X = A\b;
% return column vector [ed; edd; ed; edd]
dz = [z(2);
X(1);
z(4);
X(2)];