I am attempting to write a function which expands another into a Fourier series. However for some functions the integral() function keeps spitting out warnings claiming it has reached minimum step size which is likely due to a singularity at x = -1. My code is as follows:
H = #(t) 1 * (t >= 0) + 0; % Heaviside step function
x_a = #(t) 2*(H(mod(t+1, 4)) - H(mod(t+1, 4) - 2)) - 1;
time = linspace(-8, 8, 25);
plot(time, x_a(time))
ylim([-1.5 1.5])
xlim([-8 8])
% This is where it starts spitting out warnings if the next line is uncommented
%x_a_fourier = fourier(x_a, time, 4, 10);
function x = fourier(F, I, T, m)
a_0 = (1/T) * integral(#(x) F(x), -T/2, T/2);
x = a_0 * ones(1, length(I));
w_0 = (2*pi) / T;
a_n = #(n) (2/T) * integral(#(x) F(x) .* cos(n*w_0*x), -T/2, T/2);
b_n = #(n) (2/T) * integral(#(x) F(x) .* sin(n*w_0*x), -T/2, T/2);
for k = 1:length(I)
for l = 1:m
x(k) = x(k) + a_n(l) * cos(l*w_0*I(k)) + b_n(l) * sin(l*w_0*I(k));
end
end
end
From the plot() statement it should be obvious that the integral() function shouldn't run into any singularities. Any ideas as to what may be the problem?
Related
My following code generates the plot of V and D values in figure 1. In the graph, the parabolas and straigh lines intersect, and I need to find the roots from the plot from a loop. So I tried to use fzero function, but the error appeared:
Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values. Use the ANY or ALL functions to reduce operands to logical scalar values.
Error in fzero (line 326)
elseif ~isfinite(fx) || ~isreal(fx)
Error in HW1 (line 35)
x=fzero(fun,1);
My code is:
clear all; close all
W = 10000; %[N]
S = 40; %[m^2]
AR = 7;
cd0 = 0.01;
k = 1 / pi / AR;
clalpha = 2*pi;
Tsl=800;
figure(1);hold on; xlabel('V');ylabel('D')
for h=0:1:8;
i=0;
for alpha = 1:0.25:12
i=i+1;
rho(i)=1.2*exp(-h/10.4);
cl(i) = clalpha * alpha * pi/180;
V(i) = sqrt(2*W/rho(i)/S/cl(i));
L(i) = 0.5 * rho(i) * V(i) * V(i) * S * cl(i);
cd(i) = cd0 + k * cl(i) * cl(i);
D(i) = 0.5 * rho(i) * V(i) * V(i) * S * cd(i);
clcd(i) = cl(i)/cd(i);
p(i) = D(i)*V(i);
ang(i) = alpha;
T(i)=Tsl*(rho(i)/1.2).^0.75;
end
figure(1); plot(V,D); hold on
plot(V,T);
end
fun = #(V) 0.5*V.*V.*rho.*S.*cd-T;
x=fzero(fun,1);
Probably, I should not use the fzero function, but the task is to find the roots of V from a plot (figure 1). There are parabolas and straight lines respectively.
From the documentation for fzero(fun,x)
fun: Function to solve, specified as a handle to a scalar-valued function or the name of such a function. fun accepts a scalar x and returns a scalar fun(x).
Your function does not return a scalar value for a scalar input, it always returns a vector which is not valid for a function which is being used with fzero.
1.- Your codes doesn't plot V and D: Your code plots D(V) and T(V)
2.- T is completely flat, despite taking part in the inner for loop calculations with T(i)=Tsl*(rho(i)/1.2).^0.75; as it had to be somehow modified.
But in fact it remains constant for all samples of V, constant temperature (°C ?), and for all laps of the outer for loop sweeping variable h within [0:1:8].
The produced T(V) functions are the flat lines.
3.- Then you try building a 3rd function f that you put as if f(V) only but in fact it's f(V,T) with the right hand side of the function with a numerical expression, without a symbolic expression, the symbolic expression that fzero expects to attempt zero solving.
In MATLAB Zero finding has to be done either symbolically or numerically.
A symbolic zero-finding function like fzero doesn't work with numerical expressions like the ones you have calculated throughout the 2 loops for h and for alpha .
Examples of function expressions solvable by fzero :
3.1.-
fun = #(x)sin(cosh(x));
x0 = 1;
options = optimset('PlotFcns',{#optimplotx,#optimplotfval});
x = fzero(fun,x0,options)
3.2.-
fun = #sin; % function
x0 = 3; % initial point
x = fzero(fun,x0)
3.3.- put the following 3 lines in a separate file, call this file f.m
function y = f(x)
y = x.^3 - 2*x - 5;
and solve
fun = #f; % function
x0 = 2; % initial point
z = fzero(fun,x0)
3.4.- fzeros can solve parametrically
myfun = #(x,c) cos(c*x); % parameterized function
c = 2; % parameter
fun = #(x) myfun(x,c); % function of x alone
x = fzero(fun,0.1)
4.- So since you have already done all the numerical calculations and no symbolic expression is supplied, it's reasonable to solve numerically, not symbolically.
To this purpose there's a really useful function called intersections.m written by Douglas Schwarz available here
clear all; close all;clc
W = 10000; %[N]
S = 40; %[m^2]
AR = 7;
cd0 = 0.01;
k = 1 / pi / AR;
clalpha = 2*pi;
Tsl=800;
figure(1);
ax1=gca
hold(ax1,'on');xlabel(ax1,'V');ylabel(ax1,'D');grid(ax1,'on');
title(ax1,'1st graph');
reczeros={}
for h=0:1:8;
i=0;
for alpha = 1:0.25:12
i=i+1;
rho(i)=1.2*exp(-h/10.4);
cl(i) = clalpha * alpha * pi/180;
V(i) = sqrt(2*W/rho(i)/S/cl(i));
L(i) = 0.5 * rho(i) * V(i) * V(i) * S * cl(i);
cd(i) = cd0 + k * cl(i) * cl(i);
D(i) = 0.5 * rho(i) * V(i) * V(i) * S * cd(i);
clcd(i) = cl(i)/cd(i);
p(i) = D(i)*V(i);
ang(i) = alpha;
T(i)=Tsl*(rho(i)/1.2).^0.75;
end
plot(ax1,V,D); hold(ax1,'on');
plot(ax1,V,T);
[x0,y0]=intersections(V,D,V,T,'robust');
reczeros=[reczeros [x0 y0]];
for k1=1:1:numel(x0)
plot(ax1,x0,y0,'r*');hold(ax1,'on')
end
end
Because at each pair D(V) T(V) there may be no roots, 1 root or more than 1 root, it makes sense to use a cell, reczeros, to store whatever roots obtained.
To read obtained roots in let's say laps 3 and 5:
reczeros{3}
=
55.8850 692.5504
reczeros{5}
=
23.3517 599.5325
55.8657 599.5325
5.- And now the 2nd graph, the function that is defined in a different way as done in the double for loop:
P = 0.5*V.*V.*rho.*S.*cd-T;
figure(2);
ax2=gca
hold(ax2,'on');xlabel(ax2,'V');ylabel(ax2,'P');grid(ax2,'on');
title(ax2,'2nd graph')
plot(ax2,V,P)
plot(ax2,V,T)
[x0,y0]=intersections(V,T,V,P,'robust');
for k1=1:1:numel(x0)
plot(ax2,x0,y0,'r*');hold(ax2,'on')
end
format short
V0=x0
P0=y0
V0 =
86.9993
P0 =
449.2990
For the purpose of generalization, I hope Matlab can automatically compute the 1st & 2nd derivatives of the associated function f(x). (in case I change f(x) = sin(6x) to f(x) = sin(8x))
I know there exists built-in commands called diff() and syms, but I cannot figure out how to deal with them with the index i in the for-loop. This is the key problem I am struggling with.
How do I make changes to the following set of codes? I am using MATLAB R2019b.
n = 10;
h = (2.0 * pi) / (n - 1);
for i = 1 : n
x(i) = 0.0 + (i - 1) * h;
f(i) = sin(6 * x(i));
dfe(i) = 6 * cos(6 * x(i)); % first derivative
ddfe(i) = -36 * sin(6 * x(i)); % second derivative
end
You can simply use subs and double to do that. For your case:
% x is given here
n = 10;
h = (2.0 * pi) / (n - 1);
syms 'y';
g = sin(6 * y);
for i = 1 : n
x(i) = 0.0 + (i - 1) * h;
f(i) = double(subs(g,y,x(i)));
dfe(i) = double(subs(diff(g),y,x(i))); % first derivative
ddfe(i) = double(subs(diff(g,2),y,x(i))); % second derivative
end
By #Daivd comment, you can vectorize the loop as well:
% x is given here
n = 10;
h = (2.0 * pi) / (n - 1);
syms 'y';
g = sin(6 * y);
x = 0.0 + ((1:n) - 1) * h;
f = double(subs(g,y,x));
dfe = double(subs(diff(g),y,x)); % first derivative
ddfe = double(subs(diff(g,2),y,x)); % second derivative
I am new to Matlab and trying to find a solution to the error of my code:
Not enough input arguments.
Error in F9>f (line 42)
y = (2 - 2*t*x) / (x^2 + 1) ;
Error in F9 (line 18)
e = euler(f, trange(1), y0_value, h, trange(end));
function [] = F9()
% Euler's Method to solve given functions
% Set initial values
hi = [1/2, 1/4];
trange = [0, 2];
y0_value = 1;
% Set functions' and exact functions' handles
% Calculate and show results
% Loop for functions
for i = 1:2
fprintf('###########\n');
fprintf('Function #%d\n', i)
fprintf('###########\n');
exact_value = f_exact(trange(end));
% Loop for h
for h = hi
% Euler calculations
e = euler(f, trange(1), y0_value, h, trange(end));
fprintf('\nh: %f\n', h);
fprintf('\nEuler: %f \n', e(end));
fprintf('Error: %f\n\n', abs((e(end)-exact_value)/exact_value));
end
fprintf('Exact: %f\n\n', exact_value);
end
end
% Euler's Method
function y = euler(f, t0, y0, h, tn)
n = (tn-t0)/h;
% Initialize t, y
[t, y] = deal(zeros(n, 1));
% Set t0, y0
t(1) = t0;
y(1) = y0;
for i = 1:n
t(i+1) = t(i) + h;
y(i+1) = y(i) + h/2 * (f(t(i), y(i))+ f(t(i+1) , y(i) + h * f(t(i), y(i))));
end
end
% Functions to solve
function y = f(t, x)
y = (2 - 2*t*x) / (x^2 + 1) ;
end
function y = f_exact(x)
y = (2*x + 1) / (x^2 + 1);
end
When you pass f to euler you need to pass it as a handle, i.e. precede it with a #:
e = euler(#f, trange(1), y0_value, h, trange(end));
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)];
I'm using ode45 to solve second order differential equation. the time span is determined based on how many numbers in txt file, therefore, the time span is defined as follows
i = 1;
t(i) = 0;
dt = 0.1;
numel(theta_d)
while ( i < numel(theta_d) )
i = i + 1;
t(i) = t(i-1) + dt;
end
Now the time elements should not exceed the size of txt (i.e. numel(theta_d)). In main.m, I have
x0 = [0; 0];
options= odeset('Reltol',dt,'Stats','on');
[t, x] = ode45('ODESolver', t, x0, options);
and ODESolver.m header is
function dx = ODESolver(t, x)
If I run the code, I'm getting this error
Attempted to access theta_d(56); index out of bounds because numel(theta_d)=55.
Error in ODESolver (line 29)
theta_dDot = ( theta_d(i) - theta_dPrev ) / dt;
Why the ode45 is not being fixed with the time span?
Edit: this is the entire code
main.m
clear all
clc
global error theta_d dt;
error = 0;
theta_d = load('trajectory.txt');
i = 1;
t(i) = 0;
dt = 0.1;
numel(theta_d)
while ( i < numel(theta_d) )
i = i + 1;
t(i) = t(i-1) + dt;
end
x0 = [pi/4; 0];
options= odeset('Reltol',dt,'Stats','on');
[t, x] = ode45(#ODESolver, t, x0, options);
%e = x(:,1) - theta_d; % Error theta
plot(t, x(:,2), 'r', 'LineWidth', 2);
title('Tracking Problem','Interpreter','LaTex');
xlabel('time (sec)');
ylabel('$\dot{\theta}(t)$', 'Interpreter','LaTex');
grid on
and ODESolver.m
function dx = ODESolver(t, x)
persistent i theta_dPrev
if isempty(i)
i = 1;
theta_dPrev = 0;
end
global error theta_d dt ;
dx = zeros(2,1);
%Parameters:
m = 0.5; % mass (Kg)
d = 0.0023e-6; % viscous friction coefficient
L = 1; % arm length (m)
I = 1/3*m*L^2; % inertia seen at the rotation axis. (Kg.m^2)
g = 9.81; % acceleration due to gravity m/s^2
% PID tuning
Kp = 5;
Kd = 1.9;
Ki = 0.02;
% theta_d first derivative
theta_dDot = ( theta_d(i) - theta_dPrev ) / dt;
theta_dPrev = theta_d(i);
% u: joint torque
u = Kp*(theta_d(i) - x(1)) + Kd*( theta_dDot - x(2)) + Ki*error;
error = error + (theta_dDot - x(1));
dx(1) = x(2);
dx(2) = 1/I*(u - d*x(2) - m*g*L*sin(x(1)));
i = i + 1;
end
and this is the error
Attempted to access theta_d(56); index out of bounds because numel(theta_d)=55.
Error in ODESolver (line 28)
theta_dDot = ( theta_d(i) - theta_dPrev ) / dt;
Error in ode45 (line 261)
f(:,2) = feval(odeFcn,t+hA(1),y+f*hB(:,1),odeArgs{:});
Error in main (line 21)
[t, x] = ode45(#ODESolver, t, x0, options);
The problem here is because you have data at discrete time points, but ode45 needs to be able to calculate the derivative at any time point in your time range. Once it solves the problem, it will interpolate the results back onto your desired time points. So it will calculate the derivative many times more than at just the time points you specified, thus your i counter will not work at all.
Since you have discrete data, the only way to proceed with ode45 is to interpolate theta_d to any time t. You have a list of values theta_d corresponding to times 0:dt:(dt*(numel(theta_d)-1)), so to interpolate to a particular time t, use interp1(0:dt:(dt*(numel(theta_d)-1)),theta_d,t), and I turned this into an anonymous function to give the interpolated value of theta_p at a given time t
Then your derivative function will look like
function dx = ODESolver(t, x,thetaI)
dx = zeros(2,1);
%Parameters:
m = 0.5; % mass (Kg)
d = 0.0023e-6; % viscous friction coefficient
L = 1; % arm length (m)
I = 1/3*m*L^2; % inertia seen at the rotation axis. (Kg.m^2)
g = 9.81; % acceleration due to gravity m/s^2
% PID tuning
Kp = 5;
Kd = 1.9;
Ki = 0.02;
% theta_d first derivative
dt=1e-4;
theta_dDot = (thetaI(t) - theta(I-dt)) / dt;
%// Note thetaI(t) is the interpolated theta_d values at time t
% u: joint torque
u = Kp*(thetaI(t) - x(1)) + Kd*( theta_dDot - x(2)) + Ki*error;
error = error + (theta_dDot - x(1));
dx=[x(2); 1/I*(u - d*x(2) - m*g*L*sin(x(1)))];
end
and you will have to define thetaI=#(t) interp1(0:dt:(dt*(numel(theta_d)-1)),theta_d,t); before calling ode45 using [t, x] = ode45(#(t,x) ODESolver(t,x,thetaI, t, x0, options);.
I removed a few things from ODESolver and changed how the derivative was computed.
Note I can't test this, but it should get you on the way.