First time post here. Pretty frustrated right now working on this assignment for class.
Basically, the idea is to use Euler's method to simulate and graph an equation of motion. The equation of motion is in the form of an ODE.
My professor has already put down some code for slightly similar system and would like us to derive the equation of motion using Lagrange. I believe that I have derived the EOM correctly, however I am running into problems on the Matlab side of things.
What's weird is that using a similar technique on another, seperate EOM, I have no issues. So I am unsure what I am doing wrong.
Here's the code for the part that is working correctly:
close all; clear all; clc;
% System parameters
w = 2*pi;
c = 0.02;
% Time vectors
dt = 1e-5;
t = 0:dt:4;
theta = zeros(size(t));
thetadot = zeros(size(t));
% Initial conditions
theta(1)=pi/2; %theta(0)
thetadot(1)=0; %thetadot(0)
for I = 1 : length(t)-1;
thetaddot = -c*thetadot(I)-w^2*sin(theta(I));
thetadot(I+1)=thetadot(I)+thetaddot*dt;
theta(I+1)=theta(I)+thetadot(I)*dt ;
end
figure(1);
plot(t,theta,'b');
xlabel('time(s)');
ylabel('theta');
title('Figure 1');
zoom on;
% Output the plot to a pdf file, and make it 6 inches by 4 inches
printFigureToPdf('fig1.pdf', [6,4],'in');
% Open the pdf for viewing
open fig1.pdf
Everything runs fine, except Matlab complains about the printFigureToPdf command.
Now, here is the code for the problem that I am having issues with.
close all; clear all; clc; clf
% System parameters
m=0.2;
g=9.81;
c=.2;
d=0.075;
L=0.001; %L is used for Gamma
B=0.001; %B is used for Beta
W=210*pi; %W is used for Omega
%Time vectors
dt = 1e-6; %Time Step
t=0:dt:10; %Range of times that simulation goes through
x=zeros(size(t));
xdot=zeros(size(t));
%Initialconditions
x(1)=0;%x(0)
xdot(1)=0; %xdot(0)
for I = 1 : length(t)-1;
xddot =-1/m*(c*xdot(I)-c*L*W*cos(W)+m*g-3*B*((d+x-L*W*sin(W*t)).^(-4)-(d-x-L*W*sin(W*t)).^(-4)));
xdot(I+1)=xdot(I)+xddot*dt;
x(I+1)=x(I)+xdot(I+1)*dt ;
end
figure(1);
plot(t,x,'b');
xlabel('time(s)');
ylabel('distance(m)');
title('Figure 2');
zoom on;
% Output the plot to a pdf file, and make it 6 inches by 4 inches
printFigureToPdf('fig1.pdf', [6,4],'in');
% Open the pdf for viewing
open fig1.pdf
With this code, I followed the same procedure and is giving an error on line 23: "In an assignment A(I) = B, the number of elements in B and I must be the same."
Like I said, I am confused because the other code worked okay, and this second set of code gives an error.
If anyone could give me a hand with this, I would greatly appreciate it.
Thanks in advance,
Dave
Edit: As suggested, I changed x(I+1)=x(I)+xdot(I+1)*dt to x(I+1)=x(I)+xdot(I)*dt. However, I am still getting an error for line 23: "In an assignment A(I) = B, the number of elements in B and I must be the same."
Line 23 is: xdot(I+1)=xdot(I)+xddot*dt;
So, I tried adjusting the code as suggested for the other line to xdot(I+1)=xdot(I)+xddot(I)*dt;
After making this change, Matlab gets stuck, I tried letting it run for a few minutes but won't execute. I ended up having to close and reopen the application.
The error In an assignment A(I) = B, the number of elements in B and I must be the same. is something you should understand because it may pop up frequently in Matlab if you are not careful.
In your case, you are trying to assign 1 element value xdot(I+1) with something which has more than 1 element xdot(I)+xddot*dt.
Indeed, if you step through the code line by line and observe your workspace, you will notice that xddot is not a scalar value as intended, but a full blown vector the size of t. This is because in the precedent line where you define xddot:
xddot =-1/m*(c*xdot(I)-c*L*W*cos(W)+m*g-3*B*((d+x-L*W*sin(W*t)).^(-4)-(d-x-L*W*sin(W*t)).^(-4)));
you still have many references to x (full vector) and t (full vector). You have to replace all these references to full vectors to only one index of them, i.e use x(I) and t(I). The line becomes:
xddot =-1/m*(c*xdot(I)-c*L*W*cos(W)+m*g-3*B*((d+x(I)-L*W*sin(W*t(I))).^(-4)-(d-x(I)-L*W*sin(W*t(I))).^(-4)));
With that your code runs just fine. However, it is far from optimized and it runs relatively slow. I have a powerful machine and it still takes a long time to run for me. I suggest you reduce your time step to something more sensible, at least when you are still trying your code. If you really need that kind of precision, first make sure your code runs fine then when it is ready let it run at full precision and go have a coffee while your computer is doing the work.
The snippet below is the loop part of your code with the correct assignment for xddot. I also added a simple progress bar so you can see that your code is doing something.
hw = waitbar(0,'Please wait...') ;
npt = length(t)-1 ;
for I = 1 : npt
xddot =-1/m*(c*xdot(I)-c*L*W*cos(W)+m*g-3*B*((d+x(I)-L*W*sin(W*t(I))).^(-4)-(d-x(I)-L*W*sin(W*t(I))).^(-4)));
xdot(I+1) = xdot(I)+xddot*dt;
x(I+1) = x(I)+xdot(I+1)*dt ;
pcdone = I / npt ;
waitbar(pcdone,hw,[num2str(pcdone*100,'%5.2f') '% done'])
end
close(hw)
I strongly suggest you reduce your time step to dt = 1e-3; until you are satisfied with everything else.
In the final version, you can remove or comment the calls to the waitbar as it slows down things too.
Related
I wanted to plot the load voltage across the resistor in series with a diode using matlab. This is a rather simple example involving piecewise functions, however I ran into an unexpected error.
t=[0:0.001:10]
vs=4*sin(pi * t)
for i =1:length(vs)
if(vs(i)<=0.7)
v(i)=0;
else
v(i)=vs(i)-0.7;
end
end
plot(t,v)
t is the time, vs is the source voltage, and v is the load voltage. Now, running this gave me an error saying "error, t and v are of different sizes..". Using length() I found out that while t and vs are of lengths 10001, v is somehow of length 1000001.
This is truly baffling to me. How can v and vs possibly differ in size? every element of vs was mapped to an element of v, and yet the size of v comes out to be about 100 times the size of vs.
Being new to matlab, I still am not very comfortable with not explicitly declaring the array v before using it in the for loop. However, I went through with it, because the example I worked on prior to this, used the same thing and it worked without any problems. We simply had a plot a piecewise function:
x=[-2 : 0.00001 : 20];
for i=1: length(x)
if(x(i)>=-2 && x(i)<0)
y(i)=sqrt(x(i)^2+1);
else if(x(i)>=0 && x(i)<10)
y(i)=3*x(i)+1;
else
y(i)=9*sin(5*x(i)-50);
end
end
end
plot(x,y)
This code worked flawlessly, and in my opinion the initial code is fundamentally doing the same thing, so again, I'm clueless as to why it failed.
The original code works if you initialise v to be of the same size as t (and therefore, vs), but still, I want to know why the code involving x,y worked (where y wasn't initialised) and the code involving (t,v) failed.
Also, my friend copy pasted the entire code into the command window of matlab 2016, and it worked. (and I'm using the 2021 version).
Its good practice to initialize variables before entering a loop. It will help avoid undefinied behaviour when you run the script multiple times. If you run the script with different lengths for t, it would fail the second run. One solution would be:
t=0:0.001:10;
vs=4*sin(pi * t);
v=nan(size(t));
for i =1:length(vs)
if(vs(i)<=0.7)
v(i)=0;
else
v(i)=vs(i)-0.7;
end
end
figure;
plot(t,v);
You could also avoid the for loop and use matrix operations instead:
t=0:0.001:10;
vs=4*sin(pi * t);
v=vs-0.7;
v(vs<=0.7)=0;
figure;
plot(t,v);
I need to obtain the values of a, b and c in the following equations so that the step response of the system matches that of the figure below.
x_dot = a*x + b+u;
y = c*x;
Where x_dot is the first derivative of x.
I have been trying to achieve this through Matlab and have so far achieved the following, using just arbitrary values for a, b and c for testing purposes:
clc;
close all;
clear all;
a=1;
b=2;
c=3;
tspan = [0:0.01:12];
x_dot = a*x+b*xu;
x = (a*x^2)/2 + b*u*x;
y = c*x;
f = #(t,x) [a*x(1)+b*x(2); c*x(1)];
[t, xa] = ode45(f,tspan,[0,0]);
plot(t,xa(:,1));
This certainly sounds like a parameter estimation problem as already hinted. You want to minimise the error between the outcome modelled using your ode and the values in your graph (fitting your three parameters a,b and c to the data).
A first step is to write an error function that takes the ode output values and compares how close it is to the data values (sum of least squares error for instance).
Then you have to search through a range of a,b,c values (which may be a large search space) and you pick the set of (a,b,c) which minimise your error function (i.e. get as close to your graph as possible).
Many search/optimisation strategies exist (.. e.g. genetic algorithm/etc.).
Please Note that the parameters are elements of real numbers(which includes negative values and extremely large or small values), the large search space is usually what makes these problems difficult to solve.
Also I think you have to be careful of initial conditions e.g. [0,0] doesn't seem to lead to interesting results. (try a = -0.5,b=0.2 and c=-0.00000001, with IC of [0,10] as below)
clc;
close all;
clear all;
a=-0.5;
b=0.2;
c=-0.00000001;
tspan = [0:0.01:12];
f = #(t,x) [a*x(1)+b*x(2); c*x(1)];
[t, xa] = ode45(f,tspan,[0,10]);
plot(t,xa);
hold on
plot(t,4)
Here 10 is the starting point of the green line and blue line starts at 0. What I would also note is that the IC change the results.. so there any many possible solutions for a,b,c given IC.
Looks interesting.. good luck.
I am writing a code that generates a function f in a loop. This function f changes in every loop, for example from f = x + 2x to f = 3x^2 + 1 (randomly), and I want to evaluate f at different points in every loop. I have tried using subs, eval, matlabFunction etc but it is still running slowly. How would you tackle a problem like this in the most efficient way?
This is as fast as I have been able to do it. ****matlabFunction and subs go slower than this.
The code below is my solution and it is one loop. In my larger code the function f and point x0 change in every loop so you can imagine why I want this to go as fast as possible. I would greatly appreciate it if someone could go through this, and give me any pointers. If my coding is crap feel free to tell me :D
x = sym('x',[2,1]);
f = [x(1)-x(1)cos(x(2)), x(2)-3x(2)^2*cos(x(1))];
J = jacobian(f,x);
x0 = [2,1];
N=length(x0); % Number of equations
%% Transform into string
fstr = map2mat(char(f));
Jstr = map2mat(char(J));
% replace every occurence of 'xi' with 'x(i)'
Jstr = addPar(Jstr,N);
fstr = addPar(fstr,N);
x = x0;
phi0 = eval(fstr)
J = eval(Jstr)
function str = addPar(str,N)
% pstr = addPar(str,N)
% Transforms every occurence of xi in str into x(i)
% N is the maximum value of i
% replace every occurence of xi with x(i)
% note that we do this backwards to avoid x10 being
% replaced with x(1)0
for i=N:-1:1
is = num2str(i);
xis = ['x' is];
xpis = ['x(' is ')'];
str = strrep(str,xis,xpis);
end
function r = map2mat(r)
% MAP2MAT Maple to MATLAB string conversion.
% Lifted from the symbolic toolbox source code
% MAP2MAT(r) converts the Maple string r containing
% matrix, vector, or array to a valid MATLAB string.
%
% Examples: map2mat(matrix([[a,b], [c,d]]) returns
% [a,b;c,d]
% map2mat(array([[a,b], [c,d]]) returns
% [a,b;c,d]
% map2mat(vector([[a,b,c,d]]) returns
% [a,b,c,d]
% Deblank.
r(findstr(r,' ')) = [];
% Special case of the empty matrix or vector
if strcmp(r,'vector([])') | strcmp(r,'matrix([])') | ...
strcmp(r,'array([])')
r = [];
else
% Remove matrix, vector, or array from the string.
r = strrep(r,'matrix([[','['); r = strrep(r,'array([[','[');
r = strrep(r,'vector([','['); r = strrep(r,'],[',';');
r = strrep(r,']])',']'); r = strrep(r,'])',']');
end
There are several ways to get huge boosts in speed for this sort of problem:
The java GUI front end slows everything down. Go back to version 2010a or earlier. Go back to when it was based on C or fortran. The MATLAB script runs as fast as if you had put it into the MATLAB "compiler".
If you have MatLab compiler (or builder, I forget which) but not the coder, then you can process your code and have it run a few times faster without modifying the code.
write it to a file, then call it as a function. I have done this for changing finite-element expressions, so large ugly math that makes $y = 3x^2 +1$ look simple. In that it gave me solid speed increase.
vectorize, vectorize, vectorize. It used to reliably give 10x to 12x speed increase. Pull it out of loops. The java, I think, obscures this some by making everything slower.
have you "profiled" your function to make sure that "eval" or such are the problem? If you fix "eval" and your bottleneck is elsewhere then you will have problems.
If you have the choice between eval and subs, stick with eval. subs gives you a symbolic solution, not a numeric one.
If there is a clean way to have multiple instances of MatLab running, especially if you have a decently core-rich cpu that MatLab does not fully utilize, then get several of them going. If you are at an educational institution you might try running several different versions (2010a, 2010b, 2009a,...) on the same system. I (fuzzily) recall they didn't collide when I did it. Running more than about 8 started slowing things down more than it improved them. Make sure they aren't colliding on file access if you are using files to share control.
You could write your program in LabVIEW (not MathScript, not MatLab) and because it is a compiled language, there are times that code can run 1000x faster.
You could go all numeric and make it a matrix activity. This depends on your code, but if you could randomly populate the columns in the matrix then matrix multiply it to a matrix $ \left[ 1, x, x^{2}, ...\right] $, that would likely be several hundreds or thousands of times faster than your current level of equation handling and still in MatLab.
About your coding:
don't redeclare "x" as a symbol every loop, that is expensive.
what is this "map2mat" then "addPar" stuff?
the string handling functions are horrible for runtime. Stick to one language. The symbolic toolbox IS maple, and you don't have to get goofy hand-made parsing to make it work with the rest of MatLab.
I must to use angle = atan2(norm(cross(a,b)),dot(a,b)), for calculating the angle between two vectors a,b and these are double type and norm is undefined for this type. How do I resolve this problem? I need to calculate the angle between two vectors this way.
In your comments, you have shown us how you are actually writing out the angle calculation and it is not the same as how you have put it in your post.
atan2(norm(cross(I(i,j,:),I_avg)),dot(I(i,j,:),I_avg));
I is an image you are loading in. I'm assuming it's colour because of the way you are subsetting I. Because I is a 3D matrix, doing I(i,j,:) will give you a 1 x 1 x 3 vector when in fact this has to be a 1D vector. norm does not recognize this structure which is why you're getting this error. Therefore, you need to use squeeze to remove the singleton dimensions so that this will become a 3 x 1 vector, rather than a 1 x 1 x 3 vector. As such, you need to rewrite your code so that you're doing this instead. Bear in mind that in your comments, angle is always overwritten inside the for loop, so you probably want to save the results of each pixel. With this, you probably want to create a 2D array of angles that will store these results. In other words:
I=imread('thesis.jpg');
I = double(I);
angles = zeros(m,n);
I_avg = squeeze(I_avg); %// Just in case
for i=1:m
for j=1:n
pixels = squeeze(I(i,j,:)); %// Add this statement and squeeze
angles(i,j) = atan2(norm(pixels,I_avg)),dot(pixels,I_avg)); %// Change
end
end
Minor note
MATLAB has a built-in function called angle that determines the angle from the origin to a complex number in the complex plane. It is not recommended you call your variable angle as this will unintentionally shadow over the angle function, and any other code that you create from this point onwards may rely on that actual angle function, and you will get unintended results.
Another minor note
Using i and j as loop variables is not recommended. These letters are reserved for the complex number, and this can produce unintentional results. Take a look at this question and post by Shai here - Using i and j as variables in Matlab. As such, it is suggested you use other variable names instead.
As #rayryeng has successfully answered this question, I would like to turn my post into a more general one by sharing my experience in debugging in Matlab. I hope anyone who somehow managed to find this post get more or less thinking about the habits a good programmer should have.
The question goes like: "How would I do if I get errors?"
Here's an excellent article by Eric in which he lists the rule-of-thumbs when you encounter a bug and wish to get rid of it. It's originally been cited by Stackoverflow, and that's the reason I read it.
If you still get no clue / idea how you can play with your code, see how this person does:
Pin-point the buggy line
(The number should start with 0) Make sure before running a script, you clear out any previously stored variables, including the notorious i and j's (you should never see them in any workspace). If any one is needed for the buggy code to run, save('buggy.mat','importantvar') before clear and load('buggy.mat') after clear.
By doing so, you can isolate your buggy code from anything else, which could have bad influences. For example, in a previously called script, there is a line
double = [2,4,6]; % you should never name a variable `double`
and in the next script, you have
>> e = str2num('uint8(200)')
e =
200
>> double(e)
Index exceeds matrix dimensions.
>>
>> f = single(2.36)
f =
2.3600
>> double(f)
Subscript indices must either be real positive integers or
logicals.
>>
The reason is double is no longer an inbuild function, but a user-defined variable. Too bad to pick up a name carelessly!
....anyway, let's clear the workspace and get rid of double.
>> clear
Read the error message, thoroughly.
Now let's begin with OP's problem. The original code (trimmed) goes like this -
img = imread('peppers.png');
a = img(300,200,:);
b = img(200,300,:);
d = norm(cross(a,b));
.... hence the error
Undefined function 'norm' for input arguments of type 'uint8'.
Error in untitled (line 6)
d = norm(cross(a,b));
Most beginners are only interested in the first line of the error message, which by it alone usually doesn't provide any useful help, or only in the red color, which leads to the famous question "my code does not work!"
But think twice. You still have another 2 lines unread! Error in untitled (line 6) says I'm running a script named untitled and the (first) error lies in line 6, and the code in that line is d = norm(cross(a,b));.
Now, at least you know a little more about your code - "My code d = norm(cross(a,b)); doesn't work!"
Although most likely we may also vote this kind of question to get closed, it's still much much better than a simply "It does not work!".
Now we can pin-point the buggy line
try
% this line will raise an error
d = norm(cross(a,b));
catch err
disp(err.message)
end
Look into the functions
First, make sure the inner function cross works as expected -
>> cross(a,b)
ans(:,:,1) =
0
ans(:,:,2) =
255
ans(:,:,3) =
0
>>
Good. So now we can even narrow down the error to the outer norm.
One more thing to mention. You can always find Mathworks' documentation for any in-build function, by typing "matlab function", such as "matlab norm" in Google (or any other search engine) and clicking on the first result. If you prefer, you can also type in Matlab command window doc _function_ such as doc norm and read the doc in Matlab. It's of course a pleasure of us on Stackoverflow to give you the reference by doing the same thing, but it takes a longer time because a human is, in this aspect, always slower than a search engine.
The error reads Undefined function 'norm' for input arguments of type 'uint8'.. So the input for norm should not be uint8, unsigned 8-bit integer. But what should it be?
% why `norm` "does not work"?
% this line runs perfectly well
norm(cross([1,2,3], [4,5,6]))
% so what is working?
class([1,2,3]) % so `norm` works for `double`
One thing we can do now is convert a and b to double precision. Let's try it now.
% try fixing 'uint8' error
a2 = double(a);
b2 = double(b);
whos a b % now they are double, which `norm` should work for
try
% this line will raise an error
d = norm(cross(a2,b2));
catch err
disp(err.message)
end
Now the error becomes Input must be 2-D.. What's wrong with the input?
% what is "must be 2-D" error?
size(a2) % a2 is 3-D
disp(b2) % b2 is also 3-D
This gives output in command window
ans =
1 1 3
(:,:,1) =
255
(:,:,2) =
150
(:,:,3) =
0
In OP's problem, he/she is trying to calculate something about color difference (to the best of my knowledge) which involves the angle between two color vectors in RGB space. So the vectors are needed. With imread, each pixel of the image is stored as 3 elements in the matrix, first 2 dimension being its physical position, the 3 dimension being RGB channel components. Hence pixel(200,300) with color rgb[255,150,0] is stored by us in variable b wihch is a 3-D vector.
By understanding what we need and what Matlab can do, we can combine these two points into one. We need the norm of the cross product of a and b, while the useful information (the 3 component values) is stored in the 3rd dimension. Matlab can calculate the norm of the cross product of a vector with all its information in the 1st dimension. (Here, "dimension" refers to that of the Matlab variable; a vector with 3 elements in its 1st dimension is physically a 3-D vector).
After thinking twice, we are now able to debug our code - just put all 3 elements into the 1st dimension.
% so we want the 3 elements in the 3rd dimension become in the 1st dim
a3 = squeeze(a2);
b3 = reshape(b2,numel(b2),[]);
try
d = norm(cross(a3,b3));
catch err
disp(err.message)
end
d
Bonus: If by default Matlab treats a 3-D vector as a "1-D array", then most probably the cross function has not been working correctly. Let's make a check -
>> clear
>> a = [1,2,3]
a =
1 2 3
>> b=[4,5,6]
b =
4 5 6
>> cross(a,b)
ans =
-3 6 -3
>>
The result should be the same as the one we can get by calculating by hand.
Now if we put the components into the 3rd dimension of the variable -
>> clear
>> a(1,1,:)=[1,2,3]
a(:,:,1) =
1
a(:,:,2) =
2
a(:,:,3) =
3
>> b(1,1,:)=[4,5,6]
b(:,:,1) =
4
b(:,:,2) =
5
b(:,:,3) =
6
>> cross(a,b)
ans(:,:,1) =
-3
ans(:,:,2) =
6
ans(:,:,3) =
-3
>>
.... seems OK. cross also puts the result in the 3rd dimension. In fact, Mathworks' documentation says
If A and B are vectors, then they must have a length of 3.
If A and B are matrices or multidimensional arrays, then they must
have the same size. In this case, the cross function treats A and B as
collections of three-element vectors. The function calculates the
cross product of corresponding vectors along the first array dimension
whose size equals 3.
At last, one thing is always correct to anyone who wants to do something with programming - be cautious and prudent when writing your code.
Hey, I've got a problem plotting a function in Matlab.
I first run this:
format long
f = inline('-x.^2');
for i = 0:10
[I(i+1) h(i+1) tid(i+1)] = trapets(f,0,1,2^i);
end
trunk = I - log(2);
hold on
grid on
plot(log(h),log(trunk),'r+')
t = -7:0;
c = polyfit(log(h),log(trunk),1);
yy = polyval(c,t);
plot(t,yy)
grid off
hold off
koefficienter = real(c)
and after that I run this file:
hold on
plot(h,trunk,'r+:','linewidth',2)
axis([0 0.6 0 0.0014])
Thing is, I don't get any errors, and the plot windows pops up with axes and all, but there is no graph to be found. It's just an empty window with two axes.
Anyone got any ideas?
Edit:
Okay, so I'm new to this site and couldn't find the reply button, so I add a reply here instead.
#woodchips :
I just realized that I hadn't given you all the information for this problem.. Sorry about that, anyhow I would really appreciate it if someone had the time to help me with this, it would seriously make my week.
This is the part I accidentally left out:
function [ I,h,tid ] = trapets(
f,a,b,n )
h=(b-a)/n;
tic; I=(f(a)+f(b));
for k=2:2:n-2
I = I+2*f(a+k*h);
end
for k = 1:2:n-1
I = I + 4*f(a+k*h);
end
I = I * h/3;
tid = toc;
end
Edit 2: So, I think that the graph I'm seeking is actually getting plotted in the first code that I wrote, the problem is that the variabe 'I' is not changing, which I expect it to do, although the variabels 'n' and 'h' do change. If 'I' was working correctly, I would probably get the right graph (hopefully). Any ideas, anyone?
Unfortunately the home computer I had with Matlab on it died the other day so I can't test anything. First thing I can think of if to simply run step by step through the code and see if the results of the math are what you are expecting. For instance Matlab was primarily made and runs as a matrix calculator if I recall correctly. As such most of the simple math doesn't function as it would punching it in a calculator. An example would be that 2^i needs to be 2.^i to operate correctly in some cases. Same with .* and ./ to use the singular scalar verses the matrices math.
The best way to find out what is going wrong is to iterate through the math a few times to ensure that it is being performed as expected. Once that is verified then you can move on to looking at plotting formatting.