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.
Related
I have a following lab where i was asked to write the command matlab lines for these questions:
If initial Conditions are: x(0)=[2;0].๐๐๐๐ ๐กโ๐ ๐๐๐ ๐๐๐๐ ๐ ๐ฆ(๐ก).
Find the response y(t) due to step input with Amplitude of
Find the Transfer Function for the above state space model
Derive back the state space model from (3).
a = [0,1;0,4];
b = [0;1];
c = [0 -5];
x0=[2;0];
sys = ss(a,b,c,2);
initial(sys,x0); % to get 1
[n,d]=ss2tf(a,b,c,0);
mySys_tf=tf(n,d) % to get 3
[num den] = tfdata(mySys_tf, 'v')
tf2ss(num,den) % to get 4
I have written this code but it seems like its not giving me any results in the response graph and thus i can't also solve 2 and it get error in 4 if you can help me out to check what is wrong
I believe that the error comes from the fact that the system is unstable. If you were to plot the system's reaction to a step input using step() then you will see how it goes to infinity. I also don't know how far you are into your controls course and if you've seen the root locus yet, but you can plot the root locus of the system via rlocus(sys) and you'll see that the real portion of the root is on the right half of the plane and therefore letting you know that the system is unstable.
The response is 0 and will stay zero as x(2) = 0. It requires an input u to get x(2) off zero. So the graph is totally fine.
use step(sys) and you will see the drop to -Inf. Optionally you can define the end-time. Call step(sys,1) to see a reasonable range.
You solved 3 & 4 yourself.
To check stability you simply need to ask MATLAB isstable(sys) (isn't it continent? Well, there is a danger that people will forget the theory behind it and how it is connected...)
To check observability: rank(obsv(sys)) and make sure that it is the same as the system matrix
assert(rank(obsv(sys)) == length(sys.A), 'System is not observable!')
I am trying to solve a very large non-linear problem using the MatCont package. Due to the large number of dimensions, and the non-linear nature, I believe that supplying the Jacobian for my system to the MatCont algorithm will speed things up immensely. However, I cannot get it to recognise that it has a Jacobian to use!
As a minimal working example, I have modified the circle finder from the help documentation to include a Jacobian:
function out = curve()
out{1}=#curvefunc;
out{2}=#defaultprocessor;
out{3}=#options;
out{4}=#jacobian;
out{13}=#adapt;
end
function f=curvefunc(x)
f=x(1)^2+x(2)^2-1;
end
function J=jacobian(x)
disp('USE JACOBIAN')
J=[2*x(1) , 2*x(2)];
end
function varargout=defaultprocessor(varargin)
if nargin>2
varargout{3}=varargin{3};
end
varargout{2}=[0];
varargout{1}=0;
end
function option=options()
option=contset;
end
function [res,x,v]=adapt(x,v)
res=[];
end
I then try to run this program from the command line using
[x,v,s,h,f] = cont(#curve,[1;0]);
However, the response is
first point found
tangent vector to first point found
elapsed time = 0.2 secs
npoints curve = 300
Since I told it to output 'USE JACOBIAN' every time the Jacobian function was called, it is clear that MatCont is not using it.
How do I get the Jacobian to be used?
I solved my own problem! Seems I was quite close to getting it working. The below is a bit of a botch, so if anyone knows how to do it with options please post an answer also.
Firstly I edit the options settings so that when it performs the continuation it only locates the first point:
function option=options()
option = contset;
option = contset(option,'MaxNumPoints',1);
end
its fine for it to do this using numerical Jacobians, the first point is known very well in most problems. This is then called from a script or a function using the following:
[x,v,s,h,f] = cont(#curve,[1;0]);
global cds
cds.options.MaxNumPoints=[];
cds.symjac=1;
[x,v,s,h,f] = cont(x,v,s,h,f,cds);
The first line finds the initial point using numerical Jacobians, as it was set up to do. The continuer is then manually adjusted to firstly have no limit on the maximal number of points (this can be set to any appropriate number) and then the use of the user provided Jacobian is set to 1 (true). The continuer is then resumed with the new settings and uses the Jacobain properly.
I want to draw a line on a graph to find the intersection point with another line. However, there's no response after I executed the script below. May I know what is the problem and how can I solve it?
x=1:2^20;
y2=2^24;
plot(x,y2);
Thanks!
What you want is to plot a line on 2^24. However, there are too many points for you computer probably, and you run out of memory
I am guessing that you'll need to plot your other inequality as well.
Something like
x=1:100:2^20;
% As Zoran and others suggested, You may not want all the points!
% It is too much memory
y2=2^24*ones(size(x)); % This ones is optional, but its good to know what you are doing (personal opinion)
plot(x,y2);
hold on
y1=(x+1).*log(x);
plot(x,y1);
However, you are still not there!
Another solution, which does not rely on plotting:
>> f = #(x) (x+1)*log(x)-2^24;
>> soln = fzero(f,1e6)
soln = 1.1987e+006
>> f(soln)
ans = 3.7253e-009
So your intersection point is at 1.1987e6.
Apparently,you have too many points for x, 2^20
Have to wait program to calculate, or plot,for example, every 100th point
This solution works for Matlab
x=1:100:2^20;
y2=2^2;
plot(x,y2,'o');
There is one more and maybe a bit smarter way: if you want to solve ((k+1)(ln k)<2^24) as you've commented above, use fsolve function to get just solution of an equation(!). Then use that solution to specify the area of your interest, so you won't have to plot the domain of 2^20.
(All functions are continuous so you don't have to worry about any wild singularities. Just examine the neigborhood of ks for which (k+1)(ln k)-2^24=0.)
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.
I have a Matlab project in which I need to make a GUI that receives two mathematical functions from the user. I then need to find their intersection point, and then plot the two functions.
So, I have several questions:
Do you know of any algorithm I can use to find the intersection point? Of course I prefer one to which I can already find a Matlab code for in the internet. Also, I prefer it wouldn't be the Newton-Raphson method.
I should point out I'm not allowed to use built in Matlab functions.
I'm having trouble plotting the functions. What I basically did is this:
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
cla % To clear axes when plotting new functions
ezplot(fun_f);
hold on
ezplot(fun_g);
axis ([-20 20 -10 10]);
The problem is that sometimes, the axes limits do not allow me to see the other function. This will happen, if, for example, I will have one function as log10(x) and the other as y=1, the y=1 will not be shown.
I have already tried using all the axis commands but to no avail. If I set the limits myself, the functions only exist in certain limits. I have no idea why.
3 . How do I display numbers in a static text? Or better yet, string with numbers?
I want to display something like x0 = [root1]; x1 = [root2]. The only solution I found was turning the roots I found into strings but I prefer not to.
As for the equation solver, this is the code I have so far. I know it is very amateurish but it seemed like the most "intuitive" way. Also keep in mind it is very very not finished (for example, it will show me only two solutions, I'm not so sure how to display multiple roots in one static text as they are strings, hence question #3).
function [Sol] = SolveEquation(handles)
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
f = inline(fun_f);
g = inline(fun_g);
i = 1;
Sol = 0;
for x = -10:0.1:10;
if (g(x) - f(x)) >= 0 && (g(x) - f(x)) < 0.01
Sol(i) = x;
i = i + 1;
end
end
solution1 = num2str(Sol(1));
solution2 = num2str(Sol(2));
set(handles.roots1,'string',solution1);
set(handles.roots2,'string',solution2);
The if condition is because the subtraction will never give me an absolute zero, and this seems to somewhat solve it, though it's really not perfect, sometimes it will give me more than two very similar solutions (e.g 1.9 and 2).
The range of x is arbitrary, chosen by me.
I know this is a long question, so I really appreciate your patience.
Thank you very much in advance!
Question 1
I think this is a more robust method for finding the roots given data at discrete points. Looking for when the difference between the functions changes sign, which corresponds to them crossing over.
S=sign(g(x)-f(x));
h=find(diff(S)~=0)
Sol=x(h);
If you can evaluate the function wherever you want there are more methods you can use, but it depends on the size of the domain and the accuracy you want as to what is best. For example, if you don't need a great deal of accurac, your f and g functions are simple to calculate, and you can't or don't want to use derivatives, you can get a more accurate root using the same idea as the first code snippet, but do it iteratively:
G=inline('sin(x)');
F=inline('1');
g=vectorize(G);
f=vectorize(F);
tol=1e-9;
tic()
x = -2*pi:.001:pi;
S=sign(g(x)-f(x));
h=find(diff(S)~=0); % Find where two lines cross over
Sol=zeros(size(h));
Err=zeros(size(h));
if ~isempty(h) % There are some cross-over points
for i=1:length(h) % For each point, improve the approximation
xN=x(h(i):h(i)+1);
err=1;
while(abs(err)>tol) % Iteratively improve aproximation
S=sign(g(xN)-f(xN));
hF=find(diff(S)~=0);
xN=xN(hF:hF+1);
[~,I]=min(abs(f(xN)-g(xN)));
xG=xN(I);
err=f(xG)-g(xG);
xN=linspace(xN(1),xN(2),15);
end
Sol(i)=xG;
Err(i)=f(xG)-g(xG);
end
else % No crossover points - lines could meet at tangents
[h,I]=findpeaks(-abs(g(x)-f(x)));
Sol=x(I(abs(f(x(I))-g(x(I)))<1e-5));
Err=f(Sol)-g(Sol)
end
% We also have to check each endpoint
if abs(f(x(end))-g(x(end)))<tol && abs(Sol(end)-x(end))>1e-12
Sol=[Sol x(end)];
Err=[Err g(x(end))-f(x(end))];
end
if abs(f(x(1))-g(x(1)))<tol && abs(Sol(1)-x(1))>1e-12
Sol=[x(1) Sol];
Err=[g(x(1))-f(x(1)) Err];
end
toc()
Sol
Err
This will "zoom" in to the region around each suspected root, and iteratively improve the accuracy. You can tweak the parameters to see whether they give better behaviour (the tolerance tol, the 15, number of new points to generate, could be higher probably).
Question 2
You would probably be best off avoiding ezplot, and using plot, which gives you greater control. You can vectorise inline functions so that you can evaluate them like anonymous functions, as I did in the previous code snippet, using
f=inline('x^2')
F=vectorize(f)
F(1:5)
and this should make plotting much easier:
plot(x,f(x),'b',Sol,f(Sol),'ro',x,g(x),'k',Sol,G(Sol),'ro')
Question 3
I'm not sure why you don't want to display your roots as strings, what's wrong with this:
text(xPos,yPos,['x0=' num2str(Sol(1))]);