why when i use ezplot in for example [1 1.5] interval, a discontinuity will appear in some pieces of lines but when i use a closer interval like [1.3 1.5], the discontinuity will be annihilated?
EZPLOT is a general purpose plotting function that will automatically select a set of points at which to evaluate and plot the function you pass to it. Most of the time, things work fine. But there are some special cases where EZPLOT can have some trouble. It may not render well near discontinuities or points where there are rapid changes in the function (which it may mistake for a discontinuity).
That's the drawback of a function that is designed to be general enough to accept any function you give it: it's hard to make it general enough to handle everything exactly right, so some special edge cases look a little funny. In such cases, you should avoid functions like EZPLOT (which make a lot of choices for you) and plot things yourself by evaluating your functions at points you choose and plotting those points using the PLOT function. Here's a helpful link for this.
The problem is that ezplot() is useful, but not that robust.
A better option for plotting a function without discrete points is fplot(). Check out the documentation for it.
Here is an example of how to use it compared with ezplot():
lowerBound = 0;
upperBound = 1;
%# The ezplot way:
ezplot('y=sin(1/x)',[lowerBound,upperBound,-1,1])
%# The fplot way:
fplot('sin(1/x)',[lowerBound,upperBound])
fplot() will evaluate more points where the function changes more rapidly. Thus discontinuities will still cause problems in the graph if you look closely, but it will try harder to plot them accurately.
To plot a level curve of a function with three variables requires a little more typing:
%# First create a grid where you want the function to be drawn
[x,y]=meshgrid(-2:.01:2);
%# Remember that -2:.01:2 creates a vector with values from -2 to 2
%# in steps of .01
%# Then define your function
z=-3*y./(x.^2+y.^2+1);
%# Now graph the level curve of the function. I chose the level z=0.5:
contour(x,y,z,[0.5])
Related
Here is the given system I want to plot and obtain the vector field and the angles they make with the x axis. I want to find the index of a closed curve.
I know how to do this theoretically by choosing convenient points and see how the vector looks like at that point. Also I can always use
to compute the angles. However I am having trouble trying to code it. Please don't mark me down if the question is unclear. I am asking it the way I understand it. I am new to matlab. Can someone point me in the right direction please?
This is a pretty hard challenge for someone new to matlab, I would recommend taking on some smaller challenges first to get you used to matlab's conventions.
That said, Matlab is all about numerical solutions so, unless you want to go down the symbolic maths route (and in that case I would probably opt for Mathematica instead), your first task is to decide on the limits and granularity of your simulated space, then define them so you can apply your system of equations to it.
There are lots of ways of doing this - some more efficient - but for ease of understanding I propose this:
Define the axes individually first
xpts = -10:0.1:10;
ypts = -10:0.1:10;
tpts = 0:0.01:10;
The a:b:c syntax gives you the lower limit (a), the upper limit (c) and the spacing (b), so you'll get 201 points for the x. You could use the linspace notation if that suits you better, look it up by typing doc linspace into the matlab console.
Now you can create a grid of your coordinate points. You actually end up with three 3d matrices, one holding the x-coords of your space and the others holding the y and t. They look redundant, but it's worth it because you can use matrix operations on them.
[XX, YY, TT] = meshgrid(xpts, ypts, tpts);
From here on you can perform whatever operations you like on those matrices. So to compute x^2.y you could do
x2y = XX.^2 .* YY;
remembering that you'll get a 3d matrix out of it and all the slices in the third dimension (corresponding to t) will be the same.
Some notes
Matlab has a good builtin help system. You can type 'help functionname' to get a quick reminder in the console or 'doc functionname' to open the help browser for details and examples. They really are very good, they'll help enormously.
I used XX and YY because that's just my preference, but I avoid single-letter variable names as a general rule. You don't have to.
Matrix multiplication is the default so if you try to do XX*YY you won't get the answer you expect! To do element-wise multiplication use the .* operator instead. This will do a11 = b11*c11, a12 = b12*c12, ...
To raise each element of the matrix to a given power use .^rather than ^ for similar reasons. Likewise division.
You have to make sure your matrices are the correct size for your operations. To do elementwise operations on matrices they have to be the same size. To do matrix operations they have to follow the matrix rules on sizing, as will the output. You will find the size() function handy for debugging.
Plotting vector fields can be done with quiver. To plot the components separately you have more options: surf, contour and others. Look up the help docs and they will link to similar types. The plot family are mainly about lines so they aren't much help for fields without creative use of the markers, colours and alpha.
To plot the curve, or any other contour, you don't have to test the values of a matrix - it won't work well anyway because of the granularity - you can use the contour plot with specific contour values.
Solving systems of dynamic equations is completely possible, but you will be doing a numeric simulation and your results will again be subject to the granularity of your grid. If you have closed form solutions, like your phi expression, they may be easier to work with conceptually but harder to get working in matlab.
This kind of problem is tractable in matlab but it involves some non-basic uses which are pretty hard to follow until you've got your head round Matlab's syntax. I would advise to start with a 2d grid instead
[XX, YY] = meshgrid(xpts, ypts);
and compute some functions of that like x^2.y or x^2 - y^2. Get used to plotting them using quiver or plotting the coordinates separately in intensity maps or surfaces.
I want to plot some equations and inequalities like x>=50, y>=0,4x-5y>=8,x=40,x=60,y=25, y=45 in matlab and want to get the area produced by intersecting these equations and inequalities. Is it possible using matlab? If yes can someone provide me some manual? If not, is there some other software that can do this?
Integrals would work for your purposes, provided you know the points at which the curves intersect (something Matlab is also able to compute). Take a look at the documentation on the integral function.
q = integral(fun,xmin,xmax) approximates the integral of function fun
from xmin to xmax using global adaptive quadrature and default error
tolerances.
EDIT: As an additional resource, take a look at the code provided by user Grzegorz Konz on the Mathworks blog.
EDIT #2: I'm not familiar with any Matlab functions that'll take a vector of functions and return the points of intersection (if any) between all the curves. Users have produced functions that return the set of intersection points between two curves. You could run this function for each pair of equations in your list and use a function like polyarea to compute the area of the enclosed region if the curves are all straight lines.
I have some 3D parametric equations that I plot in matlab/octave and would like to find out where they intersect how can I go about doing this. Please note this is just a simple example I plan on having multiple parametric equations that will intersect.
What I'm trying to do is begin and end each plot at an intersection point.
My first thought was putting all the intersection points found at (t) into an array and plotting the ones I want I just couldn't figure out how to get the intersections at (t)
Example matlab/octave code below:
clear all, clf
t = 0:pi/60:2.45*pi;
plot3 ((t).*cos(t), (t).*sin(t), (t),'b*');
hold on
plot3 ((t).*sin((t)), (t).*cos((t)), (t),'r');
Here's an image with the arrows pointing to the intersecting points
I'd say that it is better to solve this system of equations analytically if possible.
Look, with your original code (just the plotted line styles were changed)
clear all, clf
t = 0:pi/60:2.45*pi;
plot3 ((t).*cos(t), (t).*sin(t), (t),'*b-');
hold on
plot3 ((t).*sin((t)), (t).*cos((t)), (t),'or-');
I have this picture at one of the intersections:
Its nice because the two plots intersect perfectly with each other in one of the plotted points.
Now consider the following situation. I changed your second line to
t = 0:0.05:2.45*pi;
Now I see this at the intersection point:
Now the two 3D polygonal lines generated for different values of t parameter in general case will not have any common points. So simple check for intersection of those linear segments will fail.
Same thing will happed if you update your equations to more complicated ones with more complicated dependencies over t (otherwise you are really lucky).
Analytical solution (if possible) will be unbeatable here.
As a numerical way to do this, consider distance D from an arbitrary point P1(x1(t1),y1(t1),z1(t1)) on one line to an arbitrary point P2(x2(t2),y2(t2),z2(t2)) on another one. Then find minimum of D as a function of two independent parameters t1 and t2. If D at minimum is zero (or very close to zero), then these two lines intersect. Otherwise they just pass each other.
If I have an RC circuit with transfer function 1/(1+sRC) how do I draw the transfer function using MATLAB?
Num2=[1];
Den2=[R*C 1];
RCcirc=tf(Num2,Den2);
How do I declare the R and the C so that there are no errors?
tf is the wrong tool for plotting the transfer function. Try these instead:
Use linspace to generate a range of values for s. Give R and C reasonable values of your choice.
Read up on arithmetic operations in MATLAB, especially ./
Look at how to use plot and familiarize yourself with the command using some simple examples from the docs.
With these you should be able to plot the transfer function in MATLAB :)
First of all you need to understand what transfer function you want. Without defined values of R and C you won't get any transfer function. Compare it to this, you want to plot a sine wave: x = sin(w*t), I hope you can agree with me that you cannot plot such a function (including axes) unless I specifically say e.g. t is the time, ranging from 0 seconds to 10 seconds and w is a pulsation of 1 rad/s. It's exactly the same with your RC network: without any values, it is impossible for numerical software such as MATLAB to come up with a plot.
If you fill in those values, you can use th tf function to display the transfer function in whatever way you like (e.g. a bode plot).
On the other hand, if you just want the expression 1/(1+s*R*C), take a look at the symbolic toolbox, you can do such things there. But to make a plot, you will still have to fill in the R and C value (and even a value for your Laplace variable in this case).
I found that if I use griddata Method with Cubic Interpolation method, for certain values of x, y, it will return NaN. One post says that this is because the x and y data are very near to convex hull.
Any idea how to fix this?
Edit: Note that I can't make sure that my inputs are monotonously increasing ( thus, gridfit doesn't work). The reason is because I would have to mesh my area ( which could be an irregular polygon in 2D), get all the points before generating the coressponding Z values for each points. My code is as follows:
function ZI=Interpolate3D(scatteredData, boundary)
%scatteredData is the scattered points, boundary is the area that I want to generate 3D surface.
% Given the boundaries, generate mesh first
[element,points]= GenMesh(boundary);
ZI = griddata(scatteredData(:,1),scatteredData(:,2),scatteredData(:,3),points(:,1),points(:,2), 'cubic',{'QJ'});
If your points are outside of the convex hull, you CANNOT get a result other than NaN from griddata using the cubic option. If the point is right on the line, then a NaN may result, depending upon what happens in the least significant bits of the computation.
The issue is that the cubic method uses a triangulation. If your point is outside of the convex hull, then the triangulation fails on that point.
Of course, you CAN use the -v4 method, but there are tremendously good reasons why it has largely been superceded. It uses a distance based interpolation method, where for n data points, a full nxn matrix must be generated. Then a system of equations is solved using that matrix. This will be quite slow for even moderately large problems.
The virtue of the -v4 method is it will extrapolate smoothly without producing nans. This is why it was left in there.
For larger problems where you need a smooth result, and you still wish to extrapolate outside of the convex hull, you can use my gridfit tool. It does do smoothing though, not pure interpolation.
All such methods have tradeoffs that you must resolve for your particular problem.
Since the release of Matlab R2013a you can use scatteredInterpolant instead of griddata.
This has several advantages:
Matlab can perform interpolation as well as extrapolation on a scatteredInterpolant object. You can specify a point outside the convex hull of your scattered data and will still not get a NaN.
Once created, the scatteredInterpolant object can be evaluated multiple times, thus saving computational time compared to calling griddata several times.
On the down side: While you can specify the interpolation and extrapolation methods, cubic is not available but only linear, nearest and natural.
Using scatteredInterpolant your code could look like
F = scatteredInterpolant(scatteredData(:,1),scatteredData(:,2),scatteredData(:,3));
ZI=F(points(:,1),points(:,2));
Are you sure you want cubic interpolation? For some input data the calculated z-nodes could have extreme large values!
I always use -v4 option like the post in your link mentions. You can also play with options used in Qhull via delaunayn, some (but not all) are {'Qt','Qbb','Qc'} http://www.qhull.org/html/qhull.htm