It seems to be very basic question, but I wonder when I plot x values against y values, what interpolation technique is used behind the scene to show me the discrete data as continuous? Consider the following example:
x = 0:pi/100:2*pi;
y = sin(x);
plot(x,y)
My guess is it is a Lagrangian interpolation?
No, it's just a linear interpolation. Your example uses a quite long dataset, so you can't tell the difference. Try plotting a short dataset and you'll see it.
MATLAB's plot performs simple linear interpolation. For finer resolution you'd have to supply more sample points or interpolate between the given x values.
For example taking the sinus from the answer of FamousBlueRaincoat, one can just create an x vector with more equidistant values. Note, that the linear interpolated values coincide with the original plot lines, as the original does use linear interpolation as well. Note also, that the x_ip vector does not include (all) of the original points. This is why the do not coincide at point (~0.8, ~0.7).
Code
x = 0:pi/4:2*pi;
y = sin(x);
x_ip = linspace(x(1),x(end),5*numel(x));
y_lin = interp1(x,y,x_ip,'linear');
y_pch = interp1(x,y,x_ip,'pchip');
y_v5c = interp1(x,y,x_ip,'v5cubic');
y_spl = interp1(x,y,x_ip,'spline');
plot(x,y,x_ip,y_lin,x_ip,y_pch,x_ip,y_v5c,x_ip,y_spl,'LineWidth',1.2)
set(gca,'xlim',[pi/5 pi/2],'ylim',[0.5 1],'FontSize',16)
hLeg = legend(...
'No Interpolation','Linear Interpolation',...
'PChip Interpolation','v5cubic Interpolation',...
'Spline Interpolation');
set(hLeg,'Location','south','Fontsize',16);
By the way..this does also apply to mesh and others
[X,Y] = meshgrid(-8:2:8);
R = sqrt(X.^2 + Y.^2) + eps;
Z = sin(R)./R;
figure
mesh(Z)
No, Lagrangian interpolation with 200 equally spaced points would be an incredibly bad idea. (See: Runge's phenomenon).
The plot command simply connects the given (x,y) points by straight lines, in the order given. To see this for yourself, use fewer points:
x = 0:pi/4:2*pi;
y = sin(x);
plot(x,y)
Related
I am using MATLAB to try to interpolate data for an object that is moving in 3D space at variable speed. I am starting with an array with columns: time, x, y, z, velocity. I want to interpolate the xyz path, but I can't figure out how to interpolate time and velocity at the same points that the xyz interpolation produces.
My problem is that if I interpolate using the cscvn function and fnval, then the interpolated points are not evenly spaced (yellow stars in the below image) and I'm not sure how to interpolate the times and speeds for those points.
Alternatively if I interpolate with interp1, it does produce evenly spaced points, but the interpolation is not better with cscvn.
I tried doing a 5 dimensional interpolation, and that didn't produce the desired results. I'm not sure how to deal with this problem.
How can I interpolate the xyz path, and then interpolate the time and velocities at those unevenly spaced points?
Here is the code I am using:
% Generate some fake data
flightPathRate = 1;
x = (-5:flightPathRate:10)';
y = sin(4*x);
z = linspace(3,5, length(x))';
t = ((0:length(x)-1)*flightPathRate)';
vel = x.^2 + 10;
pathData = [t x y z vel];
% Interpolate with cscvn
curve = cscvn(pathData(:,2:4)');
plot3(x, y, z, 'ob-')
hold on
fnplt(curve)
% Evaluate the spline curve created with cscvn at finer points.
splinePoints = fnval(curve, 0:0.1:16);
plot3(splinePoints(1,:), splinePoints(2,:), splinePoints(3,:), '*')
%% Interpolate with interp1
cs = cat(1,0,cumsum(sqrt(sum(diff([x, y, z], [], 1).^2, 2))));
dd = interp1(cs, [x, y, z], unique([cs(:)' linspace(0,cs(end),100)]), 'spline')
hold on
plot3(dd(:,1), dd(:,2), dd(:,3), '.r-')
axis image, view(3), legend({'Original', 'Spline Curve with cscvn/fnplt', 'Interp. Points with cscvn/fnval', 'Interp. Spline with interp1'})
Here is the plot produced.
I ran your code successfully.
Q.1: Please explain the role of velocity in your code. You define vel as fake data, at every fake data time point. But it is totally unrelated to the x,y,z fake data.
Answer to your question about how to determine the times associated with the x,y,z points which you get from cscvn(): there is no reasonable proven way to do that. Therefore I recommend that you not use cscvn(). Its main advantage is to make periodic splines, i.e. splines for cirves that return to their origin. But this problem is not like that.
Your interpolated data from interp1() looks good to me. I would go with it. It is a simple matter to esitmate velocity (|v| and vx, vy, vz) from the interpolated points. When you do, I recommend using the interpolated points before and after each point, to get an un-shifted estimate in the middle. Try the 'makima' or 'pchip' method with interp1(), instead of the 'spline' method, if you think 'spline' overshoots too much.
I'm trying to fit a shape perserving interpolation curve to my angular data (r/phi).
However, as I have repeated x-values when I transform the datapoints to (x/y), I can not simply use pchip.
I know for spline interpolation, there is cscvn and fnplt, is there anything similar for pchip?
Furthermore, there is an example of spline fitting to angular data in the matlab documentation of "spline", but I don't quite get it how I could adapt it to pchip and different data points.
I also found the interparc-function by John d'Errico, but I would like to keep my datapoints instead of having equally spaced ones.
To make it clearer, here a figure of my datapoints with linear (blue) and spline interpolation (black). The curve I'd like to get would be something in between this two, without the steep edges in the linear case but with less overshoot than in the spline case....
Thanks for your help!
use 1D parametric interpolation:
n = 20;
r = 1 + rand(n-1,1)*0.01;%noisy r's
theta = sort(2*pi*rand(n-1,1));
% closing the circle
r(end+1) = r(1);
theta(end+1) = theta(1);
% convert to cartesian
[x,y] = pol2cart(theta,r);
% interpolate with parameter t
t = (1:n)';
v = [x,y];
tt = linspace(1,n,100);
X = interp1(t,v,tt,'pchip');
% plot
plot(x,y,'o');
hold on
plot(X(:,1),X(:,2));
MATLAB's surf command allows you to pass it optional X and Y data that specify non-cartesian x-y components. (they essentially change the basis vectors). I desire to pass similar arguments to a function that will draw a line.
How do I plot a line using a non-cartesian coordinate system?
My apologies if my terminology is a little off. This still might technically be a cartesian space but it wouldn't be square in the sense that one unit in the x-direction is orthogonal to one unit in the y-direction. If you can correct my terminology, I would really appreciate it!
EDIT:
Below better demonstrates what I mean:
The commands:
datA=1:10;
datB=1:10;
X=cosd(8*datA)'*datB;
Y=datA'*log10(datB*3);
Z=ones(size(datA'))*cosd(datB);
XX=X./(1+Z);
YY=Y./(1+Z);
surf(XX,YY,eye(10)); view([0 0 1])
produces the following graph:
Here, the X and Y dimensions are not orthogonal nor equi-spaced. One unit in x could correspond to 5 cm in the x direction but the next one unit in x could correspond to 2 cm in the x direction + 1 cm in the y direction. I desire to replicate this functionality but drawing a line instead of a surf For instance, I'm looking for a function where:
straightLine=[(1:10)' (1:10)'];
my_line(XX,YY,straightLine(:,1),straightLine(:,2))
would produce a line that traced the red squares on the surf graph.
I'm still not certain of what your input data are about, and what you want to plot. However, from how you want to plot it, I can help.
When you call
surf(XX,YY,eye(10)); view([0 0 1]);
and want to get only the "red parts", i.e. the maxima of the function, you are essentially selecting a subset of the XX, YY matrices using the diagonal matrix as indicator. So you could select those points manually, and use plot to plot them as a line:
Xplot = diag(XX);
Yplot = diag(YY);
plot(Xplot,Yplot,'r.-');
The call to diag(XX) will take the diagonal elements of the matrix XX, which is exactly where you'll get the red patches when you use surf with the z data according to eye().
Result:
Also, if you're just trying to do what your example states, then there's no need to use matrices just to take out the diagonal eventually. Here's the same result, using elementwise operations on your input vectors:
datA = 1:10;
datB = 1:10;
X2 = cosd(8*datA).*datB;
Y2 = datA.*log10(datB*3);
Z2 = cosd(datB);
XX2 = X2./(1+Z2);
YY2 = Y2./(1+Z2);
plot(Xplot,Yplot,'rs-',XX2,YY2,'bo--','linewidth',2,'markersize',10);
legend('original','vector')
Result:
Matlab has many built-in function to assist you.
In 2D the easiest way to do this is polar that allows you to make a graph using theta and rho vectors:
theta = linspace(0,2*pi,100);
r = sin(2*theta);
figure(1)
polar(theta, r), grid on
So, you would get this.
There also is pol2cart function that would convert your data into x and y format:
[x,y] = pol2cart(theta,r);
figure(2)
plot(x, y), grid on
This would look slightly different
Then, if we extend this to 3D, you are only left with plot3. So, If you have data like:
theta = linspace(0,10*pi,500);
r = ones(size(theta));
z = linspace(-10,10,500);
you need to use pol2cart with 3 arguments to produce this:
[x,y,z] = pol2cart(theta,r,z);
figure(3)
plot3(x,y,z),grid on
Finally, if you have spherical data, you have sph2cart:
theta = linspace(0,2*pi,100);
phi = linspace(-pi/2,pi/2,100);
rho = sin(2*theta - phi);
[x,y,z] = sph2cart(theta, phi, rho);
figure(4)
plot3(x,y,z),grid on
view([-150 70])
That would look this way
I want to plot the area above and below a particular value in x axis.
The problem i am facing is with discrete values. The code below for instance has an explicit X=10 so i have written it in such a way that i can find the index and calculate the values above and below that particular value but if i want to find the area under the curve above and below 4 this program will now work.
Though in the plot matlab does a spline fitting(or some sort of fitting for connecting discrete values) there is a value for y corresponding to x=4 that matlab computes i cant seem to store or access it.
%Example for Area under the curve and partial area under the curve using Trapezoidal rule of integration
clc;
close all;
clear all;
x=[0,5,10,15,20];% domain
y=[0,25,50,25,0];% Values
LP=log2(y);
plot(x,y);
full = trapz(x,y);% plot of the total area
I=find(x==10);% in our case will be the distance value up to which we want
half = trapz(x(1:I),y(1:I));%Plot of the partial area
How can we find the area under the curve for a value of ie x = 2 or 3 or 4 or 6 or 7 or ...
This is an elaboration of patrik's comment, "first interpolate and then integrate".
For the purpose of this answer I'll assume that the area in question is the area that can be seen in the plot, and since plot connects points by straight lines I assume that linear interpolation is adequate. Moreover, since the trapezoidal rule itself is based on linear interpolation, we only need interpolated values at the beginning and end of the interval.
Starting from the given points
x = [0, 5, 10, 15, 20];
y = [0, 25, 50, 25, 0];
and the integration interval limits, say
xa = 4;
xb = 20;
we first select the data points within the limits
ind = (x > xa) & (x < xb);
xw = x(ind);
yw = y(ind);
and then complete them by interpolation values at the edges:
ya = interp1(x, y, xa);
yb = interp1(x, y, xb);
xw = [xa, xw, xb];
yw = [ya, yw, yb];
Now we can simply apply trapezoidal integration:
area = trapz(xw, yw);
I think that you either need more samples, or to interpolate the data. Another alternative is to use a function handle. Then you need to know the function though. Example using linear interpolation follows.
x0 = [0;5;10;15;20];
y0 = [0,25,50,25,0];
x1 = 0:20;
y1 = interp1(x0,y0,x1,'linear');
xMax = 4;
partInt = trapz(x1(x1<=xMax),y1(x1<=xMax));
Some other kind of interpolation may be suitable, but that is hard to say without more information. Also, this interpolates from the beginning to x. However, I guess figuring out how to change the limits should be easy from here. This solution is different than the former, since it is less depending on the pyramid shape of the data. So to say, it is more general.
I have a function z = f(x, y), where z is the value at point (x, y). How may I integrate z over the x-y plane in MATLAB?
By function above, I actually mean I have something similar to a hash table. That is, given a (x, y) pair, I can look up the table to find the corresponding z value.
The problem would be rather simple, if the points were uniformly distributed over x-y plane, in which case I can simply sum up all the z values, multiply it with the bottom area, and finally divide it by the number of points I have. However, the distribution is not uniform as shown below. So I am actually asking for the computation method that minimises the error.
The currently accepted answer will only work for gridded data. If your data is scattered you can use the following approach instead:
scatteredInterpolant + integral2:
f = scatteredInterpolant(x(:), y(:), z(:), 'linear');
int = integral2(#(x,y) f(x,y), xmin, xmax, ymin, ymax);
This defines the linear interpolant f of the data z(i) = f(x(i),y(i)) and uses it as an argument to integral2. Note that ymin and ymax, instead of doubles, can be function handles depending on x. So usually you will be integrating rectangles, but this could be used for integration regions a bit more complicated.
If your integration area is rather complicated or has holes, you should consider triangulating your data.
DIY using triangulation:
Let's say your integration area is given by the triangulation trep, which for example could be obtained by trep = delaunayTriangulation(x(:), y(:)). If you have your values z corresponding to z(i) = f(trep.Points(i,1), trep.Points(i,2)), you can use the following integration routine. It computes the exact integral of the linear interpolant. This is done by evaluating the areas of all the triangles and then using these areas as weights for the midpoint(mean)-value on each triangle.
function int = integrateTriangulation(trep, z)
P = trep.Points; T = trep.ConnectivityList;
d21 = P(T(:,2),:)-P(T(:,1),:);
d31 = P(T(:,3),:)-P(T(:,1),:);
areas = abs(1/2*(d21(:,1).*d31(:,2)-d21(:,2).*d31(:,1)));
int = areas'*mean(z(T),2);
If you have a discrete dataset for which you have all the x and y values over which z is defined, then just obtain the Zdata matrix corresponding to those (x,y) pairs. Save this matrix, and then you can make it a continuous function using interp2:
function z_interp = fun(x,y)
z_interp = interp2(Xdata,Ydata,Zdata,x,y);
end
Then you can use integral2 to find the integral:
q = integral2(#fun,xmin,xmax,ymin,ymax)
where #fun is your function handle that takes in two inputs.
I had to integrate a biavariate normal distribution recently in MatLab. The idea is very simple. Matlab defines a surface through a meshgrid, so from x, y you need to do this:
x = -10:0.05:10;
y = x;
[X,Y] = meshgrid(x',y');
...for example. Then, let's call FX the function that defines the value at each point of the surface. To calculate the integral you just need to do this:
surfint = zeros(length(X),1);
for a = 1:length(X)
surfint(a,1) = trapz(x,FX(:,a));
end
trapz(x, surfint)
For me, this is the simplest way.