Suppose we have five vertices:
X = [0 1;
2 1;
4 1;
1 0;
3 0];
a triangulation:
T = [1 4 2;
4 5 2;
5 3 2];
and function values defined on the vertices:
Fx = [1;
2;
3;
4;
-5];
then we can easily compute the function value for any point inside the triangle by using the barycentric coordinates. For point P = [1 .5], which lies in the first triangle, the barycentric coordinates are B = [.25 .5 .25], so the function evaluates to Fxi = 1/4 + 4/2 + 2/4 = 2.75.
However, I have difficulty to see how one would extrapolate this surface. We could find the closest triangle and extrapolate from that. The problem is that this results in a discontinuous function. Consider e.g. point P = [2 2]. According to triangle 1, its value would be -0.5, whereas according to triangle 3 its value would be 9.5.
Is there a "standard" or generally accepted approach to extrapolate from piecewise linear functions? Any pointers to existing material also greatly appreciated.
A possibility is Shepard's method:
https://en.wikipedia.org/wiki/Inverse_distance_weighting
The resulting function interpolates the input values defined at the vertices and is non-linear but continuous everywhere else.
The choice p=2 usually gives decent results.
Another technique to look for are "Barycentric coordinates for non-convex polygons".
The following publication shows (page 8 etc.) how the weight functions behave outside the polygons
https://www.in.tu-clausthal.de/fileadmin/homes/techreports/ifi0505hormann.pdf
However, even this solution does not behave piecewise-linear on your given triangulation.
Related
I created a circle in MATLAB using following code.
I need to find the points inside the circle in 3D space
radius = 5;
theta=linspace(0,2*pi);
rho=ones(1,100).*radius;
[x,z]=pol2cart(theta,rho);
y=center(2)*ones(1,length(x))
fill3(x,y,z,'yellow')
How can I find Cartesian co-ordinates of points inside this circle?
Not sure I'm understanding well your question. Obviously there are infinite points inside the circle so I guess you want to check whether a point (or a set of points) is inside or not. If you loop through a list of points, those who meet the following criteria are inside (or in the perimeter of the circle):
norm([xi,zi]) <= radius
yi = 0 (same plane)
Was this what you were asking?
Edit: you can do it pretty quickly in matlab without a loop. Lets imagine you have x = [1 2 3] and z = [4 5 6]. To check all combinations you can use repmat with x and z' (transverse) to obtain: xr = [ 1 2 3; 1 2 3; 1 2 3] and zr = [ 4 4 4 ; 5 5 5 ; 6 6 6]. So you have 2 matrixes with the coordinates of all possible points. Now you can calculate the norm as: N = sqrt(xr.^2+zr.^2). All i,j with Nij <= radius are inside your circle (considering all your x and z are <= radius of course)
I am to fit planes through various points in an image, but I am having issues with forcing the line through a particular point in the image. This happens particularly when the line is 90 degrees.
My code is as follows:
I = [3 3 3 3 3 2 2
3 3 3 3 2 2 2
3 3 3 3 2 2 2
3 3 1 2 2 2 2
1 1 1 2 2 2 2
1 1 1 1 1 2 2
1 1 1 1 1 1 1];
% force the line through point p
p = [3,3];
% points to fit plane through
edgeA = [3,3.5; 3,4; 2.5,4; 2,4; 1.5,4];
edgeB = [3.5,3; 4,3; 4.5,3; 5,3];
% fit a plane through p and edgeA
xws = [p(2), edgeA(:,2)']';
yws = [p(1), edgeA(:,1)']';
Cws = [xws ones(size(xws))];
dws = yws;
Aeqws = [p(2) 1];
beqws = [p(1)];
planefitA = lsqlin(Cws ,dws,[],[],Aeqws, beqws);
% fit a plane through p and edgeB
xwn = [p(2), edgeB(:,2)']';
ywn = [p(1), edgeB(:,1)']';
Cwn = [xwn ones(size(xwn))];
dwn = ywn;
Aeqwn = [p(2) 1];
beqwn = [p(1)];
planefitB = lsqlin(Cwn ,dwn,[],[],Aeqwn, beqwn);
%%%%% plot the fitted planes:
xAxis = linspace(0, size(I, 2), 12);
%obtain linear curve
fA = planefitA(1)*xAxis + planefitA(2);
fB = planefitB(1)*xAxis + planefitB(2);
%plot the fitted curve
RI = imref2d(size(I),[0 size(I, 2)],[0 size(I, 1)]);
figure, imshow(I, RI, [], 'InitialMagnification','fit')
grid on;
hold on;
plot(xAxis,fA, 'Color', 'b', 'linewidth', 2);
plot(xAxis,fB, 'Color', 'r', 'linewidth', 2);
All the points in edgeB fall on a 90 degrees line. However, the function ends up fitting a wrong line through those points. I know this because using
planefitB = polyfit([p(2), edgeB(:,2)'], [p(1), edgeB(:,1)'], 1);
works for this particular line but the problem is that i have these process repeated so many times at different locations in my image, hence i do not know how to suggest polyfit when the line would be 90 degrees.
Please, any ideas/suggestions on how i could make this work? Many thanks.
This amounts to the least squares solution of only the angle of the line. The offset is fixed by the fact that it has to go through (3,3). The easiest way to express this is by offsetting your data points by the known crossing. That is, subtract (3,3) from your data points, and fit the best m for y=mx, the b being fixed to 0.
For the non-vertical case, you can use a classic least-squares formulation, but don't augment the constant 1 into the Vandermonde matrix:
slope = (edgeA(:,2) - p(2)) \ (edgeA(:,1) - p(1));
This gives exactly the same answer as your constrained lsq solution.
Now for the vertical line: A non-vertical line can be expressed in the standard functional form of y=mx, where the least squares formulation implicitly assumes an independent and a dependent variable. A vertical line doesn't follow that, so the only general choice is a "Total Least Squares" formulation, where errors in both variables are considered, rather than just the residuals in the dependent (y) variable.
The simplest way to write this is to choose a and b to minimize ax - by, in the least squares sense. [x_k -y_k]*[a b].' should be as close to a zero vector as possible. This is the vector closest to the null space of the [x -y] matrix, which can be computed with the svd. Swapping columns and fudging signs lets us just use svd directly:
[u s v] = svd(bsxfun(#minus, edgeA, p));
The last column of v is the closest to the null space, so mapping back to your x/y definitions, (edgeA-p)*v(:,2) is the line, so the y multiplier is in the top position, and the x in the lower, with a sign flip. To convert to y=mx form, just divide:
slope = -v(2,2)/v(1,2);
Note that this answer will be quite a bit different than the normal least squares answer, since you are treating the residuals differently. Also, the final step of computing "slope" won't work in the vertical case for the reasons we've already discussed (it produces Inf), so you are probably better off leaving the line as a normalized 2-vector, which won't have any corner cases.
I have two datasets, one of which is a target position, and the other is the actual position. I would like to plot the target with a +/- acceptable range and then overlay with the actual. This question is only concerning the target position however.
I have unsuccessfully attempted the built in area, fill, and rectangle functions. Using code found on stackoverflow here, it is only correct in certain areas.
For example
y = [1 1 1 2 1 1 3 3 1 1 1 1 1 1 1]; % Target datum
y1 = y+1; %variation in target size
y2 = y-1;
t = 1:15;
X=[t,fliplr(t)]; %create continuous x value array for plotting
Y=[y1,fliplr(y2)]; %create y values for out and then back
fill(X,Y,'b');
The figure produced looks like this:
I would prefer it to be filled within the red boxes drawn on here:
Thank you!
If you would just plot a function y against x, then you could use a stairs plot. Luckily for us, you can use the stairs function like:
[xs,ys] = stairs(x,y);
to create the vectors xs, ys which generate a stairs-plot when using the plot function. We can now use these vectors to generate the correct X and Y vectors for the fill function. Note that stairs generates column vectors, so we have to transpose them first:
y = [1 1 1 2 1 1 3 3 1 1 1 1 1 1 1]; % Target datum
y1 = y+1; %variation in target size
y2 = y-1;
t = 1:15;
[ts,ys1] = stairs(t,y1);
[ts,ys2] = stairs(t,y2);
X=[ts.',fliplr(ts.')]; %create continuous x value array for plotting
Y=[ys1.',fliplr(ys2.')]; %create y values for out and then back
fill(X,Y,'b');
Again, thank you hbaderts. You answered my question perfectly, however when I applied it to the large data set I needed for, I obtained this image
https://dl.dropboxusercontent.com/u/37982601/stair%20fill.png
I think it is because the fill function connects vertices to fill?
In any case, for the potential solution of another individual, combined your suggested code with the stair function and used the area function.
By plotting them on top of one another and setting the color of the lower area to be white, it appears as the rectangular figures I was after.
%sample code. produces image similar to o.p.
y = [1 1 1 2 1 1 3 3 1 1 1 1 1 1 1];
y1 = y+1;
y2 = y-1;
t = 1:15;
[ts,ys1] = stairs(t,y1);
[ts,ys2] = stairs(t,y2);
area(ts,ys1,'FaceColor','b','EdgeColor','none')
hold on
area(ts,ys2,'FaceColor','w','EdgeColor','none')
https://dl.dropboxusercontent.com/u/37982601/stair%20area.png
Thanks again for your help and for pointing me in the right direction!
How can I plot amplitude of transfer function in three dimension (for instance to check poles and zeros on graph) ?
Suppose this is my transfer function:
My code:
b = [6 -10 2];
a = [1 -3 2];
[x, y] = meshgrid(-3:0.1:3);
z = x+y*j;
res = (polyval(b, z))./(polyval(a,z));
surf(x,y, abs(res));
Is it correct? I'd also like to know is it possible to mark unit circle on plot?
I think it's correct. However, you're computing H(z^-1), not H(z). Is that you want to do? For H(z), just reverse the entries in a from left to right (with fliplr), and do the same to b:
res = (polyval(fliplr(b), z))./(polyval(fliplr(a),z));
To plot the unit circle you can use rectangle. Seriously :-) It has a 'Curvature' property which can be set to generate a circle.
It's best if you use imagesc instead of surf to make the circle clearly visible. You will get a view from above, where color represents height (value of abs(H)):
imagesc(-3:0.1:3,-3:0.1:3, abs(res));
hold on
rectangle('curvature', [1 1], 'position', [-1 -1 2 2], 'edgecolor', 'w');
axis equal
I have never in my whole life heard of a 3D transfer function, it doesn't make sense. I think you are completely wrong: z does not represent a complex number, but the fact that your transfer function is a discrete one, rather than a continuous one (see the Z transform for more details).
The correct way to do this in MATLAB is to use the tf function, which requires the Control System Toolbox (note that I have assumed your discrete sample time to be 0.1s, adjust as required):
>> b = [6 -10 2];
a = [1 -3 2];
>> sys = tf(b,a,0.1,'variable','z^-1')
sys =
6 - 10 z^-1 + 2 z^-2
--------------------
1 - 3 z^-1 + 2 z^-2
Sample time: 0.1 seconds
Discrete-time transfer function.
To plot the transfer function, use the bode or bodeplot function:
bode(sys)
For the poles and zeros, simply use the pole and zero functions.
I have two vectors of the same size. The first one can have any different numbers with any order, the second one is decreasing (but can have the same elements) and consists of only positive integers. For example:
a = [7 8 13 6];
b = [5 2 2 1];
I would like to plot them in the following way: on the x axis I have points from a vector and on the y axis I have the sum of elements from vector b before this points divided by the sum(b). Therefore I will have points:
(7; 0.5) - 0.5 = 5/(5+2+2+1)
(8; 0.7) - 0.7 = (5+2)/(5+2+2+1)
(13; 0.9) ...
(6; 1) ...
I assume that this explanation might not help, so I included the image
Because this looks to me as a cumulative distribution function, I tried to find luck with cdfplot but with no success.
I have another option is to draw the image by plotting each line segment separately, but I hope that there is a better way of doing this.
I find the values on the x axis a little confusing. Leaving that aside for the moment, I think this does what you want:
b = [5 2 2 1];
stairs(cumsum(b)/sum(b));
set(gca,'Ylim',[0 1])
And if you really need those values on the x axis, simply rename the ticks of that axis:
a = [7 8 13 6];
set(gca,'xtick',1:length(b),'xticklabel',a)
Also grid on will add grid to the plot