I am trying to plot a zplot in Matlab that displays a unit circle, centered at 0 along with the poles and zeros of the plot. I am not allowed to use any other matlab function such as zplane or pzplot to do this. So far I am able to plot a unit circle just fine but I am having trouble getting my plot to display more of the axis without warping my circle. I also am having a heard time finding the poles and zeros of my function and also how to display the poles as little x's and the zeros as little o's on my plot. Any help would be greatly appreciated! My assignment looks like this and must correctly handle cases such as
zplot([0 1 1], [0 1]);
zplot([0 1 1], [0 0 1]);
function zplot(b, a)
% ZPLOT Plot a zero-pole plot.
-1 -nb
B(z) b(1) + b(2)z + .... + b(nb+1)z
H(z) = ---- = ---------------------------------
-1 -na
A(z) a(1) + a(2)z + .... + a(na+1)z
% zplot(b, a) plots the zeros and poles which determined by vectors b and a
% The plot includes the unit circle and axes for reference, plotted in black.
% Each zero is represented with a blue 'o' and each pole with a red 'x' on the
%plot.
xmin;
xmax;
ymin;
ymax;
% vector of angles at which points are drawn
angle = 0:2*pi/100:2*pi;
% Unit radius
R = 1;
% Coordinates of the circle
x = R*cos(angle);
y = R*sin(angle);
% Plot the circle
plot(x,y);
axis ([xmin, xmax, ymin, ymax]);
grid on;
end
If you can't use pzplot() it is not hard. Here is a hint:
num = [1 4 1];%numerator coefficients of transfer function
den = [1 2 1];%denominator coefficients
z = roots(num)%zeros
p = roots(den)%poles
angle = 0:2*pi/100:2*pi;
xp = cos(angle);
yp = sin(angle);
figure(1)
scatter(z,zeros(length(z),1),'o');
hold on
scatter(p,zeros(length(p),1),'x');
plot(xp,yp);
axis equal
The output
Note that I haven't dealt with imaginary poles/zeros in this example. You'll need to calculate the proper x,y coordinates for a given imaginary pole or zero. (all the poles/zeros in this example are real, not imaginary)
Given a transfer function G, you can use the pzplot() command and add a circle to it.
G = tf([1 4 1],[1 2 1]);
angle = [0:0.1:2*pi+0.1];
xp = cos(angle);
yp = sin(angle);
figure(1)
hold on
pzplot(G);
plot(xp,yp);
axis equal;
This should give you the pole-zero plot with x's for poles, o's for zeros, and the unit circle.
Here is the result.
Related
I try to solve the following 2D elliptic PDE electrostatic problem by fixing the Parallel plate Capacitors code. But I have problem to plot the circle region. How can I plot a circle region rather than the square?
% I use following two lines to label the 50V and 100V squares
% (but it should be two circles)
V(pp1-r_circle_small:pp1+r_circle_small,pp1-r_circle_small:pp1+r_circle_small) = 50;
V(pp2-r_circle_big:pp2+r_circle_big,pp2-r_circle_big:pp2+r_circle_big) = 100;
% Contour Display for electric potential
figure(1)
contour_range_V = -101:0.5:101;
contour(x,y,V,contour_range_V,'linewidth',0.5);
axis([min(x) max(x) min(y) max(y)]);
colorbar('location','eastoutside','fontsize',10);
xlabel('x-axis in meters','fontsize',10);
ylabel('y-axis in meters','fontsize',10);
title('Electric Potential distribution, V(x,y) in volts','fontsize',14);
h1=gca;
set(h1,'fontsize',10);
fh1 = figure(1);
set(fh1, 'color', 'white')
% Contour Display for electric field
figure(2)
contour_range_E = -20:0.05:20;
contour(x,y,E,contour_range_E,'linewidth',0.5);
axis([min(x) max(x) min(y) max(y)]);
colorbar('location','eastoutside','fontsize',10);
xlabel('x-axis in meters','fontsize',10);
ylabel('y-axis in meters','fontsize',10);
title('Electric field distribution, E (x,y) in V/m','fontsize',14);
h2=gca;
set(h2,'fontsize',10);
fh2 = figure(2);
set(fh2, 'color', 'white')
You're creating a square due to the way you're indexing (see this post on indexing). You've specified the rows to run from pp1-r_circle_small to pp1+r_circle_small and similar for the columns. Given that Swiss cheese is not an option, you're creating a complete square.
From geometry we know that all points within distance sqrt((X-X0)^2 - (Y-Y0)^2) < R from the centre of the circle at (X0,Y0) with radius R are within the circle, and the rest outside. This means that you can simply build a mask:
% Set up your grid
Xsize = 30; % Your case: 1
Ysize = 30; % Your case: 1
step = 1; % Amount of gridpoints; use 0.001 or something
% Build indexing grid for circle search, adapt as necessary
X = 0:step:Xsize;
Y = 0:step:Ysize;
[XX,YY] = meshgrid(X, Y);
V = zeros(numel(X), numel(Y));
% Repeat the below for both circles
R = 10; % Radius of your circle; your case 0.1 and 0.15
X0 = 11; % X coordinate of the circle's origin; your case 0.3 and 0.7
Y0 = 15; % Y coordinate of the circle's origin; your case 0.3 and 0.7
% Logical check to see whether a point is inside or outside
mask = sqrt( (XX - X0).^2 + (YY - Y0).^2) < R;
V(mask) = 50; % Give your circle the desired value
imagesc(V) % Just to show you the result
axis equal % Use equal axis to have no image distortion
mask is a logical matrix containing 1 where points are within your circle and 0 where points are outside. You can then use this mask to logically index your potential grid V to set it to the desired value.
Note: This will, obviously, not create a perfect circle, given you cannot plot a perfect circle on a square grid. The finer the grid, the more circle-like your "circle" will be. This shows the result with step = 0.01
Note 2: You'll need to tweek your definition of X, Y, X0, Y0 and R to match your values.
Applying 3D rotation matrix to the x,y,z values obtained from surface function object. The error I get is due to the matrix not being nonconforment but how can I adjust the matrix correctly?
I know hgtransform / makehgtform can do rotations but I need to use rotation matrices since I plan on testing it using matrices created from quaternions.
I've created a little plane out of cylinders and the surface functions.
See code below:
clear all,clf
ax=axes('XLim',[-2 2],'YLim', [-2 10],'ZLim',[-1.5 1.5]);
grid on;
%axis equal;
xlabel('x');
ylabel('y');
zlabel('z');
ax
% rotate around
rot_mat = [.707 -.707 0;.707 .707 0; 0 0 1] %rotation matrix
[xc yc zc] = cylinder([0.1 0.0]); %cone
[x y z]= cylinder([0.2 0.2]);
h(1) = surface(xc,zc,-yc,'FaceColor', 'red'); %noise cone
h(2) = surface(z,y,0.5*x,'FaceColor', 'blue'); %right wing
h(3) = surface(-z,y,0.5*x,'FaceColor', 'yellow');%left wing
h(4) = surface(x,-1.5*z,0.5*y,'FaceColor', 'green'); %main body
h(5) = surface(xc,(1.5*yc)-1.3,z*.5,'FaceColor', 'red'); %tail
view(3);
x_temp = get(h(1),'xdata'); % get x values
y_temp = get(h(1),'ydata');
z_temp =get(h(1),'zdata');
xc_new=x_temp.*rot_mat;
%zc_new=
%yc_new=
I can get the x,y, and z value by using the commands
x_temp = get(h(1),'xdata');
y_temp = get(h(1),'ydata');
z_temp = get(h(1),'zdata');
The error I get is due to the matrix being nonconforment but how can I adjust the matrix correctly?
error: test_object_matrix_rot: product: nonconformant arguments (op1 is 2x21, op2 is 3x3).
The error is with the line xc_new=x_temp.*rot_mat;
PS: I'm using Octave 5.0.91 which is like Matlab
YOu are messing up a lot of things......in fact I would say, you have made your work complex. YOu should straight away work on matrices to rotate to new positons instead of arrays and picking them from the figure.
This line:
x_temp = get(h(1),'xdata'); % get x values
giving you a 2*21 array and your rot_mat is 3X3.....you cannot multiply them. YOu need to pick (x,y,z) and multiply this point with rotation matrix to get the point shifted. Check the below pseudo code.....yo can develop your logic with the below example code.
t = 0:0.1:1;
[X,Y,Z] = cylinder((t));
%% Rotation
th = pi/2 ;
Rx = [1 0 0 ; 0 cos(th) -sin(th) ; 0 sin(th) cos(th)] ;
P0 = [X(:) Y(:) Z(:)] ;
P1 = P0*Rx ;
X1 = reshape(P1(:,1),size(X)) ;
Y1 = reshape(P1(:,2),size(X)) ;
Z1 = reshape(P1(:,3),size(X)) ;
figure
hold on
surf(X,Y,Z)
surf(X1,Y1,Z1)
view(3)
I have about four series of data on a Matlab plot, two of them are quite close and can only be differentiated with a zoom. How do I depict the zoomed part within the existing plot for the viewer. I have checked similar posts but the answers seem very unclear.
I look for something like this:
Here is a suggestion how to do this with MATLAB. It may need some fine tuning, but it will give you the result:
function pan = zoomin(ax,areaToMagnify,panPosition)
% AX is a handle to the axes to magnify
% AREATOMAGNIFY is the area to magnify, given by a 4-element vector that defines the
% lower-left and upper-right corners of a rectangle [x1 y1 x2 y2]
% PANPOSTION is the position of the magnifying pan in the figure, defined by
% the normalized units of the figure [x y w h]
%
fig = ax.Parent;
pan = copyobj(ax,fig);
pan.Position = panPosition;
pan.XLim = areaToMagnify([1 3]);
pan.YLim = areaToMagnify([2 4]);
pan.XTick = [];
pan.YTick = [];
rectangle(ax,'Position',...
[areaToMagnify(1:2) areaToMagnify(3:4)-areaToMagnify(1:2)])
xy = ax2annot(ax,areaToMagnify([1 4;3 2]));
annotation(fig,'line',[xy(1,1) panPosition(1)],...
[xy(1,2) panPosition(2)+panPosition(4)],'Color','k')
annotation(fig,'line',[xy(2,1) panPosition(1)+panPosition(3)],...
[xy(2,2) panPosition(2)],'Color','k')
end
function anxy = ax2annot(ax,xy)
% This function converts the axis unites to the figure normalized unites
% AX is a handle to the figure
% XY is a n-by-2 matrix, where the first column is the x values and the
% second is the y values
% ANXY is a matrix in the same size of XY, but with all the values
% converted to normalized units
pos = ax.Position;
% white area * ((value - axis min) / axis length) + gray area
normx = pos(3)*((xy(:,1)-ax.XLim(1))./range(ax.XLim))+ pos(1);
normy = pos(4)*((xy(:,2)-ax.YLim(1))./range(ax.YLim))+ pos(2);
anxy = [normx normy];
end
Note that the units of areaToMagnify are like the axis units, while the units of panPosition are between 0 to 1, like the position property in MATLAB.
Here is an example:
x = -5:0.1:5;
subplot(3,3,[4 5 7 8])
plot(x,cos(x-2),x,sin(x),x,-x-0.5,x,0.1.*x+0.1)
ax = gca;
area = [-0.4 -0.4 0.25 0.25];
inlarge = subplot(3,3,3);
panpos = inlarge.Position;
delete(inlarge);
inlarge = zoomin(ax,area,panpos);
title(inlarge,'Zoom in')
So I have already implemented every part of a Hough Transform on my own, except for actually plotting the lines back onto the original image.
I can set up my array of data that I have like this.
points | theta | rho
-------|-------|----
[246,0] -90 -246
[128,0] -90 -128
[9,0] -90 -9
[0,9] 0 9
[0,128] 0 128
[0,246] 0 246
The points are the points that were converted from the peaks in Polar Coordinates. So now I need to draw all six of these lines and I have had no luck.
Edit
So I tried to change my code based off suggestions. This is what I came up with.
function help(img, outfile, peaks, rho, theta)
imshow(img);
x0 = 1;
xend = size(img,2);
peaks_len=length(peaks);
for i=1:peaks_len
peak=peaks(i,:);
r_ind=peak(1);
t_ind=peak(2);
r=rho(r_ind);
th=theta(t_ind);
%display([r,th,peak]);
%// if a vertical line, then draw a vertical line centered at x = r
% display([r, th]);
if (th == 0)
display('th=0');
display([1, size(img,1)]);
line([r r], [1 size(img,1)], 'Color', 'green');
else
%// Compute starting y coordinate
y0 = abs((-cosd(th)/sind(th))*x0 + (r / sind(th)))+11;%-25;
%// Compute ending y coordinate
yend = abs((-cosd(th)/sind(th))*xend + (r / sind(th)))+11;%-25;
display('y');
display([y0, yend]);
display('x');
display([x0 xend]);
%// Draw the line
line([x0 xend], [y0 yend], 'Color', 'green');
end
end
end
I had to change from r==0 to th==0 because th=0 would give NAN errors when r was not 0.
Based off the peaks, I then used that to get the data I needed to then calculate some values... but for some reason this isn't plotting well.
If you notice the + 11 for both y values. I had to do that to get the lines to be where they need to. I have a feeling something else went wrong.
I did change it so that my Rho values are all now positive.
If you recall from the parameterization of the Hough space, the direct relation between rho,theta to x,y is:
rho = x*cos(theta) + y*sin(theta)
Bear in mind that x,y represent the column and row location respectively. In addition, the origin is defined at the top-left corner of the image. Now that you want to plot the equation of the line, you have your rho and theta. Simply re-arrange the equation so that you can solve for an equation of the line of the form y = mx + b:
As such , simply loop over each rho and theta you have and draw a line that starts from the origin at x = 0 up to the limit of your image x = width-1. However, because MATLAB is 1-indexed, we need to go from x = 1 to x = width. Supposing that your rho and theta are stored in separate arrays of the same lengths and you have your edge image stored in im, you can do something like this:
imshow(im); %// Show the image
hold on; %// Hold so we can draw lines
numLines = numel(rho); %// or numel(theta);
%// These are constant and never change
x0 = 1;
xend = size(im,2); %// Get the width of the image
%// For each rho,theta pair...
for idx = 1 : numLines
r = rho(idx); th = theta(idx); %// Get rho and theta
%// Compute starting y coordinate
y0 = (-cosd(th)/sind(th))*x0 + (r / sind(th)); %// Note theta in degrees to respect your convention
%// Compute ending y coordinate
yend = (-cosd(th)/sind(th))*xend + (r / sind(th));
%// Draw the line
line([x0 xend], [y0 yend], 'Color', 'blue');
end
The above code is pretty simple. First, show the image using imshow in MATLAB. Next, use hold on so we can draw our lines in the image that will go on top of the image. Next, we calculate how many rho,theta pairs there are, and then we define the two x coordinates to be 1 and width as we will use these to determine where the starting and ending y coordinates are, given these x coordinates. Next, for each rho,theta pair we have, determine the corresponding y coordinates, then use line to draw a line from the starting and ending (x,y) coordinates in blue. We repeat this until we run out of lines.
Don't be alarmed if the y coordinates that are produced go out of bounds in the image. line will be intelligent enough to simply cap the result.
When theta = 0
The above code works assuming that you have no vertical lines detected in the Hough Transform, or when theta = 0. If theta = 0 (like in your case), this means that we have a vertical line which would thus produce an infinite slope and our formulation of y = mx + b is invalid. Should theta = 0, the equation of the line becomes x = rho. As such, you will need an additional if statement inside your loop that will detect this:
imshow(im); %// Show the image
hold on; %// Hold so we can draw lines
numLines = numel(rho); %// or numel(theta);
%// These are constant and never change
x0 = 1;
xend = size(im,2); %// Get the width of the image
%// For each rho,theta pair...
for idx = 1 : numLines
r = rho(idx); th = theta(idx); %// Get rho and theta
%// if a vertical line, then draw a vertical line centered at x = r
if (th == 0)
line([r r], [1 size(im,1)], 'Color', 'blue');
else
%// Compute starting y coordinate
y0 = (-cosd(th)/sind(th))*x0 + (r / sind(th)); %// Note theta in degrees to respect your convention
%// Compute ending y coordinate
yend = (-cosd(th)/sind(th))*xend + (r / sind(th));
%// Draw the line
line([x0 xend], [y0 yend], 'Color', 'blue');
end
end
In order to draw the vertical line, I need to know how high the image is so that we can draw a vertical line from the top of the image (y = 1) down to the bottom of the image (y = height) which is anchored at x = rho. As such, the above code should now properly handle any line, as well as the degenerate case when the slope is infinite. Therefore, this second version of the code is what you're after.
Good luck!
I am stuck with this problem for a long time now. I have a polygonal region (lets say, a hexagon). I can calculate the values of certain function at any point inside the polygon. Now I need to create a filled contour (using contourf in MATLAB) of this data. How do I go about it. I found some discussion on this topic at the link below (page 121)
http://www-personal.umich.edu/~jpboyd/eng403_chap4_contourplts.pdf
This works somewhat ok but it still produces jagged edges which I don't want. Anyone has any suggestion on this problem? Thanks. Here is my code
close all
Node = [ 1.0 0
0.5 0.8660
-0.5 0.8660
-1.0 0
-0.5 -0.8660
0.5 -0.8660];
[x,y] = meshgrid(-1:0.1:1,-1:0.1:1);
N = zeros(size(x));
for i=1:size(x,2)
for j=1:size(y,2)
p = [x(i,j) y(i,j)];
IN = inpolygon(p(1),p(2),Node(:,1),Node(:,2));
if IN
N(i,j)= rand;
else
N(i,j)= NaN;
end
end
end
figure
contourf(x,y,N,'LineStyle','none'), hold on;
xlabel('X'), ylabel('Y'), axis equal; axis off; colorbar;
line([Node(:,1);Node(1,1)],[Node(:,2);Node(1,2)],'Color',[1 1 1],'LineWidth',2.0)
clear IN i j p x y
You're using a square grid to sample a hexagonal area. This will indeed lead to boundary problems.
A better solution (but still not quite optimal) is to use a hexagonal grid:
%# define hexagon
Node = [ 1.0 0
0.5 0.8660
-0.5 0.8660
-1.0 0
-0.5 -0.8660
0.5 -0.8660];
%# Generate hexagonal grid
Rad3Over2 = sqrt(3) / 2;
[x, y] = meshgrid(0:51);
n = size(y,1);
x = Rad3Over2 * x;
y = y + repmat([0 0.5],[n,n/2]);
%# re-scale to -1..1
x = 2*x/max(x(:))-1;
y = 2*y/max(y(:))-1;
%# insode-polygon check
N = zeros(size(x));
for i=1:size(x,2)
for j=1:size(y,2)
p = [x(i,j) y(i,j)];
IN = inpolygon(p(1),p(2),Node(:,1),Node(:,2));
if IN
N(i,j)= rand;
else
N(i,j)= NaN;
end
end
end
%# make contour plot
figure(1)
contourf(x,y,N,'LineStyle','none'), hold on;
xlabel('X'), ylabel('Y'), axis equal; axis off; colorbar;
line([Node(:,1);Node(1,1)],[Node(:,2);Node(1,2)],'Color',[1 1 1],'LineWidth',2.0)
If you want even better coverage, you'll have to devise a grid that better covers your area and/or up the number of sample points.
For arbitrary irregular areas, you might want to experiment with irregular/random grids, distributed such that there are more points close to the boundaries, than there are in the middle of the area.
Suppose you have a function defined on a hexagon. Then take some square that contains the hexagon.
A simple first test could be to extend your function on the square to take the value = 0 for points outside of the hexagon.
Depending on the range of your function this may, or may not, give you a good answer.
If this doesn't work, then find the min of the function on the hexagon using min(min(A))) for an nxn matrix A. Then define the function to be min(min(A)) - 1.0 outside of the hexagon but in the square.
Then use contourf(x, y, z, v) where v is a vector of values for the output, so take v(1) = min(min(A)) - 1.0, and v(2:n) = linspace(min(min(A)), max(max(A)), n) for some integer n. You can probably specify the color of the level set min(min(A)) - 1.0 to be white, but I've never done this. I've had to do something like this before, and setting the function = 0 outside of he hexagon was enough.
If your function can only be evaluated inside the polygonal region, then my hexagonal grid answer still stands. However, if your function can be (or modified to be) evaluated outside the polygon, you might want to use a cheat:
figure(1), clf, hold on
%# External contour, square.
x1 = [-3 -3 +3 +3 -3]/2;
y1 = [-3 +3 +3 -3 -3]/2;
%# internal contour, some polygon
x2 = [1.0 0.5 -0.5 -1.0 -0.5 0.5];
y2 = [0 0.8660 0.8660 0 -0.8660 -0.8660];
%# convert to patches
[f, v] = poly2fv({x1, x2}, {y1, y2});
patch(...
'Faces' , f,...
'Vertices' , v,...
'FaceColor', get(0, 'defaultuicontrolbackgroundcolor'), ...
'EdgeColor', 'none'...
);
%# Generate function for contourplot
[x, y] = meshgrid(-2.8/2:0.1:2.8/2);
N = rand(size(x));
%# make contour plot
contourf(x,y,N,'LineStyle','none'), hold on;
xlabel('X'), ylabel('Y'), axis equal;
axis off; colorbar;
It basically creates your contour plot on a square region, and overlays a mask with a hexagonal hole in it, so it looks like it's only a contour of the hexagonal. I suspect it is a lot easier to tweak your function to allow function evaluations outside the region, then it is to (re-)invent some sort of grid that contains the boundaries.