I am trying to simulate a network of mobile robots that uses artificial potential fields for movement planning to a shared destination xd. This is done by generating a series of m-files (one for each robot) from a symbolic expression, as this seems to be the best way in terms of computational time and accuracy. However, I can't figure out what is going wrong with my gradient computation: the analytical gradient that is being computed seems to be faulty, while the numerical gradient is calculated correctly (see the image posted below). I have written a MWE listed below, which also exhibits this problem. I have checked the file generating part of the code, and it does return a correct function file with a correct gradient. But I can't figure out why the analytic and numerical gradient are so different.
(A larger version of the image below can be found here)
% create symbolic variables
xd = sym('xd',[1 2]);
x = sym('x',[2 2]);
% create a potential function and a gradient function for both (x,y) pairs
% in x
for i=1:size(x,1)
phi = norm(x(i,:)-xd)/norm(x(1,:)-x(2,:)); % potential field function
xvector = reshape(x.',1,size(x,1)*size(x,2)); % reshape x to allow for gradient computation
grad = gradient(phi,xvector(2*i-1:2*i)); % compute the gradient
gradx = grad(1);grady=grad(2); % split the gradient in two components
% create function file names
gradfun = strcat('GradTester',int2str(i),'.m');
phifun = strcat('PotTester',int2str(i),'.m');
% generate two output files
matlabFunction(gradx, grady,'file',gradfun,'outputs',{'gradx','grady'},'vars',{xvector, xd});
matlabFunction(phi,'file',phifun,'vars',{xvector, xd});
end
clear all % make sure the workspace is empty: the functions are in the files
pause(0.1) % ensure the function file has been generated before it is called
% these are later overwritten by a specific case, but they can be used for
% debugging
x = 0.5*rand(2);
xd = 0.5*rand(1,2);
% values for the Stackoverflow case
x = [0.0533 0.0023;
0.4809 0.3875];
xd = [0.4087 0.4343];
xp = x; % dummy variable to keep x intact
% compute potential field and gradient for both (x,y) pairs
for i=1:size(x,1)
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
% preallocate the gradient and potential matrices
gradx = zeros(length(xGrid),length(yGrid));
grady = zeros(length(xGrid),length(yGrid));
phi = zeros(length(xGrid),length(yGrid));
% generate appropriate function handles
fun = str2func(strcat('GradTester',int2str(i)));
fun2 = str2func(strcat('PotTester',int2str(i)));
% compute analytic gradient and potential for each position in the xGrid and
% yGrid vectors
for ii = 1:length(yGrid)
for jj = 1:length(xGrid)
xp(i,:) = [xGrid(ii) yGrid(jj)]; % select the position
Xvec = reshape(xp.',1,size(x,1)*size(x,2)); % turn the input into a vector
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd); % compute gradients
phi(jj,ii) = fun2(Xvec,xd); % compute potential value
end
end
[FX,FY] = gradient(phi); % compute the NUMERICAL gradient for comparison
%scale the numerical gradient
FX = FX/0.005;
FY = FY/0.005;
% plot analytic result
subplot(2,2,2*i-1)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-gradx,-grady)
contour(xGrid,yGrid,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-FX,-FY)
contour(xGrid,yGrid,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
end
The potential field I am trying to generate is defined by an (x,y) position, in my code called xd. x is the position matrix of dimension N x 2, where the first column represents x1, x2, and so on, and the second column represents y1, y2, and so on. Xvec is simply a reshaping of this vector to x1,y1,x2,y2,x3,y3 and so on, as the matlabfunction I am generating only accepts vector inputs.
The gradient for robot i is being calculated by taking the derivative w.r.t. x_i and y_i, these two components together yield a single derivative 'vector' shown in the quiver plots. The derivative should look like this, and I checked that the symbolic expression for [gradx,grady] indeed looks like that before an m-file is generated.
To fix the particular problem given in the question, you were actually calculating phi in such a way that meant you doing gradient(phi) was not giving the correct results compared to the symbolic gradient. I'll try and explain. Here is how you created xGrid and yGrid:
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
But then in the for loop, ii and jj were used like phi(jj,ii) or gradx(ii,jj), but corresponding to the same physical position. This is why your results were different. Another problem you had was you used gradient incorrectly. Matlab assumes that [FX,FY]=gradient(phi) means that phi is calculated from phi=f(x,y) where x and y are matrices created using meshgrid. You effectively had the elements of phi arranged differently to that, an so gradient(phi) gave the wrong answer. Between reversing the jj and ii, and the incorrect gradient, the errors cancelled out (I suspect you tried doing phi(jj,ii) after trying phi(ii,jj) first and finding it didn't work).
Anyway, to sort it all out, on the line after you create xGrid and yGrid, put this in:
[X,Y]=meshgrid(xGrid,yGrid);
Then change the code after you load fun and fun2 to:
for ii = 1:length(xGrid) %// x loop
for jj = 1:length(yGrid) %// y loop
xp(i,:) = [X(ii,jj);Y(ii,jj)]; %// using X and Y not xGrid and yGrid
Xvec = reshape(xp.',1,size(x,1)*size(x,2));
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd);
phi(ii,jj) = fun2(Xvec,xd);
end
end
[FX,FY] = gradient(phi,0.005); %// use the second argument of gradient to set spacing
subplot(2,2,2*i-1)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))]) %// use axis rather than xlim/ylim
quiver(X,Y,gradx,grady)
contour(X,Y,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))])
quiver(X,Y,FX,FY)
contour(X,Y,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
I have some other comments about your code. I think your potential function is ill-defined, which is causing all sorts of problems. You say in the question that x is an Nx2 matrix, but you potential function is defined as
norm(x(i,:)-xd)/norm(x(1,:)-x(2,:));
which means if N was three, you'd have the following three potentials:
norm(x(1,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(2,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(3,:)-xd)/norm(x(1,:)-x(2,:));
and I don't think the third one makes sense. I think this could be causing some confusion with the gradients.
Also, I'm not sure if there is a reason to create the .m file functions in your real code, but they are not necessary for the code you posted.
Related
I tried everything and looked everywhere but can't find any solution for my question.
clc
clear all
%% Solving the Ordinary Differential Equation
G = 6.67408e-11; %Gravitational constant
M = 10; %Mass of the fixed object
r = 1; %Distance between the objects
tspan = [0 100000]; %Time Progression from 0 to 100000s
conditions = [1;0]; %y0= 1m apart, v0=0 m/s
F=#(t,y)var_r(y,G,M,r);
[t,y]=ode45(F,tspan,conditions); %ODE solver algorithm
%%part1: Plotting the Graph
% plot(t,y(:,1)); %Plotting the Graph
% xlabel('time (s)')
% ylabel('distance (m)')
%% part2: Animation of Results
plot(0,0,'b.','MarkerSize', 40);
hold on %to keep the first graph
for i=1:length(t)
k = plot(y(i,1),0,'r.','MarkerSize', 12);
pause(0.05);
axis([-1 2 -2 2]) %Defining the Axis
xlabel('X-axis') %X-Axis Label
ylabel('Y-axis') %Y-Axis Label
delete(k)
end
function yd=var_r(y,G,M,r) %function of variable r
g = (G*M)/(r + y(1))^2;
yd = [y(2); -g];
end
this is the code where I'm trying to replace the ode45 with the runge kutta method but its giving me errors. my runge kutta function:
function y = Runge_Kutta(f,x0,xf,y0,h)
n= (xf-x0)/h;
y=zeros(n+1,1);
x=(x0:h:xf);
y(1) = y0;
for i=1:n
k1 = f(x(i),y(i));
k2= f(x(i)+ h/2 , y(i) +h*(k1)/2);
y(i+1) = y(i)+(h*k2);
end
plot(x,y,'-.M')
legend('RKM')
title ('solution of y(x)');
xlabel('x');
ylabel('y(x)')
hold on
end
Before converting your ode45( ) solution to manually written RK scheme, it doesn't even look like your ode45( ) solution is correct. It appears you have a gravitational problem set up where the initial velocity is 0 so a small object will simply fall into a large mass M on a line (rectilinear motion), and that is why you have scalar position and velocity.
Going with this assumption, r is something you should be calculating on the fly, not using as a fixed input to the derivative function. E.g., I would have expected something like this:
F=#(t,y)var_r(y,G,M); % get rid of r
:
function yd=var_r(y,G,M) % function of current position y(1) and velocity y(2)
g = (G*M)/y(1)^2; % gravity accel based on current position
yd = [y(2); -g]; % assumes y(1) is positive, so acceleration is negative
end
The small object must start with a positive initial position for the derivative code to be valid as you have it written. As the small object falls into the large mass M, the above will only hold until it hits the surface or atmosphere of M. Or if you model M as a point mass, then this scheme will become increasingly difficult to integrate correctly because the acceleration becomes large without bound as the small mass gets very close to the point mass M. You would definitely need a variable step size approach in this case. The solution becomes invalid if it goes "through" mass M. In fact, once the speed gets too large the whole setup becomes invalid because of relativistic effects.
Maybe you could explain in more detail if your system is supposed to be set up this way, and what the purpose of the integration is. If it is really supposed to be a 2D or 3D problem, then more states need to be added.
For your manual Runge-Kutta code, you completely forgot to integrate the velocity so this is going to fail miserably. You need to carry a 2-element state from step to step, not a scalar as you are currently doing. E.g., something like this:
y=zeros(2,n+1); % 2-element state as columns of the y variable
x=(x0:h:xf);
y(:,1) = y0; % initial state is the first 2-element column
% change all the scalar y(i) to column y(:,i)
for i=1:n
k1 = f(x(i),y(:,i));
k2= f(x(i)+ h/2 , y(:,i) +h*(k1)/2);
y(:,i+1) = y(:,i)+(h*k2);
end
plot(x,y(1,:),'-.M') % plot the position part of the solution
This is all assuming the f that gets passed in is the same F you have in your original code.
y(1) is the first scalar element in the data structure of y (this counts in column-first order). You want to generate in y a list of column vectors, as your ODE is a system with state dimension 2. Thus you need to generate y with that format, y=zeros(length(x0),n+1); and then address the list entries as matrix columns y(:,1)=x0 and the same modification in every place where you extract or assign a list entry.
Matlab introduce various short-cuts that, if used consequently, lead to contradictions (I think the script-hater rant (german) is still valid in large parts). Essentially, unlike in other systems, Matlab gives direct access to the underlying data structure of matrices. y(k) is the element of the underlying flat array (that is interpreted column-first in Matlab like in Fortran, unlike, e.g., Numpy where it is row-first).
Only the two-index access is to the matrix with its dimensions. So y(:,k) is the k-th matrix column and y(k,:) the k-th matrix row. The single-index access is nice for row or column vectors, but leads immediately to problems when collecting such vectors in lists, as these lists are automatically matrices.
I have a spectral data (1000 variables on xaxis, and peak intensities as y) and a list of peaks of interest at various specific x locations (a matrix called Peak) which I obtained from a function I made. Here, I would like to draw a line from the maximum value of each peaks to the xaxis - or, eventually, place a vertical arrow above each peaks but I read it is quite troublesome, so just a vertical line is welcome. However, using the following code, I get "Error using line Value must be a vector of numeric type". Any thoughts?
X = spectra;
[Peak,intensity]=PeakDetection(X);
nrow = length(Peak);
Peak2=Peak; % to put inside the real xaxis value
plot(xaxis,X);
hold on
for i = 1 : nbrow
Peak2(:,i) = round(xaxis(:,i)); % to get the real xaxis value and round it
xline = Peak2(:,i);
line('XData',xline,'YData',X,'Color','red','LineWidth',2);
end
hold off
Simple annotation:
Here is a simple way to annotate the peaks:
plot(x,y,x_peak,y_peak+0.1,'v','MarkerFaceColor','r');
where x and y is your data, and x_peak and y_peak is the coordinates of the peaks you want to annotate. The add of 0.1 is just for a better placing of the annotation and should be calibrated for your data.
For example (with some arbitrary data):
x = 1:1000;
y = sin(0.01*x).*cos(0.05*x);
[y_peak,x_peak] = PeakDetection(y); % this is just a sketch based on your code...
plot(x,y,x_peak,y_peak+0.1,'v','MarkerFaceColor','r');
the result:
Line annotation:
This is just a little bit more complicated because we need 4 values for each line. Again, assuming x_peak and y_peak as before:
plot(x,y);
hold on
ax = gca;
ymin = ax.YLim(1);
plot([x_peak;x_peak],[ymin*ones(1,numel(y_peak));y_peak],'r')
% you could write instead:
% line([x_peak;x_peak],[ymin*ones(1,numel(y_peak));y_peak],'Color','r')
% but I prefer the PLOT function.
hold off
and the result:
Arrow annotation:
If you really want those arrows, then you need to first convert the peak location to the normalized figure units. Here how to do that:
plot(x,y);
ylim([-1.5 1.5]) % only for a better look of the arrows
peaks = [x_peak.' y_peak.'];
ax = gca;
% This prat converts the axis unites to the figure normalized unites
% AX is a handle to the figure
% PEAKS is a n-by-2 matrix, where the first column is the x values and the
% second is the y values
pos = ax.Position;
% NORMPEAKS is a matrix in the same size of PEAKS, but with all the values
% converted to normalized units
normpx = pos(3)*((peaks(:,1)-ax.XLim(1))./range(ax.XLim))+ pos(1);
normpy = pos(4)*((peaks(:,2)-ax.YLim(1))./range(ax.YLim))+ pos(2);
normpeaks = [normpx normpy];
for k = 1:size(normpeaks,1)
annotation('arrow',[normpeaks(k,1) normpeaks(k,1)],...
[normpeaks(k,2)+0.1 normpeaks(k,2)],...
'Color','red','LineWidth',2)
end
and the result:
I haven't used MATLAB in a while and I am stuck on a small detail. I would really appreciate it if someone could help me out!
So I am trying to plot a transfer function using a specific function called freqs but I can't figure out how I can label specific points on the graph.
b = [0 0 10.0455]; % Numerator coefficients
a = [(1/139344) (1/183.75) 1]; % Denominator coefficients
w = logspace(-3,5); % Frequency vector
freqs(b,a,w)
grid on
I want to mark values at points x=600 Hz and 7500 Hz with a marker or to be more specific, points (600,20) and (7500,-71), both of which should lie on the curve. For some reason, freqs doesn't let me do that.
freqs is very limited when you want to rely on it plotting the frequency response for you. Basically, you have no control on how to modify the graph on top of what MATLAB generates for you.
Instead, generate the output response in a vector yourself, then plot the magnitude and phase of the output yourself so that you have full control. If you specify an output when calling freqs, you will get the response of the system.
With this, you can find the magnitude of the output by abs and the phase by angle. BTW, (600,20) and (7500,-71) make absolutely no sense unless you're talking about magnitude in dB.... which I will assume is the case for the moment.
As such, we can reproduce the plot that freqs gives by the following. The key is to use semilogx to get a semi-logarithmic graph on the x-axis. On top of this, declare those points that you want to mark on the magnitude, so (600,20) and (7500,-71):
%// Your code:
b = [0 0 10.0455]; % Numerator coefficients
a = [(1/139344) (1/183.75) 1]; % Denominator coefficients
w = logspace(-3,5); % Frequency vector
%// New code
h = freqs(b,a,w); %// Output of freqs
mag = 20*log10(abs(h)); %// Magnitude in dB
pha = (180/pi)*angle(h); %// Phase in degrees
%// Declare points
wpt = [600, 7500];
mpt = [20, -71];
%// Plot the magnitude as well as markers
figure;
subplot(2,1,1);
semilogx(w, mag, wpt, mpt, 'r.');
xlabel('Frequency');
ylabel('Magnitude (dB)');
grid;
%// Plot phase
subplot(2,1,2);
semilogx(w, pha);
xlabel('Frequency');
ylabel('Phase (Degrees)');
grid;
We get this:
If you check what freqs generates for you, you'll see that we get the same thing, but the magnitude is in gain (V/V) instead of dB. If you want it in V/V, then just plot the magnitude without the 20*log10() call. Using your data, the markers I plotted are not on the graph (wpt and mpt), so adjust the points to whatever you see fit.
There are a couple issues before we attempt to answer your question. First, there is no data-point at 600Hz or 7500Hz. These frequencies fall between data-points when graphed using the freqs command. See the image below, with datatips added interactively. I copy-pasted your code to generate this data.
Second, it does not appear that either (600,20) or (7500,-71) lie on the curves, at least with the data as you entered above.
One solution is to use plot a marker on the desired position, and use a "text" object to add a string describing the point. I put together a script using your data, to generate this figure:
The code is as follows:
b = [0 0 10.0455];
a = [(1/139344) (1/183.75) 1];
w = logspace(-3,5);
freqs(b,a,w)
grid on
figureHandle = gcf;
figureChildren = get ( figureHandle , 'children' ); % The children this returns may vary.
axes1Handle = figureChildren(1);
axes2Handle = figureChildren(2);
axes1Children = get(axes1Handle,'children'); % This should be a "line" object.
axes2Children = get(axes2Handle,'children'); % This should be a "line" object.
axes1XData = get(axes1Children,'xdata');
axes1YData = get(axes1Children,'ydata');
axes2XData = get(axes2Children,'xdata');
axes2YData = get(axes2Children,'ydata');
hold(axes1Handle,'on');
plot(axes1Handle,axes1XData(40),axes1YData(40),'m*');
pointString1 = ['(',num2str(axes1XData(40)),',',num2str(axes1YData(40)),')'];
handleText1 = text(axes1XData(40),axes1YData(40),pointString1,'parent',axes1Handle);
hold(axes2Handle,'on');
plot(axes2Handle,axes2XData(40),axes2YData(40),'m*');
pointString2 = ['(',num2str(axes2XData(40)),',',num2str(axes2YData(40)),')'];
handleText2 = text(axes2XData(40),axes2YData(40),pointString2,'parent',axes2Handle);
Hi to all,
When I run a Matlab code which generate two plot, these are overplotted (the second over the first).
I would like obtain a result as this figure, where the two plots are like in a subplot(211) and subplot(212), the first and the second in two colons, but without using subplot.
It is possible?
UPDATE
I generate this two plot with two subfuncion:
function create_figure(X1, YMatrix1, p)
%CREATE_FIGURE(X1, YMATRIX1)
% X1: vector of x data
% YMATRIX1: matrix of y data
% P: parameters used in legend
% Create figure
figure1 = figure('Name','Acceleration Power vs. Velocity LPF 1st order');
...
and
function create_figure_gamma(X1, YMatrix1, p)
%CREATE_FIGURE_GAMMA(X1, YMATRIX1, P)
% X1: vector of x data
% YMATRIX1: matrix of y data
% P: parameters used in legend
% Create figure
figure1 = figure('Name','gamma trend vs. Velocity');
...
Of course, I can give in output the parameter figure1, writing:
function figure1 = create_figure(X1, YMatrix1, p)
I think that which this parameter is possible set the position of the two plots, but I do not know the procedure respect the generic window size.
You could also set the Units to 'normalized' and enter relative positions for the figures:
set(h1,'Units','normalized');
set(h2,'Units','normalized');
set(h1,'Position',[0.1021 0.1708 0.2917 0.3500]);
set(h2,'Position',[0.4021 0.1700 0.2917 0.3508]);
This way you are independent of the current screen resolution.
This would produce two figures with plots side by side:
x = 0:0.1:2*pi;
y1 = sin(x);
y2 = cos(x);
h1=figure
plot(x,y1);
h2=figure
plot(x,y2);
% x, y, width, height
set(h1,'Position',[20 616,560,420])
set(h2,'Position',[20+560 616,560,420])
for an implicit equation(name it "y") of lambda and beta-bar which is plotted with "ezplot" command, i know it is possible that by a root finding algorithm like "bisection method", i can find solutions of beta-bar for each increment of lambda. but how to build such an algorithm to obtain the lines correctly.
(i think solutions of beta-bar should lie in an n*m matrix)
would you in general show the methods of plotting such problem? thanks.
one of my reasons is discontinuity of "ezplot" command for my equation.
ok here is my pic:
alt text http://www.mojoimage.com/free-image-hosting-view-05.php?id=5039TE-beta-bar-L-n2-.png
or
http://www.mojoimage.com/free-image-hosting-05/5039TE-beta-bar-L-n2-.pngFree Image Hosting
and my code (in short):
h=ezplot('f1',[0.8,1.8,0.7,1.0]);
and in another m.file
function y=f1(lambda,betab)
n1=1.5; n2=1; z0=120*pi;
d1=1; d2=1; a=1;
k0=2*pi/lambda;
u= sqrt(n1^2-betab^2);
wb= sqrt(n2^2-betab^2);
uu=k0*u*d1;
wwb=k0*wb*d2 ;
z1=z0/u; z1_b=z1/z0;
a0_b=tan(wwb)/u+tan(uu)/wb;
b0_b=(1/u^2-1/wb^2)*tan(uu)*tan(wwb);
c0_b=1/(u*wb)*(tan(uu)/u+tan(wwb)/wb);
uu0= k0*u*a; m=0;
y=(a0_b*z1_b^2+c0_b)+(a0_b*z1_b^2-c0_b)*...
cos(2*uu0+m*pi)+b0_b*z1_b*sin(2*uu0+m*pi);
end
fzero cant find roots; it says "Function value must be real and finite".
anyway, is it possible to eliminate discontinuity and only plot real zeros of y?
heretofore,for another function (namely fTE), which is :
function y=fTE(lambda,betab,s)
m=s;
n1=1.5; n2=1;
d1=1; d2=1; a=1;
z0=120*pi;
k0=2*pi/lambda;
u = sqrt(n1^2-betab^2);
w = sqrt(betab^2-n2^2);
U = k0*u*d1;
W = k0*w*d2 ;
z1 = z0/u; z1_b = z1/z0;
a0_b = tanh(W)/u-tan(U)/w;
b0_b = (1/u^2+1/w^2)*tan(U)*tanh(W);
c0_b = -(tan(U)/u+tanh(W)/w)/(u*w);
U0 = k0*u*a;
y = (a0_b*z1_b^2+c0_b)+(a0_b*z1_b^2-c0_b)*cos(2*U0+m*pi)...
+ b0_b*z1_b*sin(2*U0+m*pi);
end
i'd plotted real zeros of "y" by these codes:
s=0; % s=0 for even modes and s=1 for odd modes.
lmin=0.8; lmax=1.8;
bmin=1; bmax=1.5;
lam=linspace(lmin,lmax,1000);
for n=1:length(lam)
increment=0.001; tolerence=1e-14; xstart=bmax-increment;
x=xstart;
dx=increment;
m=0;
while x > bmin
while dx/x >= tolerence
if fTE(lam(n),x,s)*fTE(lam(n),x-dx,s)<0
dx=dx/2;
else
x=x-dx;
end
end
if abs(real(fTE(lam(n),x,s))) < 1e-6 %because of discontinuity some answers are not correct.%
m=m+1;
r(n,m)=x;
end
dx=increment;
x=0.99*x;
end
end
figure
hold on,plot(lam,r(:,1),'k'),plot(lam,r(:,2),'c'),plot(lam,r(:,3),'m'),
xlim([lmin,lmax]);ylim([1,1.5]),
xlabel('\lambda(\mum)'),ylabel('\beta-bar')
you see i use matrix to save data for this plot.
![alt text][2]
because here lines start from left(axis) to rigth. but if the first line(upper) starts someplace from up to rigth(for the first figure and f1 function), then i dont know how to use matrix. lets improve this method.
[2]: http://www.mojoimage.com/free-image-hosting-05/2812untitled.pngFree Image Hosting
Sometimes EZPLOT will display discontinuities because there really are discontinuities or some form of complicated behavior of the function occurring there. You can see this by generating your plot in an alternative way using the CONTOUR function.
You should first modify your f1 function by replacing the arithmetic operators (*, /, and ^) with their element-wise equivalents (.*, ./, and .^) so that f1 can accept matrix inputs for lambda and betab. Then, run the code below:
lambda = linspace(0.8,1.8,500); %# Create a vector of 500 lambda values
betab = linspace(0.7,1,500); %# Create a vector of 500 betab values
[L,B] = meshgrid(lambda,betab); %# Create 2-D grids of values
y = f1(L,B); %# Evaluate f1 at every point in the grid
[c,h] = contour(L,B,y,[0 0]); %# Plot contour lines for the value 0
set(h,'Color','b'); %# Change the lines to blue
xlabel('\lambda'); %# Add an x label
ylabel('$\overline{\beta}$','Interpreter','latex'); %# Add a y label
title('y = 0'); %# Add a title
And you should see the following plot:
Notice that there are now additional lines in the plot that did not appear when using EZPLOT, and these lines are very jagged. You can zoom in on the crossing at the top left and make a plot using SURF to get an idea of what's going on:
lambda = linspace(0.85,0.95,100); %# Some new lambda values
betab = linspace(0.95,1,100); %# Some new betab values
[L,B] = meshgrid(lambda,betab); %# Create 2-D grids of values
y = f1(L,B); %# Evaluate f1 at every point in the grid
surf(L,B,y); %# Make a 3-D surface plot of y
axis([0.85 0.95 0.95 1 -5000 5000]); %# Change the axes limits
xlabel('\lambda'); %# Add an x label
ylabel('$\overline{\beta}$','Interpreter','latex'); %# Add a y label
zlabel('y'); %# Add a z label
Notice that there is a lot of high-frequency periodic activity going on along those additional lines, which is why they look so jagged in the contour plot. This is also why a very general utility like EZPLOT was displaying a break in the lines there, since it really isn't designed to handle specific cases of complicated and poorly behaved functions.
EDIT: (response to comments)
These additional lines may not be true zero crossings, although it is difficult to tell from the SURF plot. There may be a discontinuity at those lines, where the function shoots off to -Inf on one side of the line and Inf on the other side of the line. When rendering the surface or computing the contour, these points on either side of the line may be mistakenly connected, giving the false appearance of a zero crossing along the line.
If you want to find a zero crossing given a value of lambda, you can try using the function FZERO along with an anonymous function to turn your function of two variables f1 into a function of one variable fcn:
lambda_zero = 1.5; %# The value of lambda at the zero crossing
fcn = #(x) f1(lambda_zero,x); %# A function of one variable (lambda is fixed)
betab_zero = fzero(fcn,0.94); %# Find the value of betab at the zero crossing,
%# using 0.94 as an initial guess