If I have a function f(x) which is defined as
f(x) = x^2; x>0 & x<=1
= x^3; x>1 & x<2
= 2*x; elsewhere
How do I plot this in Matlab in the same graph?
I would do this without fplot :
x = 0:0.1:3;
x1 = x(x>0 & x<=1);
x2 = x(x>1 & x<2);
x3 = x(x>2);
y1 = (x1).^2;
y2 = (x2).^3;
y3 = 2*(x3);
plot([x1 x2 x3], [y1 y2 y3])
You can define a set of x values (say between 4 <= x <= 4), then apply 2*x to every value within this interval. After, search for those x values that are within the intervals of the other functions and set those values to what they should be within those intervals. As such, try something like this:
x = -4 : 0.001 : 4;
y = 2*x;
y(x > 0 & x <= 1) = x(x > 0 & x <= 1).^2;
y(x > 1 & x < 2) = x(x > 1 & x < 2).^3;
plot(x,y);
grid;
This is what I get:
Related
I'm trying to plot a function subject to three constraints (see code)
Now I tried the following
function value = example(x1, x2)
if x1 < 0 || x2 < 0 || x1+2*x2 > 6
value = NaN;
else
value = abs(x1 - x2) + exp(-x1 - x2);
end
[X, Y] = meshgrid(-10:10, -10:10);
Z = example(X,Y);
surf(X, Y, Z)
Now, this raises an error since the if clause cannot be evaluated for inputs X and Y. Any idea how to make this work?
As #Cris mentioned, use logical indexing.
The basic idea is (x1 < 0 | x2 < 0 | x1+2*x2 > 6) will gives you a matrix (same size as value) of zeros and ones. The positions of ones correspond to the true condition. Try this:
function value = example(x1, x2)
value = abs(x1 - x2) + exp(-x1 - x2);
value(x1 < 0 | x2 < 0 | x1+2*x2 > 6) = NaN;
Output:
I'm trying to solve a piecewise function, but I am getting an error. The following is the code.
script:
syms x
y_intercept = 2;
answerr = solve(pw_f(x) == y_intercept, x);
piecewise function (in a separate file within the same folder):
function y = pw_f(x)
if x < 0
y = x;
elseif (x >=0) && (x <= 20)
y = 2*x;
elseif x > 20
y = 4*x - 40;
else
end
end
The error I'm getting after running the script is:
Conversion to logical from sym is not possible.
Error in pw_f (line 3)
if x < 0
Error in solve_test
answerr = fsolve(pw_f(x) == y_intercept, x);
I know that the error is because Matlab can't perform the comparison x < 0 because x is a symbolic variable, so it does not know what x is yet. I also tried using fsolve, and vpasolve but I'm still getting the same error. Do you know how to solve this in Matlab or get around this error?
Of course, this is an easy problem that I can do in my head (x = 1 is the solution) so Matlab should be able to do this!! However, I want to make this generic for any y-intercept (maybe some random number that is not such a nice whole number) that I choose. PLEASE HELP!!!! Thanks :)
FYI, I am using Matlab R2013a.
In file called pw_f.m
function y = pw_f(x)
if x < 0
y = x;
elseif (x >=0) && (x <= 20)
y = 2*x;
elseif x > 20
y = 4*x - 40;
else
end
end
In command window
>> y_intercept = 2; % set object value
>> x0 = 0; % initial guess
>> answerr = fzero(#(x)pw_f(x) - y_intercept, x0) % solve
answerr =
1
>> pw_f(answerr) % test solution
ans =
2
I have a function like this:
f(x) = { x if 0 < x < n
{ n-x if n < x < 2*n
How to enter this function in MATLAB?
Best way is to put this in a sub-function or nested function, or in a separate m-file:
function y = f(x)
n = 4; %// Or whatever your N is
if x <= 0 || x >= 2*n
y = 0;
elseif x < n
y = x;
else
y = n-x;
end
end
or, more generally, when x is a vector/matrix,
function y = f(x)
y = x;
y(x >= n) = n-x(x >= n);
y(x <= 0 | x >= 2*n) = 0;
end
Alternatively, you can of course pass the n as an argument:
function y = f(x, n)
...
end
Alternatively, you can use this anonymous function:
f = #(x) (x>0 & x<n).*x + (x>=n & x<=2*n).*(n-x);
again, optionally, pass the n:
f = #(x,n) ...
This question already has answers here:
Draw 3d Inequality on Matlab
(1 answer)
matlab - plot inequality in 3d with surf
(1 answer)
Closed 1 year ago.
I want to plot a 3d region in MATLAB bounded from a set of inequalities.
For example:
0 <= x <= 1
sqrt(x) <= y <= 1
0 <= z <= 1 - y
I found a 2d example that someone has done on this site but I'm not sure how to convert that to 3d. How to plot inequalities.
Edit:
From #Tobold's help I modified the code to restrict the points that are plotted to those that are defined by all three regions, but it plots only 2 or 3 points. It looks like the points in the vectors X1, Y1 and Z1 are right but for some reason its plotting only a few. Any ideas why it is only plotting a few points from the X1, Y1 and Z1 vectors instead of all of them?
[X,Y,Z]=meshgrid(0:0.1:1,0:0.1:1,0:0.1:1); % Make a grid of points between 0 and 1
p1=0.1; p2=0.2; % Choose some parameters
X1 = (X >= 0 & X <= 1) & (Y >= sqrt(X) & Y <= 1) & (Z >= 0 & Z <= 1 - Y);
Y1 = (X >= 0 & X <= 1) & (Y >= sqrt(X) & Y <= 1) & (Z >= 0 & Z <= 1 - Y);
Z1 = (X >= 0 & X <= 1) & (Y >= sqrt(X) & Y <= 1) & (Z >= 0 & Z <= 1 - Y);
ineq1 = (X >= 0 & X <= 1) * 2;
ineq2 = (Y >= sqrt(X) & Y <= 1) * 4;
ineq3 = (Z >= 0 & Z <= 1 - Y) * 8;
all = ineq1 & ineq2 & ineq3;
colors = zeros(size(X))+ineq1+ineq2+ineq3;
scatter3(X1(:),Y1(:),Z1(:),3,colors(:)','filled')
You can do almost the same thing as in the 2d case that you linked to. Just write down your three inequalities, use a 3d meshgrid, multiply each inequality with a number from a set of three numbers that has unique subset sums (e.g. 2, 4, 8) and use scatter3:
[X,Y,Z]=meshgrid(0:0.1:1,0:0.1:1,0:0.1:1); % Make a grid of points between 0 and 1
p1=0.1; p2=0.2; % Choose some parameters
ineq1 = (X >= 0 & X <= 1) * 2;
ineq2 = (X >= sqrt(X) & Y <= 1) * 4;
ineq3 = (Z >= 0 & Z <= 1 - Y) * 8;
colors = zeros(size(X))+ineq1+ineq2+ineq3;
scatter3(X(:),Y(:),Z(:),3,colors(:),'filled')
I've been trying to figure out the same thing, and the trick is to make the size of everything not in the intersection 0. Tobold's scatter3 line uses '3' as the option for size, meaning all points will show up as point 3. This can be substituted for a matrix of equal size to X1 with the set of sizes. The easiest way to do this is just make s = 3*all:
all = ineq1 & ineq2 & ineq3;
colors = zeros(size(X))+all;
sizes = 3 * all;
scatter3(X1(:),Y1(:),Z1(:),sizes,colors(:)','filled')
That should get you just the area in the intersection.
--
edit: The color variable needs to change too. You just want the intersection, not the other inequalities.
I don't understand several things in the code that you wrote as modification of #Tobold's help. For example what are the parameters p1 and p2 doing in your code?
Anyway, The code plotting only the points of your grid satisfying all of your inequalities;
[X,Y,Z]=meshgrid(0:0.1:1,0:0.1:1,0:0.1:1);
ineq1 = (X >= 0 & X <= 1);
ineq2 = (Y >= sqrt(X) & Y <= 1);
ineq3 = (Z >= 0 & Z <= 1 - Y);
all = ineq1 & ineq2 & ineq3;
scatter3(X(all),Y(all),Z(all),'b','filled')
The result is brought in the following image.
I have the following function:
f(t) = 0 if t < 0
f(t) = 2*t^2 - 4*t +3 if 1 <= t < 2
f(t) = Cos(t) if 2 <= t
I am a new MATLAB user, and I do not how to plot the function on a single figure over the range 0<=t<=5.
Any ideas about What I have to do?
Write a function for your Laplace formula.
Something like this
function [ft] = func(t)
if t <= 0
ft = 0;
elseif t > 0 && t < 2
ft = 2 * t^2 - 4 * t + 3;
elseif t >= 2
ft = cos(t);
end
You can then plot the function with fplot, the second parameter defines the plotting range.
fplot('func', [0, 5])
thanks for your help but I could not implement any code or commands to get the answer. Instead of, I was lucky and I found an example and the MATLAB commands are as follow:
x=linspace(0,5,3000);
y=(0*x).*(x<1) + (2*(x.^2)-(4.*x)+3).*((1<=x) & (x<2))
+ (cos(x)).*(2<=x);
plot(x,y, '.'), grid
axis([0 5 -2 4])
title ('Plot of f(t)'), xlabel('t'), ylabel('f(t)')
If you mean limiting x axis, then after using plot use
xlim([xmin xmax])
In your case
xlim([0 5])
Use ylim for limiting y axis
Ok, I think I misunderstood you
Also I think, you've made mistake in your formulas
f(t) = 0 if 0<= t < 1
f(t) = 2*t^2 - 4*t +3 if 1 <= t < 2
f(t) = Cos(t) if 2 <= t
figure;
hold on;
x = 0:0.1:0.9; y = 0 * x; plot( x, y );
x = 1:0.1:1.9; y = 2 * x * x - 4 * x + 3; plot( x, y );
x = 2:0.1:5; y = cos( x ); plot( x, y );