a = 10:100
b = 10:100
c = power(a,b)
surf(a,b,c)
=> Error using surf (line 78)
Z must be a matrix, not a scalar or vector
any clue ?
Here's a vectorized way using bsxfun:
a = 10:100;
b = 1:.1:10; %// changed b to avoid very large numbers
c = bsxfun(#power, a, b.');
surf(a,b,c)
c=power(a,b) does not give you the combinations of all a power b, unfortunately.
Here there is a way of doing it (though most likely there is a vectorizedd way of doing it)
a = 10:100;
b = linspace(1,10,length(a));
% I changed the values of b, because 100^100 is that a big number that Matlab will not plot it, it is too big for storing in a double
%loop and save
for ii=1:length(b)
c(ii,:)=a.^(b(ii));
end
surf(a,b,c)
Related
I am using this command in Matlab:
grazAng = grazingang(H,R)
If i fix H, I can treat R as a vector:
z=[];
for i=1:1000
z(i)=abs(grazingang(1,i));
end
Now I would like to have both H and R to by dynamic. For example:
H=[0,0.25,0.5]
R=[1,2,3]
And I would like my loop to run three times, each time selecting a pair of (H,R) values with the same indexes, i.e. (0,1),(0.25,2),(0.5,3) and then store the result in z. Could anyone help me out with this?
Remember, everything in MATLAB is an array. To do this with a loop, you need to index into the arrays:
H = [0,0.25,0.5];
R = [1,2,3];
z = zeros(size(H)); % Pre-allocation is generally advised
for i = 1:1000
z(i) = abs(grazingang(H(i),R(i)));
end
But MATLAB functions generally accept vectors and do this for you, so all you need to do is:
H=[0,0.25,0.5];
R=[1,2,3];
z = abs(grazingang(H,R));
If you define a range from A1 to A10 in Excel using VBA you cam use Range("A1:A10"). On the other hand side it's possible to write down Range("A1:A"&10). How can I use the second way in MATLAB, please?
I have some matrices M1, M2, M3, ..., and I wish to define them by using iterator FOR that helps me stop writing the matrix names completely.
Notwithstanding the comment that this approach is frowned upon, below is a way to do it programmatically:
% define cell with matrix names
number_of_matrices_I_want = 5;
my_matrix_names = repmat({'NA'}, 1, number_of_matrices_I_want);
for ii = 1:length(my_matrix_names)
my_matrix_names(ii) = {strcat('M',num2str(ii))};
end
% example for how to populate a matrix from "my_matrix_names"
x = rand(5,1); % fake numbers
y = rand(5,1);
eval([my_matrix_names{1} '= [x y]']);
Can you please tell me what's wrong with the following code?
function [n]=calculate_n(p,delta)
e = 1.6*power(10,-19);
k = 1.38*power(10,-23);
T = 298;
co = 3.25*power(10,13)*e*power(10,4);
er=12.5;
eo=1.0;
Nv=3*power(10,13);
us = log((p*e)/sqrt(2*k*T*er*eo*Nv))*2*k*T;
tmp = delta+(e*e*p)/co+us;
n = 1/(exp((tmp))+1);
end
I am getting matrix dimension error while calculating n. Please help me.
Caller:
e = 1.6*power(10,-19);
x = logspace(13,18);
y=calculate_n(x,0.2*e);
semilogx(x,y,'-s');
grid on;
Just replace n = 1/(exp((tmp))+1); with n = 1./(exp(tmp)+1);. But beware, tmp is so small for these values that exp(tmp) will always be 1. Also, there is a surplus bracket around tmp, you might want to check if you put them correctly.
Edit:
The reason is that A/B tries to solve the system of linear equations A*x = B for x which is not what you wanted. It threw an error because it requires both variables to have the same number of columns. A./B performs element-wise matrix division which is what you wanted. However, if A and B are singular A/B = A./B. See the documentation for more info.
I have a cubic expression here
I am trying to determine and plot δ𝛿 in the expression for P values of 0.0 to 5000. I'm really struggling to get the expression for δ in terms of the pressure P.
clear all;
close all;
t = 0.335*1e-9;
r = 62*1e-6;
delta = 1.2*1e+9;
E = 1e+12;
v = 0.17;
P = 0:100:5000
P = (4*delta*t)*w/r^2 + (2*E*t)*w^3/((1-v)*r^4);
I would appreciate if anyone could provide pointers.
I suggest two simple methods.
You evaluate P as a function of delta then you plot(P,delta). This is quick and dirty but if all you need is a plot it will do. The inconvenience is that you may to do some guess-and-trial to find the correct interval of P values, but you can also take a large enough value of delta_max and then restrict the x-axis limit of the plot.
Your function is a simple cubic, which you can solve analytically (see here if you are lost) to invert P(delta) into delta(P).
What you want is the functional inverse of your expression, i.e., δ𝛿 as a function of P. Since it's a cubic polynomial, you can expect up to three solutions (roots) for a give value of P. However, I'm guessing that you're only interested in real-valued solutions and nonnegative values of P. In that case there's just one real root for each value of P.
Given the values of your parameters, it makes most sense to solve this numerically using fzero. Using the parameter names in your code (different from equations):
t = 0.335*1e-9;
r = 62*1e-6;
delta = 1.2*1e9;
E = 1e12;
v = 0.17;
f = #(w,p)2*E*t*w.^3/((1-v)*r^4)+4*delta*t*w/r^2-p;
P = 0:100:5000;
w0 = [0 1]; % Bounded initial guess, valid up to very large values of P
w_sol = zeros(length(P),1);
for i = 1:length(P)
w_sol(i) = fzero(#(w)f(w,P(i)),w0); % Find solution for each P
end
figure;
plot(P,w_sol);
You could also solve this using symbolic math:
syms w p
t = 0.335*sym(1e-9);
r = 62*sym(1e-6);
delta = 1.2*sym(1e9);
E = sym(1e12);
v = sym(0.17);
w_sol = solve(p==2*E*t*w^3/((1-v)*r^4)+4*delta*t*w/r^2,w);
P = 0:100:5000;
w_sol = double(subs(w_sol(1),p,P)); % Plug in P values and convert to floating point
figure;
plot(P,w_sol);
Because of your numeric parameter values, solve returns an answer in terms of three RootOf objects, the first of which is the real one you want.
I need to generate two chaotic sequences based on chen's hyperchaotic system.It has to be generated from the following four formulas
X=ay-x;
Y=-xz+dx+cy-q;
Y=xy-bz;
Q=x+k;
where a,b,c,d,x,y,z,q are all initialised as follows.
I need only X and Y
where
X=[x1,x2,...x4n]
Y=[y1,y2,...y4n]
a=36 ;
b=3 ;
c=28 ;
d=16 ;
k=0.2 ;
x=0.3 ;
y=-0.4 ;
z=1.2 ;
q=1 ;
n=256 ;
I tried the following code but i'm not able to get it properly.
clc
clear all
close all
w=imread('C:\Users\Desktop\a.png');
[m n]=size(w)
a=36;
b=3;
c=28;
d=16;
k=0.2;
x(1)=0.3;
y(1)=-0.4;
z(1)=1.2;
q(1)=1;
for i=1:1:4(n)
x(i+1)=(a*(y(i)-x(i)));
y(i+1)=-(x(i)*z(i))+(d*x(i))+(c*y(i))-q(i);
z(i+1)=(x(i)*y(i))-(b*z(i));
q(i+1)=x(i)+k;
end
disp(x);
disp(y);
pls help. thanks in advance.
Your code isn't even close to doing what you want it to. Fortunately, I'm vaguely interested in the problem and I have a bunch of spare time, so I thought I'd try and implement it step by step to show you what to do. I've left a few gaps for you to fill in.
It sounds like you want to integrate the hyperchaotic chen system, which has various definitions online, but you seem to be focusing on
So let's write a matlab function that defines that system
function vdot = chen(t, v, a, b, c, d, k)
% Here you unpack the input vector v -
x = v(1); y = v(2); z = v(3); q = v(4);
% Here you need to implement your equations as xdot, ydot etc.
% xdot = ...
% ydot = ...
% I'll leave that for you to do yourself.
% Then you pack them up into an output vector -
vdot = [xdot; ydot; zdot; qdot];
end
Save that in a file called chen.m. Now you need to define the values of the parameters a, b, c, d and k, as well as your initial condition.
% You need to define the values of a, b, c, d, k here.
% a = ...
% b = ...
% You also need to define the vector v0, which is a 4x1 vector of your
% initial conditions
% v0 = ...
%
This next line creates a function that can be used by Matlab's integration routines. The first parameter t is the current time (which you don't actually use) and the second parameter is a 4x1 vector containing x, y, z, q.
>> fun = #(t,v) chen(t,v,a,b,c,d,k)
Now you can use ode45 (which does numerical integration using a 4th order runge-kutta scheme) to integrate it and plot some paths. The first argument to ode45 is the function you want to be integrated, the second argument is the timespan to be integrated over (I chose to integrate from 0 to 100, maybe you want to do something different) and the third argument is your initial condition (which hopefully you already defined).
>> [t, v] = ode45(fun, [0 100], v0);
The outputs are t, a vector of times, and v, which will be a matrix whose columns are the different components (x, y, z, q) and whose rows are the values of the components at each point in time. So you can pull out a column for each of the x and y components, and plot them
>> x = v(:,1);
>> y = v(:,2);
>> plot(x,y)
Which gives a reasonably chaotic looking plot:
#Abirami Anbalagan and Sir #Chris Taylor, I have also studied hyperchaotic system up to some extent. According to me, for system to be chaotic, values should be like
a= 35; b= 3; c= 12; d= 7;
v(n) = [-422 -274 0 -2.4]transpose
where v(n) is a 4*1 Matrix.