I'm trying to use the bisection method in Matlab to find the root of an equation as q varies from 2000-3000 in 10 step intervals. My code however does not print out a graph even though I have a plot statement and I think it creates an infinite loop since when I run it matlab says busy and I can't close the program unless I force close. I can't see anything in my code that would cause this though, could someone help me out?
function myFunction
a = 20;
b = 40;
tol = 1e-4;
q = 2000:10:3000;
t = zeros(101,1);
for i=(1:length(q))
f = #(x) (((1800).*log((160000)./(160000 - (x.*q(i)))) - (9.812).*x)./750) - 1;
t(i) = bisect(f,a,b,tol);
end
figure(1)
plot(q,t)
function c=bisect(f,a,b,tol)
k=0;
while b-a > tol
c = (a-b)/2;
if sign(f(c)) == sign(f(b))
b=c;
else
a=c;
end
k=k+1;
end
end
end
It should also be noted that I have used this bisect method before and it does work so I don't think the problem is with that function.
Your error is here:
c = (a-b)/2;
You initialize a=20 and b=40. c is initially set to -10. But you really want c to be halfway between a and b, which means you want:
c = (a+b)/2;
Also, add a drawnow right after you plot statement to force MATLAB to draw graphics.
Related
I have a two-track model implemented in simulink. To control the velocity I use a PID-controller, so that the the output of the velocity looks like this:
now I want to implement a MATLAB function or simulink block that tracks the time when the velocity reaches a steady-state-behaviour and puts it into some kind of storage. I tried to implement something like this via the following MATLAB function with MATLAB-function-block:
function y = fcn(t,v,dv,tv)
%#codegen
if (v==tv+0.01) & (dv<0)
y=t
end
t is the clock-signal, v the velocity, dv the first derivation of the velocity and tv is the targetvelocity. The problem about this function is that there "y is not defined on some execution paths". do you have any ideas how to make this work?
I solved the problem without a MATLAB function using the simulink blocks data store memory and its read & write blocks. The signal that is coming in from bottom right side is the momentary velocity. The if statement is
(u1 >= 22.2) & (u1<=22.3) & (u2<0)
Since simulink is using time steps and the momentary velocity will never be exactly 22.2, you can not use u1==22.2
In SimuLink add a second output and a fifth input to your function. Then use that new output as a feedback to the function.
function [y, output] = fcn(t,v,dv,tv,input)
y = 0;
output = input;
if (v == tv + 0.01) && (dv < 0)
y = t;
if (input == -1)
output = t;
end
end
Attach the output to an IC block where you set the input initial value to -1 or whatever value you want to use. Then attach the IC block to the input of the function. output will be feedback constantly through the function. Once it's set it will reatin it's value forever.
function y = fcn(t,v,dv,tv)
%#codegen
y = zeros(length(t),1); % Initialise the array
for ii = 1:length(t)
if (v==tv+0.01) & (dv<0)
y(ii)=t;
else
y(ii)=0;
end
end
y(y==0)=[];
end
Two changes: added a semicolon after y=t to force it to not print it every time it is set. Second, your question, else y=[];, which means that y will be an empty matrix if you do not adhere to your if statement.
It now stores a 0 each time you do not adhere to the if statement. The line y(y==0)=[]; deletes all zeros, comment this line if you want your y to be the same length as the input variables.
function y = fcn(t,v,dv,tv)
%#codegen
y = zeros(length(t),1); % Initialise the array
ii=1;
while exist(t)
if (v==tv+0.01) & (dv<0)
y(ii)=t;
else
y(ii)=0;
end
ii = ii+1;
end
y(y==0)=[];
end
I am attempting to write a function that, if called before any animation, will handle the close events without extra code in the animation file.
function matlabStopFunction(varargin)
persistent runs
if runs==2
runs = [];
end
if isempty(runs)
evalin('base','figure(''DeleteFcn'',#matlabStopFunction);');
runs = 1;
else
assignin('base','play',false);
pause(1);
runs = 2;
end
end
Here's a sample animation code that I've been using:
function sampleAnimation
matlabStopFunction;
r = 5;
th = 0;
play = true;
while play
x = r * cosd(th);
y = r * -sind(th);
plot(x,y,'r*');
axis([-10 10 -10 10]);
th = th + 45;
pause(0.25);
end
end
The stop function works fine creating the figure, and when I close the figure, it calls the same function as expected (including the assignin on line 10). However, when stepping through, when I first get back to the base function (sampleAnimation), play is false as would be expected:
But when I step one more line, play is reset to true
Am I incorrectly assigning the value of play to false in the stop function, and if so, how could I correct this so that the animation stops when the figure is closed while keeping the code inside the animation to a minimum? I am trying to replicate the method on this blog except with all the code contained in a separate file.
I am running Matlab 2014b on Windows 8.1.
To answer your question - you are modifying the value play in the base workspace - where as the loop is in the workspace of the sampleAnimation function -> so you are not changing the required value to stop the animation. To verify this clear your variables in the base workspace clear before you run your code and you will see that the variable play is created and set to false.
By the way there is a much simpler way to do this, you animation can create a figure and then you can stop when it is deleted:
function sampleAnimation
h = figure;
r = 5;
th = 0;
while ishandle ( h )
x = r * cosd(th);
y = r * -sind(th);
plot(x,y,'r*');
axis([-10 10 -10 10]);
th = th + 45;
pause(0.25);
end
end
I am trying to make my program work. It is for prime numbers.
Below is my function then my main program
I believe this is the right formatting, (obviously I'm probably wrong cause its not working) and I have been trying to fix it to no avail. Any help will be most appreciated and an explanation of what I am doing wrong (I am pretty sure it has to do with subtle formatting) would be great too.
function [answer,primefactors ] = primey1 (N)
for i=2:(N-1)
A(i-1)=rem(N,i);
end
A;
if(all(A)==1)
answer=['Yes']
primefactors=[1,N]
elseif(all(A)==0)
answer=['No']
fac=[]
for i=2:N
q=0;
while N/i==floor(N/i)
N=N/i;
q=q+1;
end
if q>0
fac=[fac,i]
if N==1
break
primefactors=[fac]
end
end
end
end
endfunction
As noted by Magla, MATLAB comes with the primes and factor functions, which you can at least use to compare your implementation with and/or check your outcomes.
Anyway, as for your code, try this:
function [answer, primefactors] = primey1(N)
% Use vectorization for simple cases such as these:
A = rem(N,2:N-1);
if all(A)
answer = 'Yes';
primefactors = [1,N];
% Re-wrote this entire section. There were a bunch of things wrong with it:
% the order of commands was wrong, variables were not assigned for some N,
% etc. Just compare the two implementations
else
answer = 'No';
primefactors = [];
for ii = 2:N
q = 0;
while N/ii == floor(N/ii)
N = N/ii;
q = q+1;
end
if q > 0
primefactors = [primefactors,ii]; %#ok<AGROW>
if N==1
break;
end
end
end
end
end
Matlab has a factor function that does what your code is trying to do
p = factor(10)
returns 2, 5
and
p = factor(11)
returns 11.
All you have is to test for length
if length(p) == 1
is true for prime numbers.
The solution as offered by #Rody should do the job, in theory even more efficient than this, however to illustrate the concept of initialization, I would recommend you to initialize the output variables of your function right after the function start. In your case this would mean that I recommend starting the code like this:
function [answer, primefactors] = primey1(N)
% Function to determine whether a number is prime and which prime factors it has
% Assign default values
answer = 'No';
primefactors = [];
I am trying to using a while loop inside a for loop in Matlab. The while loop will repeat the same action until it satifies some criteria. The outcome from the while loop is one iteration in the for loop. I am having a problem to get that correctly.
n=100;
for i=1:n
while b<0.5
x(i)=rand;
b=x(i);
end
end
I am not sure what i am doing wrongly.
Thanks
Approach the problem differently. There's no need to try again if rand doesn't give you the value you want. Just scale the result of rand to be in the range you want. This should do it:
x = 0.5 + 0.5*rand(1, 100);
With the example you showed, you have to initialize b or the while-statement cannot be evaluated when it is first called.
Do it inside the for-loop to avoid false positives after the first for-iteration:
n=100;
for ii=1:n
b = 0;
while b<0.5
x(ii)=rand;
b=x(ii);
end
end
Or, without b:
n=100;
x = zeros(1,100);
for ii=1:n
while x(ii)<0.5
x(ii)=rand;
end
end
I have two vectors and I want to get their dor product without for loop. It would save a lot of time.
Here is sample code.
p=1;
C=zeros(100,1);
r_a_= rand(1,3);
r_b= rand(100,3);
for it=1:size(r_b,1)
C(p,1) = dot(r_a_,r_b(it,:)) ;
p=p+1
end
To avoide for loop and to call dot function only once, I thought like this
r_a=repmat(r_a_,100,1);
C = dot(r_a,r_b) %%%%%%%% But obviously it gives wrong answer
Any idea?
You can do it with a matrix multiply, like so:
r_a_ = rand(1,3);
r_b = rand(100,3);
C = r_a_ * r_b.'