When I run this code that I've written to simulate a heat flow model in MATLAB i get an error that says 'Subscript indices must either be real positive integers or logicals.' I think this is probably something to do with my linspace command generating a different type of variable not integers and so it's not working properly but I'm not sure how to amend my script to correct for this.
Cp = 400;
p = 8960;
k = 400;
a = k/(p*Cp);
dt = 0.01;
dx = sqrt(5*a*dt); %% 5 as 1/5 is smaller than 1/4 for stability
T = zeros(20000,10000);
for x = linspace(1,10000,10000);
T(x,:) = 1000;
end
for x = linspace(10001,20000,10000);
T(x,:) = 25;
end
for t = linspace(1,10000,10000);
for x = linspace(1,20000,20000);
T(x,t+1) = T(x,t)+a*dt*((T(x-1,t)-2*T(x,t)+ T(x+1,t))/(dx*dx));
end
end
The line that blows up is:
T(x,t+1) = T(x,t)+a*dt*((T(x-1,t)-2*T(x,t)+ T(x+1,t))/(dx*dx));
Specifically T(x-1,t) triggers the error because x starts as 1, hence x - 1 = 0 and 0 is not a valid index.
On a more general Matlab coding note, I would write x = 1:10000 instead of x = linspace(1,10000,10000), but this is not causing the error. Note that I'm only addressing the Matlab error message. I have no idea whether your overall code works.
Related
I am trying to use the Metropolis Hastings algorithm with a random walk sampler to simulate samples from a function $$ in matlab, but something is wrong with my code. The proposal density is the uniform PDF on the ellipse 2s^2 + 3t^2 ≤ 1/4. Can I use the acceptance rejection method to sample from the proposal density?
N=5000;
alpha = #(x1,x2,y1,y2) (min(1,f(y1,y2)/f(x1,x2)));
X = zeros(2,N);
accept = false;
n = 0;
while n < 5000
accept = false;
while ~accept
s = 1-rand*(2);
t = 1-rand*(2);
val = 2*s^2 + 3*t^2;
% check acceptance
accept = val <= 1/4;
end
% and then draw uniformly distributed points checking that u< alpha?
u = rand();
c = u < alpha(X(1,i-1),X(2,i-1),X(1,i-1)+s,X(2,i-1)+t);
X(1,i) = c*s + X(1,i-1);
X(2,i) = c*t + X(2,i-1);
n = n+1;
end
figure;
plot(X(1,:), X(2,:), 'r+');
You may just want to use the native implementation of matlab mhsample.
Regarding your code, there are a few things missing:
- function alpha,
- loop variable i (it might be just n but it is not suited for indexing since it starts at zero).
And you should always allocate memory in matlab if you want to fill it dynamically, i.e. X in your case.
To expand on the suggestions by #max, the code appears to work if you change the i indices to n and replace
n = 0;
with
n = 2;
X(:,1) = [.1,.1];
It would probably be better to assign X(:,1) to random values within your accept region (using the same code you use later), and/or include a burn-in period.
Depending upon what you are going to do with this, it may also make things cleaner to evaluate the argument to sin in the f function to keep it within 0 to 2 pi (likely by shifting the value by 2 pi if it exceeds those bounds)
numberOfTrials = 10;
numberOfSizes = 6;
sizesArray = zeros(numberOfSizes, 1);
randomMAveragesArray = zeros(numberOfSizes, 1);
for i=1:numberOfSizes
N = 2^i;
%x=rand(N,1);
randomMTimesArray = zeros(numberOfTrials, 1);
for j=1:numberOfTrials
tic;
for k=1:N^3
x = .323452345e-999 * .98989898989889e-953;
end
randomMTimesArray(j) = toc;
end
sizesArray(i) = N;
end
randomMPolyfit = polyfit(log10(sizesArray), log10(randomMAveragesArray), 1);
randomMSlope = randomMPolyfit(1);
That is my Matlab script. I was originally timing a NxN random matrix using '\' to solve. The runtime on this is O(n^3). But my slope for the log graph was always about 1.8.
My understanding from this is that the timing results are O(n^k) where k is the slope from the log/log graph. So therefore the slope I should get should be around 3.
The code I posted above I have made an arbitrary loop that is N^3 with a floating point operation to test if this works.
However with the for loop I'm getting a slope of 2.5.
Why is this?
Since the O() behavior is asymptotic, you sometimes cannot see the behavior for small values of N. For example, if I set numberOfSizes = 9 and discard the first 3 points for the polynomial fit, the slope is much closer to 3:
randomMPolyfit = polyfit(log10(sizesArray(4:end)), log10(randomMAveragesArray(4:end)), 1);
randomMSlope = randomMPolyfit(1)
randomMSlope =
2.91911869082081
If you plot the timing array this behavior is clearer.
So I am trying to go through a for loop that will increment .1 every time and will do this until the another variable h is less than or equal to zero. Then I am suppose to graph this h variable along another variable x. The code that I wrote looks like this:
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
h(t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
The Error that I keep getting when running this code says "Attempted to access h(0); index must be a positive integer or logical." I don't understand what exactly is going on in order for this to happen. So my question is why is this happening and is there a way I can solve it, Thank you in advance.
You're using t as your loop variable as well as your indexing variable. This doesn't work, because you'll try to access h(0), h(0.1), h(0.2), etc, which doesn't make sense. As the error says, you can only access variables using integers. You could replace your code with the following:
t = 0:0.1:12;
for i = 1:length(t)
% use t(i) instead of t now
end
I will also point out that you don't need to use a for loop to do this. MATLAB is optimised for acting on matrices (and vectors), and will in general run faster on vectorised functions rather than for loops. For instance, your equation for h could be replaced with the following:
O = 20;
v = 200;
g = 32.2;
t = 0:0.1:12;
h = v * t * sin(O) - 0.5 * g * t.^2;
The only difference is that you have to use the element-wise square (.^2) rather than the normal square (^2). This means that MATLAB will square each element of the vector t, rather than multiplying the vector t by itself.
In short:
As the error says, t needs to be an integer or logical.
But your t is t=0:0.1:12, therefore a decimal value.
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
idx_t = 1:numel(t);
h(idx_t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(idx_t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
Look this question's answer for more options: Subscript indices must either be real positive integers or logical error
I was trying to write an algorithm to approximate integrals with Simpson's method. When I try to plot it in a loglog plot, however, I don't get the correct order of accuracy which is O(h^4) (I get O(n)). I can't find any errors though. This is my code:
%Reference solution with Simpson's method (the reference solution works well)
yk = 0;
yj = 0;
href = 0.0001;
mref = (b-a)/href;
for k=2:2:mref-1
yk=yk+y(k*href+a);
end
for j=1:2:mref
yj=yj+y(href*j+a);
end
Iref=(href/3)*(y(a)+y(b)+2*yk+4*yj);
%Simpson's method
iter = 1;
Ehmatrix = 0;
for n = 0:nmax
h = b/(2^n+1);
x = a:h:b;
xodd = x(2:2:end-1);
xeven = x(3:2:end);
yodd = y(xodd);
yeven = y(xeven);
Isimp = (h/3)*(y(x(1))+4*sum(yodd)+2*sum(yeven)+y(b));
E = abs(Isimp-Iref);
Ehmatrix([iter],1) = [E];
Ehmatrix([iter],2) = [h];
iter = iter + 1;
end
figure
loglog(Ehmatrix(:,2),Ehmatrix(:,1))
a and b are the integration limits and y is the integrand that we want to approximate.
Djamillah - your code looks fine though the initialization of h is probably valid only for the case where a==0, so you may want to change this line of code to
h = (b-a)/(2^n+1);
I wonder if x = a:h:b; will always be valid - sometimes b may be included in the list, and sometimes it might not be, depending upon h. You may want to reconsider and use linspace instead
x = linspace(a,b,2^n+1);
which will guarantee that x has 2^n+1 points distributed evenly in the interval [a,b]. h could then be initialized as
h = x(2)-x(1);
Also, when determining the even and odd indices, we need to ignore the last element of x for both even and odd. So instead of
xodd = x(2:2:end-1);
xeven = x(3:2:end);
do
xodd = x(2:2:end-1);
xeven = x(3:2:end-1);
Finally, rather than using a vector y (how is this set?) I might just use the function handle to the function that I'm integrating instead and replace the calculation above as
Isimp = delta/3*(func(x(1)) + 4*sum(func(xodd)) + 2*sum(func(xeven)) + ...
func(x(end)));
Other than these tiny things (which are probably negligible), there is nothing in your algorithm to indicate a problem. It produced similar results to a version that I have.
As for the order of convergence, should it be O(n^4) or O(h^4)?
Taking into account Geoff's suggestions, and making a few other changes, it all works as expected.
%Reference solution with Simpson's method (the reference solution works well)
a=0;
b=1;
y=#(x) cos(x);
nmax=10;
%Simpson's method
Ehmatrix = [];
for n = 0:nmax
x = linspace(a,b,2^n+1);
h = x(2)-x(1);
xodd = x(2:2:end-1);
xeven = x(3:2:end-1);
yodd = y(xodd);
yeven = y(xeven);
Isimp = (h/3)*(y(x(1))+4*sum(yodd)+2*sum(yeven)+y(b));
E = abs(Isimp-integral(y,0,1));
Ehmatrix(n+1,:) = [E h];
end
loglog(Ehmatrix(:,2),Ehmatrix(:,1))
P=polyfit(log(Ehmatrix(:,2)),log(Ehmatrix(:,1)),1);
OrderofAccuracy=P(1)
You were getting O(h) accuracy because xeven=x(3:2:end) was wrong. Replacing it by xeven=x(3:e:end-1) fixes the code, and thus the accuracy.
I'm trying to set up a piecewise transfer function of a filter in matlab to get its impulse response. I have the code below:
function H = H(w)
H = zeros(size(w)); % Preallocating enough memory for y
nd = 0;
region1 = (abs(w)<(pi/4)) & (abs(w)>(pi/8)) ; % First interval
H(region1) = exp((-(w(region1))*1i*nd));
region2 = (abs(w)<(7*pi/8)) & (abs(w)>(5*pi/8)); % Second interval
H(region2) = exp((-0.5*(w(region1))*1i*nd));
region3 = ~(abs(w)<(pi/4)) & (abs(w)>(pi/8)) & ~(abs(w)<(7*pi/8)) & (abs(w)>(5*pi/8)) ; % Third interval
H(region3) = 0;
But it gives me this error, when I try to run:
In an assignment A(I) = B, the number of elements in B and I must be the same.Error
in H (line 9)
H(region2) = exp((-0.5*(w(region1))*1i*nd));
Am I going about this the right way or is there an easier way to do something like this?
I think the problem is that:
H(region2) = exp((-0.5*(w(region1))*1i*nd));
Should be:
H(region2) = exp((-0.5*(w(region2))*1i*nd));
Where region1 is corrected as region2.
Also, nd is always 0.
You ask if you're going about it the right way, seems decent enough to me as long as you realize the frequency response between the points you specify could be all over the place, or not depending on the transitions.