I am trying to generate an array from some starting values using this formula in MATLAB:
yt = a0 + ∑i=1p (ai ⋅ yt-i), t ≥ p
p is some small number compared to T (max t). I have been able to make this using two for cycles but it is really slow. Is there some easy way to do it?
First p values of y are provided and vector a (its length is p+1) is provided too...
This is what I have so far, but now when I tried it, it doesn't work 100% (I think it's because of indexing from 1 in MATLAB):
y1 = zeros(T+1, 1);
y1(1:p) = y(1:p);
for t = p+1:T+1
value = a1(1);
for j = 2:p+1
value = value + a1(j)*y1(t-j+1);
end
y1(t) = value;
end
EDIT: I solved it, I am just not used to Matlab indexing from 1...
This statement
if(p>=t)
looks odd inside a loop whose index expression is
for t = p+1:T+1
which seems to guarantee that t>p for the entire duration of the loop. Is that what you meant to write ?
EDIT in response to comment
Inside a loop indexed with this statement
for j = 2:p
how does the reference you make to a(j) ever call for a(0) ?
y1 = zeros(T+1, 1);
y1(1:p) = y(1:p);
for t = p+1:T+1
value = a1(1);
for j = 2:p+1
value = value + a1(j)*y1(t-j+1);
end
y1(t) = value;
end
Related
I'm trying to implement my own fft in MATLAB the following way:
function z=FastFourierTransform(x)
N=length(x);
if N <= 1
z = x;
else
range = (0:N/2-1);
e = exp(-2i*pi/N).^range;
odd = FastFourierTransform(x(1:2:N-1));
even = e.*FastFourierTransform(x(2:2:N));
z = [even + odd, even - odd];
end
return
Turns out, there seems to be somthing wrong with it since it does not give the same result as the built in function provided by MATLAB.
I'm calling the function the following way:
N = 128;
x = 32*pi*(1:N)'/N;
u = cos(x/16).*(1+sin(x/16));
actualSolution = fft(u);
actualSolution
mySolution = FastFourierTransform(u)';
mySolution
actualSolution
mySolution
The numbers are always the same but they sometimes differ in their sign.
You have swapped odd and even.
Using this line to compute z will produce the correct FFT:
z = [odd + even, odd - even];
My guess is that the source of confusion is that Matlab uses 1-based indices, and the pseudocode you used to implement the function uses 0-based indices.
I've run an experiment where a machine exerts a force on a bridge until it breaks. I need to cycle through my entire data set and calculate the toughness until I've hit the breaking point (fdValue1).
The toughness is calculated by summing up all the rectangles under my load vs. distance curve. (Basically the integral)
However, I have not been able to find a reasonable way of doing so and my current loop is an infinite loop and I'm not sure why.
%Initializing variables
bridge1Data = xlsread('Bridge1Data.xlsx', 'A2:C2971');
bridge2Data = xlsread('Bridge2Data.xlsx', 'A2:C1440');
bridge1Load = bridge1Data(:, 2);
bridge2Load = bridge2Data(:, 2);
bridge1Dist = bridge1Data(:, 3);
bridge2Dist = bridge2Data(:, 3);
[row1, col1] = size(bridge1Dist);
[row2, col2] = size(bridge2Dist);
bridge1Disp = zeros(row1, col1);
bridge2Disp = zeros(row2, col2);
fdValue1 = 0.000407350000000029;
&Main code
%Displacement
for k = 2:length(bridge1Dist)
bridge1Disp(k-1, 1) = bridge1Dist(k, 1) - bridge1Dist(k-1, 1);
end
%Max Load Bridge 1
maxLoad1 = 0;
for n = 1:length(bridge1Load)
for k = 1
if bridge1Load(n, k) > maxLoad1
maxLoad1 = bridge1Load(n, k);
end
end
end
%Cycle through data till failure, change proj data
totalRect1 = 0;
for j = 2:length(bridge1Disp)
while bridge1Disp(j, 1) ~= fdValue1
rectangle = (bridge1Disp(j, 1) - bridge1Disp(j-1, 1))*...
((bridge1Load(j, 1) + bridge1Load(j-1, 1))/2);
totalRect1 = totalRect1 + rectangle;
end
end
Basically, I make an array for the load and distance the machine pushes down on the bridge, assign a 'Failure Distance' value (fdValue) which should be used to determine when we stop calculating toughness. I then calculate displacement, calculate the maximum load. Then through the variable 'rectangle', calculate each rectangle and sum them all up in 'totalRect1', and use that to calculate the toughness by finding the area under the curve. Is anyone able to see why the loop is an infinite loop?
Thanks
The problem with the condition while bridge1Disp(j, 1) ~= fdValue1 is that you need to check for <= and not for (in)equality, since double numbers will almost never evaluate to be equal even if they seem so. To read more about this you can look here and also google for matlab double comparison. Generally it has something to do with precision issues.
Usually when checking for double equality you should use something like if abs(val-TARGET)<1E-4, and specify some tolerance which you are willing to permit.
Regardless,
You don't need to use loops for what you're trying to do. I'm guessing this comes from some C-programming habits, which are not required in MATLAB.
The 1st loop (Displacement), which computes the difference between every two adjacent elements can be replaced by the function diff like so:
bridge1Disp = diff(bridge1Dist);
The 2nd loop construct (Max Load Bridge 1), which retrieves the maximum element of bridge1Load can be replaced by the command max as follows:
maxLoad1 = max(bridge1Load);
For the last loop construct (Cycle ...) consider the functions I've mentioned above, and also find.
In the code section
%Cycle through data till failure, change proj data
totalRect1 = 0;
for j = 2:length(bridge1Disp)
while bridge1Disp(j, 1) ~= fdValue1
rectangle = (bridge1Disp(j, 1) - bridge1Disp(j-1, 1))*...
((bridge1Load(j, 1) + bridge1Load(j-1, 1))/2);
totalRect1 = totalRect1 + rectangle;
end
end
the test condition of the while loop is
bridge1Disp(j, 1) ~= fdValue1
nevertheless, in the while loop the value of bridge1Disp(j, 1) does not change so if at the first iteratiion of the while loop bridge1Disp(j, 1) is ~= fdValue1 the loop will never end.
Hope this helps.
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 am trying to rewrite part of a vector given that:
t = -10:.1:10
x = exp((-3.*t);
The length of x will be 201, and I want to rewrite the first 100 values.
The only way I've gotten to work is by doing this:
EDIT Fixed typo.
t = 0:.1:10;
x = exp((-3.*t); % EDIT: THERE WAS A TYPO HERE
z = zeros(1,100);
for k = 1 : 100
x(k) = z(k);
end
There are three questions. First: What is a faster and more efficient way of doing this? Second: What do I do if I don't want to overwrite the first part of the code but rather the middle or the second part? Third: Is there a way of utilizing the full range of t where t = -10:.1:10 and just ignoring the first half instead of writing a whole new variable for it?
First: Nothing else I've tried has been successful.
Second: The only way I can think to do that is to append the two vectors together, but then it doesn't overwrite the data, so that is a no go.
Third:I have tried an if statement and that didn't work at all.
Your code appears to assign something to y, then changes the value of x. I assume that is a typo - and not the problem you actually want to fix.
In general, if you have
t = -10:0.1:10; % my preference: t = linspace(-10,10,201);
and
y = exp(-3 * t );
but you want to set the first 100 elements of y to zero, you can then do
y(1:100) = 0;
If you wanted never to compute y(1:100) in the first place you might do
y = zeros(size(t));
y(101:end) = exp(-3 * t(101:end));
There are many variations on this. I think the above code samples address all three of your questions.
change your
for k = 1 : 100
x(k) = z(k); % i think it should be y(k) though
end
to
x(1:100) = 0;
You could use logical indexing; that is, you can use a logical statement to select elements of a vector/matrix:
t = -10:0.1:10;
x = exp((-3.*t);
x(t < 0) = 0;
This works for the middle of a matrix too:
x(t > -5 & t < 5) = whatever;