I am facing some problem during calculation of Numerical Integration with two data set.
For integration i am using simpsons 1/3 rule.
function I = Simpsons(f,a,b,n)
if numel(f)>1 % If the input provided is a vector
n=numel(f)-1; h=(b-a)/n;
I= h/3*(f(1)+2*sum(f(3:2:end-2))+4*sum(f(2:2:end))+f(end));
else
h=(b-a)/n; xi=a:h:b;
I= h/3*(f(xi(1))+2*sum(f(xi(3:2:end-2)))+4*sum(f(xi(2:2:end)))+f(xi(end)));
end
This code correctly calculates the integration.
Now the problem occurs during the calculation of multiplied values.
for example I have two functions f and g both are depending on same variable.
the variables is in the same ranges. SO lower Limit and Upper Limit is same.
$\int_a^b \! f(x) *g(x) \, \mathrm{d} x.$
here the resolution of x is different. Means for f(x) we have 1000 data where as for g(x) we have 1700 data points. so element by element multiplication cant be done.
How to Solve this integration ..
you'll need to interpolate one of your functions, f or g, to the other function points, for 1D functions that is achievable using interp1.
For example:
% x1 an x2 have the same limits but different # of elements
x1 = linspace(-10,10,100);
x2 = sort(rand(1,170)*20-10); # non-unifrom points from -10 to 10
f1 = sin(x1);
f2 = cos(x2);
now say we want to multiply f1*f2, we need them to have the # of elements, so
f2i= interp1(x2,f2,x1,'spline');
will make f2 to have the same # of elements as f1, or instead
f1i= interp1(x1,f1,x2,'spline');
will make f1 to have the same # of elements as f2.
Related
I'm trying to plot a probability mass function for the probability of certain sums when rolling three dices and I found this example in MathCad and wondered if there is anything like it in MatLab?
I imagine you’d build that as a lookup table rather than a series of if/else statements. It is easy to compute the probabilities using convolution:
f1 = ones(1,6);
f2 = conv(f1,f1);
f = conv(f2,f1);
% because f contains values from x=3 to 18,
% rather than starting at 1 as MATLAB arrays do,
% we add two zeros to the front:
f = [0,0,f];
Now f(x) returns the probability of throwing x.
Working in MATLAB. I have the following equation:
S = aW + bX + cY + dZ
where S,W,X,Y, and Z are all known n x 1 vectors. I am trying to fit the data of S with a linear combination of the basis vectors W,X,Y, and Z with the constraint of the constants (a,b,c,d) being greater than 0. I have managed to do this in Excel's solver, and have attempted to figure it out on MATLAB, being directed towards functions like fmincon, but I am not overly familiar with MATLAB and feel I am misunderstanding the use of fmincon.
I am looking for help with understanding fmincon's use towards my problem, or redirection towards a more efficient method for solving.
Currently I have:
initials = [0.2 0.2 0.2 0.2];
fun = #(x)x(1)*W + x(2)*X + x(3)*Y + x(4)*Z;
lb = [0,0,0,0];
soln = fmincon(fun,initials,data,b,Aeq,beq,lb,ub);
I am receiving an error stating "A must have 4 column(s)". Where A is referring to my variable data which corresponds to my S in the above equation. I do not understand why it is expecting 4 columns. Also to note the variables that are in my above snippet that are not explicitly defined are defined as [], serving as space holders.
Using fmincon is a huge overkill in this case. It's like using big heavy microscope to crack nuts... or martian rover to tow a pushcart or... anyway :) May be it's OK if you don't have to process large sets of vectors. If you need to fit hundreds of thousands of such vectors it can take hours. But this classic solution will be faster by several orders of magnitude.
%first make a n x 4 matrix of your vectors;
P=[W,X,Y,Z];
% now your equation looks like this S = P*m where m is 4 x 1 vectro
% containing your coefficients ( m = [a,b,c,d] )
% so solution will be simply
m_1 = inv(P'*P)*P'*S;
% or you can use this form
m_2 = (P'*P)\P'*S;
% or even simpler
m_3 = (S'/P')';
% all three solutions should give exactly same resul
% third solution is the neatest but may not work in every version of matlab
% Your modeled vector will be
Sm = P*m_3; %you can use any of m_1, m_2 or m_3;
% And your residual
R = S-Sm;
If you need to procees many vectors don't use for cycle. For cycles are very slow in Matlab and you should use matrices instead, if possible. S can also be nk matrix, where k is number vectors you want to process. In this case m will be 4k matrix.
What you are trying to do is similar to the answer I gave at is there a python (or matlab) function that achieves the minimum MSE between given set of output vector and calculated set of vector?.
The procedure is similar to what you are doing in EXCEL with solver. You create an objective function that takes (a, b, c, d) as the input parameters and output a measure of fit (mse) and then use fmincon or similar solver to get the best (a, b, c, d) that minimize this mse. See the code below (no MATLAB to test it but it should work).
function [a, b, c, d] = optimize_abcd(W, X, Y, Z)
%=========================================================
%Second argument is the starting point, second to the last argument is the lower bound
%to ensure the solutions are all positive
res = fmincon(#MSE, [0,0,0,0], [], [], [], [], [0,0,0,0], []);
a=res(1);
b=res(2);
c=res(3);
d=res(4);
function mse = MSE(x_)
a_=x_(1);
b_=x_(2);
c_=x_(3);
d_=x_(4);
S_ = a_*W + b_*X + c_*Y + d_*Z
mse = norm(S_-S);
end
end
I'm kind've new to Matlab and stack overflow to begin with, so if I do something wrong outside of the guidelines, please don't hesitate to point it out. Thanks!
I have been trying to do convolution between two functions and I have been having a hard time trying to get it to work.
t=0:.01:10;
h=exp(-t);
x=zeros(size(t)); % When I used length(t), I would get an error that says in conv(), A and B must be vectors.
x(1)=2;
x(4)=5;
y=conv(h,x);
figure; subplot(3,1,1);plot(t,x); % The discrete function would not show (at x=1 and x=4)
subplot(3,1,2);plot(t,h);
subplot(3,1,3);plot(t,y(1:length(t))); %Nothing is plotted here when ran
I commented my issues with the code. I don't understand the difference of length and size in this case and how it would make a difference.
For the second comment, x=1 should have an amplitude of 2. While x=4 should have an amplitude of 5. When plotted, it only shows nothing in the locations specified but looks jumbled up at x=0. I'm assuming that's the reason why the convoluted plot won't be displayed.
The original problem statement is given if it helps to understand what I was thinking throughout.
Consider an input signal x(t) that consists of two delta functions at t = 1 and t = 4 with amplitudes A1 = 5 and A2 = 2, respectively, to a linear system with impulse response h that is an exponential pulse (h(t) = e ^−t ). Plot x(t), h(t) and the output of the linear system y(t) for t in the range of 0 to 10 using increments of 0.01. Use the MATLAB built-in function conv.
The initial question regarding size vs length
length yields a scalar that is equal to the largest dimension of the input. In the case of your array, the size is 1 x N, so length yields N.
size(t)
% 1 1001
length(t)
% 1001
If you pass a scalar (N) to ones, zeros, or a similar function, it will create a square matrix that is N x N. This results in the error that you see when using conv since conv does not accept matrix inputs.
size(ones(length(t)))
% 1001 1001
When you pass a vector to ones or zeros, the output will be that size so since size returns a vector (as shown above), the output is the same size (and a vector) so conv does not have any issues
size(ones(size(t)))
% 1 1001
If you want a vector, you need to explicitly specify the number of rows and columns. Also, in my opinion, it's better to use numel to the number of elements in a vector as it's less ambiguous than length
z = zeros(1, numel(t));
The second question regarding the convolution output:
First of all, the impulses that you create are at the first and fourth index of x and not at the locations where t = 1 and t = 4. Since you create t using a spacing of 0.01, t(1) actually corresponds to t = 0 and t(4) corresponds to t = 0.03
You instead want to use the value of t to specify where to put your impulses
x(t == 1) = 2;
x(t == 4) = 5;
Note that due to floating point errors, you may not have exactly t == 1 and t == 4 so you can use a small epsilon instead
x(abs(t - 1) < eps) = 2;
x(abs(t - 4) < eps) = 5;
Once we make this change, we get the expected scaled and shifted versions of the input function.
I'm new to programming in Matlab. I'm trying to figure out how to calculate the following function:
I know my code is off, I just wanted to start with some form of the function. I have attempted to write out the sum of the function in the program below.
function [g] = square_wave(n)
g = symsum(((sin((2k-1)*t))/(2k-1)), 1,n);
end
Any help would be much appreciated.
Update:
My code as of now:
function [yout] = square_wave(n)
syms n;
f = n^4;
df = diff(f);
syms t k;
f = 1; %//Define frequency here
funcSum = (sin(2*pi*(2*k - 1)*f*t) / (2*k - 1));
funcOut = symsum(func, v, start, finish);
xsquare = (4/pi) * symsum(funcSum, k, 1, Inf);
tVector = 0 : 0.01 : 4*pi; %// Choose a step size of 0.01
yout = subs(xsquare, t, tVector);
end
Note: This answer was partly inspired by a previous post I wrote here: How to have square wave in Matlab symbolic equation - However, it isn't quite the same, which is why I'm providing an answer here.
Alright, so it looks like you got the first bit of the question right. However, when you're multiplying things together, you need to use the * operator... and so 2k - 1 should be 2*k - 1. Ignoring this, you are symsuming correctly given that square wave equation. The input into this function is only one parameter only - n. What you see in the above equation is a Fourier Series representation of a square wave. A bastardized version of this theory is that you can represent a periodic function as an infinite summation of sinusoidal functions with each function weighted by a certain amount. What you see in the equation is in fact the Fourier Series of a square wave.
n controls the total number of sinusoids to add into the equation. The more sinusoids you have, the more the function is going to look like a square wave. In the question, they want you to play around with the value of n. If n becomes very large, it should start approaching what looks like to be a square wave.
The symsum will represent this Fourier Series as a function with respect to t. What you need to do now is you need to substitute values of t into this expression to get the output amplitude for each value t. They define that for you already where it's a vector from 0 to 4*pi with 1001 points in between.
Define this vector, then you'll need to use subs to substitute the time values into the symsum expression and when you're done, cast them back to double so that you actually get a numeric vector.
As such, your function should simply be this:
function [g] = square_wave(n)
syms t k; %// Define t and k
f = sin((2*k-1)*t)/(2*k-1); %// Define function
F = symsum(f, k, 1, n); %// Define Fourier Series
tVector = linspace(0, 4*pi, 1001); %// Define time points
g = double(subs(F, t, tVector)); %// Get numeric output
end
The first line defines t and k to be symbolic because t and k are symbolic in the expression. Next, I'll define f to be the term inside the summation with respect to t and k. The line after that defines the actual sum itself. We use f and sum with respect to k as that is what the summation calls for and we sum from 1 up to n. Last but not least, we define a time vector from 0 to 4*pi with 1001 points in between and we use subs to substitute the value of t in the Fourier Series with all values in this vector. The result should be a 1001 vector which I then cast to double to get a numerical result and we get your desired output.
To show you that this works, we can try this with n = 20. Do this in the command prompt now:
>> g = square_wave(20);
>> t = linspace(0, 4*pi, 1001);
>> plot(t, g);
We get:
Therefore, if you make n go higher... so 200 as they suggest, you'll see that the wave will eventually look like what you expect from a square wave.
If you don't have the Symbolic Math Toolbox, which symsum, syms and subs relies on, we can do it completely numerically. What you'll have to do is define a meshgrid of points for pairs of t and n, substitute each pair into the sequence equation for the Fourier Series and sum up all of the results.
As such, you'd do something like this:
function [g] = square_wave(n)
tVector = linspace(0, 4*pi, 1001); %// Define time points
[t,k] = meshgrid(tVector, 1:n); %// Define meshgrid
f = sin((2*k-1).*t)./(2*k-1); %// Define Fourier Series
g = sum(f, 1); %// Sum up for each time point
end
The first line of code defines our time points from 0 to 4*pi. The next line of code defines a meshgrid of points. How this works is that for t, each column defines a unique time point, so the first column is 200 zeroes, up to the last column which is a column of 200 4*pi values. Similarly for k, each row denotes a unique n value so the first row is 1001 1s, followed by 1001 2s, up to 1001 1s. The implications with this is now each column of t and k denotes the right (t,n) pairs to compute the output of the Fourier series for each time that is unique to that column.
As such, you'd simply use the sequence equation and do element-wise multiplication and division, then sum along each individual column to finally get the square wave output. With the above code, you will get the same result as above, and it'll be much faster than symsum because we're doing it numerically now and not doing it symbolically which has a lot more computational overhead.
Here's what we get when n = 200:
This code with n=200 ran in milliseconds whereas the symsum equivalent took almost 2 minutes on my machine - Mac OS X 10.10.3 Yosemite, 16 GB RAM, Intel Core i7 2.3 GHz.
I am trying to integrate the sine-function. My goal is to get not just the value of the area inbetween a certain distance but the specific values of the integrated course.
One way to achieve this is by using cumtrapz. I want to get the same result using integral or quad. So I am wondering if there is something like cumquad?
I tried to write something for myself but it works very slow and seems to be even worse than cumtrapz. Later on I want to integrate measured data. So it won't be as simple as a sine.
Here is my current code:
a = 0; b = 10;
x = a:0.1:b;
y = 2*sin(3*x);
pp = spline(x,y);
y2=zeros(1,length(y));
y3=zeros(1,length(y));
y2(1)=integral(#(x)ppval(pp,x),x(1),x(2));
y3(1)=integral(#(x)ppval(pp,x),x(1),x(2));
for a=2:(length(y)-1)
y2(a) = y2(a-1)+integral(#(x)ppval(pp,x),x(a-1),x(a));
y3(a) = y3(a-1)+quad(#(x)ppval(pp,x),x(a-1),x(a));
end
y4=cumtrapz(x,y);
% y5=cumsum(y);
plot(x,y)
hold on
plot(x,y2,'-ro')
plot(x,y3,'-kx')
plot(x,y4,'g')
syms x % compare with analytical result
ya=2*sin(3*x);
ya5=int(ya)+(2/3);
ezplot(x,ya5)
Using integral
I don't think there is a way to have MATLAB return the integral along the path, so you are correct in performing the integration one Δx at a time.
The slowness comes from the loop and subsequent restart of every integral call.
You can avoid the loop by posing the integral over every interval as a vector-valued function.
The Math
Suppose we divide x into N-1 intervals with N total boundaries and denote an interval boundary as xn where n ∈{1,2,3...,N} such that x1 ≤ x2 ≤ x3 ... ≤ xN.
Then any integral over the interval would be
Using the u-substitution:
The integral becomes:
where Δxn = xn - xn-1
The Code
So now, we can pose the interval integration of any function by specifying the lower bound xn-1, specifying the interval width Δx, and integrating from 0 to 1.
The best part is that if the lower bound and interval widths are vectors, we can create a vector-valued function in terms of u and have integral integrate with the option 'ArrayValued' = true.
x = a:0.1:b;
xnm1 = x(1:end-1);
dx = x(2:end) - xnm1;
fx = #(x) 2*sin(3*x);
f = #(u) dx .* fx(dx*u+xnm1);
y = cumsum([0,integral(#(u)f(u),0,1,'ArrayValued',true)]);
The cumsum accounts for the fact that each integral over a given interval needs to have the value of the previous interval added to it.
On my machine, this is at least order of magnitude faster than the loop version and gets better as the interval count increases.
Using ode45
Use can also use ode45 to perform the integration.
It is not nearly as efficient as the integral method, but it may be easier conceptually and look cleaner.
In fact, ode45 is about 10 times slower than the integral method above when required to return an absolute error on par with that of integral.
a = 0;
b = 10;
% These options are necessary to approach the accuracy of integral
opt = odeset('RelTol',100*eps(),'AbsTol',eps());
sol = ode45(#(x,y) 2*sin(3*x),[a,b],0,opt);
x = a:0.01:b;
yint = deval(sol,x);