I have here an equation that is in the image
I'm trying to make the plot where y-axis is this equation, and x-axis is time - just a vector.
All the initials I have:
%Initials
Beta=[24 123 117 262 108 45]*10^-5; %pcm
Lambda=[0.0127 0.0317 0.1160 0.3106 1.4006 3.8760]; %1/s
LAMBDA=10^-4 ; %s
W=[ 0.376 -0.0133 -0.0426 -0.153 -0.972 -3.38 -29.5]
Rho=400*10^-5
t=linspace(1,30,7)
This is the code I'm using:
for n=1:7
for j=1:6
S1=Rho*sum(exp(W(n)'.*t)/(W(n)'.*(LAMBDA+(sum(Beta(j).*Lambda(j)./(W(n)+Lambda(j))^2)))))
end
end
semilogy(t,S1,'b','linewidth',2);
And S1 returns too much answers, and as I undestand it should give only 7...
And I am new to matlab and coding in general, so if the answer is obviuos I still don't know how to make it work :D
Let's clarify a few things first.
In order to do a 2D plot (of any kind), you need two vectors in Matlab. They should be of equal length. One for all the x-coordinates. Another for all the y-coordinates.
You get the x-coordinates in t=linspace(1,30,7). However, you do not have the corresponding y-coordinates.
In your case, it is best to phrase your formula as a function of t. And let's break down the sums for clarity. For example,
function num = oscillation_modes_of_sort(t)
outer_sum = 0;
for j=1:numel(W)
inner_sum = 0;
for i=1:numel(Beta)
inner_sum = inner_sum + Beta(i)*Lambda(i)/(W(j)+Lambda(i))^2;
end
outer_sum = out_sum + exp(W(j)*t)/(W(j)*(LAMBDA+inner_sum));
end
num = Rho * outer_sum;
end
Now, your y-coordinates will be oscillation_modes_of_sort(t).
There are ways to either make the code more brief or make it more friendly if W and Beta are much much longer. But let's do those at a future time.
Related
I am trying to make a model of planets' movement plot it in 3d using Matlab.
I used Newton's law with the gravitational force between two objects and I got the differential equation below:
matlab code:
function dy=F(t,y,CurrentPos,j)
m=[1.98854E+30 3.302E+23 4.8685E+24 5.97219E+24 6.4185E+23 1.89813E+27 5.68319E+26 8.68103E+25 1.0241E+26 1.307E+22];
G=6.67E-11;
dy = zeros(6,1);
dy(1) = y(4);
dy(2) = y(5);
dy(3) = y(6);
for i=1:10
if i~=j
deltaX=(CurrentPos(j,1)-CurrentPos(i,1));
deltaY=(CurrentPos(j,2)-CurrentPos(i,2));
deltaZ=(CurrentPos(j,3)-CurrentPos(i,3));
ray=sqrt((deltaX^2)+(deltaY^2)+(deltaZ^2));
dy(4) = dy(4) + G*m(i)*(deltaX/(ray^3));
dy(5) = dy(5) + G*m(i)*(deltaY/(ray^3));
dy(6) = dy(6) + G*m(i)*(deltaZ/(ray^3));
end
end
where the 'm' array is the planet masses.
then I used the numerical method Runge-Kutta-4 to solve it, and here's the code:
function [y,t]=RK4(F,intPos,a,b,N)
h=(b-a)/N;
t=zeros(N,1);
y = zeros(10*N,6);
y(1,:)=intPos(1,:);
y(2,:)=intPos(2,:);
y(3,:)=intPos(3,:);
y(4,:)=intPos(4,:);
y(5,:)=intPos(5,:);
y(6,:)=intPos(6,:);
y(7,:)=intPos(7,:);
y(8,:)=intPos(8,:);
y(9,:)=intPos(9,:);
y(10,:)=intPos(10,:);
t(1)=a;
for i=1:N
t(i+1)=a+i*h;
CurrentPos=y((i*10)-9:i*10,:);
% CurrentPos(1,:)=intPos(1,:);
y((i*10)+1,:)=intPos(1,:);
for j=2:10
k1=F(t(i),y(((i-1)*10)+j,:),CurrentPos,j);
k2=F(t(i)+h/2,y(((i-1)*10)+j,:)+(h/2).*k1',CurrentPos,j);
k3=F(t(i)+h/2,y(((i-1)*10)+j,:)+(h/2).*k2',CurrentPos,j);
k4=F(t(i)+h,y(((i-1)*10)+j,:)+h.*k3',CurrentPos,j);
y((i*10)+j,:)=y(((i-1)*10)+j,:)+(h/6)*(k1+2*k2+2*k3+k4)';
end
end
Finally applied the function for the Initial States from JPL HORIZONS System:
format short
intPos=zeros(10,6);
intPos(1,:)=[1.81899E+08 9.83630E+08 -1.58778E+07 -1.12474E+01 7.54876E+00 2.68723E-01];
intPos(2,:)=[-5.67576E+10 -2.73592E+10 2.89173E+09 1.16497E+04 -4.14793E+04 -4.45952E+03];
intPos(3,:)=[4.28480E+10 1.00073E+11 -1.11872E+09 -3.22930E+04 1.36960E+04 2.05091E+03];
intPos(4,:)=[-1.43778E+11 -4.00067E+10 -1.38875E+07 7.65151E+03 -2.87514E+04 2.08354E+00];
intPos(5,:)=[-1.14746E+11 -1.96294E+11 -1.32908E+09 2.18369E+04 -1.01132E+04 -7.47957E+02];
intPos(6,:)=[-5.66899E+11 -5.77495E+11 1.50755E+10 9.16793E+03 -8.53244E+03 -1.69767E+02];
intPos(7,:)=[8.20513E+10 -1.50241E+12 2.28565E+10 9.11312E+03 4.96372E+02 -3.71643E+02];
intPos(8,:)=[2.62506E+12 1.40273E+12 -2.87982E+10 -3.25937E+03 5.68878E+03 6.32569E+01];
intPos(9,:)=[4.30300E+12 -1.24223E+12 -7.35857E+10 1.47132E+03 5.25363E+03 -1.42701E+02];
intPos(10,:)=[1.65554E+12 -4.73503E+12 2.77962E+10 5.24541E+03 6.38510E+02 -1.60709E+03];
[yy,t]=RK4(#F,intPos,0,1e8,1e3);
x=zeros(101,1);
y=zeros(101,1);
z=zeros(101,1);
for i=1:1e3
x(i,:)=yy((i-1)*10+4,1);
y(i,:)=yy((i-1)*10+4,2);
z(i,:)=yy((i-1)*10+4,3);
end
plot3(x,y,z)
Finally, the result wasn't satisfying at all and I got many 'NAN', then I did some adjustment on the RK4 method and started to get numbers, but when I plotted them it turned out I'm plotting a line instead of an orbit.
Any help would be appreciated.
Thanks in advance.
Two errors: One physical: The alpha in the formula is the j in the code, the running index j in the formulas is the loop index i in the formula. In total this makes a sign error, transforming the attracting gravity force into a repelling force like between electrons. Thus the physics dictates that the bodies move away from each other almost linearly, as long as their paths don't cross.
Second, you are applying the RK4 method in such a way that in total it is an order 1 method. These also tend to behave un-physically rather quickly. You need to update first all positions to the first stage in a temporary StagePos variable, then use that to compute all position updates for the second stage etc. The difference to the current implementation may be small in each step, but such systematic errors quickly sum up.
Hi there I’ve been working on trying to plot the convergence of 'e' through using the equation 1/N! with limits from 0 to 9.
clc,clear
terms=[1];
x=10;
for i=2:x
terms(i,1)=terms(i-1,1) + 1/factorial(i);
end
disp(terms)
xplotrange = 0:9;
plot(xplotrange,terms,'b-')
With the code I intened to plot the number of terms in the 'x' axis and the result of the series in the 'y' axis. But I am confused as to why the array of numbers outputted in the for loop converges at 1.718 instead of 2.718?
As #Daniel stated, Euler's number via Taylor expansion should start from x=0 . Thus you can adjust your code to something like below
terms=[1];
x=10;
for i=2:x
terms(i,1)=terms(i-1,1) + 1/factorial(i-1);
end
disp(terms)
xplotrange = 0:9;
plot(xplotrange,terms,'b-')
or a method using cumsum, e.g.,
terms=[1];
x=10;
terms = cumsum(1./factorial(0:x));
disp(terms)
xplotrange = 0:x;
plot(xplotrange,terms,'b-');
Initializing terms with 1 and starting your for loop at 2, you effectively start at i=1, but the sum has to start at i=0. 1/0! is the 1 you are missing.
I need help finding an integral of a function using trapezoidal sums.
The program should take successive trapezoidal sums with n = 1, 2, 3, ...
subintervals until there are two neighouring values of n that differ by less than a given tolerance. I want at least one FOR loop within a WHILE loop and I don't want to use the trapz function. The program takes four inputs:
f: A function handle for a function of x.
a: A real number.
b: A real number larger than a.
tolerance: A real number that is positive and very small
The problem I have is trying to implement the formula for trapezoidal sums which is
Δx/2[y0 + 2y1 + 2y2 + … + 2yn-1 + yn]
Here is my code, and the area I'm stuck in is the "sum" part within the FOR loop. I'm trying to sum up 2y2 + 2y3....2yn-1 since I already accounted for 2y1. I get an answer, but it isn't as accurate as it should be. For example, I get 6.071717974723753 instead of 6.101605982576467.
Thanks for any help!
function t=trapintegral(f,a,b,tol)
format compact; format long;
syms x;
oldtrap = ((b-a)/2)*(f(a)+f(b));
n = 2;
h = (b-a)/n;
newtrap = (h/2)*(f(a)+(2*f(a+h))+f(b));
while (abs(newtrap-oldtrap)>=tol)
oldtrap = newtrap;
for i=[3:n]
dx = (b-a)/n;
trapezoidsum = (dx/2)*(f(x) + (2*sum(f(a+(3:n-1))))+f(b));
newtrap = trapezoidsum;
end
end
t = newtrap;
end
The reason why this code isn't working is because there are two slight errors in your summation for the trapezoidal rule. What I am precisely referring to is this statement:
trapezoidsum = (dx/2)*(f(x) + (2*sum(f(a+(3:n-1))))+f(b));
Recall the equation for the trapezoidal integration rule:
Source: Wikipedia
For the first error, f(x) should be f(a) as you are including the starting point, and shouldn't be left as symbolic. In fact, you should simply get rid of the syms x statement as it is not useful in your script. a corresponds to x1 by consulting the above equation.
The next error is the second term. You actually need to multiply your index values (3:n-1) by dx. Also, this should actually go from (1:n-1) and I'll explain later. The equation above goes from 2 to N, but for our purposes, we are going to go from 1 to N-1 as you have your code set up like that.
Remember, in the trapezoidal rule, you are subdividing the finite interval into n pieces. The ith piece is defined as:
x_i = a + dx*i; ,
where i goes from 1 up to N-1. Note that this starts at 1 and not 3. The reason why is because the first piece is already taken into account by f(a), and we only count up to N-1 as piece N is accounted by f(b). For the equation, this goes from 2 to N and by modifying the code this way, this is precisely what we are doing in the end.
Therefore, your statement actually needs to be:
trapezoidsum = (dx/2)*(f(a) + (2*sum(f(a+dx*(1:n-1))))+f(b));
Try this and let me know if you get the right answer. FWIW, MATLAB already implements trapezoidal integration by doing trapz as #ADonda already pointed out. However, you need to properly structure what your x and y values are before you set this up. In other words, you would need to set up your dx before hand, then calculate your x points using the x_i equation that I specified above, then use these to generate your y values. You then use trapz to calculate the area. In other words:
dx = (b-a) / n;
x = a + dx*(0:n);
y = f(x);
trapezoidsum = trapz(x,y);
You can use the above code as a reference to see if you are implementing the trapezoidal rule correctly. Your implementation and using the above code should generate the same results. All you have to do is change the value of n, then run this code to generate the approximation of the area for different subdivisions underneath your curve.
Edit - August 17th, 2014
I figured out why your code isn't working. Here are the reasons why:
The for loop is unnecessary. Take a look at the for loop iteration. You have a loop going from i = [3:n] yet you don't reference the i variable at all in your loop. As such, you don't need this at all.
You are not computing successive intervals properly. What you need to do is when you compute the trapezoidal sum for the nth subinterval, you then increment this value of n, then compute the trapezoidal rule again. This value is not being incremented properly in your while loop, which is why your area is never improving.
You need to save the previous area inside the while loop, then when you compute the next area, that's when you determine whether or not the difference between the areas is less than the tolerance. We can also get rid of that code at the beginning that tries and compute the area for n = 2. That's not needed, as we can place this inside your while loop. As such, this is what your code should look like:
function t=trapintegral(f,a,b,tol)
format long; %// Got rid of format compact. Useless
%// n starts at 2 - Also removed syms x - Useless statement
n = 2;
newtrap = ((b-a)/2)*(f(a) + f(b)); %// Initialize
oldtrap = 0; %// Initialize to 0
while (abs(newtrap-oldtrap)>=tol)
oldtrap = newtrap; %//Save the old area from the previous iteration
dx = (b-a)/n; %//Compute width
%//Determine sum
trapezoidsum = (dx/2)*(f(a) + (2*sum(f(a+dx*(1:n-1))))+f(b));
newtrap = trapezoidsum; % //This is the new sum
n = n + 1; % //Go to the next value of n
end
t = newtrap;
end
By running your code, this is what I get:
trapezoidsum = trapintegral(#(x) (x+x.^2).^(1/3),1,4,0.00001)
trapezoidsum =
6.111776299189033
Caveat
Look at the way I defined your function. You must use element-by-element operations as the sum command inside the loop will be vectorized. Take a look at the ^ operations specifically. You need to prepend a dot to the operations. Once you do this, I get the right answer.
Edit #2 - August 18th, 2014
You said you want at least one for loop. This is highly inefficient, and whoever specified having one for loop in the code really doesn't know how MATLAB works. Nevertheless, you can use the for loop to accumulate the sum term. As such:
function t=trapintegral(f,a,b,tol)
format long; %// Got rid of format compact. Useless
%// n starts at 3 - Also removed syms x - Useless statement
n = 3;
%// Compute for n = 2 first, then proceed if we don't get a better
%// difference tolerance
newtrap = ((b-a)/2)*(f(a) + f(b)); %// Initialize
oldtrap = 0; %// Initialize to 0
while (abs(newtrap-oldtrap)>=tol)
oldtrap = newtrap; %//Save the old area from the previous iteration
dx = (b-a)/n; %//Compute width
%//Determine sum
%// Initialize
trapezoidsum = (dx/2)*(f(a) + f(b));
%// Accumulate sum terms
%// Note that we multiply each term by (dx/2), but because of the
%// factor of 2 for each of these terms, these cancel and we thus have dx
for n2 = 1 : n-1
trapezoidsum = trapezoidsum + dx*f(a + dx*n2);
end
newtrap = trapezoidsum; % //This is the new sum
n = n + 1; % //Go to the next value of n
end
t = newtrap;
end
Good luck!
i would like to understand one thing and please help me to clarify it,sometimes it is necessary to represent given data by sum of complex exponentials with additive white noise,let us consider following model using sinusoidal model
clear all;
A1=24;
A2=23;
A3=23;
A4=23;
A5=10;
f1=11.01;
f2=11.005;
f3= 10;
f4=10.9
phi=2*pi*(rand(1,4)-0.5);
t=0:0.01:2.93;
k=1:1:294;
x=rand([1,length(t)]);
y(k)=A1.*sin(2*pi*f1*t+phi(1))+A2.*cos(2*pi*f2*t+phi(2))+A3.*sin(2*pi*f3*t+phi(3))+A4.*cos(2*pi*f4*t+phi(4))+A5.*x;
[pxx,f]=periodogram(y,[],[],100);
plot(f,pxx)
there phases are distributed uniformly in range of [-pi pi],but my main question is related following fact.to represent data as linear combination of complex exponentials with phases uniformly distributed in [-pi pi] interval, should we generate these phases outside of sampling or at each sampling process we should generate new list of phases?please help me to clarify this things
As I stated in the my comment, I don't really understand what you're asking. But, I will answer this as if you had asked it on codereview.
The following is not good practice in MATLAB:
A1=24;
A2=23;
A3=23;
A4=23;
A5=10;
There are very few cases (if any), where you actually need such variable names. Instead, the following would be much better:
A = [24 23 23 23 10];
Now, if you want to use A1, you do A(1) instead.
These two lines:
t=0:0.01:2.93;
k=1:1:294;
They are of course the same size (1x294), but when you do it that way, it's easy to get it wrong. You will of course get errors later on if they're not the same size, so it's nice to make sure that you have it correct on the first try, thus using linspace might be a good idea. The following line will give you the same t as the line above. This way it's easier to be sure you have exactly 294 elements, not 293, 295 or 2940 (it is sometimes easy to miss).
t = linspace(0,2.93,294);
Not really important, but k = 1:1:294 can be simplified to k = 1:294, as the default step size is 1.
The syntax .*, is used for element-wise operations. That is, if you want to multiply each element of a vector (or matrix) with the corresponding element in another one. Using it when multiplying vectors with scalars is therefore unnecessary, * is enough.
Again, not an important point, but x=rand([1,length(t)]); is simpler written x=rand(1, length(t)); (without brackets).
You don't need the index k in y(k) = ..., as k is continuous, starting at 1, with increments of 1. This is the default behavior in MATLAB, thus y = ... is enough. If, however, you only wanted to fill in every other number between 1 and 100, you could do y(1:2:100).
This is far from perfect, but in my opinion big step in the right direction:
A = [24 23 23 23 10];
f = [11.01 11.005 10 10.9]; % You might want to use , as a separator here
phi = 2*pi*(rand(1,4)-0.5);
t = linspace(0,2.93,294);
x = rand(1, length(t));
w = 2*pi*f; % For simplicity
y = A(1)*sin(w(1)*t+phi(1)) + A(2)*cos(w(2)*t+phi(2)) + ...
A(3)*sin(w(3)*t+phi(3)) + A(4)*cos(w(4)*t+phi(4))+A(5)*x;
Another option would be:
z = [sin(w(1)*t+phi(1)); cos(w(2)*t+phi(2)); sin(w(3)*t+phi(3)); ...
cos(w(4)*t+phi(4)); x];
y = A.*z;
This will give you the same y as the first one. Having the same w, t and phi as above, the following will also give you the same results:
c = bsxfun(#times,w,t') + kron(phi,ones(294,1));
y = sum(bsxfun(#times,A,[sin(c(:,1)), cos(c(:,2)), sin(c(:,3)), cos(c(:,4)), x']),2)';
I hope something in here might help you some in your further work. And maybe I actually answered your question. =)
I need to produce a signal x=-2*cos(100*pi*n)+2*cos(140*pi*n)+cos(200*pi*n)
So I put it like this :
N=1024;
for n=1:N
x=-2*cos(100*pi*n)+2*cos(140*pi*n)+cos(200*pi*n);
end
But What I get is that the result keeps giving out 1
I tried to test each values according to each n, and I get the same results for any n
For example -2*cos(100*pi*n) with n=1 has to be -1.393310473. Instead of that, Matlab gave the result -2 for it and it always gave -2 for any n
I don't know how to fix it, so I hope someone could help me out! Thank you!
Not sure where you get the idea that -2*cos(100*pi) should be anything other than -2. Maybe you are not aware that Matlab works in radians?
Look at your expression. Each term can be factored to contain 2*pi*(an integer). And you should know that cos(2*pi*(an integer)) = 1.
So the results are exactly as expected.
What you are seeing is basically what happens when you under-sample a waveform. You may know that the Nyquist criterion says that you need to have a sampling rate that is at least two times greater than the highest frequency component present; but in your case, you are sampling one point every 50, 70, 100 complete cycles. So you are "far beyond Nyquist". And that can only be solved by sampling more closely.
For example, you could do:
t = linspace(0, 1, 1024); % sample the waveform 1024 times between 0 and 1
f1 = 50;
f2 = 70;
f3 = 100;
signal = -2*cos(2*pi*f1*t) + 2*cos(2*pi*f2*t) + cos(2*pi*f3*t);
figure; plot(t, signal)
I think you are using degrees when you are doing your calculations, so do this:
n = 1:1024
x=-2*cosd(100*pi*n)+2*cosd(140*pi*n)+cosd(200*pi*n);
cosd uses degrees instead of radians. Radians is the default for cos so matlab has a separate function when degree input is used. For me this gave:
-2*cosd(100*pi*1) = -1.3933
The first term that I got using:
x=-2*cosd(100*pi*1)+2*cosd(140*pi*1)+cosd(200*pi*1)
x = -1.0693
Also notice that I defined n as n = 1:1024; this will give all integers from 1,2,...,1024,
there is no need to use a for loop since many of Matlab's built in functions are vectorized. Meaning you can just input a vector and it will calculate the function for every element in the vector.