MATLAB Discretizing Sine Function with +/- - matlab

Hello I am relatively new to MATLAB and have received and assignment in which we could use any programming language. I would like to continue MATLAB and have decided to use it for this assignment. The questions has to do with the following formula:
x(t) = A[1+a1*E(t)]*sin{w[1+a2*E(t)]*t+y}(+/-)a3*E(t)
The first question we have is to develop an appropriate discretization of x(t) with a time step h. I think i understand how to do this using step but because there is a +/- in the end I am running into errors. Here is what I have (I have simplified the equation by assigning arbitrary values to each variable):
A = 1;
E = 1;
a1 = 1;
a2 = 2;
a3 = 3;
w = 1;
y = 0;
% ts = .1;
% t = 0:ts:10;
t = 1:1:10;
x1(t) = A*(1+a1*E)*sin(w*(1+a2*E)*t+y);
x2(t) = a3*E;
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
plot(y)
The problem is I keep getting the following error because of the +/-:
In an assignment A(I) = B, the number of elements in B and I must be the same.
Error in Try1 (line 21)
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
Any help?? Thanks!


You can remove the (t) from the left-hand side of all three assignments.
y = [x1+x2, x1-x2]
MATLAB knows what to do with vectors and matrices.
Or, if you want to write it out the long way, tell MATLAB there will be two columns:
y(t, 1:2) = [x1(t)'+x2(t)', x1(t)'-x2(t)']
or two rows:
y(1:2, t) = [x1(t)+x2(t); x1(t)-x2(t)]
But this won't work when you have fractional values of t. The value in parentheses is required to be the index, not a dependent variable. If you want the whole vector, just leave it out.

Related

SIR model using fsolve and Euler 3BDF

Hi i've been asked to solve SIR model using fsolve command in MATLAB, and Euler 3 point backward. I'm really confused on how to proceed, please help. This is what i have so far. I created a function for 3BDF scheme but i'm not sure how to proceed with fsolve and solve the system of nonlinear ODEs. The SIR model is shown as and 3BDF scheme is formulated as
clc
clear all
gamma=1/7;
beta=1/3;
ode1= #(R,S,I) -(beta*I*S)/(S+I+R);
ode2= #(R,S,I) (beta*I*S)/(S+I+R)-I*gamma;
ode3= #(I) gamma*I;
f(t,[S,I,R]) = [-(beta*I*S)/(S+I+R); (beta*I*S)/(S+I+R)-I*gamma; gamma*I];
R0=0;
I0=10;
S0=8e6;
odes={ode1;ode2;ode3}
fun = #root2d;
x0 = [0,0];
x = fsolve(fun,x0)
function [xs,yb] = ThreePointBDF(f,x0, xmax, h, y0)
% This function should return the numerical solution of y at x = xmax.
% (It should not return the entire time history of y.)
% TO BE COMPLETED
xs=x0:h:xmax;
y=zeros(1,length(xs));
y(1)=y0;
yb(1)=y0+f(x0,y0)*h;
for i=1:length(xs)-1
R =R0;
y1(i+1,:) = fsolve(#(u) u-2*h/3*f(t(i+1),u) - R, y1(i-1,:)+2*h*F(i,:))
S = S0;
y2(i+1,:) = fsolve(#(u) u-2*h/3*f(t(i+1),u) - S, y2(i-1,:)+2*h*F(i,:))
I= I0;
y3(i+1,:) = fsolve(#(u) u-2*h/3*f(t(i+1),u) - I, y3(i-1,:)+2*h*F(i,:))
end
end

You have an implicit equation
y(i+1) - 2*h/3*f(t(i+1),y(i+1)) = G = (4*y(i) - y(i-1))/3
where the right-side term G is constant in the call to fsolve, that is, during the solution of the implicit step equation.
Note that this is for the vector valued system y'(t)=f(t,y(t)) where
f(t,[S,I,R]) = [-(beta*I*S)/(S+I+R); (beta*I*S)/(S+I+R)-I*gamma; gamma*I];
To solve this write
G = (4*y(i,:) - y(i-1,:))/3
y(i+1,:) = fsolve(#(u) u-2*h/3*f(t(i+1),u) - G, y(i-1,:)+2*h*F(i,:))
where a midpoint step is used to get an order 2 approximation as initial guess, F(i,:)=f(t(i),y(i,:)). Add solver options for error tolerances as necessary, you want the error in the implicit equation smaller than the truncation error O(h^3) of the step. One can also keep only a short array of function values, then one has to be careful for the correspondence of the position in the short array to the time index.
Using all that and a reference solution by a higher order standard solver produces the following error graphs for the components
where one can see that the first order error of the constant first step results in a first order global error, while with a second order error in the first step using the Euler method results in a clear second order global error.
Implement the method in general terms
from scipy.optimize import fsolve
def BDF2(f,t,y0,y1):
N, h = len(t)-1, t[1]-t[0];
y = (N+1)*[np.asarray(y0)];
y[1] = y1;
for i in range(1,N):
t1, G = t[i+1], (4*y[i]-y[i-1])/3
y[i+1] = fsolve(lambda u: u-2*h/3*f(t1,u)-G, y[i-1]+2*h*f(t[i],y[i]), xtol=1e-3*h**3)
return np.vstack(y)
Set up the model to be solved
gamma=1/7;
beta=1/3;
print beta, gamma
y0 = np.array([8e6, 10, 0])
P = sum(y0); y0 = y0/P
def f(t,y): S,I,R = y; trns = beta*S*I/(S+I+R); recv=gamma*I; return np.array([-trns, trns-recv, recv])
Compute a reference solution and method solutions for the two initialization variants
from scipy.integrate import odeint
tg = np.linspace(0,120,25*128)
yg = odeint(f,y0,tg,atol=1e-12, rtol=1e-14, tfirst=True)
M = 16; # 8,4
t = tg[::M];
h = t[1]-t[0];
y1 = BDF2(f,t,y0,y0)
e1 = y1-yg[::M]
y2 = BDF2(f,t,y0,y0+h*f(0,y0))
e2 = y2-yg[::M]
Plot the errors, computation as above, but embedded in the plot commands, could be separated in principle by first computing a list of solutions
fig,ax = plt.subplots(3,2,figsize=(12,6))
for M in [16, 8, 4]:
t = tg[::M];
h = t[1]-t[0];
y = BDF2(f,t,y0,y0)
e = (y-yg[::M])
for k in range(3): ax[k,0].plot(t,e[:,k],'-o', ms=1, lw=0.5, label = "h=%.3f"%h)
y = BDF2(f,t,y0,y0+h*f(0,y0))
e = (y-yg[::M])
for k in range(3): ax[k,1].plot(t,e[:,k],'-o', ms=1, lw=0.5, label = "h=%.3f"%h)
for k in range(3):
for j in range(2): ax[k,j].set_ylabel(["$e_S$","$e_I$","$e_R$"][k]); ax[k,j].legend(); ax[k,j].grid()
ax[0,0].set_title("Errors: first step constant");
ax[0,1].set_title("Errors: first step Euler")

Solve for independent variable between data points in MATLAB

I have many sets of data over the same time period, with a timestep of 300 seconds. Sets that terminate before the end of the observation period (here I've truncated it to 0 to 3000 seconds) have NaNs in the remaining spaces:
x = [0;300;600;900;1200;1500;1800;2100;2400;2700;3000];
y(:,1) = [4.65;3.67;2.92;2.39;2.02;1.67;1.36;1.07;NaN;NaN;NaN];
y(:,2) = [4.65;2.65;2.33;2.18;2.03;1.89;1.75;1.61;1.48;1.36;1.24];
y(:,3) = [4.65;2.73;1.99;1.49;1.05;NaN;NaN;NaN;NaN;NaN;NaN];
I would like to know at what time each dataset would reach the point where y is equal to a specific value, in this case y = 2.5
I first tried finding the nearest y value to 2.5, and then using the associated time, but this isn't very accurate (the dots should all fall on the same horizontal line):
ybreak = 2.5;
for ii = 1:3
[~, index] = min(abs(y(:,ii)-ybreak));
yclosest(ii) = y(index,ii);
xbreak(ii) = x(index);
end
I then tried doing a linear interpolation between data points, and then solving for x at y=2.5, but wasn't able to make this work:
First I removed the NaNs (which it seems like there must be a simpler way of doing?):
for ii = 1:3
NaNs(:,ii) = isnan(y(:,ii));
for jj = 1:length(x);
if NaNs(jj,ii) == 0;
ycopy(jj,ii) = y(jj,ii);
end
end
end
Then tried fitting:
for ii = 1:3
f(ii) = fit(x(1:length(ycopy(:,ii))),ycopy(:,ii),'linearinterp');
end
And get the following error message:
Error using cfit/subsasgn (line 7)
Can't assign to an empty FIT.
When I try fitting outside the loop (for just one dataset), it works fine:
f = fit(x(1:length(ycopy(:,1))),ycopy(:,1),'linearinterp');
f =
Linear interpolant:
f(x) = piecewise polynomial computed from p
Coefficients:
p = coefficient structure
But I then still can't solve f(x)=2.5 to find the time at which y=2.5
syms x;
xbreak = solve(f(x) == 2.5,x);
Error using cfit/subsref>iParenthesesReference (line 45)
Cannot evaluate CFIT model for some reason.
Error in cfit/subsref (line 15)
out = iParenthesesReference( obj, currsubs );
Any advice or thoughts on other approaches to this would be much appreciated. I need to be able to do it for many many datasets, all of which have different numbers of NaN values.

As you mention y=2.5 is not in your data set so the value of x which corresponds to this depends on the interpolation method you use. For linear interpolation, you could use something like the following
x = [0;300;600;900;1200;1500;1800;2100;2400;2700;3000];
y(:,1) = [4.65;3.67;2.92;2.39;2.02;1.67;1.36;1.07;NaN;NaN;NaN];
y(:,2) = [4.65;2.65;2.33;2.18;2.03;1.89;1.75;1.61;1.48;1.36;1.24];
y(:,3) = [4.65;2.73;1.99;1.49;1.05;NaN;NaN;NaN;NaN;NaN;NaN];
N = size(y, 2);
x_interp = NaN(N, 1);
for i = 1:N
idx = find(y(:,i) >= 2.5, 1, 'last');
x_interp(i) = interp1(y(idx:idx+1, i), x(idx:idx+1), 2.5);
end
figure
hold on
plot(x, y)
scatter(x_interp, repmat(2.5, N, 1))
hold off
It's worth keeping in mind that the above code is assuming your data is monotonically decreasing (as your data is), but this solution could be adapted for monotonically increasing as well.

Create a variable number of terms in an anonymous function that outputs a vector

I'd like to create an anonymous function that does something like this:
n = 5;
x = linspace(-4,4,1000);
f = #(x,a,b,n) a(1)*exp(b(1)^2*x.^2) + a(2)*exp(b(2)^2*x.^2) + ... a(n)*exp(b(n)^2*x.^2);
I can do this as such, without passing explicit parameter n:
f1 = #(x,a,b) a(1)*exp(-b(1)^2*x.^2);
for j = 2:n
f1 = #(x,a,b) f1(x,a,b) + a(j)*exp(b(j)^2*x.^2);
end
but it seems, well, kind of hacky. Does someone have a better solution for this? I'd like to know how someone else would treat this.

Your hacky solution is definitely not the best, as recursive function calls in MATLAB are not very efficient, and you can quickly run into the maximum recursion depth (500 by default).
You can introduce a new dimension along which you can sum up your arrays a and b. Assuming that x, a and b are row vectors:
f = #(x,a,b,n) a(1:n)*exp((b(1:n).^2).'*x.^2)
This will use the first dimension as summing dimension: (b(1:n).^2).' is a column vector, which produces a matrix when multiplied by x (this is a dyadic product, to be precise). The resulting n * length(x) matrix can be multiplied by a(1:n), since the latter is a matrix of size [1,n]. This vector-matrix product will also perform the summation for us.
Mini-proof:
n = 5;
x = linspace(-4,4,1000);
a = rand(1,10);
b = rand(1,10);
y = 0;
for k=1:n
y = y + a(k)*exp(b(k)^2*x.^2);
end
y2 = a(1:n)*exp((b(1:n).^2).'*x.^2); %'
all(abs(y-y2))<1e-10
The last command returns 1, so the two are essentially identical.

Numerical Integral in MatLab using integral command

I am trying to compute the value of this integral using Matlab
Here the other parameters have been defined or computed in the earlier part of the program as follows
N = 2;
sigma = [0.01 0.1];
l = [15];
meu = 4*pi*10^(-7);
f = logspace ( 1, 6, 500);
w=2*pi.*f;
for j = 1 : length(f)
q2(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(2));
q1(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(1));
C2(j)= 1/(q2(j));
C1(j)= (q1(j)*C2(j) + tanh(q1(j)*l))/(q1(j)*(1+q1(j)*C2(j)*tanh(q1(j)*l)));
Z(j) = sqrt(-1)*2*pi*f(j)*C1(j);
Apprho(j) = meu*(1/(2*pi*f(j))*(abs(Z(j))^2));
Phi(j) = atan(imag(Z(j))/real(Z(j)));
end
%integration part
c1=w./(2*pi);
rho0=1;
fun = #(x) log(Apprho(x)/rho0)/(x.^2-w^2);
c2= integral(fun,0,Inf);
phin=pi/4-c1.*c2;
I am getting an error like this
could anyone help and tell me where i am going wrong.thanks in advance

Define Apprho in a separate *.m function file, instead of storing it in an array:
function [ result ] = Apprho(x)
%
% Calculate f and Z based on input argument x
%
% ...
%
meu = 4*pi*10^(-7);
result = meu*(1/(2*pi*f)*(abs(Z)^2));
end
How you calculate f and Z is up to you.
MATLAB's integral works by calling the function (in this case, Apprho) repeatedly at many different x values. The x values called by integral don't necessarily correspond to the 1: length(f) values used in your original code, which is why you received errors.

MATLAB Loop error when adding noise function

One of the tasks for my assignment is to add an appropriate noise function to the following equation:
x = A*(1+a1*E)*sin(w*(1+a2*E)*t+y)+ a3*E
We must then plot the noise function vs. time as well as the original function with the noise added. I have asked the professor if a random number generator between -1 and 1 will suffice and he has agreed. I have the following code so far:
t = 0:0.1:6.5;
A = 2;
a1 = 2;
a2 = 4;
a3 = 3;
w = 1;
y = 2;
for i=1:length(t)
E(i) = random('unif', -1, 1, 1, 1);
x(i) = A*(1+a1*E(i))*sin(w*(1+a2*E(i))*t+y)+ a3*E(i);
i=i+1;
end
plot(t,E)
figure
stem(t,x)
I keep getting the following error In an assignment A(I) = B, the number of elements in B and I must
be the same.
Error in Try1 (line 58)
x(i) = A*(1+a1*E(i))*sin(w*(1+a2*E(i))*t+y)+ a3*E(i);
I don't understand the error because the E is just one number.
Any help appreciated!! Thanks!!

You're correct that E is just one number, but you're multiplying by t, which is not just one number - it's an array. I think you meant to multiply by t(i).
here------v
x(i) = A*(1+a1*E(i))*sin(w*(1+a2*E(i))*t(i)+y)+ a3*E(i);