while using the lsim command of matlab I found out that the initial condition in my program doesn't affect the simulation's output.
y = lsim(F,input,time,x0);
Where F is a transfer function, and x0 the initial condition that I calculate with the state-space model.
The value of x0 doesn't affect y, I even replaced it with different numbers and the simulation's output y is always the same.
I'm actually trying to get the parameters of the tf from a real measured output. So that's the main part of the code:
tsmpl = 1e-2;
Sizeof_y = length(y_real);
R = zeros(Sizeof_y,3);
R(1,:) = [y_real(1) input(1) 0];
for i=2:Sizeof_y
R(i,:)=[y_real(i-1) input(i) input(i-1)];
end
p = pinv(R)*y_real;
z = tf('z',tsmpl);
num = p(2)*z+p(3);
den = z-p(1);
F = num/den;
sys = ss(F);
x0 = (y_real(1) - (sys.d*sig(1)))*pinv(sys.c);
y = lsim(F,input,time,x0);
y_real is the measured output. It's a vector of complexe numbers.
time is the time vector that represents the duration of the process. (given by the measurment) time = 6:0.01:24
input represents the test signal which is a vector defined like this:
Size_input = length(time);
Size_sine = length(halftime) ; %halftime is the duration of the exitation also known from the measurment
input = zeros(Size_input,1);
input(1:Size_sine) = complex(30);
Vectors y_real, time, and input have the same length.
I would be thankful for any idea :))
Related
I want to find the maximum value using the second derivative of the the expression when x is between 0 and 1. In other words I am taking the derivative of cox(x^2) twice to get the second derivative resulting in - 2*sin(x^2) - 4*x^2*cos(x^2), then I want to evaluate this second derivative at x = 0 to x = 1, and display the maximum value of the populated values.
I have:
syms x
f = cos(x^2);
secondD = diff(diff(f));
for i = 0:1
y = max(secondD(i))
end
Can someone help?
You can do it easily by subs and double:
syms x
f = cos(x^2);
secondD = diff(diff(f));
% instead of the for loop
epsilon = 0.01;
specified_range = 0:epsilon:1;
[max_val, max_ind] = max(double(subs(secondD, specified_range)));
Please note that it is a numerical approach to find the maximum and the returned answer is not completely correct all the time. However, by increasing the epsilon, you can expect a better result in general (again in some cases it is not completely correct).
I'm trying to verify if my implementation of Logistic Regression in Matlab is good. I'm doing so by comparing the results I get via my implementation with the results given by the built-in function mnrfit.
The dataset D,Y that I have is such that each row of D is an observation in R^2 and the labels in Y are either 0 or 1. Thus, D is a matrix of size (n,2), and Y is a vector of size (n,1)
Here's how I do my implementation:
I first normalize my data and augment it to include the offset :
d = 2; %dimension of data
M = mean(D) ;
centered = D-repmat(M,n,1) ;
devs = sqrt(sum(centered.^2)) ;
normalized = centered./repmat(devs,n,1) ;
X = [normalized,ones(n,1)];
I will be doing my calculations on X.
Second, I define the gradient and hessian of the likelihood of Y|X:
function grad = gradient(w)
grad = zeros(1,d+1) ;
for i=1:n
grad = grad + (Y(i)-sigma(w'*X(i,:)'))*X(i,:) ;
end
end
function hess = hessian(w)
hess = zeros(d+1,d+1) ;
for i=1:n
hess = hess - sigma(w'*X(i,:)')*sigma(-w'*X(i,:)')*X(i,:)'*X(i,:) ;
end
end
with sigma being a Matlab function encoding the sigmoid function z-->1/(1+exp(-z)).
Third, I run the Newton algorithm on gradient to find the roots of the gradient of the likelihood. I implemented it myself. It behaves as expected as the norm of the difference between the iterates goes to 0. I wrote it based on this script.
I verified that the gradient at the wOPT returned by my Newton implementation is null:
gradient(wOP)
ans =
1.0e-15 *
0.0139 -0.0021 0.2290
and that the hessian has strictly negative eigenvalues
eig(hessian(wOPT))
ans =
-7.5459
-0.0027
-0.0194
Here's the wOPT I get with my implementation:
wOPT =
-110.8873
28.9114
1.3706
the offset being the last element. In order to plot the decision line, I should convert the slope wOPT(1:2) using M and devs. So I set :
my_offset = wOPT(end);
my_slope = wOPT(1:d)'.*devs + M ;
and I get:
my_slope =
1.0e+03 *
-7.2109 0.8166
my_offset =
1.3706
Now, when I run B=mnrfit(D,Y+1), I get
B =
-1.3496
1.7052
-1.0238
The offset is stored in B(1).
I get very different values. I would like to know what I am doing wrong. I have some doubt about the normalization and 'un-normalization' process. But I'm not sure, may be I'm doing something else wrong.
Additional Info
When I tape :
B=mnrfit(normalized,Y+1)
I get
-1.3706
110.8873
-28.9114
which is a rearranged version of the opposite of my wOPT. It contains exactly the same elements.
It seems likely that my scaling back of the learnt parameters is wrong. Otherwise, it would have given the same as B=mnrfit(D,Y+1)
I'm writing a program in matlab to observe how a function evolves in time. I'd like to set up a matrix that fills its first row with the initial function, and progressively fills more rows based off of a time derivative (that's dependent on the spatial derivative). The function is arbitrary, the program just needs to 'evolve' it. This is what I have so far:
xleft = -10;
xright = 10;
xsampling = 1000;
tmax = 1000;
tsampling = 1000;
dt = tmax/tsampling;
x = linspace(xleft,xright,xsampling);
t = linspace(0,tmax,tsampling);
funset = [exp(-(x.^2)/100);cos(x)]; %Test functions.
funsetvel = zeros(size(funset)); %The functions velocities.
spacetimevalue1 = zeros(length(x), length(t));
spacetimevalue2 = zeros(length(x), length(t));
% Loop that fills the first functions spacetime matrix.
for j=1:length(t)
funsetvel(1,j) = diff(funset(1,:),x,2);
spacetimevalue1(:,j) = funsetvel(1,j)*dt + funset(1,j);
end
This outputs the error, Difference order N must be a positive integer scalar. I'm unsure what this means. I'm fairly new to Matlab. I will exchange the Euler-method for another algorithm once I can actually get some output along the proper expectation. Aside from the error associated with taking the spatial derivative, do you all have any suggestions on how to evaluate this sort of process? Thank you.
I have a MATLAB code that solves a large scale ODE system of following type
function [F] = myfun(t,C,u)
u here is a time dependent vector used as a input in the function myfun. My current code reads as:-
u = zeros(4,1);
u(1) = 0.1;
u(2) = 0.1;
u(3) = 5.01/36;
u(4) = 0.1;
C0 = zeros(15*4*20+12,1);
options = odeset('Mass',Mf,'RelTol',1e-5,'AbsTol',1e-5);
[T,C] = ode15s(#OCFDonecolumnA, [0,5000] ,C0,options,u);
I would like to have the entire time domain split into 5 different sections and have the elements of u take different values at different times(something like a multiple step function). what would be the best way to code it considering that I later want to solve a optimization problem to determine values of u for each time interval in the domain [0,5000].
Thanks
ode15s expects your model function to accept a parameter t for time and a vector x (or C as you named it) of differential state variables. Since ode15s will not let us pass in another vector u of control inputs, I usually wrap the ODE function ffcn inside a model function:
function [t, x] = model(x0, t_seg, u)
idx_seg = 0;
function y = ffcn(t, x)
% simple example of exponential growth
y(1) = u(idx_seg) * x(1)
end
...
end
In the above, the parameters of model are the initial state values x0, the vector of switching times t_seg, and the vector u of control input values in different segments. You can see that idx_seg is visible from within ffcn. This allows us to integrate the model over all segments (replace ... in the above with the following):
t_start = 0;
t = t_start;
x = x0;
while idx_seg < length(t_seg)
idx_seg = idx_seg + 1;
t_end = t_seg(idx_seg);
[t_sol, x_sol] = ode15s(#ffcn, [t_start, t_end], x(end, :));
t = [t; t_sol(2 : end)];
x = [x; x_sol(2 : end, :)];
t_start = t_end;
end
In the first iteration of the loop, t_start is 0 and t_end is the first switching point. We use ode15s now to only integrate over this interval, and our ffcn evaluates the ODE with u(1). The solution y_sol and the corresponding time points t_sol are appended to our overall solution (t, x).
For the next iteration, we set t_start to the end of the current segment and set the new t_end to the next switching point. You can also see that the last element of t_seg must be the time at which the simulation ends. Importantly, we pass to ode15s the current tail y(end, :) of the simulated trajectory as the initial state vector of the next segment.
In summary, the function model will use ode15s to simulate the model segment by segment and return the overall trajectory y and its time points t. You could convince yourself with an example like
[t, x] = model1(1, [4, 6, 12], [0.4, -0.7, 0.3]);
plot(t, x);
which for my exponential growth example should yield
For your optimization runs, you will need to write an objective function. This objective function can pass u to model, and then calculate the merit associated with u by looking at x.
Is there some simple way of calculating of p-value of t-Test in MATLAB.
I found something like it however I think that it does not return correct values:
Pval=2*(1-tcdf(abs(t),n-2))
I want to calculate the p-value for the test that the slope of regression is equal to 0. Therefore I calculate the Standard Error
$SE= \sqrt{\frac{\sum_{s = i-w }^{i+w}{(y_{s}-\widehat{y}s})^2}{(w-2)\sum{s=i-w}^{i+w}{(x_{s}-\bar{x}})^2}}$
where $y_s$ is the value of analyzed parameter in time period $s$,
$\widehat{y}_s$ is the estimated value of the analyzed parameter in time period $s$,
$x_i$ is the time point of the observed value of the analysed parameter,
$\bar{x}$ is the mean of time points from analysed period and then
$t_{score} = (a - a_{0})/SE$ where $a_{0}$ where $a_{0} = 0$.
I checked that p values from ttest function and the one calculated using this formula:
% Let n be your sample size
% Let v be your degrees of freedom
% Then:
pvalues = 2*(1-tcdf(abs(t),n-v))
and they are the same!
Example with Matlab demo dataset:
load accidents
x = hwydata(:,2:3);
y = hwydata(:,4);
stats = regstats(y,x,eye(size(x,2)));
fprintf('T stat using built-in function: \t %.4f\n', stats.tstat.t);
fprintf('P value using built-in function: \t %.4f\n', stats.tstat.pval);
fprintf('\n\n');
n = size(x,1);
v = size(x,2);
b = x\y;
se = diag(sqrt(sumsqr(y-x*b)/(n-v)*inv(x'*x)));
t = b./se;
p = 2*(1-tcdf(abs(t),n-v));
fprintf('T stat using own calculation: \t\t %.4f\n', t);
fprintf('P value using own calculation: \t\t %.4f\n', p);