i want to know how to convert ARMA (Autoregression moving average) process to AR(Autoregression) process by PARAMETRIC METHOD.
i.e. i have a transfer function H(z) = (a + b*z)/(c +d*z) e.g. H(z) = (0.26 + 0.073*z^-1)/(1 - z^-1) i.e. Autoregression(ARMA)
and i want to convert it to AR process i.e. H(z) = 1/(p + q*z + r*z^2 + ...)(i.e only pole system).
Kindly Give some hints.
Thanks in advance!
if you are looking for partial partial fraction expansion commands, I suggest you try
[r, p, k] = residue(b,a)
http://www.mathworks.com/help/matlab/ref/residue.html
Related
I have this function:
x[n] = (1/2) ^ n * u[n] + (-1/3) ^ n * u[n]
I need to do two things with this using MATLAB:
Find it's z-transform.
Plot it's poles and zeros.
I am using the following code:
syms n;
f = (1/2)^n + (-1/3)^n;
F = ztrans(f);
I get the z-transform in the F variable, but I can't see how to create it's pole-zero plot. I am using the built-in function pzmap (pzmap(F);), but it doesn't seem to work with the output of ztrans(f).
What am I doing wrong? Do I need to change the z-transform into some other form like like a transfer function model or a zero-pole gain model? If so, can someone explain how that can be done using the output of ztrans(f)?
The first bit of code you gave uses symbolic math to solve for the z-transform. You'll need to convert the output to a discrete-time model supported by the Control System toolbox.
syms n;
f = (1/2)^n + (-1/3)^n;
F = ztrans(f)
returns z/(z - 1/2) + z/(z + 1/3). You can optionally use collect to convert this
F2 = collect(F)
to (12*z^2 - z)/(6*z^2 - z - 1). Then you'll want to find the coefficients of the polynomials in the numerator and denominator and create a discrete-time transfer function object with tf for a particular sampling period:
[num,den] = numden(F2);
Ts = 0.1; % Sampling period
H = tf(sym2poly(num),sym2poly(den),Ts)
Then pzmap(H) will produce a plot like this:
I'm trying to calculate the phase2 angle/value in the y2 equation of a signal given at a specific frequency if I know the other values. Is this possible? Example below: along with picture example:
y1=A1*cos*(2*pi*f1*t+phase1) we know A1,f1,t=1,phase1
y1=0.00720858*cos*(2*pi*6+6.33)
y2=A2*cos*(2*pi*f2*t+phase2) we know A2,f2,t=1, trying to find **phase2**
y2=.4*cos*(2*pi*6.4951+phase2)
y3=A3*cos*(2*pi*f3*t+phase3) we know A3,f3,t=1,phase3
y3=0.0135274*cos(2*pi*7+.786473)
I'm using maxima 13.04.2, octave 3.8.1.
I tried to solve the y2 equation for phase2 in maxima but it got rid of the cos function
kill(all);
A:A; phase:phase; solve(A*cos*(2*pi*t+phase)=0,phase);
the answer came back as phase=-2pi*t
Is this possible? or should I go about this another way?
Thanks
The weird result might stem from the fact that you multiply the cos function with what is supposed to be its argument (by the way, this is mathematically unsound). What you might want is to apply the cos function to the argument. To illustrate what I mean, compare:
A*cos*(2*pi*t+phase)
with:
A*cos(2*pi*t+phase)
On another hand, why not solve the equation pen-on-paper style?
y2 = A2×cos(2πf2t + φ2) ⇒
y2/A2 = cos(2πf2t + φ2) ⇒
arccos(y2/A2) = 2πf2t + φ2 ⇒
arccos(y2/A2) - 2πf2t = φ2
With the values that you provided:
A2 = 0.4, f2 = 6.4951, t = 1.
you can calculate the phase φ2 as function of your level y2 (left as exercise to you).
I plotted the responses of two difference equation obtained from a Z transform transfer function. I used two methods and get two different results. Why?
tf = y(output)/u(input) = z/ (z^2 - 3z + 3)
Method 1, using Matlab, taking the inverse Z transform
tf_difference = iztrans(tf, z, k);
yields: y = 2^k - 1, for timesteps 'k'. This is an exponential.
Method 2, algebraic rearranging by hand, turning z^n into (k-n):
y(k+2) = 3y(k+1) - 3y(k) + u(k+1)
or equivalently
y(k+1) = 3y(k) - 3y(k-1) + u(k)
Also, for the second method, i need to specify a 'u', whereas in the first method that's not needed, only k is needed. Why does the first method not need an input signal u, only a time step?
I'm fitting custom functions to my data.
After obtaining the fit I would like to get something like a function handle of my fit function including the parameters set to the ones found by the fit.
I know I can get the model with
formula(fit)
and I can get the parameters with
coeffvalues(fit)
but is there any easy way to combine the two in one step?
This little loop will do the trick:
x = (1:100).'; %'
y = 1*x.^5 + 2*x.^4 + 3*x.^3 + 4*x.^2 + 5*x + 6;
fitobject = fit(x,y,'poly5');
cvalues = coeffvalues(fitobject);
cnames = coeffnames(fitobject);
output = formula(fitobject);
for ii=1:1:numel(cvalues)
cname = cnames{ii};
cvalue = num2str(cvalues(ii));
output = strrep(output, cname , cvalue);
end
output = 1*x^5 + 2*x^4 + 3*x^3 + 4*x^2 + 5*x + 6
The loop needs to be adapted to the number of coefficients of your fit.
Edit: two slight changes in order to fully answer the question.
fhandle = #(x) eval(output)
returns a function handle. Secondly output as given by your procedure doesn't work, as the power operation reads .^ instead of x, which can obviously be replaced by
strrep(output, '^', '.^');
You can use the Matlab curve fitting function, polyfit.
p = polyfit(x,y,n)
So, p contains the coefficients of the polynomial, x and y are the coordinates of the function you're trying to fit. n is the order of the polynomial. For example, n=1 is linear, n=2 is quadratic, etc. For more info, see this documentation centre link. The only issue is that you may not want a polynomial fit, in which case you'll have to use different method.
Oh, and you can use the calculated coefficients p to to re-evaluate the polynomial with:
f = polyval(p,x);
Here, f is the value of the polynomial with coefficients p evaluated at points x.
I’m currently a Physics student and for several weeks have been compiling data related to ‘Quantum Entanglement’. I’ve now got to a point where I have to plot my data (which should resemble a cos² graph - and does) to a sort of “best fit” cos² graph. The lab script says the following:
A more precise determination of the visibility V (this is basically how 'clean' the data is) follows from the best fit to the measured data using the function:
f(b) = A/2[1-Vsin(b-b(center)/P)]
Granted this probably doesn’t mean much out of context, but essentially A is the amplitude, b is an angle and P is the periodicity. Hence this is also a “wave” like the experimental data I have found.
From this I understand, as previously mentioned, I am making a “best fit” curve. However, I have been told that this isn’t possible with Excel and that the best approach is Matlab.
I know intermediate JavaScript but do not know Matlab and was hoping for some direction.
Is there a tutorial I can read for this? Is it possible for someone to go through it with me? I really have no idea what it entails, so any feed back would be greatly appreciated.
Thanks a lot!
Initial steps
I guess we should begin by getting a representation in Matlab of the function that you're trying to model. A direct translation of your formula looks like this:
function y = targetfunction(A,V,P,bc,b)
y = (A/2) * (1 - V * sin((b-bc) / P));
end
Getting hold of the data
My next step is going to be to generate some data to work with (you'll use your own data, naturally). So here's a function that generates some noisy data. Notice that I've supplied some values for the parameters.
function [y b] = generateData(npoints,noise)
A = 2;
V = 1;
P = 0.7;
bc = 0;
b = 2 * pi * rand(npoints,1);
y = targetfunction(A,V,P,bc,b) + noise * randn(npoints,1);
end
The function rand generates random points on the interval [0,1], and I multiplied those by 2*pi to get points randomly on the interval [0, 2*pi]. I then applied the target function at those points, and added a bit of noise (the function randn generates normally distributed random variables).
Fitting parameters
The most complicated function is the one that fits a model to your data. For this I use the function fminunc, which does unconstrained minimization. The routine looks like this:
function [A V P bc] = bestfit(y,b)
x0(1) = 1; %# A
x0(2) = 1; %# V
x0(3) = 0.5; %# P
x0(4) = 0; %# bc
f = #(x) norm(y - targetfunction(x(1),x(2),x(3),x(4),b));
x = fminunc(f,x0);
A = x(1);
V = x(2);
P = x(3);
bc = x(4);
end
Let's go through line by line. First, I define the function f that I want to minimize. This isn't too hard. To minimize a function in Matlab, it needs to take a single vector as a parameter. Therefore we have to pack our four parameters into a vector, which I do in the first four lines. I used values that are close, but not the same, as the ones that I used to generate the data.
Then I define the function I want to minimize. It takes a single argument x, which it unpacks and feeds to the targetfunction, along with the points b in our dataset. Hopefully these are close to y. We measure how far they are from y by subtracting from y and applying the function norm, which squares every component, adds them up and takes the square root (i.e. it computes the root mean square error).
Then I call fminunc with our function to be minimized, and the initial guess for the parameters. This uses an internal routine to find the closest match for each of the parameters, and returns them in the vector x.
Finally, I unpack the parameters from the vector x.
Putting it all together
We now have all the components we need, so we just want one final function to tie them together. Here it is:
function master
%# Generate some data (you should read in your own data here)
[f b] = generateData(1000,1);
%# Find the best fitting parameters
[A V P bc] = bestfit(f,b);
%# Print them to the screen
fprintf('A = %f\n',A)
fprintf('V = %f\n',V)
fprintf('P = %f\n',P)
fprintf('bc = %f\n',bc)
%# Make plots of the data and the function we have fitted
plot(b,f,'.');
hold on
plot(sort(b),targetfunction(A,V,P,bc,sort(b)),'r','LineWidth',2)
end
If I run this function, I see this being printed to the screen:
>> master
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the function tolerance.
A = 1.991727
V = 0.979819
P = 0.695265
bc = 0.067431
And the following plot appears:
That fit looks good enough to me. Let me know if you have any questions about anything I've done here.
I am a bit surprised as you mention f(a) and your function does not contain an a, but in general, suppose you want to plot f(x) = cos(x)^2
First determine for which values of x you want to make a plot, for example
xmin = 0;
stepsize = 1/100;
xmax = 6.5;
x = xmin:stepsize:xmax;
y = cos(x).^2;
plot(x,y)
However, note that this approach works just as well in excel, you just have to do some work to get your x values and function in the right cells.