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From: Gwen Farrow on 5 Aug 2010 18:16
Hi everyone, I'm new here, and a begginer at matlab; I am working in a project using the ZAD technique to control a boost converter. I've already figured out what are the equations for my problem, and I have put them into a function that calculates what i need; this function is presented below:

function X=switchinboostzad(xi,A1,A2,B,K,T,gamma,x1ref,x2ref)
Xref=[x1ref; x2ref],
s=K*(xi-Xref);
s1=K*(A1*xi+B);
s2=K*(A2*xi+B);
d=((2*s+T*s2)/(s2-s1))/T;
if 0<d & d<1
phi1=expm(A1*d/2*T);
psi1=B*d/2*T;
Y0=phi1*xi+psi1;
phi2=expm(A2*(T-d*T));
psi2=A2\(expm(A2*(T-d*T))-eye(2))*B;
Z0=phi2*Y0+psi2;
X=phi1*Z0+psi1;
elseif d<=0
d=0;
phi2=expm(A2*T);
psi2=A2\(expm(A2*T)-eye(2))*B;
X=phi2*xi+psi2;
elseif d>=1
d=1;
phi1=expm(A1*T);
psi1=B*T;
X=phi1*xi+psi1;
end

So, starting from an xi point (as a column vector), This function gives the value of X after passing through the control action: computes a duty cycle d for the pwm centered signal, and depending of this value computes the next x after a period of the pwmc signal has passed, using what is called a Poincare Map.
So, this is f(xi)=X
The other parameters such as A1,gamma etc., are constants:

gamma=.35; T=0.18; k2=-1; x1ref=2.5; x2ref=(x1ref)^(2)*gamma;
A1= [-gamma 0; 0 0]; B=[0;1]; A2= [-gamma 1;-1 0];
k1 can take any value between 0.1 and 1.

The value I took for xi was [x1ref;x2ref], and X is delivered in a column vector form too.

Well, I needed to find out what was the value for xi, that was equal to X, i.e. I need to find the roots for f(xi)=xi i.e. X=xi ; X-xi=0 or xi-X=0 and to do that, I did the next change to the function above:

function out=switchinboostzad(xi,A1,A2,B,K,T,gamma,x1ref,x2ref)
Xref=[x1ref; x2ref],
s=K*(xi-Xref);
s1=K*(A1*xi+B);
s2=K*(A2*xi+B);
d=((2*s+T*s2)/(s2-s1))/T;
if 0<d & d<1
phi1=expm(A1*d/2*T);
psi1=B*d/2*T;
Y0=phi1*xi+psi1;
phi2=expm(A2*(T-d*T));
psi2=A2\(expm(A2*(T-d*T))-eye(2))*B;
Z0=phi2*Y0+psi2;
X=phi1*Z0+psi1;
elseif d<=0
d=0;
phi2=expm(A2*T);
psi2=A2\(expm(A2*T)-eye(2))*B;
X=phi2*xi+psi2;
elseif d>=1
d=1;
phi1=expm(A1*T);
psi1=B*T;
X=phi1*xi+psi1;
end
out=[xi-X];
It is a system of nonlinear equations with 2 variables, so I used the fsolve function to find the zeros of this equation (xi-X=0, where X is a function of xi) from an starting point xi:
Xref=[x1ref; x2ref];
xi=Xref;
[x0,fval,exitflag,output,jacobian]=fsolve(@switchinboostzad,xi,[],A1,A2,B,K,T,gamma,x1ref,x2ref);
I get the next result for x0:
2.4993
2.1873
It seems to work.
Now, I need to know the solution for the second iteration of this function:
f(f(xi))=xi
And the third iteration:
f(f(f(xi)))=xi
and the fourth:
f(f(f(f(xi))))=xi
and the nth:
f(f(f(f(.....f(xi)....))))=xi n-times

(note: this iterations are not the same as the iterations of fsolve.)

And I just dont know how to do it.
I tried to do the next change to the function above:

function [out]=switchinboostzad(xi,A1,A2,B,K,T,gamma,x1ref,x2ref,numo)
Xref=[x1ref; x2ref];
for i=1:numo
if i==1
s=K*(xi-Xref);
s1=K*(A1*xi+B);
s2=K*(A2*xi+B);
d=((2*s+T*s2)/(s2-s1))/T;
if 0<d & d<1
phi1=expm(A1*d/2*T);
psi1=B*d/2*T;
Y0=phi1*xi+psi1;
phi2=expm(A2*(T-d*T));
psi2=A2\(expm(A2*(T-d*T))-eye(2))*B;
Z0=phi2*Y0+psi2;
X=phi1*Z0+psi1;
elseif d<=0
d=0;
phi2=expm(A2*T);
psi2=A2\(expm(A2*T)-eye(2))*B;
X=phi2*xi+psi2;
elseif d>=1
d=1;
phi1=expm(A1*T);
psi1=B*T;
X=phi1*xi+psi1;
end
else
s=K*(X-Xref);
s1=K*(A1*X+B);
s2=K*(A2*X+B);
d=((2*s+T*s2)/(s2-s1))/T;
if 0<d & d<1
phi1=expm(A1*d/2*T);
psi1=B*d/2*T;
Y0=phi1*X+psi1;
phi2=expm(A2*(T-d*T));
psi2=A2\(expm(A2*(T-d*T))-eye(2))*B;
Z0=phi2*Y0+psi2;
X=phi1*Z0+psi1;
elseif d<=0
d=0;
phi2=expm(A2*T);
psi2=A2\(expm(A2*T)-eye(2))*B;
X=phi2*X+psi2;
elseif d>=1
d=1;
phi1=expm(A1*T);
psi1=B*T;
X=phi1*X+psi1;
end
end
end
out=[xi-X];

where numo is the number of iterations of the function, and in the script, it goes like this:
options = optimset('Jacobian','off');
[x0,F,exitflag,output,JAC]=fsolve(@switchinboostzad,xi,options,A1,A2,B,K,T,gamma,x1ref,x2ref,numo);

but it gives the same outcome for x0 for any value of numo, although it is a bit different with format long:

for numo=1 x0=
2.499308123908802
2.187384146706227
for numo=2 x0=
2.499308014143268
2.187384088662747
for numo=3 x0=
2.499308036596760
2.187384135574587
for numo=4 x0=
2.499308026058130
2.187384092188981
for numo=10 x0=
2.499308049304966
2.187384099869770
And I'm getting very simmilar results.
The questions are: ¿Am I using this function fsolve the wrong way?
¿It is ok to iterate the function switchinboostzad this way (with a for)?
¿What happens if I want to compute the Jacobian for the Poincare map of this function and his iterations; it is ok to do it with the default function in matlab?
If I should not use fsolve this way, or cannot iterate the poincare map the way I have done it, ¿Is there any way to use fsolve to achieve this? or ¿What function or process or algorithm should I use?
¿if I use 'TolFun',1e-17 in the options of fsolve, is there a chance to get more suitable results?
I appreciate any help that anyone can give to me.
From: Alan Weiss on 6 Aug 2010 12:24
On 8/5/2010 6:13 PM, Gwen Farrow wrote:
> Hi everyone, I'm new here, and a begginer at matlab; I am working in a
> project using the ZAD technique to control a boost converter. I've
> already figured out what are the equations for my problem, and I have
> put them into a function that calculates what i need; this function is
> presented below:
>
***SNIP***
>
> Well, I needed to find out what was the value for xi, that was equal to
> X, i.e. I need to find the roots for f(xi)=xi i.e. X=xi ; X-xi=0 or
> xi-X=0 and to do that, I did the next change to the function above:
>
***SNIP***
> I get the next result for x0:
> 2.4993
> 2.1873
> It seems to work.
> Now, I need to know the solution for the second iteration of this function:
> f(f(xi))=xi
> And the third iteration:
> f(f(f(xi)))=xi
> and the fourth:
> f(f(f(f(xi))))=xi
> and the nth:
> f(f(f(f(.....f(xi)....))))=xi n-times
>
> (note: this iterations are not the same as the iterations of fsolve.)
>
> And I just dont know how to do it.

The solution to your problem f(x) = x is also the solution to all the
problems
f(f(x)) = x
f(f(f(x))) = x
etc.

Why? Suppose you have a point x such that f(x) = x. Then f(f(x)) = f(x)
(by substituting x for the value of f(x)) = x.
etc.

Alan Weiss
MATLAB mathematical toolbox documentation
From: Gwen Farrow on 6 Aug 2010 16:45
Hi, Thanks for your reply, but still there is something:

Alan Weiss <aweiss(a)mathworks.com> wrote in message <i3hd0d$ptq$1(a)fred.mathworks.com>...
> On 8/5/2010 6:13 PM, Gwen Farrow wrote:
> > Hi everyone, I'm new here, and a begginer at matlab; I am working in a
> > project using the ZAD technique to control a boost converter. I've
> > already figured out what are the equations for my problem, and I have
> > put them into a function that calculates what i need; this function is
> > presented below:
> >
> ***SNIP***
> >
> > Well, I needed to find out what was the value for xi, that was equal to
> > X, i.e. I need to find the roots for f(xi)=xi i.e. X=xi ; X-xi=0 or
> > xi-X=0 and to do that, I did the next change to the function above:
> >
> ***SNIP***
> > I get the next result for x0:
> > 2.4993
> > 2.1873
> > It seems to work.
> > Now, I need to know the solution for the second iteration of this function:
> > f(f(xi))=xi
> > And the third iteration:
> > f(f(f(xi)))=xi
> > and the fourth:
> > f(f(f(f(xi))))=xi
> > and the nth:
> > f(f(f(f(.....f(xi)....))))=xi n-times
> >
> > (note: this iterations are not the same as the iterations of fsolve.)
> >
> > And I just dont know how to do it.
>
> The solution to your problem f(x) = x is also the solution to all the
> problems
> f(f(x)) = x
> f(f(f(x))) = x
> etc.
>
> Why? Suppose you have a point x such that f(x) = x. Then f(f(x)) = f(x)
> (by substituting x for the value of f(x)) = x.
> etc.

when I have:
f(f(x))=x
f(f(x)) - x = 0
The fixed points of this function are more than the function f(x), see:
f(x)=f1(x) ;
f(f(x))=f2(x);
and
f1(x) is different from f2(x)

Then I have , for example:
f(x)=f1(x)=x^2
f1(f1(x))=f2(x)=(x^2)^2=x^4

if I do the next:
solve('x^2-x','x')
I get 0 and 1 as the fixed points.
If I do:
solve('x^4-x','x')
I get:
ans =
0
1
-1/2+1/2*i*3^(1/2)
-1/2-1/2*i*3^(1/2)
So, is not only 0 and 1 the solutions, there are more fixed points; so how can I obtain all the solutions without using symbolic variables nor solve, but instead using fsolve?
Tnx
> Alan Weiss
> MATLAB mathematical toolbox documentation
From: Alan Weiss on 9 Aug 2010 07:34
If you think there are more real-valued solutions for your iterated
function, the way to find them is to start fsolve from many different
initial values. For a theoretical discussion with a few suggestions on
practical steps, see
http://www.mathworks.com/access/helpdesk/help/toolbox/optim/ug/br44i40.html#brhkghv-65

Alan Weiss
MATLAB mathematical toolbox documentation

On 8/6/2010 4:45 PM, Gwen Farrow wrote:
> Hi, Thanks for your reply, but still there is something:
>
> Alan Weiss <aweiss(a)mathworks.com> wrote in message
> <i3hd0d$ptq$1(a)fred.mathworks.com>...
>> On 8/5/2010 6:13 PM, Gwen Farrow wrote:
>> > Hi everyone, I'm new here, and a begginer at matlab; I am working in a
>> > project using the ZAD technique to control a boost converter. I've
>> > already figured out what are the equations for my problem, and I have
>> > put them into a function that calculates what i need; this function is
>> > presented below:
>> >
>> ***SNIP***
>> >
>> > Well, I needed to find out what was the value for xi, that was equal to
>> > X, i.e. I need to find the roots for f(xi)=xi i.e. X=xi ; X-xi=0 or
>> > xi-X=0 and to do that, I did the next change to the function above:
>> >
>> ***SNIP***
>> > I get the next result for x0:
>> > 2.4993
>> > 2.1873
>> > It seems to work.
>> > Now, I need to know the solution for the second iteration of this
>> function:
>> > f(f(xi))=xi
>> > And the third iteration:
>> > f(f(f(xi)))=xi
>> > and the fourth:
>> > f(f(f(f(xi))))=xi
>> > and the nth:
>> > f(f(f(f(.....f(xi)....))))=xi n-times
>> >
>> > (note: this iterations are not the same as the iterations of fsolve.)
>> >
>> > And I just dont know how to do it.
>>
>> The solution to your problem f(x) = x is also the solution to all the
>> problems
>> f(f(x)) = x
>> f(f(f(x))) = x
>> etc.
>>
>> Why? Suppose you have a point x such that f(x) = x. Then f(f(x)) =
>> f(x) (by substituting x for the value of f(x)) = x.
>> etc.
>
> when I have:
> f(f(x))=x
> f(f(x)) - x = 0
> The fixed points of this function are more than the function f(x), see:
> f(x)=f1(x) ; f(f(x))=f2(x);
> and f1(x) is different from f2(x)
>
> Then I have , for example: f(x)=f1(x)=x^2
> f1(f1(x))=f2(x)=(x^2)^2=x^4
>
> if I do the next:
> solve('x^2-x','x')
> I get 0 and 1 as the fixed points.
> If I do:
> solve('x^4-x','x')
> I get:
> ans =
> 0
> 1
> -1/2+1/2*i*3^(1/2)
> -1/2-1/2*i*3^(1/2)
> So, is not only 0 and 1 the solutions, there are more fixed points; so
> how can I obtain all the solutions without using symbolic variables nor
> solve, but instead using fsolve?
> Tnx
>> Alan Weiss
>> MATLAB mathematical toolbox documentation

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