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bowser
Posted: Thu Jul 06, 2006 11:43 am
Guest
Hello, I have a problem in a non-linear equation analysis.

I have a non-linear equation:
x'' + kx(1+ax^2)+bx' = A*sin(wt)

that I already transformed into the following system
x' = y
y' = -kx(1+ax^2) - by + A*sin(wt)

Now I have to obtain the transfer function, so I suppose the only way
to do it is to first linearize (approximating) the second equation. Do
I have to use Taylor series to linearize around the zero (which I
calculated to be (0,0))? If so, how?

My attempt was to posticipate the "linearizing": I let a particular
solution to be H*sin(w1t) (where w1 <> w),
then I substitute it in the main equation (the first that I wrote), but
I have a term
ka*H^3*(sin(w1t))^3
that blocks me from calculating H as a function of the other terms. I
guess I have to linearize that term,
but now what is the sense in linearizing around the zero??

Than you in advance for your help.

Alessandro
Lou Pecora
Posted: Thu Jul 06, 2006 3:44 pm
Guest
In article <1152204229.963759.318370@m38g2000cwc.googlegroups.com>,
"bowser" <alessandro.denardi@gmail.com> wrote:

Quote:
Hello, I have a problem in a non-linear equation analysis.

I have a non-linear equation:
x'' + kx(1+ax^2)+bx' = A*sin(wt)

that I already transformed into the following system
x' = y
y' = -kx(1+ax^2) - by + A*sin(wt)

Now I have to obtain the transfer function, so I suppose the only way
to do it is to first linearize (approximating) the second equation. Do
I have to use Taylor series to linearize around the zero (which I
calculated to be (0,0))? If so, how?

The equations of motion are nonlinear. I'm not sure a transfer function
makes sense. That's usually only for the linear systems. You might get
an approximate transfer function by linearizing especially if the
parameter 'a' is small.

Quote:
My attempt was to posticipate the "linearizing": I let a particular
solution to be H*sin(w1t) (where w1 <> w),
then I substitute it in the main equation (the first that I wrote), but
I have a term
ka*H^3*(sin(w1t))^3
that blocks me from calculating H as a function of the other terms. I
guess I have to linearize that term,
but now what is the sense in linearizing around the zero??

Yes, that's the nonlinearity showing itself. That's the problem. I
think you are stuck with the case 'a' << 1. That might give results
that are good for short times. I think the above equation might better
be solved by other methods. Given that it's 2D there may be closed-form
solutions for it. It looks similar to a Van der Pol type, except
driven. You need to start searching the literature and, probably, the
web (Google Scholar, for example).

Perhaps someone in this group will recognize the equation.

-- Lou Pecora (my views are my own) REMOVE THIS to email me.
bowser
Posted: Fri Jul 07, 2006 6:32 am
Guest
Quote:
I think you are stuck with the case 'a' << 1. That might give results
that are good for short times.

Yes, sorry, I forgot to say that I can assume a small. But what does it
imply?
May I remove the whole term kaH^3(sin(wt))^3 ? I guess not..

Quote:
I think the above equation might better
be solved by other methods. Given that it's 2D there may be closed-form
solutions for it. It looks similar to a Van der Pol type, except
driven. You need to start searching the literature and, probably, the
web (Google Scholar, for example).

I don't know what you mean by closed-form solutions, but the system is
actually 3d since time has its influence in the dynamic, due to the
presence of the drive force. Indeed the trajectory in the phase space
can overlap to itself.

I'll look for something about the Van der Pol pendulum. Thank you!
Lou Pecora
Posted: Fri Jul 07, 2006 8:35 am
Guest
In article <1152271940.216281.115630@b28g2000cwb.googlegroups.com>,
"bowser" <alessandro.denardi@gmail.com> wrote:

Quote:
I think you are stuck with the case 'a' << 1. That might give results
that are good for short times.

Yes, sorry, I forgot to say that I can assume a small. But what does it
imply?
May I remove the whole term kaH^3(sin(wt))^3 ? I guess not..

I think the above equation might better
be solved by other methods. Given that it's 2D there may be closed-form
solutions for it. It looks similar to a Van der Pol type, except
driven. You need to start searching the literature and, probably, the
web (Google Scholar, for example).

I don't know what you mean by closed-form solutions,

Closed form means you can write a mathematical expression for the
solution in terms of calculable functions (e.g. sines, hyperbolic tan,
etc.).

Quote:
but the system is
actually 3d since time has its influence in the dynamic, due to the
presence of the drive force. Indeed the trajectory in the phase space
can overlap to itself.

The equation is really motion on a circle (sin wt) driving another
system, so 4D is probably more accurate. But 1/2 of the system is
already integrated (the drive expressed in closed form sin wt). A lot
of driven 2D systems of this form have been solved in closed form.
That's what I meant.

-- Lou Pecora (my views are my own) REMOVE THIS to email me.
 
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