laplace transform of a product

[robxx17]

Hello,

I realize that this is a silly question but I'm currently working on a control system for a robotic arm and I'm trying to compute a laplace transform for the following function:

cos(x)*x" + (x')^2 = y

in particular I'm having a problem with the laplace transform of a product of cos(x) and the 2nd derivative of x. also, I'm a bit puzzled by the square of the 1st derivative of x. Thank you very much for any help.

r.

 

[electricpete]

Laplace transforms are easiest to apply in linear situations... you have unfortunately a very nonlinear equation in x.

Sometimes you can get rid of the product through partial fraction expansion, but I don't think that will work here.

By the way, it's not clear to me the context of the equation. Is this a differential equation in x(t), where y(t) is an input/independent function?



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[robxx17]

sorry about the ambiguity, x(t) is the input and y(t) is the output.

I'm writing the Langrangian equation for the motion of the arm that's where I get the product of the cosine and the 2nd derivative of the variable:

M(q)q” + V(q,q’) + G(q)= ?
? = d/dt(?L/??’) - (?L/??)

where '?' is the partial derivative. Also, here I have 'q' as my input and '?' as the output. But, for the simplicity, I originally used 'x' and 'y' and my input and output variables.

I tried just about everything, including MathCad, MatLab and even trying to derive the expression myself using the definition of Laplace Transform. But, so far, I am failing miserably. Thanks for your help.

r.

 

[zekeman]

Ordinarily, you shouldn't get a term in x'^2 when you perform the La Grangian on the arm. And forget about using LaPlace for a nonlinear equation (it's for linear equations only) if that is what you got.
I strongly suggest that you present the problem as it appears so you can get a verification of that equation first. Then we can talk about a solution.
For example, for an arm falling freely ( a simple pendulum), I get
T=I/2*@'^2
V=Mgl/2*(1-cos(@))
L=T-V
And
d/dt part(L)/d@'-part(L)/d@=I@"+1/2Mglsin(@)=0