Hi there,

Please can you help - I need to find the Laplace transform of this equation:

(dy(t)/dt)^3 + y(t) - y(t)^3 = u(t) + 2u(t)^2

Thanks

 

Thats pretty standard. Use the tables and properties.

What's the problem? If you show your work, we can help... but we won't do your work for you

 

I would post an attempt if I could Maybe I'm missing something, but it doesn't seem standard at all...

On all the tables there is nothing that helps with (y')^3. y' and y would be ok, but when raised to the power of 3, I'm stuck...

 

Ohhh i thought at first that it was simply the third derivative.

Anyway, i think you can still use the tables/properties.

Hint: you can combine the following two properties:

Use the table to find F{t^3} = 3!/s^4

Derivative property: sY(s) - y'(t)

 

Do you mean, Derivative property: sY(s) - y(0) ?

I don't think you can use F{t^3} = 3!/s^4, since this is for a specific case of y(t). We are working with y(t) as an arbitrary function of t.

Even Wolfram Alpha can't get it Type:

LaplaceTransform[(f'[t])^3, t, s]

in http://www.wolframalpha.com/

Hmm.. I still don't know...

 

Ohhh wait a second. I was getting confused.

This is a non-linear differential equation. You cannot take the laplace transform of it.

You'd need to do some SERIOUS maths to find the direct transfer function of that using laplace methods (volterra kernel).

Alternatively, you could linearize the system. This involves writing the state space form, finding the equilibrium points of the system, evaluating the Jacobian of the system at those equilibrium points...

What class is this for, might i ask?

 

Your equation is a non-linear equation. The Laplace transform approach to dealing with differential equations, is tricky and often not useful for nonlinear equations. Still, it sometimes works out in special cases and it's worth a try when you don't have a better method. However, this takes considerable mathematical skill. Note that multiplication in the time domain will become convolution in the Laplace domain, so you can try to convert your powers in this way. It's not clear to me what this will yield in the end, without going through all the math.

How have you obtained this equation? What is your motivation for taking the Laplace transform?

EDIT: I just noticed guitarguy's post. He's thinking along the same lines as I am.

 

Thanks for the replies. I should have explained the context and it would have cleared things up more.

The equation relates to a system. I was asked 2 questions, given the equation:

1) i) Determine the state space model of the system.
ii) Determine the system stability about each of the equilibrium points for u(t) = 0.

and then

2) Determine the linearised state space model of the system about the equilibrium set point y0...etc, etc...

So I thought you should be able to find the pure transform then find the [linear] state space model, since it only asks for the linearised model in question 2. I guess you can't though. But then what is Q1 asking for if we're only asked for the linearised model in Q2?

By the way, when is it linear and when is it not?

 

State space is NOT the same thing as S-domain. You should research state space first.

In short, what is happening is that this equation of a system (again, non-linear). You want to describe that system. That is, you wish to model it for a local space around the equilibrium points. The reason you do this is because the local system behaves linearly for points *near* the equilibrium points. So you are asked to give the linear 'approximation' about those points.

Question 2 is asking you to write the equations for the new linearized system. Whereas, question 1 is the state space model of the original non-linear system.

See my above tip for how to linearize a system. State space, find equilibrium, find jacobian, evaluate jacobian at equilibrium points to give you your new (linearized) system.

y = mx+b is linear. Anything with a polynomial, exponential, sinusoidal, etc. function is nonlinear.

 

Thanks again Problem solved. You guys are awesome

 

No problemo! Good luck.