How Is Equilibrium Achieved in a Magnetic Levitation System?

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Equilibrium in a magnetic levitation system occurs when the vertical displacement of a metallic object is described by the equation m(d^2H/dt^2) = mg - k(I^2/H^2), leading to the condition Ho = Io * sqrt(k/mg). To analyze the system's behavior near this equilibrium, the differential equation can be linearized using a Taylor series expansion around the equilibrium point. This results in a transfer function that can be expressed as (deltaH(S)/deltaI(s)) = -a/(s^2 - b^2). The discussion emphasizes the importance of understanding the algebra involved in proving the equilibrium condition and performing the Taylor expansion. Overall, mastering these concepts simplifies the problem significantly.
erahartman
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Here is a doozy

In a magnetic levatation experiment a metallic object is help up in the air suspended under an electromagnet. The vertical displacement of the object can be described by the following nonlinear differential equation.

m(d^2H/dt^2)=mg-k(I^2/H^2)

m=mass of object
g=gravity acc const
k=const
H=distance between electromagnet and object (output signal)
I=electromagnet current ( input signal)

a) show that the system is in equilibrium when
Ho=Io*sqrt(k/mg)

b) Linearize the equation about the equilibrium point found in part a and show that the resulting transfer function obtained from the linearized differential equation can be expressed as

(deltaH(S)/deltaI(s))=-a/(s^2-b^2)

a>0
 
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erahartman said:
Here is a doozy

In a magnetic levatation experiment a metallic object is help up in the air suspended under an electromagnet. The vertical displacement of the object can be described by the following nonlinear differential equation.

m(d^2H/dt^2)=mg-k(I^2/H^2)

m=mass of object
g=gravity acc const
k=const
H=distance between electromagnet and object (output signal)
I=electromagnet current ( input signal)

a) show that the system is in equilibrium when
Ho=Io*sqrt(k/mg)

b) Linearize the equation about the equilibrium point found in part a and show that the resulting transfer function obtained from the linearized differential equation can be expressed as

(deltaH(S)/deltaI(s))=-a/(s^2-b^2)

a>0

In equilibrium, the time derivative is equal to zero. This should make sense to you because equilibrium means that nothing is changing, and anything that is constant, has a derivative of zero.

The second part is to linearize the differential equation around the equilibrium point which is simply done by doing a Taylor series expansion around the operating point and keeping only the linear terms. You can then transform the linearized equation.

The problem goes from being a doozy to being trivial, if you take this approach.
 
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If we are being assigned problems like this it should be "obvious" that we know to do a taylor expansion. If this problem is "simply done" and "trivial" how about you show us how to do it instead of reitterating what the book says.

Thanks.
 
What would the first couple steps be?
 
Fr3nch said:
If we are being assigned problems like this it should be "obvious" that we know to do a taylor expansion. If this problem is "simply done" and "trivial" how about you show us how to do it instead of reitterating what the book says.

Thanks.

How do I know what your book says? How do I know what your professor has taught you?

How about if you guys make an attempt at the solution according to the rules in this forum?

I would expect that at least the first part is doable considering it's a trivial algebra problem once I told you that the time derivative is zero.

Perhaps if you show that you can solve the first part and then explain what is confusing about doing the Taylor expansion, you might find that I, or someone else, would be inclined to help further.
 
I can't get started on either I am stuck that's why I posted the problem. I know that you set the first part equal to zero but i don't know how to prove that Ho=Io* sqrt(k/mg)
 
erahartman said:
I can't get started on either I am stuck that's why I posted the problem. I know that you set the first part equal to zero but i don't know how to prove that Ho=Io* sqrt(k/mg)

OK, it's just simple algebra. Write out the equation for the time derivative, as you already did. Then set it equal to zero, which you claim you already know that you should do.

Now, the next step may be your issue. Re-label H and I as Ho and Io so that it is clear that these are the equilibrium values and not the time dependent values.

Then rearrange the equation so that you have Ho as a function of Io. This part is just simple algebra.
 
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Thanks
 
erahartman said:
Thanks

OK, it sounds like you understand the first part. The second part is simpler than it sounds, but I agree that it will look daunting the first time you do it. I would ask that if you have any trouble with the second part and want help, post your attempt, and then we can step through the solution.
 
  • #10
Got the first part now thanks. For the second part what to values would I do the taylor expansion between?
 
  • #11
Sorry I meant 2 not to.
 
  • #12
erahartman said:
Got the first part now thanks. For the second part what two values would I do the taylor expansion between?

I wouldn't word the question this way. I would rather say that the Taylor expansion will be done around a point with two values specifying the point (Ho,Io). So the two values are Ho and Io.

To help get you started, I would recommend writing the differential equation as follows.

{{d^2\; H}\over{dt^2}}=g-{{k\;I^2}\over{m\; H^2}}=f(H,I)

Next, the idea is going to be that the time dependent input and output signals (I and H) can be represented as their equalibrium positions (Io and Ho), plus small deviations around those values, as follows.

I=I_o+\Delta I
H=H_o+\Delta H

Now, you can remember your first order Taylor expansion (in two variables) and express it as follows.

f(H,I)\approx f(H_o, I_o) + {{\partial\; f}\over{\partial H}}\Bigg\vert_{(Ho,Io)} \Delta H + {{\partial\; f}\over{\partial I}}\Bigg\vert_{(Ho,Io)} \Delta I

From there, you just need to work out the math.
 
  • #13
Got it thanks Steven for the help, it is much appreciated.

Era
 
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