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Homework Help: Transfer equation of a Voice coil motor for dissertation

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm currently creating a PID control system for a Voice coil motor, my issue is obtaining the transfer equation that relates the voltage to the displacement. Ignoring gravity it is easy however once gravity is included I can't figure out how to rearrange it.

    2. Relevant equations

    In the time domain the equation that I have come up with is
    http://mathhelpforum.com/attachments/differential-equations/27771-transfer-equation-voice-coil-motor-equation.png

    the derivation from fist principles can be provided if needed

    3. The attempt at a solution

    I've applied the laplace transform to the equation to get:

    http://mathhelpforum.com/attachments/differential-equations/27772d1364923937-transfer-equation-voice-coil-motor-equation.png

    I could really use some help, and I would be very grateful for any advice offered.

    Thanks in advance
     
  2. jcsd
  3. Apr 4, 2013 #2

    rude man

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    I think including gravity is both extremely difficult and pointless. The cone will not be materially deformed or otherwise affected by gravity.

    BTW all I get for your images is red x's.
     
  4. Apr 5, 2013 #3
    I can seem to edit the original post so I'm reposting fixing the equations and trying to clear a few thing up. Thanks again for taking the time to look at this

    1. The problem statement, all variables and given/known data

    I'm currently creating a PID control system for a Voice coil motor. The end goal of the project is to create a tactile simulation display that mimics softness. The basic priciple it is modeling is if you take a samle of silicone and press on it the material will defom and the top will bow around you finger.

    The device will mimic this by measuring the force that a person presses on a load cell with and then the voice coil motor will move the frame causing a strip of polyimide film to wrap around the participants finger simulating softness.

    My issue is obtaining the transfer equation that relates the voltage to the displacement. Ignoring gravity it is easy however once gravity is included I can't figure out how to rearrange it.

    I have actually developed a PID controller and I'm now writing up my report, as such I know that the effects of gravity cause a steady state error of about 3mm (this is compare to when it was moving horizontally when there was no error)

    2. Relevant equations

    In the time domain the equation that I have come up with is

    [tex]v = \frac{R.m.\ddot{x}+K_b.K_E.\dot{x}+ \frac{mg}{S} }{K_b}[/tex]

    the derivation from fist principles can be provided if needed

    3. The attempt at a solution

    I've applied the laplace transform to the equation to get:

    [tex]V(s) = \frac{R.m.X(s).S^2+K_b.K_E.X(s).S+ \frac{mg}{S} }{K_b}[/tex]

    I could really use some help, and I would be very grateful for any advice offered.

    Thanks in advance
     
    Last edited: Apr 5, 2013
  5. Apr 5, 2013 #4

    rude man

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    How about this:

    When the speaker is directing sound upwards or downwards there is indeed a deformation of the cone and voice coil. However, if the cone deflection-to-current transfer function is linear, operating about a finite cone deflection makes no difference.

    The analogy is to a mass dangling at the end of a spring vertically. The mass stretches the spring by an amount mg/k but applying an additional external force F will stretch or compress the spring an equal amount F/k.

    Of course, any transfer function nonlinearity will show up to some extent as distortion.
     
  6. Apr 5, 2013 #5
    We are using the voice coil motors as normal linear motors, so the weight due to the frame is much more than a speaker cone (unless I'm mistaken). For clarification this is the device in question.

    dhtxlQ6.png

    As I know from testing gravity does have a significat effect on the model meaning that any simulations run aould be pointless unless I include gravity. My equation comes from the following derivation (I'm actually working on VCM1, but the equation is the same for 2&3 the constant values will be different).

    ---------------------------------------------------------------------------------------------

    The first step to defining the transfer equation is to create a free body diagram this is shown below

    JYYxx9b.png

    It can therefore be stated that the sum of the forces is;
    [itex]∑F = F_{mag} - F_{grav}[/itex]
    Where Fmag is the force due to the magnetic field which is defined as
    [itex]F_{mag} = BIL[/itex]
    And Fgrav is the force due to gravity defined as
    [itex]F_{grav} = mg[/itex]
    Using the standard equation [itex]F=m\ddot{X}[/itex], it can be said that
    [itex]m\ddot{X}= F_{mag} - F_{grav}[/itex]
    [itex]m\ddot{X} = BIL-mg[/itex]
    [itex]\ddot{X}=\frac{(BIL-mg)}{m}[/itex]
    The next step is to define the relationship between the voltage and the current, this is achieved by analysing the equivalent circuit.

    XXEx7sj.png

    From the figure the equation below was derived
    [itex]V=IR+E+L\frac{di}{dt}[/itex]
    As the inductance for the motors that will be used is negligiable in comparison to the other variable taking this and rearranging with respect to current, I, we get
    [itex]I=\frac{(V-E)}{R}[/itex]
    As the back emf is proportional to the velocity of the voice coil motor ([itex]E=K_E\ddot{X}[/itex]) equation XX can be derived.
    [itex]I=\frac{V-K_E \ddot{X}}{R}[/itex]
    where KE is the back emf constant. Substituting equation XX into equation XX gives
    [itex]\ddot{X}=\frac{B \frac{V-K_E\dot{X}}{R} L-mg}{m}[/itex]
    Rearranging (and using the magnetic field constant [itex]K_b = BL[/itex] gives
    [itex]V=\frac{RmX ̈+K_b K_E\dot{X}+mg)}{K_b} [/itex]
    Applying the Laplace transform gives
    [itex]V(s)=\frac{RmX(s)s^2+K_b K_E X(s)s+\frac{mg}{s}}{K_b}[/itex]
    Rearranging gives the transform function
    [itex]\frac{X(s)}{V(s)}=[/itex]

    and that is where I am stuck
     
  7. Apr 5, 2013 #6

    rude man

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    E = KE dx/dt, not d2x/dt2.

    May I suggest you use lower case for the time variables and upper case for the transformed variables: x(t) and X(s), i(t) and I(s), v(t) and V(s) etc.


    I will work on this and see what I come up with, now that I have a better idea of what you're trying to do.
     
  8. Apr 5, 2013 #7
    I might be wrong with E = KE, I got that equation from some where I'll check it, I may be wrong though. I did get the feeling you might be getting the wrong idea about what I was trying to do, the main thing I'm stuck on is the rearrangement but that may well be because it is wrong.

    Im also not sue about how mg should be converted, I feel that maybe it should be convertd into a form of displacement, possibly using SUVAT? that would give me the X(s) term that I am looking for.

    Thanks for taking the time to look it over I've been wracking my head for weeks, luckily I was able to tune it without simulation, but my report will need me to include some nice pretty graphs and simulation results.
     
  9. Apr 5, 2013 #8

    berkeman

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    Sorry if I'm missing the obvious, but why are there 3 actuators, and what does a "participant finger" have to do with all of this? What is that mechanism used for?
     
  10. Apr 5, 2013 #9

    rude man

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    Why do this? You have a linear, constant-coefficient, 2nd order ODE in x(t) with v(t) (BL/mR)
    - g as your driving function. So transform this and get your X(s) and then x(t).
     
  11. Apr 5, 2013 #10
    the two smaller VCMs will vibrate and hopefully simulate wetness. The larger one simulates softness.

    I don't quite follow what you mean sorry..
     
  12. Apr 5, 2013 #11

    berkeman

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    :confused:
     
  13. Apr 5, 2013 #12
    Lol, if you are interested the work we are doing is based of the research done by some Japanese researchers, their paper explains the priciples about how softness can be displayed, we are creating a similar device and then also testing for wetness which is based on work done by a 3rd year student who determied vibrating surfaces can mimic wetness. The original paper is called "A Softness Feeling Display with an Active Tensioner Controlling Contact Pressure Distribution on a Fingertip" if you want to look it up.
     
  14. Apr 5, 2013 #13

    rude man

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    Don't you want to know x(t), given voltage v(t)?

    BTW I believe the expression for E = constant * dx/dt is correct. The motor is acting like a generator, as all motors and generators do unless they're not spinning.
     
  15. Apr 5, 2013 #14

    rude man

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    Oh, maybe you want voltage as a function of x. That makes more sense from your 'softness' viewpoint.

    Here's the problem: you started with BLi = mg = m d2x(t)/dt2 which says that if i = 0 then x(t) increases without limit.

    You need to revise your basic system definition.
     
  16. Apr 5, 2013 #15
    so do you mean take the equation in the form

    [itex]\ddot{x} = \frac{BL\frac{v-K_E\dot{x}}{r}-mg}{m}[/itex]

    rearrange to get

    [itex]\ddot{x} = \frac{K_bv}{mr}-\frac{K_E\dot{x}}{mr}-g[/itex]

    and then transform that?

    Don't I still get the same issue that the gravity term will mess up any rearrangement?
     
  17. Apr 5, 2013 #16
    The plant equation of the voice coil motor should take voltage as an input and give my a displacement as an output, the displacement modelling the deflection of a material with the youngs modules that is being simulated.

    where did I have "BLi = mg = m d2x(t)/dt2" I came up with BLi - mg = m d2x(t)/dt2. thus when i is 0 then the only force is gravity thus it cancels to -g=d2x(t)/dt2, which is correct as it falls to the ground (or table :P)
     
  18. Apr 5, 2013 #17
    edit:bad post
     
  19. Apr 5, 2013 #18

    rude man

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    Sorry,I meant BLi - mg = m d2x(t)/dx2, same as yours.

    My point is you will get infinite output displacement without any input voltage with your equation above.

    So this brings me back to my question why you did not leave your equation as
    d2x/dt2 + (BLKE)/mR) dx/dt = (BL/mR) v - g where v = voltage input v(t) and solved for x(t) for a given input voltage v(t)?

    Still doesn' change the fact that you don't have a system that makes sense, with output x(t) unlimited negative for any finite input v(t). Unless there's another feedback path you haven't told us about.
     
  20. Apr 5, 2013 #19

    rude man

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    PS the thing that makes a speaker voice coil give a displacement output for a given current (or voltage) input is the restoring force of the speaker cone which acts like a spring. You have no restoring force increasing with x in some manner, that's your problem.
     
    Last edited: Apr 5, 2013
  21. Apr 5, 2013 #20

    rude man

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    Yes you do. See my previous post.
     
  22. Apr 6, 2013 #21
    There's another force? I thought gravity was the restoring force. Is this right or is there another force I've not realised??? Assuming it is gravity do I want to go along the lines of:

    [itex]s=ut+\frac{1}{2}at^2[/itex]

    or in terms of x

    [itex]x(t) = \dot{x(t-1)}t+\frac{1}{2}\ddot{x}t^2[/itex]

    rearranging to get

    [itex]g= \frac{2(x(t)- \dot{x(t-1)}t)}{t^2}[/itex]

    Not too sure on how that Laplaces but I think it would be

    [itex]g= \frac{2(X(s) - X(s)e^{-s}s^{-2})}{2S^{-3}}[/itex]

    simplifying gives

    [itex]g= X(s)S^3 - X(s)e^{-s}s[/itex]

    I can then plug that back into the main equation

    [itex]\frac{K_bV(s)}{mr} = X(s)s^2 + \frac{K_EX(s)s}{mr}+X(s)S^3 - X(s)e^{-s}s[/itex]

    then rearranging to give

    [itex]\frac{X(s)}{V(s)} = \frac{K_b}{mrs^3+mrs^2+(K_E-mre^{-s})s}[/itex]



    Edit: looking again at your post I had some more thoughts. I'm fairly sure that voltage isn't actually proportional to displacement it is proportional to acceleration. i.e If I put in 5V I wont get a fixed displacement. A fix displacement is obtained by using a fast controller, and at certain heights there are small oscillations.
     
    Last edited: Apr 6, 2013
  23. Apr 6, 2013 #22

    rude man

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    OK, so now we're looking for acceleration output a(t) for input voltage v(t)?

    So now we get, as before (at least I hope this is what you got also),
    X(s)(s2 + as) = bV(s) - g/s
    or X(s) = [bV(s) - g/s]/s(s+a)
    where
    a = BLKE/mR
    b = BL/mR

    then A(s) = s2X(s)
    where A(s) is transformed acceleration a(t)
    giving A(s) = s2X(s) = [sbV(s) - g]/(s+a)
    and now you can input any V(s) you want, such as a step 1/s, or steady-state sinusoid V(jω), or whatever.
     
  24. Apr 8, 2013 #23
    but how can I get from the form of

    [itex] X(s) = \frac{bV(s) - g/s}{s(s+a)}[/itex]

    to the transfer function G(s) where

    [itex]G(s)=X(s)/V(s)=\frac{a_0}{a_1S^3+a_2S^2+a_3S+a_4}[/itex]

    I can't just divide by V(s) as it isn't in all the terms in the numerator.
     
  25. Apr 8, 2013 #24

    rude man

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    You're right. That's because you get an output even with no voltage input V(s).

    Look at your original equation: ma = BLi - mg. What is a with i = 0? Not zero, right?

    There must be another feedback path in your overall system that balances BLi with mg. I suspect you haven't described your system fully to us.
     
  26. Apr 8, 2013 #25
    There is a negative acceleration when there is no current, it is essentially just free falling. The limiting factor is the table.

    To keep it at a fixed height I have to apply a voltage.
     
    Last edited: Apr 8, 2013
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