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Mass-two spring-damper system

  1. Jun 20, 2015 #1

    ja5

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  2. jcsd
  3. Jun 20, 2015 #2

    Hesch

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    MDS.jpg
    I assume that F is the input to the system, and that you will have to find x1, x2 ( velocities? ) as a function of F.

    If you are confident with Laplace transforms, I'd build a model by means of Laplace function blocks.
     
  4. Jun 20, 2015 #3

    ja5

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    After I gotten this equation, I tryed to excrete x1, but I'm missing one equation to make a state space model.
    I think I need to get matrix A 4x4 and I don't know how to get x2 double derivate when it doesn't exist, if you know what I mean.
     
  5. Jun 20, 2015 #4

    Hesch

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    I think that state state space is effective by multiple input/output systems, but you have a system with only one input ( F ), thus a classical model is more effective here.
    ( And, yes, I have to admit that I have only limited experience with state space ).
    When using the classical model, you can actually "see" what you are doing. In a 4x4 matrix, a certain number just disappears in the amount of numbers.
     
  6. Jun 20, 2015 #5

    ja5

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    So, I try with Laplace then?
     
  7. Jun 20, 2015 #6

    Hesch

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    Yes, I recommend that.
    Post a sketch of the ( classical ) model.
     
  8. Jun 20, 2015 #7

    ja5

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  9. Jun 20, 2015 #8

    Hesch

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    No, this is not made by function blocks, here is what I mean:
    upload_2015-6-20_15-29-43.jpeg
    You draw these blocks/loops and reduce them afterwards. For example the 4 blocks in lower left corner can be reduced to one block:

    k2/s+c2

    The loops can be reduced by Masons rule.

    At last you will have the transfer functions: x1(s)/F(s) and x2(s)/F(s). For example you put on some step function: F(s) = 1/s, and you will find x1(s) and x2(s).
     
    Last edited: Jun 20, 2015
  10. Jun 20, 2015 #9

    ja5

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    Alright, I know what you mean, but how will I draw a loop to apply mason rule, I get confused when I need to put x2
     
  11. Jun 20, 2015 #10

    Hesch

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    Say you have a closed loop (negative feed back).
    Say that the forward path has the transfer function: A(s) and that the backward path has the transfer function: B(s), then Masons rule says that the transfer function for the loop as a whole =

    A(s) / ( 1 + A(s)*B(s) ) . . . . ( It's easy to prove ).

    If the feedback is positive, you simply change the "+" to a "-".
     
  12. Jun 21, 2015 #11

    LvW

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    Perhaps not too important for solving this task - but for the sake of clarity I like to mention that the given formula for the closed-loop transfer function is known as Blacks formula; Mason`s rule is different from that.
    Questions:
    * What is x2? Input or output (according to your block diagram)?
    * Where is c1 in the diagram?
     
  13. Jun 21, 2015 #12

    Hesch

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    x2 is the velocity(?)/position(?) in the diagram shown in #2. This diagram is just a copy from one of the links in #1. I don't know, I have assumed it's an output, and that F is the input.

    c1 ( and k1 ) is coming up when the diagram is expanded in #8 to the right: ( "Finding x2" reference ).

    The diagram in #8 has just not been finished ( something left for ja5 to do ).
     
    Last edited: Jun 21, 2015
  14. Jun 21, 2015 #13

    LvW

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    OK - so I wait until it will be finished. The procedure of block diagram reduction is a standard one and shouldn`t be a problem.
     
  15. Jun 21, 2015 #14

    Hesch

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    No, maybe it should not, but it is.
     
  16. Jun 21, 2015 #15

    LvW

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    That`s why I said that I will wait - and then I can offer my help (if needed).
     
  17. Jun 25, 2015 #16

    ja5

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  18. Jun 26, 2015 #17

    Hesch

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    Capture.jpg
    This is not right. By Laplace it could be written: k2*s*x2 = k2*s*x1 + c2(x1-x2) , and that's a 1. order system. A spring and a mass will oscillate which means that the system must be a 2. order system. You may not assume a mass (A), when it is not there.

    In the (half) model in #8, you must complete the model. Then you can see what you are doing. Complete it and reduce it.

    In #8 a force, F, is induced. Dividing this force by m you will get the acceleration of m: ( dx1/dt ). If you integrate this acceleration ( divide by s ) you will get x1.
    The back-force ( as to F ) will be induced by:

    1) by c2: Fback = x1*c2.
    2) by spring: Fback = (x1/s)*k2 : Dividing x1 by s, you integrate velocity to position, thus Fback = Δposition * k2.

    But in (2) the velocity is missing: Correctly it must be: Fback = ( (x1-x2)/s ) * k2.

    The Fback is subtracted from F ( left sum block ) so that the resulting force, pushing m, is: F - Fback. ( Loop closed ).
     
    Last edited: Jun 26, 2015
  19. Jul 1, 2015 #18

    ja5

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    Can you draw me whole loop?
     
  20. Jul 1, 2015 #19

    Hesch

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    Something like this:
     

    Attached Files:

  21. Jul 1, 2015 #20

    ja5

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    but what is F2?
     
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