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Is my transfer function right?

  1. Jan 21, 2012 #1

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 21, 2012 #2

    I like Serena

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    There seem to be a couple of loops missing.
     
  4. Jan 21, 2012 #3

    Femme_physics

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    Hi ILS! Hmm you're right.

    I need to add to the R line the following:

    +G2H1 +G2H1
     
  5. Jan 21, 2012 #4

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    Yes, but that's not all (and you should only add it once and not twice ;).
     
  6. Jan 21, 2012 #5

    Femme_physics

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  7. Jan 21, 2012 #6

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    The bottom one is not a proper loop, since one of the arrows is in the wrong direction.

    Actually, you have mentioned each of the loops at least once, but I haven't seen the proper expression for it yet.

    There is also something missing from the C line.
     
  8. Jan 21, 2012 #7

    Femme_physics

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    Before I give you the proper expression (with the missing C line stuff) I'm trying to understand why the bottom one isn't a proper loop. I start from R, but I can't go through minuses?
     
  9. Jan 21, 2012 #8

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    You can and should go through the minuses, but you should not go against the arrows.
     
  10. Jan 21, 2012 #9

    Femme_physics

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    LOL - good point.

    But what's the point of this loop being there? It doesn't change anything.
     
  11. Jan 21, 2012 #10

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    Which loop?
     
  12. Jan 21, 2012 #11

    Femme_physics

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    The bottom one, the one you told me to discount.
     
  13. Jan 21, 2012 #12

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    So why is the loop there?
    Did someone tell you it should be there?
     
  14. Jan 21, 2012 #13

    Femme_physics

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    Well, nobody's told me YOU gonna be there! :wink: Sorry, I had to :smile: :biggrin:

    Anyway, that loop doesn't contain anything and just goes back to R, so I say it's pointless.

    As for the function:

    http://img819.imageshack.us/img819/7867/gggzo.jpg [Broken]

    I just hope all the signs are right, too..
     
    Last edited by a moderator: May 5, 2017
  15. Jan 21, 2012 #14

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    :)

    Looks like you have your paths and loops almost straight now.
    Where does the ##-G_1## come from in the denominator?


    Btw, last time you said you were taught:
    This was for your basic exercise at that time.
    However, I found it's not so simple for this circuit.
    I tried to apply it. The result comes close, but is not quite what it should be.

    I think you should build up your equations inside out, similar to the process of replacing resistors in series by one resistor, and replacing parallel resistors by one resistor.
     
  16. Jan 21, 2012 #15

    Femme_physics

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    http://img841.imageshack.us/img841/567/thisloop.jpg [Broken]

    Here.

    We weren't really taught how to do that, we were just told to use the fast path.
     
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  17. Jan 21, 2012 #16

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    That one accounts for the ##+G_1## in the denominator, but not for the ##-G_1##.



    Well, I'll give you a hint of what I mean.
    In your circuit you could replace the loop you just marked, by a block with ##G_1 \over 1 + G_1## in it, which is the response function of just that path/loop with the G1 block in it.
    This simplifies the circuit.
     
    Last edited by a moderator: May 5, 2017
  18. Jan 21, 2012 #17

    Femme_physics

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    You're right......

    So,


    http://img39.imageshack.us/img39/8605/g0g0.jpg [Broken]


    I really respect your methods and don't mind trying them but I have so much to study and relatively short period of time. Should I really try to understand this? I know it may not "sound" difficult but over the internet....u know how things can be.
     
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  19. Jan 21, 2012 #18

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    Yep.
    That is the proper result of the application of your path/loop formula. :)


    No, I did not expect you would want to learn this now.
    I only wanted to give you a hint that there is more to it.
    Especially since your result is not quite right now.

    I'll just give you the result I get:
    $${C(s) \over R(s)} = {2G_1G_2 + G_2 \over 1 + G_1 + 2G_1G_2H_1 + G_2H_1}$$

    As you can see it is close to what you got, except for two factors of 2 in it.
    I leave it up to you what you want to do with it.
     
    Last edited: Jan 21, 2012
  20. Jan 23, 2012 #19

    Femme_physics

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    I think I understand. There are two "reductors" (which I believe is the same as voltage sources) therefor your result makes sense. Thanks :smile:

    Edit: That "IS" why you added that factor, yes?
     
  21. Jan 23, 2012 #20

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    Sorry, but I do not know what you mean by "reductor".

    I calculated the result of the circuit with 2 sure-fire methods that are different from yours, which appears to be a rule of thumb.
    Both came up with the extra factors 2.
     
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