Transfer Function of circuit

Discussion in 'Homework Help' started by champ01, Apr 7, 2015.

  1. champ01

    Thread Starter New Member

    Aug 22, 2012
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    Trying to find the transfer func of the circuit attached, in laplace

    My answer I got was: Vo(s)/Vi(s) = R1 / (sCR1 + sCR2 + 1)(sL + R1)

    Just wondering if anyone can verify my answer?
     
  2. champ01

    Thread Starter New Member

    Aug 22, 2012
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    with verification I was just wanting to know if I got the right answer or not, not for anyone to solve it for me.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    In reality one has to solve it to verify your result anyway.

    As a comment only, perhaps it would have been more informative if you had expanded the function to show a second order polynomial (a*s^2 + b*s +c) in the denominator.

    I believe you have an error in what would be the 's' term coefficient in the aforementioned denominator expansion. The 's^2' and constant terms in the expanded denominator polynomial look OK.
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Yes i agree there is something wrong. You should not get a term with R1^2 in it.
    Try again, then post result.
     
  5. champ01

    Thread Starter New Member

    Aug 22, 2012
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    Thanks for the reply guys, attached is my new worked-answer:
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Still looks incorrect. The negative sign in the final result is puzzling.
     
  7. champ01

    Thread Starter New Member

    Aug 22, 2012
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    are my first three equations of circuit analysis correct? could be my algebra needing a brush up
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Sorry that still does not look right.

    Also, your final result should be in the form of two polynomials, one numerator and one denominator:

    Hs=N/D=(a*s^2+b*s+c)/(d*s^2+e*s+f)

    where a and/or b may be zero, and no coefficients will be negative.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Your setup equations look fine.

    But look at this equation:

    upload_2015-4-10_23-13-14.png

    Always, always, ALWAYS check your units!

    sL has units of impedance
    R has units of impedance

    what are the units on that last fraction? It's unitless. Can you add a unitless quantity to an impedance?

    This means that everything beyond this point (that uses this equation) is guaranteed to be wrong and was a waste of time. ALWAYS track and check your units as you go.
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,


    Yeah it just looks like an algebra problem when equating the far right current i3 to i2.
     
  11. champ01

    Thread Starter New Member

    Aug 22, 2012
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    I've re-worked again:
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    That may be right, but still not in the right form. You should multiply out the denominator so you have only one polynomial in the denominator and only one in the numerator, with no fractional parts in either.
    So this is an example:
    5/(s^2*2+3*s+5)

    while this is not in proper form yet:
    10/(s^2*2/R1+3*s/R1+5)

    because of the divisions.

    Sometimes you keep the divisions, but that's for a different reason.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    The units work out, now ask if the answer makes sense.

    Ask about bounding cases. You have four obvious ones, namely the various combinations of when the inductance and capacitance are zero and infinity.

    If the inductance and capacitance were both zero (i.e., if the inductor were replaced with a short and the capacitor replaced with an open), what would the transfer function be? Does your solution reduce to that result?

    Same for the other three combinations.
     
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