Transfer Function for Op-Amp Circuit

Discussion in 'Homework Help' started by tquiva, Feb 7, 2011.

  1. tquiva

    Thread Starter Member

    Oct 19, 2010
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    Could someone please assist me w/ 1a of this problem?
    I'm not sure what method to use.
    I know I would first convert it to frequency domain, but what's next?

    [​IMG]
     
  2. t_n_k

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    Mar 6, 2009
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    You could start by considering all (ideal) op-amp input terminals being at Vo/2. This is constrained by the 1Ω/1Ω voltage divider at the positive input of the lower amp.

    This looks to be a tuned circuit with the 1Ω & C values giving the resonant frequency [f0=1/2∏C] and R determining the circuit 'Q' factor.
     
    Last edited: Feb 7, 2011
  3. tquiva

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    Oct 19, 2010
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    so every terminal (+ & -) of each terminal is Vo/2 ?
     
  4. t_n_k

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    Yes - otherwise the circuit isn't operating in the linear region. The amplifiers would be considered to be ideal, for the purposes of deriving the transfer function.
     
  5. t_n_k

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    Had a bit of a closer look at the circuit response using simulation - it probably isn't a true tuned circuit. Overall, it's rather more like a highpass filter with a curious tunable peak near the corner frequency. Perhaps it has some audio function - judging by the remainder of the question.
     
  6. tquiva

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    Oct 19, 2010
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    Thank you. So for each + terminal of each op amp, the voltage is Vo/2, and for each negative terminal, the voltage will be a 0? Is that correct?
     
  7. The Electrician

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    This is the circuit of a GIC (generalized impedance converter) with one capacitor causing the GIC to simulate an inductor and the other, the input capacitor, serving to provide second order response.

    See this thread:

    http://forum.allaboutcircuits.com/showthread.php?t=15751

    for more information.
     
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  8. t_n_k

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    No - incorrect. Every input terminal, whether + or - will be at Vo/2.
     
  9. tquiva

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    Oct 19, 2010
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    Alright, but there's one part I'm a bit confused about. Say I want to take the nodal voltage at the very first node at the very top left corner. If I were to name it Va, with the input at Vs, then:

    (Vs - Va)(sC) + (0 - Va)/R = Va - ?????

    I'm not sure what to put in for the last current that is flowing out of that node. There's no element present besides another node.
    What would the equation be for this case?
     
  10. t_n_k

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    Well, call the lower opamp output node Vb, and you have

    (Vs - Va)(sC) + (0 - Va)/R = (Va - Vb)/1

    You just need to work out the relationship between Vo and Vb.

    As already discussed, Va=Vo/2
     
  11. tquiva

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    Oct 19, 2010
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    So basically... I have 4 node voltages already known which are the input terminals which is Vo/2. This leaves 2 more voltages that I need to solve for, and from then, I am able to solve for Vo.

    Is my method correct?
     
  12. t_n_k

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    Write something down and then show what you've done - then we can decide if your method is correct or not.
     
  13. tquiva

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    Oct 19, 2010
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    Alright. So you said Va = Vo/2. Are you referring to the top left node?
     
  14. t_n_k

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    Also, remember I gave you a lead in suggesting you find the relationship between the output Vo and the feedback integrator's output - the lower opamp circuit being the integrator stage.
     
  15. t_n_k

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    Yes, I am referring to the same point you named Va.
     
  16. tquiva

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    Oct 19, 2010
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    How about the point to the very right of that. Is that considered equivalent to Va to also be Vo / 2 ?
     
  17. tquiva

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    Oct 19, 2010
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    Here's my work so far:

    [​IMG]

    [​IMG]

    Did I make any mistakes?
     
  18. t_n_k

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    Everywhere along the "conductor" line, from the right-hand side of the input capacitor to the +ve input terminal of the top opamp, will be at the same potential of Vo/2.
     
  19. t_n_k

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    Points C and B on your recent diagram are one and the same node.
     
  20. tquiva

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    Oct 19, 2010
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    Ok, so that elimates VC as an unknown. Therefore, all I need are VD and VO ?
     
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