1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transfer Function Of A 2nd Order Circuit

  1. Jan 17, 2013 #1
    I'm trying to work out the Transfer function for the 2nd order RC circuit in the attachment below: I cant seem to get the right answer :(

    circuit:

    Circuit2ndO.png

    My answer:

    workings.png


    Please see attachment for my attempt and the relevant information:

    View attachment f.pdf
     
    Last edited: Jan 17, 2013
  2. jcsd
  3. Jan 17, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    When you solve for a transfer function, the assumption is that you effectively drive it with an ideal source (Vi) that has zero source impedance, and "measure" the output (Vo) with a "meter" that has infinite input impedance (no loading).

    Your problem stems from splitting the circuit into two parts and assuming that the two sections did not interact. In fact, the first section is loaded by the input impedance of the second, and the second "sees" the output impedance of the first. This interaction is lost when you divided the sections.
     
  4. Jan 17, 2013 #3

    berkeman

    User Avatar

    Staff: Mentor

    I don't think you can split and cascade the circuit like that. I think you could only treat it as two cascaded circuits if there were a buffer amp in between them (with infinite input impedance and zero output impedance).

    Try treating it as one circuit, and write the KCL equations to solve for the transfer function. Does that get you closer to the book's answer?

    EDIT -- dangit! Beat to the punch again :smile:
     
    Last edited: Jan 17, 2013
  5. Jan 17, 2013 #4
    Okay thank you, i just assumed that as when you have two transfer functions connected in series you multiply them together to get the overall function.

    What does this analogy actually represent then? Does it represent two circuits connected in series and assumes no loading effect? Or something else?
     
  6. Jan 17, 2013 #5
    This is what I mean:

    Series.png
     
  7. Jan 17, 2013 #6

    gneill

    User Avatar

    Staff: Mentor

    Yup. It means no loading effects. And it works fine if, as Berkeman suggested, you buffer them with a suitable isolation stage.

    When designing separate stages for cascading its possible to design them with a standard input impedance and assumed standard load (usually the same as the input impedance), so that the resulting transfer functions can be cascaded by multiplication (which is why you often see gear with input impedance specified, and output impedance specified). The simple solution, though, is to provide buffering amplifiers with very, very low output impedance so that loading isn't an issue.
     
  8. Jan 17, 2013 #7
    Thanks a lot man (:
     
  9. Jan 23, 2013 #8
    I'm still struggling with this question :( I don't know where to start, now I have worked out the KCL equation and KVL, I don't know what to do with them?
     
  10. Jan 23, 2013 #9

    gneill

    User Avatar

    Staff: Mentor

    You're looking to find the ratio ##V_o/V_i##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Transfer Function Of A 2nd Order Circuit
  1. 2nd order Circuits (Replies: 1)

Loading...