Connect with us

Op-Amp transfer function

Discussion in 'Electronics Homework Help' started by Dan_DG, May 18, 2014.

Scroll to continue with content
  1. Dan_DG


    May 18, 2014
    Hi everyone,

    I am stuck on a question I have to do as preliminary for a lab. I have to show that the transfer function of the following circuit is equal to the equation pictured.

    I have already related the transfer function to the gain of the amplifier (-z2/z1) where z2 is the impedance of the resistor in parallel with the capacitor. Am I correct in doing this?

    I can't figure out where the "k" or "p0" comes from, and this is ultimately what is confusing me.

    Any help will be much appreciated!

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    Hello to our forum.

    gain= -z2/z1 -> correct.

    As for the transfer function: I'd expect that the meaning of k and p0 are explained somwhere earlier in your training material. Look up the low pass filter (figure 10) in this document and relate the terms to your equation.
  3. LvW


    Apr 12, 2014
    Hello Dan,
    at first, don´t be surprised if your result has a negative sign. The opamp is in inverting configuration and, therefore, the result for T(s) must also be negative.
    Hence, the factor k must be negative.
    More than that, if you have correctly written down the expression for z2/z1 (use "s" for "jw") you have nothing to do than to re-arrange this expression with the aim to have the same form as given for T(s).
  4. Dan_DG


    May 18, 2014
    The transfer function mentioned in that PDF for the low pass filter is what I managed to derive before, which makes me feel better about the problem. I just realised K is probably used as a constant value to replace the resistor values. I am still stuck as to what p0 is. My problem is my lab supervisor and lecturers use different notations which makes this unit incredibly difficult to understand.
  5. Dan_DG


    May 18, 2014
    I figured out what I was missing. I had the right derivation, just didn't simplify 1/RC to the corner frequency.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day