# Op-Amp transfer function

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

1. ### Dan_DG

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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!

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2. ### Harald KappModeratorModerator

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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

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146
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

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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

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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.