Op-Amp Filter Circuit Transfer Function

[LordSputnik]

Hi!

I've been having some trouble deriving the transfer function of the following circuit:

Currently, I start by saying T(s) = -Zf(s)/Zi(s), where T(s) is the transfer function in the complex frequency domain, Zf(s) is the feedback impedance and Zi(s) is the input impedance.

Zf(s) = 1/sC + R = (1 + sCR) / sC = (1 + (2.63x10^-3) s) / (470x10^-9 s)

Zi(s) = 1/sC + R = (1 + sCR) / sC = (1 + (22x10^-6) s) / (22x10^-9 s)

Then:

T(s) = (1 + (2.63x10^-3) s) / (1 + (22x10^-6) s) * (22/470)

This seems to be alright to me, but when I try to work out the unit step response by splitting the numerator into two parts (one looks like 1/(1 + sT), the other is s/(1+sT), so the differentiation rule for Laplace transforms can be used), I get:

v(s) = (22/470) * (1 + 118.5 e^(-t / 22 x 10^-6) )

Which doesn't seem to match the output from my simulator exactly (about 10% difference).

Any ideas?

Thanks!

 

[Laplace]

I derived the transfer function and found the corresponding form in a table of Laplace transforms. Not sure what you mean by splitting the numerator. I just substituted 1/s for Vi as the unit step function.

Of course you realize this is only a theoretical exercise since an op amp circuit as shown would not work properly with no input bias current.

 

[LordSputnik]

Thanks for your help! It seems that substituting numbers into your solution gives the same results as mine - I expect it was just an error in the simulation.

I only had 1/(s+a) in the very simple table I was using, so by splitting the numerator, I mean:

1/s * (s+a)/(s+b) = 1/s * (s/(s+b) + a/(s+b))

Which allowed me to work out the time domain response with the one equation I had.

The circuit is part of a pre-built board, so I expect a resistor has been included on the board to bias the input, but this has been left off the diagram to reduce the complexity. I'll check for this today.

 

[duke37]

Are the feedback components connected in parallel rather than in series?