Transfer Function for Op-Amp Circuit

t_n_k

Joined Mar 6, 2009
5,455
You equation for node D doesn't look correct.

The opamp output terminal will be driving current into the circuit.

If you are not sure what to do - perhaps try deriving the transfer function of the lower feedback integrator stage which relates VD to Vo. This is quite valid.
 
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Thread Starter

tquiva

Joined Oct 19, 2010
176
Thank you so much. I really appreciate your help. In my solution above, will C also be Vo/2 or do I write it in terms of the capacitor to the left of it and the resistor to the bottom of it?
 

t_n_k

Joined Mar 6, 2009
5,455
Looking at that lower feedback integrator op-amp stage, consider the following ....

1. The 1Ω resistor into the negative terminal sees a voltage drop of Vo/2.

Why?:The negative terminal "follows" the positive terminal, the latter being at Vo/2. So the right hand side of the 1Ω has a voltage Vo and the left hand side a voltage of Vo/2. Difference across the 1Ω is therefore Vo/2.

2. Current in the 1Ω is therefore Vo/2 amp.

3. The same current flows in the feedback capacitor. So Ic=Vo/2 amp.

4. The voltage drop across the capacitor is therefore

Vcap=Icx(1/(Cs))=Vo/(2Cs)

5. The right hand side of the capacitor is at Vo/2 (being connected to the negative input terminal) so the voltage [VD] at the left hand side of the capacitor will be the voltage at the negative terminal minus the capacitor voltage drop.

Or

VD=Vo/2-Vcap=Vo/2-Vo/(2Cs)=(Vo/2)(1-1/(Cs))

Hope that's clear.
 

Thread Starter

tquiva

Joined Oct 19, 2010
176
Thank you so much for all your help. I was able to find Vo.
For part (b), it says to discuss the effect of R and C on the poles and zeroes. What exactly are the poles and zeroes?
 

t_n_k

Joined Mar 6, 2009
5,455
So having found Vo in terms of Vi you would have the 's' domain transfer function.

A zero is any complex frequency which makes the transfer function gain zero.

A pole is any complex frequency that makes the transfer function gain infinite.

So an 's+a' term in the transfer function numerator could make the transfer function gain = 0 at s=-a

An 's+b' term in the denominator could make the TF gain infinite at s=-b.

One needs to examine the individual transfer function to 'extract' the poles & zeros - sometimes by inspection, sometimes after some complex number manipulation.

Perhaps you could post your transfer function if you are not sure of how to proceed.

It's rather strange you are working on this problem if you've not yet been introduced to the concept of poles & zeros on the complex plane.
 

t_n_k

Joined Mar 6, 2009
5,455
Your transfer function is the same as what I derived.

I 'simplified' it to the form

\(H(s)=\frac{2s^2}{(s^2+\frac{1}{RC}s+\frac{1}{C^2})}\)

which makes it a bit easier for me to extract the poles and/or predict the transient behavior.
 

Thread Starter

tquiva

Joined Oct 19, 2010
176
Thank you. Would you happen to know how to go about determining the effect on the poles and zeroes through Matlab?
 
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