So, this question ask us to develop a circuit given the following:


H(s) = \(\frac{V_{out}(s)}{V_{in}(s)}\) = \(\frac{1}{as^{2}+1}\) (a>0)


part (a) asks if it's possible to realize H(s) using an op-amp and passive R, L, and C elements.

I attached my attempt to work it out(assumed R&L=1), but I've never seen an op-amp without resistors and can't seem to realize the function if they are included. Surely I am missing something important here or making a mistake?

What would happen exactly in this situation without the resistors? How can I design a circuit given H(s) when it does not have the standard s^2+s+1 form? I can't find much guidance nor does my book provide a clear picture.

part (b) then says if a=1, find the steady state response of the circuit if Vin(t) = 0.5cos(t)

I haven't started much on this one since I wasn't sure if my approach was correct for (a). Though I would like to know how to handle an input without an initial phase

Thanks for any help! I'm sure I'll stick around the website for a while, very helpful

 

"part (a) asks if it's possible to create an op-amp using passive R, L, and C elements"

Create does not make sense to me in this context.

as^2+1 actually a fully quadratic expression, just abbreviated a little. Since you want standard form, here you go: as^2+0s+1
Now you can solve it for the roots and get:
\(s=\frac{\pm\sqrt{-a}}{a}\)

Ok, I looked at the pic. I see two caps and one inductor. I don't see any resistors in there.

 

My apologies, worded it incorrectly. Meant to say possible to create a circuit USING an op-amp and passive R/L/C elements.

 

My apologies, worded it incorrectly. Meant to say possible to create a circuit USING an op-amp and passive R/L/C elements.

Possible? Yes.
People do it every day for the past 50 or 60 years.

 

The part that was confusing me was how to design an op-amp circuit with a "0s." I've never seen a transfer function with a 0s. Am I to assume my attempts are off base?

 

The part that was confusing me was how to design an op-amp circuit with a "0s." I've never seen a transfer function with a 0s. Am I to assume my attempts are off base?

You never said what you are designing. Since I don't know what you want, I am just keeping my pie hole shut and let you struggle.

 

I said I was designing a circuit using an op-amp and R/L/C elements given the transfer function, which part (a) asks you to design if possible. I was having trouble understanding how to realize H(s)=1/(s^2+1) through topology. Then I provided my two attempts at the problem as an attachment, asking if it was in the right direction. Not sure how else I can word it...

 

I said I was designing a circuit using an op-amp and R/L/C elements given the transfer function, which part (a) asks you to design if possible. I was having trouble understanding how to realize H(s)=1/(s^2+1) through topology. Then I provided my two attempts at the problem as an attachment, asking if it was in the right direction. Not sure how else I can word it...

Then it looks fine.
Ok. So you need:
* op amp
* capacitor
* inductor
* resistor

You already have:
* op amp-Check
* capacitor-Check
* inductor-Check
* resistor- NO CHECK

You need resistor. Ok. I am looking at the transfer function. It has Gain=1. Meaning your H(s)=Gain*(1/stuff)=1*(1/stuff). So use that one to introduce resistors into the circuit.
* If you use inverting topology, then your Gain=1=Rfeedback/Rinput. For the sake of exercise: Gain=1=1 kOhm/1 kOhm. There! Your circuit now has two resistors in it. But! The output signal will be inverted. Since I don't know what you want, this might be OK or it might be a bad thing. "Pilot's choice." so you decide.
* If you use none inverting topology, then your Gain=1+Rfeedback/Rinput. In this case make Rfeedback/Rinput really really small, something like 1 kOhm/ 10 or 100 kOhm, then your Gain=1.1 or 1.01 or something like that, so close to 1 that it makes no difference in grand scheme of things. There! Now you have two resistors in your circuit.

So we do checklist again:
* op amp-Check
* cap-Check
* inductor-Check
* resistor-Check
All checked. Mission accomplished.

 

Here's another approach using two equal R's and two equal C's with a potentiometer. No inductance is needed. The pot is set to give a feedback factor of 1/3, which for an ideal situation (ideal op-amp and component tolerances) eliminates the term in 's' in the 2nd order transfer function denominator.

 

Then it looks fine.
Ok. So you need:
* op amp
* capacitor
* inductor
* resistor

You already have:
* op amp-Check
* capacitor-Check
* inductor-Check
* resistor- NO CHECK

You need resistor. Ok. I am looking at the transfer function. It has Gain=1. Meaning your H(s)=Gain*(1/stuff)=1*(1/stuff). So use that one to introduce resistors into the circuit.
* If you use inverting topology, then your Gain=1=Rfeedback/Rinput. For the sake of exercise: Gain=1=1 kOhm/1 kOhm. There! Your circuit now has two resistors in it. But! The output signal will be inverted. Since I don't know what you want, this might be OK or it might be a bad thing. "Pilot's choice." so you decide.
* If you use none inverting topology, then your Gain=1+Rfeedback/Rinput. In this case make Rfeedback/Rinput really really small, something like 1 kOhm/ 10 or 100 kOhm, then your Gain=1.1 or 1.01 or something like that, so close to 1 that it makes no difference in grand scheme of things. There! Now you have two resistors in your circuit.

So we do checklist again:
* op amp-Check
* cap-Check
* inductor-Check
* resistor-Check
All checked. Mission accomplished.

I'm not clear as to what topology you are suggesting meets the design criteria. What would your proposed circuit look like? No need to show any values - just the general schematic.

 

Thanks for the reply t_n_k! this is helpful, I'm reading up on it now. Why exactly would the feedback factor of 1/3 eliminate the s value? Hope that's not a dumb question, just wasn't aware that could be done.

 

Thanks for the reply t_n_k! this is helpful, I'm reading up on it now. Why exactly would the feedback factor of 1/3 eliminate the s value? Hope that's not a dumb question, just wasn't aware that could be done.

That result is evident when one does the circuit analysis for that topology. When setting up the equations (say using nodal analysis) you assign a value to the feedback factor [e.g K] which is the fraction of the op-amp output signal fed back to the negative input terminal via the potentiometer wiper position. It so happens that when K=1/3 the term in 's' vanishes. For K less than 1/3 the latter term becomes negative (conditionally unstable situation with roots in the positive half of the complex plane). For K greater than 1/3 the term is positive, implying a conditionally stable outcome.
One must keep in mind that this for an ideal situation. Also when the term in 's' does vanish the circuit has no damping - a matter that one must also keep in mind with regard to practical circuit behaviour.