# Design Topology Given a Transfer Function H(s)

Discussion in 'Homework Help' started by kellz0r, Apr 29, 2014.

1. ### kellz0r Thread Starter New Member

Apr 28, 2014
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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

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2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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"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.

3. ### kellz0r Thread Starter New Member

Apr 28, 2014
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My apologies, worded it incorrectly. Meant to say possible to create a circuit USING an op-amp and passive R/L/C elements.

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Possible? Yes.
People do it every day for the past 50 or 60 years.

5. ### kellz0r Thread Starter New Member

Apr 28, 2014
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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?

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
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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.

7. ### kellz0r Thread Starter New Member

Apr 28, 2014
5
0
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...

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,975
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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.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

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10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.

11. ### kellz0r Thread Starter New Member

Apr 28, 2014
5
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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.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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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.