Op-amp circuit analysis using a transfer function

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SUMMARY

The discussion focuses on the analysis of op-amp circuits using transfer functions W(s) and Wk(s), where Wk(s) serves as a correction transfer function. Participants address issues related to gain discrepancies in the Bode plot, specifically the unexpected high-frequency gain of 1.47 in Wk(s) instead of the desired gain of 1.0. Key calculations involve identifying poles and zeros, with suggestions to eliminate one pole and one zero to achieve the correct frequency response. The conversation concludes with a confirmation that the task has been completed after addressing the identified issues.

PREREQUISITES
  • Understanding of transfer functions in control systems
  • Familiarity with Bode plots and frequency response analysis
  • Knowledge of op-amp circuit design and simulation
  • Ability to manipulate and analyze complex numbers in circuit equations
NEXT STEPS
  • Study the principles of transfer function design in op-amp circuits
  • Learn how to derive and analyze Bode plots for various transfer functions
  • Explore methods for correcting transfer functions in circuit simulations
  • Investigate the impact of poles and zeros on circuit gain and stability
USEFUL FOR

Electrical engineers, students in control systems, and anyone involved in op-amp circuit design and analysis will benefit from this discussion.

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Homework Statement


I have got transfer functions W(s) and Wk(s), where Wk(s) is correction transfer function.
mze1n52gh4shducqu6ss.png

So I calculate W(s) and Wk(s) in series.
Magnitude graph before correction:
325tqtj68jjtfve4xax4.png

Magnitude graph after correction:
w0u4vt6j0l6dx7gy0zb.png


The task is to make whole circuit and simulate it.

Homework Equations



The Attempt at a Solution


First of all, I tried to make it by parts:
xhzrr15rampgus2k6yn.png

e9n83phl8ipa03rslcp.png

d8kevj1wadhpmwwf5b8.png


What about 70/s?
And am I on the right way with otherones?
 
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first, you should cancel the identical pole & zero at s = -10.
then, the second ckt is incorrect. what is the high freq. gain of (0.01s+1)/(0.017s+1)? What components determine the high-freq. gain?
finally, suppose you have an inverter with gain = -Zf/Zi. Equate Zf/Zi to k/s to find your Zi and Zf components.
.
 
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It seems to me, I had a missunderstanding about the task. And there is mistake in the correction TF Wk(s), I guess.
I simulated W(s) - red colour, Wk(s) -green colour, W(s)*Wk(s) - blue colour.
Bode plot:

http://www.bildites.lv/images/5ltl503z567e84sgk3pk.png

The problem is gain of Wk(s) TF. After corner frequency W1, I expect the graphic like it is standart W(s) (red), but it isn't, because there is unexpected gain in Wk(s).

Here are calculations, how I get Wk(s) and simulation:
3iecivaigdiobzfrpbfp.png


And the problem is not because 1/60 = 0.0166666.. = ~ 0.017
There's something wrong with gain. What can be reason for wrong result?
 
I see nothing wrong with this. If Wk(s) = what you say, then your implementation is correct. The low freq. gain is 1.0, the high freq. gain is 1.47 and the time constants all check out.

What about your pole at the origin?
 
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I calculated Wk(s) by myself, it wasn't given.
I have got standart W(s) and I have to implement Wk(s) to get graphic like this (red is correction):
ifvdxt4h1auq0tf5kz13.png

on5jgaoq27ucoh6ewnju.png


But in the bode plot (in my previous post), after last corner frequency the new graphic isn't like the standart W(s), because there is this small gain. Maybe I don't understand something about it, but it suggests that there's mistake in my Wk(s) TF, because the high frequency gain is 1.47, but I want the gain to be 1.0 to get the same graphic after that corner frequency. What could be wrong there with Wk(s) in this case?

Corner frequencies: 1, 4, 10 60 rad/s were suggested by teacher.
 
Well, you're confusing me a lot.

OK, if you want to correct per your latest frequency response (the bottom graph), you need to eliminate one pole and one zero. So figure out what your W_k(s) should be to cancel those. Your Bode plot (red) tells you what to do.
 
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Thanks for the hint. Task already done.
 

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