Transfer Function and State Space Analysis of Op-Amp circuit

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SUMMARY

The discussion focuses on deriving the transfer function and state space representation of a non-inverting operational amplifier circuit. The transfer function is established using the formula \(\frac{V_{o}(s)}{V_{i}(s)} = \frac{Z_{1}(s) + Z_{2}(s)}{Z_{1}(s)}\), where \(Z_{1}(s)\) and \(Z_{2}(s)\) are the impedances defined by the circuit components. Specific component values include \(R_{4} = 110 \, KΩ\), \(C_{2} = 4 \, μF\), \(R_{3} = 600 \, KΩ\), \(R_{2} = 400 \, KΩ\), \(C_{1} = 4 \, μF\), and \(R_{1} = 600 \, KΩ\). The discussion concludes with a need for further evaluation to express the transfer function in standard form.

PREREQUISITES
  • Understanding of operational amplifier circuits
  • Familiarity with transfer functions and state space representation
  • Knowledge of impedance calculations for resistors and capacitors
  • Proficiency in Laplace transforms
NEXT STEPS
  • Learn how to derive transfer functions for different op-amp configurations
  • Study state space representation techniques in control systems
  • Explore the use of MATLAB for simulating op-amp circuits
  • Investigate the effects of varying component values on circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in control systems analysis will benefit from this discussion on operational amplifier circuit analysis.

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Homework Statement



[PLAIN]http://imagizer.imageshack.us/v2/800x600q90/690/0tab.png
I'm asked to find the transfer function and then find the state space representation and seem to be stuck.

Homework Equations





The Attempt at a Solution



My textbook states that for a noninverting operational amplifier shown below

http://imagizer.imageshack.us/v2/800x600q90/10/zmir.png

\frac{V_{o}(s)}{V_{i}(s)} = \frac{A}{1 + frac{AZ_{1}(s)}{Z_{1}(s) + Z_{2}(s)}}

and for large A

\frac{V_{o}(s)}{V_{i}(s)} = \frac{Z_{1}(s) + Z_{2}(s)}{Z_{1}(s)}

For my circuit

[PLAIN]http://imagizer.imageshack.us/v2/800x600q90/690/0tab.png

I'll make the follow substitutions

R_{4} = 110 KΩ
C_{2} = 4 μF
R_{3} = 600 KΩ
R_{2} = 400 KΩ
C_{1} = 4 μF
R_{1} = 600 KΩ

Now then

Z_{1}(s) = R_{2} + (C_{1}s + \frac{1}{R_{1}})^{-1}
Z_{2}(s) = R_{3} + (C_{2}s + \frac{1}{R_{4}})^{-1}

Hence for my circuit the transfer function is

\frac{V_{o}(s)}{V_{i}(s)} = \frac{Z_{1}(s) + Z_{2}(s)}{Z_{1}(s)} = \frac{R_{2} + (C_{1}s + \frac{1}{R_{1}})^{-1} + R_{3} + (C_{2}s + \frac{1}{R_{4}})^{-1}}{R_{2} + (C_{1}s + \frac{1}{R_{1}})^{-1}}

I can rearrange and get

V_{o}(s)(R_{2} + (C_{1}s + \frac{1}{R_{1}})^{-1}) = V_{i}(s)(R_{2} + (C_{1}s + \frac{1}{R_{1}})^{-1} + R_{3} + (C_{2}s + \frac{1}{R_{4}})^{-1})
R_{2}V_{o}(s) + \frac{R_{1}}{C_{1}R_{1}s + 1}V_{o}(s) = R_{2}V_{i}(s) + \frac{R_{1}}{C_{1}R_{1}s + 1}V_{i}(s) + R_{3}V_{i}(s) + \frac{R_{4}}{C_{2}R_{1}s + 1}V_{i}(s)
L^{-1}(R_{2}V_{o}(s)) + \frac{\frac{1}{C_{1}}}{s + \frac{1}{C_{1}R_{1}}}V_{o}(s) = L^{-1}(R_{2}V_{i}(s) + R_{3}V_{i}(s)) + \frac{\frac{1}{C_{1}}}{s + \frac{1}{C_{1}R_{1}}}V_{i}(s) + \frac{\frac{1}{C_{2}}}{s + \frac{1}{C_{2}R_{4}}}V_{i}(s)
R_{2}V_{o}(t) + \frac{1}{C_{1}}L^{-1}(\frac{V_{o}(s)}{s + \frac{1}{C_{1}R_{1}}}) = (R_{2} + R_{3})V_{i}(t) + \frac{1}{C_{1}}L^{-1}(\frac{V_{i}(s)}{s + \frac{1}{C_{1}R_{1}}}) + \frac{1}{C_{2}}L^{-1}(\frac{V_{i}(s)}{s + \frac{1}{C_{2}R_{4}}})

I'm not sure how to evaluate this further
 
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R2Vo(s)+\frac{R1}{C1R1s+1}Vo(s)=R2Vi(s)+\frac{R1}{C1R1s+1}Vi(s)+R3Vi(s)+\frac{R4}{C2R1s+1}Vi(s)

I would get everything on one side of the equation under a common denominator. I'll do the left half of an equation.

\frac{R2(C1R1s+1)}{C1R1s+1}Vo(s)+\frac{R1}{C1R1s+1}Vo(s)=...


\frac{R2(C1R1s+1)+R2}{C1R1s+1}Vo(s)=...

You do the right half. Then you can get it in the standard form for a transfer function.

We'll address the state space work when you get there
 

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