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Transfer Function and State Space Analysis of Op-Amp circuit

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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 [Broken]

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

    and for large A

    [itex]\frac{V_{o}(s)}{V_{i}(s)} = \frac{Z_{1}(s) + Z_{2}(s)}{Z_{1}(s)}[/itex]

    For my circuit

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

    I'll make the follow substitutions

    [itex]R_{4} = 110 KΩ[/itex]
    [itex]C_{2} = 4 μF[/itex]
    [itex]R_{3} = 600 KΩ[/itex]
    [itex]R_{2} = 400 KΩ[/itex]
    [itex]C_{1} = 4 μF[/itex]
    [itex]R_{1} = 600 KΩ[/itex]

    Now then

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

    Hence for my circuit the transfer function is

    [itex]\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}}[/itex]

    I can rearrange and get

    [itex]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})[/itex]
    [itex]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)[/itex]
    [itex]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)[/itex]
    [itex]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}}})[/itex]

    I'm not sure how to evaluate this further
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 27, 2014 #2

    donpacino

    User Avatar
    Gold Member

    starting here

    R2Vo(s)+[itex]\frac{R1}{C1R1s+1}[/itex]Vo(s)=R2Vi(s)+[itex]\frac{R1}{C1R1s+1}[/itex]Vi(s)+R3Vi(s)+[itex]\frac{R4}{C2R1s+1}[/itex]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.

    [itex]\frac{R2(C1R1s+1)}{C1R1s+1}[/itex]Vo(s)+[itex]\frac{R1}{C1R1s+1}[/itex]Vo(s)=....


    [itex]\frac{R2(C1R1s+1)+R2}{C1R1s+1}[/itex]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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