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Ugly xfer function

Discussion in 'Electronic Basics' started by Active8, Aug 7, 2004.

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  1. Active8

    Active8 Guest

    Being a true masochist, I thought I'd try deriving the Laplace xfer
    function for this filter and it's getting ugly. Before I go further,
    I thought I'd ask if I'm setting it up right, er, umm, ...
    efficiently?



    +------+-------+ _______
    | | | 1 | R1+R3
    | .-. | fr= ----- \|-------
    --- | |R2 | 2pi*C R1*R2*R3
    C --- | | |
    | '-' | R2
    ___ | || | |\ | gain= - -----
    o-|___|-+-+--||--+--|-\ | 2R1
    R1 | || | >-+-o
    .-. C +--|+/
    | |<-+ | |/
    R3| | | | 1
    '-' | | Bandwidth= -------
    | | | pi*R2*C
    === === ===
    GND GND GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
    view\fixed font


    I figured if I wrote the branch currents coming off the node to the
    right of R1, I'd get both e_in and e_out in the equation and go from
    there. So I set the inv op amp input to 0, and figured the
    combination of the 2 caps and R2 in || with R3 - call that Z_ugly.
    That would give me a voltage divider eq for

    e_n = e_i * Z_ugly / (R1 + Z_ugly), e_n being the voltage at the
    node to the right of R1.

    Isn't there a better way to set this up? I'd like to know before I
    try to simplify Z_ugly any further.

    TIA
     
  2. Roy McCammon

    Roy McCammon Guest

    sometimes you can use a trick.

    Vo = -Vx * R2 / (1/sC2)

    I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

    Vi = Vx + R1*I1

    solve the first equation for Vx
    Vx = -Vo/sRC2
    substitute that into the second equation
    I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
    and put that into the klast equation

    Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2 -Vo)/(1/sC1)]
     
  3. Active8

    Active8 Guest

    What's assumptions (if that's the right word) do we make to justify
    using that trick?

    In the mean time, I'll try simplifying the above and BTW, I'm trying
    to come up with those formulas, above, one way or the other.
     
  4. Ban

    Ban Guest

    Here there is the transfer function, w=omega.

    R2*R3
    - -----C*wr*P
    Vout R1+R3
    ---- = --------------------------------------
    Vin 2R1*R3 R1*R2*R3
    1 + ------C*wr*P + --------(C*wr*P)^2
    R1+R3 R1+R3

    You can then compare this with the principal bandpass equation:

    Ar
    --- P
    Vout Q
    ---- = ----------------
    Vin 1
    1 + --- P + P^2
    Q

    and derive the design equations from that.
    Well, not exactly s.e.b. :)
     
  5. Active8

    Active8 Guest

    Oh, I wasn't sure :) I'll look at that, but I was trying to figure
    out if I had the problem set up right so I could derive the
    *Laplace* xfer function, myself.

    What is wr? 2.pi.f ? What about P and Ar?
     
  6. Active8

    Active8 Guest

    I'll repost in sed.
     
  7. Ban

    Ban Guest

    wr = omega resonance = 2*pi*fr fr=resonance frequency
    P = complex frequency variable = j(w/wr) = j (f/fr)
    Ar = gain at resonance

    I gave you the solution already, sorry for that. But this way you can
    control better your effort to derive it from the knots. You are on the right
    track. Start with the first knot, and sum up the currents. Then make a loop
    and sum up the voltages. Don't forget that the -in is a virtual gnd.
     
  8. Roy McCammon

    Roy McCammon Guest

    sorry to take so long to get back to you, but it looks like you got
    plenty of good help.

    as for the trick, is a matter of recognizing that a
    sub circuit composed of the opamp, R2 and c2 is just
    a standard inverting amplifier and replacing
    that with the gain equation for the standard inverting
    amplifier. Sorry that I didn't explain that.

    But as a general rule that is the way I attack
    complicated circuits; try do identify known
    circuit blocks and replace then with an equivalent
    equation.
     
  9. Active8

    Active8 Guest

    I think the "trick" wasn't a trick at all. I've worked it all out
    days ago. Normally, I'd first try to divide output Z by in Z, but
    this had multiple feedback paths. Your "trick", IIRC was to evaluate
    the node to the right of the input R by seeing that the current in
    to and through R1 (Rs) to virtual gnd also flowed through whatever R
    it was that connected directly to Vout. That solved node Vx and thus
    the whole deal. I was looking at it ... I over complicated things.

    TNX much.
     
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