Active bandpass filter

Discussion in 'Homework Help' started by peter_morley, Apr 21, 2011.

  1. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    I have attached my work I have done so far with this problem.
    I sometimes get lost when I use op amps in circuit analysis. We didn't do much with op amps in class it was just assumed that we would learn them on our own. So let me know how this looks to you. I am trying to figure out the transfer function which is Vout/Vin and from that equation find the center frequency of the filter which happens to be the square root of A3. Well good luck helping me. I just realized the first top right equation is wrong the result should be Vout = -V*(R2)*j*w*C
     
    Last edited: Apr 21, 2011
  2. Audioguru

    New Member

    Dec 20, 2007
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    If you don't have a text book then look in Google at Multiple Feedback Bandpass filter. Each article has the formulas.
     
  3. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    alright well I don't need a random calculator I need to be able to derive an equation from the schematic using my knowledge of op amps (very little) and impedance. Can you take a look at my equation to see if I'm doing this right?
     
  4. Heavydoody

    Active Member

    Jul 31, 2009
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    Is Vout really supposed to be ground?
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    When the output of the opamp is shorted to ground like that then the electrons can't leak out. Some people do the same with a battery.;)
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    Did you mean to connect the output of the opamp to ground? If yes, can you explain your logic for doing so?

    hgmjr
     
  7. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    Vout needs to be across a load being a resistor or what have you. Lets say 500 ohms. I just left that out by mistake because im an idiot.
     
  8. beenthere

    Retired Moderator

    Apr 20, 2004
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    An ideal op amp can handle just about any load. An actual op amp has limits on the current it can source and sink. It would probably be better to choose a much higher load like 10K.
     
  9. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    Lets just pretend Vout is an open circuit at that node.
     
  10. Heavydoody

    Active Member

    Jul 31, 2009
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    I gave it a shot, but I fear I may have made an algebraic mistake somewhere...here it is.
     
  11. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I see a mistake that will have a large effect.

    At one point you have:

     \frac{V-0}{\frac{1}{\frac{1}{S C1}}} = \frac{0-Vo}{R2}

    but it should be:

     \frac{V-0}{\frac{1}{S C1}} = \frac{0-Vo}{R2}
     
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  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    It can be set up easily in admittance matrix form:

    \left[ \begin{array}{2}\frac{1}{R1}+\frac{1}{R3}+2 s C & -s C\\s C & \frac{1}{R2}\end{array}\right]\left[ \begin{array}{1}V\\Vo\\\end{array}\right]=\left[ \begin{array}{1} \frac{Vin}{R1} \\ 0 \\\end{array}\right]

    Then use a modern scientific calculator that can solve simultaneous equations expressed in matrix form, such as an HP50 or TI89, etc., or a mathematical software program such as Matlab, Scilab, etc. The individual elements of the matrix are so simple it's almost impossible to make a mistake. The computer does the final algebra and doesn't make algebraic mistakes.

    Of course, with only a 2x2 admittance matrix, the solution can be obtained by hand fairly easily, but putting the problem in matrix form to start is a good habit to acquire because it requires one to be systematic.
     
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