Bandpass Filter Transfer Function

Discussion in 'The Projects Forum' started by EEhokie13, Mar 12, 2012.

  1. EEhokie13

    Thread Starter New Member

    Oct 7, 2010
    3
    0
    I am trying to solve for the Transfer Function in this Bandpass filter. I have tried several methods. What I am trying to do is use KCL at the three essential nodes at the output of each op-amp. I have most of the equations but the one for the first op-amp(top left) I cant see to figure out what to do. I am working on this project with a few friends and we are struggling with the same spot. Any tips would be great on how to approach this question.

    [​IMG]
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    This belongs in the homework forum and you need to show your work first.

    In problems like this, you need to be clear about what OPAMP model you are using. Are you assuming ideal OPAMPs?
     
  3. EEhokie13

    Thread Starter New Member

    Oct 7, 2010
    3
    0
    Thank you for the response. This is a project my friends and I are working on and so I thought it was fitting to put under projects. As for my work and about the op-amps. We are using the ideal op-amp.

    I used KCL as mentioned above for the 3 essential nodes.

    1st Node(The upper left op-amp output node) NODE A
    2nd Node(The upper right op-amp output or Y(s)
    3rd Node(The bottom op-amp) NODE C

    1st Node KCL (NODE A)

    (V(A) - Y(s)/20k) + something...(this is where I am confused) = 0

    3rd Node KCL (NODE C)

    (V(C) - 0V/Rf) + (V(C) - Y(s)/C + Rf) = 0

    2nd Node KCL(NODE Y(s))

    (Y(s) - V(A)/20k) + (Y(s) - V(C)/C + Rf) = 0

    I am stuck on the input of this circuit and I am not sure how to deal with the short circuit. I understand the nodes connected to the op-amps are 0 and there is no entering current. I am stumped none the less. Thanks for the help.
     
  4. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    OK, if you feel it's a project, perhaps the moderators will leave it here, but you are still subject to the same rules as homework. We'll guide you, but won't do the work for you and we need to see some effort as you that show above.

    The current at the output nodes (i.e. use of KCL) is not a good constraint with ideal opamp models because there is no good way to independently know the current supplied by the output of the opamp. Basically, KCL applied to the output node allows you to calculate the OPAMP output current, but it does nothing else useful for you in this problem. You could use KCL at the negative input nodes of the opamps. You can also treat each stage as a voltage gain and apply the voltage gains around the feedback loop. This second method is what I would use because I already know how each stage behaves. However, if you don't know how each stage behaves, you can derive the voltage gain formulas for each isolated stage.

    For example, the output of the upper left opamp goes into the upper right stage which provides a gain of -1 to generate Y(s). If you don't already recognize this upper right stage as an inverting amplifier, derive it. The bottom opamp forms an integrator circuit and the upper left stage is an adder combined with a filtered gain.

    To summarize the two suggested methods.

    1. Derive the voltage gain of each of the three stages and apply the equations in the feedback loop.

    2. Derive the complete system response from fundamental KCL equations, and generalize Ohms law for caps and resistors (in the s-domain) at the negative terminal rather than the output terminals.

    Note that all derivations can be done in the Laplace domain (s-domain) which allows algebraic linear equations rather than differential equations as needed in the time domain.
     
    Last edited: Mar 13, 2012
  5. DickCappels

    Moderator

    Aug 21, 2008
    2,656
    632
    I would like to get some of those ideal opamps -been looking for them for quite some time. Please post the contact information for the supplier and the price.

    Thank you!:)
     
  6. EEhokie13

    Thread Starter New Member

    Oct 7, 2010
    3
    0
    I have been working more on the transfer function I am still stuck but I think I am making progress.

    I have been looking at each op-amp separately and I think I have determined their outputs. For the right op-amp I know it is a inverter op-amp and the output equation is (-R1/R2)(Vi) so I get (-1)(Vi) = Y(s). That means the input for the bottom op-amp is going to be Y(s). The bottom op-amp is an integrator op-amp. So (-Vi/(omega)(R)(C)) and Vi in this case will be Y(s). The top op-amp like you said is an adder with the filtered gain. I can come up with an equation for the adder but I do not know how to use the gain or incorporate it in the adder equation. I believe using this equations will get me the transfer function but maybe I am wrong. Thanks again.
     
  7. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    It looks like you are thinking along the correct lines overall. For this equation, it is better to stay in the s-domain, rather than introducing frequency omega. Hence, the impedance of a capacitor is 1/(sC), and the integrator formula is Vo=-Vi/(sRC).

    When frequency (omega) is used directly, you need to substitute in s=jω, where j is the square root of negative one (j=√-1). The imaginary value is what retains the information that current and voltage are 90 degrees out of phase. Your forumla above only considers magnitude of the impedance, but phase and magnitude are both important.

    Note the similarity of the inverting amp and the integrating amp. Both have a formula of Vo/Vi=-Z2/Z1, where Z2 is the feedback impedance and Z1 is the input impedance.

    When you get to the adder stage, you have two inputs Va and Vb, so the transfer function is Vo=-Z2*Va/Za-Z2*Vb/Zb, where the impedances are self evident. The only confusion is what is Z2, but this is the parallel combination of the resistor and the capacitor, so you can figure that out.

    These formulas you can derive using the KCL at the negative terminal of the opamp. The fact that no current flows into the inputs of an ideal opamp is very useful here.

    Note that if you ever have a coil in your circuit, then impedance of the coil is sL, and the same rules apply.
     
Loading...