Transfer function

Discussion in 'Homework Help' started by boks, Oct 26, 2008.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
  2. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    I'd start doing nodal analysis in the frequency domain then take the inverse laplace transform. Keep in mind the ideal opamp "golden rules." Are you familiar with them? Or how to solve opamp equations? Let us know what work you've gotten so far so we can give better help.

    Looks like you got a couple voltage followers (likely acting as impedance buffer) that feed an inverting amplifier.
     
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    The purpose of the circuit depends on the selection of the capacitor and resistor that make up the integrator and the differentiator at the input of the circuit. I would think that it is most likely a Band-Reject active filter.

    hgmjr
     
  4. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    boks,

    Easy, find the relationship of Vo to Vi. Let's start. Vi is feeding two op amp chains. For the top, the voltage at the + terminal of op amp 1 is (Vi)1/C1s//(R1+1/C1s) = Vi/(R1C1s+1) volts. This voltage is the same at the - terminal of op amp 1 and the output of op amp 1, since the voltages are assumed to be same across the op amp input terminals Similarly we get for the bottom chain (Vi)sR2C2/(sR2C2+1) volts at the output of op amp 2. Since the - terminal of op amp 3 is at a virtual ground, we can get the current at the - terminal of op amp 3 by dividing both voltages by R4. Now, no current can pass from one input terminal to the other of op amp 3, so it all goes through R3. Therefore the voltage Vo is (V1R3/R4)((1/(sR1C1+1))+sR2C2/(sR2C2+1)) and the transfer function is apparent by dividing by Vi.

    Ratch
     
    Last edited: Oct 28, 2008
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
    326
    I think you've got parentheses in the wrong place in this expression. If you evaluate it according to the usual rules of precedence, it doesn't give the right result.
     
  6. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    The Electrician,

    Thanks, I was missing a parenthesis, I believe. I corrected it.

    Ratch
     
  7. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    I. The output attempts to do whatever is necessary to make the voltage difference between the inputs zero.

    II. The inputs draw no current.

    We have not learnt about Laplace transforms in this course. Is it necessary to solve this problem?
     
  8. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Sorry, by "feed" i meant, feeds a voltage signal.

    Oh ok. No you don't need it. Just makes things easier. You'll just have to stay in the time domain. So the formulas that have been given won't help much.

    So just treat it as a second order circuit (cuz that's what it is). That is, find the initial and final conditions. You'll also use nodal analysis to determine the appropriate differential equation. Then solve and apply the initial conditions.

    Does that help at all? If not, i (or someone else) can be more specific if need be.
     
  9. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    Yes please. I'm quite new in this business.
     
  10. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Well i see 4 unknown nodes. Vi, V1 (inputs/output of opamp1), V2 (inputs/output of opamp2), and Vo.

    Note that the inverting input of opamp3 is ground (by opamp golden rule) and that V1 is the voltage for both inputs of opamp one (by opamp golden rule) and the output of opamp1 (because of the feedback). V2 for the same reasons as V1.

    I'll find the equation for V1, and i'll leave it to you to try the others. This is what i get:

    (V1-Vi)/R1 + C*dV1/dt = 0

    You'll get a system of diff equations which can be combined by substitution into one equation. Let us know if you get stuck!
     
  11. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    I gave it a try. First, let me know if my equations are correct.

    The voltages are the same where the color is the same here, right? So V=0 in the yellow line? Since there's zero voltage, does this mean that no current goes through R3?

    http://www.badongo.com/pic/4651107

    Here are my three equations:

    (Vi - V1)/R1 = C * dV1/dt
    C * d(Vi - V2)/dt = V2/R2
    -Vo/R3 = 0

    And how do I know that this is a secondary circuit?
     
    Last edited: Oct 28, 2008
  12. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Yup, looks good to me.

    Not necessarily. That is only the case if Vo = 0 as well. As long as there's a potential difference between Vo and the yellow node, then R3 will draw some current.

    Well you're first two equations look good to me.

    However, the third one is not necessarily true by an opamp golden rule: You cannot do nodal analysis directly at the node that is the output of an opamp because we don't know the current there.

    What i would do is do nodal analysis at the yellow node. Even though it is a known voltage, the equation there will still be able to help you find the others.

    You mean second order? Simple: there are two energy storage elements (inductor/capacitor). This just means that you'll have a circuit described by a second order differential equation.

    One energy storage element = first order, two elements = second order, three = third order, etc, etc.
     
  13. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    V1/R4 + V2/R4 = -Vo/R3, so that Vo = -2 V1

    By a lot of substitution, I end up with the equation dVi/dt + 450 000 Vo - 200 000 Vi = 0

    Can I find the transfer function from this?
     
    Last edited: Oct 28, 2008
  14. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    boks,

    No, you cannot. Where is C1,C1,R1,R2,R3,R4 ? I gave you the transfer function in post #4 of this thread. If you want to see what the Laplace transfer function represents in real time for a sinusoidal input, just substitute in s=jw . Did you read my explanation of how to derive it? Is there anything you don't understand about the explanation?

    Ratch
     
  15. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    We're not meant to use Laplace to solve this problem. I think guitarguy's node method is what I'm meant to use.
     
  16. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Well, i haven't crunched the numbers, but the first equation looks good.

    However, i'm curious why you only have a first order differential equation. I believe you should get a second order diff eqn.
     
  17. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
    326
    Have a look at this:
    http://www.me.cmu.edu/ctms/modeling/tutorial/transferfunction/mainframes.htm

    and this:
    www.uwm.edu/Course/ee474/RevisedLectureNotes/Lect7_2Up.pdf

    If you follow guitarguy's method, you will end up with differential equations, the solution of which will be time response, not the transfer function.

    To get the transfer function, you will have to find Laplace transforms of the differential equations.

    Back before about 1890, the method for solving circuits with reactive components was to set up and solve the differential equations of the circuit. But then Steinmetz (http://en.wikipedia.org/wiki/Charles_Proteus_Steinmetz) showed how to do it with simple algebra.

    To use his method, you let the impedance of a capacitor be given by 1/sC, the impedance of an inductor by sL, the impedance of a resistor by R, and then just solve the AC circuit the same way you would a DC circuit of only resistors. There must have been some mention of this in your class, because the transfer function is given in terms of the complex variable s (or, possibly with the variable s replaced with jwf). See the two references I gave above.

    The solution Ratch gave is what you want.
     
  18. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    bok,

    Then don't use Laplace. For setting up the response to a sinuoidal input, the classical and the Laplace method are virtually identical. Only when you find the inverse of the Laplace is there a difference. For example, the first equation to find the voltages at the terminals of op 1 is (Vi)1/C1s//(R1+1/C1s) = Vi/(R1C1s+1) . By simply substituting s=jw for s, we get (Vi)1/C1jw//(R1+1/C1jw) = Vi/(R1C1jw+1), which is what you would get if you did it by the classical method. You are missing a lot of good info by ignoring what I showed you in post #4. You are in effect missing the solution to your problem.

    Ratch
     
  19. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    Ok my bad possibly. I assumed that since he couldnt use laplace transforms that he wanted the solution in the time domain.

    If freq domain is ok, then heed ratch's advice.
     
  20. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    Equations:

    (1) Vi/R1 - V1/R1 = C dV1/dt

    (2) C dVi/dt - C dV2/dt = V2/R2

    (3) Vo = -2V1

    I now substitute (3) and V1 = V2 into (1) and (2):

    (4) Vi/R1 + Vo/2R1 = -(C/2) dVo/dt

    (5) C dVi/dt + (C/2) dVo/dt = -Vo/2R2

    (4) can be written

    (6) Vi = (-R1C/2)dVo/dt - Vo/2

    (6) substituted into (5) yields

    (7) (-R1C^2/2) d^2V0/dt^2 + (C/2) dV0/dt + Vo/2R2 = 0
     
    Last edited: Oct 29, 2008
Loading...