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Transfer function of Op-Amp circuit

  1. May 6, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data



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    3. The attempt at a solution
    Hello,

    I am working on this previous final exam question. I am totally stumped with getting my answer with frequency. I believe my answer is correct with impedance, but I am having a hard time changing this to ω
     

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  3. May 6, 2014 #2

    Maylis

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    Here is my attempt on the top circuit
    ImageUploadedByPhysics Forums1399359300.379890.jpg

    ImageUploadedByPhysics Forums1399359315.616875.jpg
     
  4. May 6, 2014 #3

    rude man

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    Assuming you made no math errors (your first 3 equations are correct), what's the problem? You have w in your gain expression.
     
  5. May 6, 2014 #4

    Maylis

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    I have to clean up that expression and it really sucks to do that!

    I need to put it into a standard form incase I need to make a bode plot
     
    Last edited: May 6, 2014
  6. May 6, 2014 #5

    rude man

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    Bode plot? OK, have you been introduced to the Laplace transform? If so, replace jw by s and therefore -w^2 by s^2. Do that with your last expression (I assume you did the math correctly). Then I can help you with your bode plot.
     
  7. May 6, 2014 #6

    Maylis

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    I haven't learned Laplace transforms, we always made the plots in the frequency domain
     
  8. May 6, 2014 #7

    rude man

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    Just FYI, I think teaching Bode plots without the s variable is cuckoo.

    BUT - you're stuck, so:

    1. don't multiply by comlex-conjugates as you did in the last line of your sheet 1.
    2. (1/jwC1)/R1 + 1 = 1/jwR1C1 + 1 = (jwR1C1 + 1)/jwR1C1
    Then the remaining terms are suitable for Bode plotting. Do not go (jw)^2 = -w^2. Leave everything in terms of jw. Your transfer function will always be a function of jw.

    EDIT: wait, you have to multiply numerators and denominators by jw until you don't have jw in any denominator of any fraction.
     
  9. May 6, 2014 #8

    rude man

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    Since your transfer function is always a function of jw you can substitute s=jw and get rid of all complex numbers. You don't have to have "had" the Laplace transform. It's just a simple substitution that they should have taught you.

    And BTW s is a frequency variable. A transfer function is always in terms of a frequency variable unless it's just a constant gain.
     
  10. May 10, 2014 #9

    Maylis

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    Here is where I am so far. I don't know how this is suitable for a bode plot yet.

    I am confused, to make the bode plots, is the idea that the numerator is in terms of jw, and the denominator is all real?
     
  11. May 10, 2014 #10

    Maylis

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  12. May 10, 2014 #11

    rude man

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    No.
    Example: a simple R-C low-pass network has transfer function 1/(1 + jwRC). That surely is easy to Bode-plot. But if you wrote (1 - jwRC)/[1 + (wRC)^2] you'd be all messed up.

    Leave everything in jw as much as possible.

    Again, what you should really do is change all jw to s but if they didn't teach you that I guess you'd better hold off. Personally I can't imagine doing Bode plots without that transformation. You don't even need to have had the Laplace transform.
     
  13. May 10, 2014 #12

    rude man

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    You don't want the 1/jwC2 in the numerator. Multiply num & denom by jwC2 to get rid of it.
     
  14. May 10, 2014 #13

    Maylis

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    ImageUploadedByPhysics Forums1399777817.270796.jpg

    here is my latest work. Now I am running into the problem that I don't have enough cycles on my semilog plot (given on the exam) to put everything in there!! that jω pole should go through ω=1. However, my critical values for ω are at 10^10, and 5x10^8. There are not enough cycles to fit that range of frequencies!
     
  15. May 10, 2014 #14

    Maylis

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    I just put bode plots on my regular scratch paper. I'm sure the spacing is way off

    Magnitude
    ImageUploadedByPhysics Forums1399779238.600629.jpg


    Phase
    ImageUploadedByPhysics Forums1399779359.578791.jpg
     
  16. May 10, 2014 #15

    Maylis

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    I did the bottom circuit, this one seems to be a lot easier. That one switch of a capacitor and resistor makes all the difference huh??

    ImageUploadedByPhysics Forums1399780295.562625.jpg

    ImageUploadedByPhysics Forums1399780316.833029.jpg
     
  17. May 11, 2014 #16

    rude man

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    going thru w=1 is not necessary. in fact, it makes no sense per se. you have plenty of cycles along the x axis. You now nearly have the right form for doing the Bode plots, assuming no math slipups along the way. Just one more change: multiply num. & denom. by w1*w2 to get H(jw) = (1e10*w1/w2)(jw+w1)/[(jw)(jw+w2)].

    Now, pick an w << 5e8, say w=1e7. Calculate |H(jw)| at that freq. & plot it. Note that at w=1e7 your |H(jw)| simplifies to ~ 1e10/1e7 = 1e3. Let w=1e7 be the origin along the x axis, then mark off w=1e8, 1e9, 1e10, 1e11, 1e12. That's a total of just 5 decades. Now you're ready to finish the gain Bode plot, right?

    Remember, I did not check your math. Make sure w1 and w2 are what you computed, and that they're not swapped in value.
     
  18. May 11, 2014 #17

    rude man

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    This actually looks good for the gain plot except there is no flat section on the left. There is nothing magic about w=1!

    The phase plot looks bad though. The phase angle is -90 solid until w=5e8, then goes even more negative before returning back to eventually settle at -90 for very high frequencies (say w=1e11).
    I'll let you determine the break angles. Note the correspondence between the gain slopes and the phase angles.
     
  19. May 11, 2014 #18

    NascentOxygen

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    A lot of algebra is involved. Being a sketch, perhaps the intention is that you separately sketch each first-order low-pass Bode plot, then add them graphically? Are you given the element values before being asked to sketch the Bode plot?

    Those capacitor values are in femto-Farads?
     
  20. May 11, 2014 #19

    Maylis

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    Yes those are femtofarads. We were given values to make the bode plot
     
  21. May 11, 2014 #20

    rude man

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    Since youi're only allowed 4 decades on your log paper, mark the x axis as 1e7 (origin), 1e8, 1e9, 1e10 so 1e11 is the extreme right-hand frequency.
     
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