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Homework Help: 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.
     
  22. May 11, 2014 #21

    Maylis

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    ImageUploadedByPhysics Forums1399843707.997897.jpg

    Rude man, I want to show you the table that are given in order to construct bode plots. Basically, the way we do it is that we have to get our transfer function to look like one of the factors here, and then use that to construct the plot.

    I want to take note in particular of the pole at the origin. Notice how it has a slope of -20N dB/decade at exactly 1 rad/s. That is what I am talking about in regards to that 1 rad/s that I have included in my bode plot, because I have a pole at the origin in my transfer function.
     
  23. May 11, 2014 #22

    rude man

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    I see no problem with your table. The desired form is just how I've tried to get you to arrange your transfer function - i.e. as functions of jw.
    Your post 14 clearly shows a flat segment from the left-hand end to w=1. This is wrong. The slope runs -20dB/decade all the way from the y axis to your first break point at w = 5e8 rad/s.

    Again - there is nothing sacred about w=1. Do not use it in any way unless your break frequencies are so low as to make that necessary.

    A pole at the origin means infinite gain at dc so no matter how low w goes, the slope goesd as -20 dB/decade all the way from the y axis to wherever the next break frequency is. It does not start at w=1. Even if you could plot for w=1 which you can't, as you yourself pointed out (not enough decades on the x axis!).
     
  24. May 12, 2014 #23

    Maylis

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    I don't get it, my transfer function is a function of jw, no??
     
  25. May 12, 2014 #24

    rude man

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    Yes, and a "jw" by itself (no "+1" added) IS a pole at the origin, i.e represents an integrator with a -20 dB/decade rolloff all the way to zero rad/s. You showed that correctly in an earlier post (post 14) except you straightened out the slope from the y axis to w=1. In fact, that magnitude plot was otherwise spot-on.
     
  26. May 12, 2014 #25

    Maylis

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    ImageUploadedByPhysics Forums1399928456.519430.jpg

    But look now, if I have 10^7 as my starting point, how will I know what magnitude to start its slope at???
     
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