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Homework Help: Amplification of an integrator and a differentiator

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    The amplification of an integrator is:
    A = Vout/Vin = -Zf/Zin = +j * 1/(ωRC)

    The amplification of an differentiator is:
    A = Vout/Vin = -Zf/Zin = -jωRC

    Although, that's what my book says. Not that I doubt it..

    But really, what do Zf, Zin, j, and ω stand for?
    I can't find much about it.

    The book only says the following:
    ω = 1/(RC), but why? what does this mean?
    and that Vout 90° lags (-j) at Vin, regardless the frequency.
    Thus: A is (straight?) proportional with the increasing frequency, so increases with 6dB each octave/scale and with 20dB each decade/level.

    But really, I have no idea what this actually means.

    In the classes we didn't attend this to the matter and we don't have to know this for the upcoming exam. But I'm just curious ;)

    Thanks in advance,
  2. jcsd
  3. Apr 8, 2010 #2
    I'm not sure about what you understand and what you don't... Do you know what [tex]\omega[/tex] stands for? And [tex]j[/tex]? If you don't... what do you know?
  4. Apr 8, 2010 #3
    Well... nothing about these 4 terms, really.
    We didn't attend this part of the integrators to the matter at all but I'm just curious about it! ^^

    For ω, I understand what 1/RC means. But not where it comes from, what's behind it?
    And [tex]j[/tex], 90°? Does this mean the real wavelenght is 1/2π earlier, where does this lag come from? And what has this to do with the amplification?
    And Zf and Zin, what do the Zs stand for? Zener-final and Zener-initial? Is there any Unit in the SI-system for these?

    So as you can see, I'm just guessing around.
    What do these symbols stand for?
  5. Apr 9, 2010 #4
    Hm... you need some background in electronics. Zf and Zin stand for the impedance, which is a complex value, hence the [tex]j=\sqrt{-1}[/tex]. And [tex]\omega[/tex] is the frequency of the oscillating signal. If all this confuses you, take a look at http://en.wikipedia.org/wiki/Electrical_impedance.
  6. Apr 9, 2010 #5
    Okay I think, after reading reading that link you gave me, I've got a better understanding of it now. Not 100% clear, but I was just curious about it and I'm really tired. Just had prelims/exams (what do you call it again? :smile:) the last 2 weeks.

    Anyways, Thanks a lot!
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