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Cutoff frequency in this circuit

  1. Jul 9, 2013 #1
    sOXxR.png

    I'd say that the cutoff frequency of the circuit in the following scheme is given by [tex]f_0=\frac{1}{2 \pi (R_1//R)C}[/tex] but I don't know how to prove this idea.. Do you agree with me? what would you say? Many thanks!
     
  2. jcsd
  3. Jul 9, 2013 #2

    gneill

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    Staff: Mentor

    You haven't specified what the source is connected to... We can't tell if it has additional frequency dependent components connected to its + end, or even if it's a sinusoidal source. You haven't specified where the "output" is to be measured (while we might guess that it's across "R", a guess is not the same as knowing).

    Further, you'll have to show some attempt to derive the corner frequency so that we can help. We can't just confirm or deny guesses, as this won't help you in the long run; you need to be able to plan and execute a strategy to solve this sort of problem.

    What do your course materials say about the corner (or cutoff) frequency? How is the cutoff frequency related to the transfer function? Can you determine the transfer function for your circuit?
     
  4. Jul 10, 2013 #3

    rude man

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    You're way off in your transfer function.

    This circuit has a low-frequency gain of R/11R, then climbs with increasing frequency until it reaches unity gain.

    By analysis you should be able to come up with the transfer function incl. the frequency at which the gain starts to climb and the (higher) frequency at which it starts to level off (the asymptotes). Otherwise known as a Bode plot. There is no 'cutoff frequency'. The gain is always 0.091 < gain < 1.0.

    I'm assuming the voltage source is grounded at its other end.
     
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