Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simple RC and RL Transfer Functions

  1. Aug 19, 2008 #1
    [Solved] Simple RC and RL Transfer Functions

    1. The problem statement, all variables and given/known data
    Given the transfer function:

    TF= (200)/(1 + j0.001w)

    a. Determine the cuttoff frequency in radians per second and in hertz

    2. Relevant equations

    There are a couple other equations like this in the book, but I just don't grasp the concept or the formulas that the book offers which looks like TF(w)= Vout/Vin = A < Theta degrees.

    3. The attempt at a solution

    TF= Vout/Vin = 200 / (1 + j0.001w) --> converted to rectangular form

    (200 < 0)/(1 < .057) = 200 < -.057

    Which I know is horribly wrong, but these formulas just don't make sense to me. :/
    Last edited: Aug 19, 2008
  2. jcsd
  3. Aug 19, 2008 #2
    Figured it out and did some deeper reading online, came up with:

    Wc= 1/t

    Where in the equation TF=200/(1+j.001w), t=.001

    So, 1/.001 = 1000

    Then you take Wc/(2*pi)

    Which then gives you 159.15 Hz
  4. Aug 20, 2008 #3


    User Avatar
    Science Advisor

    Looks like you've already answered your own question, but I thought I'd elaborate...

    To find the corner frequencies of a transfer function (incidentally, the transfer function you give is that of a low-pass filter--Wikipedia for it), you just have to find its poles ('w' which make the denominator go to zero).

    In this case, you just set 1+0.001j*w to zero, and determine that w_c is 1000j, which you interpret as 1000 rad/s, or with the formula that w_c = 2*pi*f_c, 159 Hz. (Technically, it should be when the magnitude of the transfer function goes to zero, which gives you the two poles at +/-1000j, and hence, a real frequency of 1000 rad/s).

    What is the physical significance of this divide by zero? (Assuming you haven't taken any complex analysis courses)--the system's 'quirks' emerge and have a significant impact (in the case of a low-pass filter, that frequencies higher than the cut-off get significantly attenuated or, cut-off, so to speak). Yes, yes, this is an idealized view of things, but it does give you significant useful information.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook