Signal Bandwidth

Discussion in 'Homework Help' started by Peytonator, Nov 11, 2010.

  1. Peytonator

    Thread Starter Active Member

    Jun 30, 2008
    105
    3
    Hi there,

    Say I have a Signal

    X(s) = 1 / [ (s/4) + 1 ] (s = jω)

    then the bandwidth is obviously √4 = 2.

    And if I have

    X(s) = 1 / [ (s^2/4) + s + 1 ]

    It's also √4 = 2.

    But what if I have

    X(s) = 1 / [ (s^3/4) + s^2 + s + 1 ]

    Is it still √4 = 2?
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    I can't nail the theoretical background right now, but I don't think the two last TFs have a knee frequency of 2rads/sec.

    Check this useful application to see what I mean: http://www.williamsonic.com/BodePlot/

    The BW is at the frequency where amplitude drops at -3dB.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The cubic can be resolved into

    s^3+4s^2+4s+4=(s+3.1304)(s^2+0.8696s+1.2778)

    So there are poles at -3.1304, -0.4348+j1.0434 & -0.4348-j1.0434

    One would expect the -3dB point to be less than √(1.2778) rad/sec - the 2nd order part ωo value.

    The underdamped second order component will also probably produce a peak in the spectrum.

    As Georacer points out the -3dB points for each case will differ. His suggestion of using Bode plot is a good one.
     
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