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Bode Plot By hand

  1. Jun 29, 2011 #1
    How would you plot the magnitude and phase Bode plots for the system transfer function of (4s^3+40s^2)/(4s^4+405s^3+504s^2+400s)
    I plugged it into MATLAB and got a result but my professor wants hand calculations.
    Thanks
     
  2. jcsd
  3. Jun 29, 2011 #2

    berkeman

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    Staff: Mentor

    Start by calculating the values of the poles and zeros of the transfer function...

    Then once you have the poles and zeros, do you have an idea of how to start plotting the gain and phase versus frequency...?
     
  4. Jun 29, 2011 #3
    I gave you the wrong transfer function its from a block diagram that looks like this
    [4+(40/s)]---->[s^2/(s+100)(4s^2+5s+4)] if i could get the transfer function from that i think i know how to get the poles and zeros
     
  5. Jun 29, 2011 #4

    berkeman

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    Staff: Mentor

    Could you please show the block diagram? Do you mean it is a traditional feedback block diagram?
     
  6. Jun 30, 2011 #5
    It is a block diagram with no feedback so the two blocks above are in series
     
  7. Jun 30, 2011 #6

    berkeman

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    So as was pointed out to you in the other thread, how do you get the overall transfer function if the two transfer functions are in series?
     
  8. Jul 1, 2011 #7
    ok so the transfer function is (4s^2+40s)/(4s^3+409s^2+540s+4) now how do i sketch the bode plot? I know i have to convert the transfer function into a different form but how do i do that
     
  9. Jul 1, 2011 #8

    berkeman

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    You need to factor the transfer function to find the poles and zeros. Those are what you use to sketch the transfer function.

    http://i.cmpnet.com/powermanagementdesignline/2008/05/BassoFig319a.jpg

    BassoFig319a.jpg
     
  10. Jul 1, 2011 #9
    instead of the 4 in the transfer function its 400 sorry about that
    Berkeman, how do you factor the transfer function?
    Thanks for your help so far
     
  11. Jul 1, 2011 #10

    berkeman

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    You factor it like you do any polynomial. You will end up with a factored polynomial in the numerator (which gives you the zeros), and a factored polynomial in the denominator (which gives you the poles).

    http://www.google.com/search?source...1T4GGLL_enUS301US302&q=factoring+a+polynomial

    .
     
  12. Jul 1, 2011 #11
    So for the numerator i get 4s(s+10) so then theres a zero at s=-10 and s=0?
     
  13. Jul 1, 2011 #12

    berkeman

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    Yes, if that's the numerator, then those are the zeros.
     
  14. Jul 1, 2011 #13
    for the denominator i but it into matlab and this is what i got

    >> p = [4 409 540 400]; % p(x) =4s^3+409s^2+540s+400
    format long; % print double-precision
    roots(p)

    ans =

    1.0e+002 *

    -1.009221534098646
    -0.006639232950677 + 0.007416660864940i
    -0.006639232950677 - 0.007416660864940i
     
  15. Jul 1, 2011 #14
    Now those are the poles with 1 real and two imaginary
     
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