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Find frequency response and circular frequency

  1. Jan 8, 2016 #1
    1. The problem statement, all variables and given/known data
    We are given that system of 1st grade
    1/2 * dy(t)/dt + y(t) = x(t)

    1) We need to find frequency response { H(w) }
    2) We need to find circular frequency -3dB
    3) We need to find the absolute value of H(w) in dB for frequency f = 3/π Hz

    2. Relevant equations
    (log is always with base 10)

    Convertion to db => db = 20log(|H(w)|)
    circular frequency = -3db => |H(w)|{w=-3db} = 1/sqrt(2) * H(w){max w}
    Y(w) = H(w) * X(w)

    3. The attempt at a solution
    I solved the 1st one with that way.
    (take Fourier) => 1/2* jw*Y(w) + Y(w) = X(w)
    (divide with X(w)) => 1/2* j*w*H(w) + H(w) = 1 => H(w) = 1 / (1/2 * j*w + 1) => H(w) = 2 / (2 + jw)


    For 2nd I dont know what is w = max, and for 3rd I did an attempt but I found wrong result.
    My book gives as answer for 2nd the value = 2rad/sec, and for 3rd the value -20dB.
    If anyone can help me with the procedure of solving I would be happy.

    Thanks in advance !
     
  2. jcsd
  3. Jan 8, 2016 #2

    BvU

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    Science Advisor
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    Gold Member

    For a first order system the "maximum response" is if dy/dt = 0 so for ##\ \omega = 0##.
     
  4. Jan 8, 2016 #3

    gneill

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    Staff: Mentor

    Hi Hepic. You should be able to recognize your transfer function H(ω) as that of a typical first order low pass filter.
    Fig1.gif

    As such you can find the -3dB frequency by inspection of H(ω) itself, or if you want to do the math, determine ω that makes |H(ω)|2 = 1/2, i.e., find the half-power point.

    I find myself disagreeing with the book's answer of -20 dB for the response at 6 rad/sec. A first order low pass filter should decline at a rate of 20 dB per decade after the corner frequency (using the straight-line curve fit approximation). Now 6 rad/sec is not a decade past 2 rad/sec. I'd expect something closer to -10 dB for the response at 6 rad/sec.
     
  5. Jan 8, 2016 #4

    LvW

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    Yes - it is exactly a value of -10 dB.
     
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