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Bode phase plots and initial angles of transfer functions

  1. Apr 29, 2017 #1
    Hello everyone. So I have a test coming up and I am struggling with the concept of figure out what the initial phase or angle of a transfer function is. For instance, consider the following transfer function:

    L(s) = 4/s(.4s+1)(s+2)

    So the initial angle for L(s) is -90 degrees. Is there a simple trick for figuring that out given any transfer function? I have heard that if there is a negative in the numerator then you add 180 degrees of phase or depending on the degree of the denominator you would subtract phase.

    Can someone help me sort this out?
  2. jcsd
  3. Apr 29, 2017 #2
    I assume you mean this function:
    L(s) = 4/(s(.4s+1)(s+2))

    Some thoughts to help you get to the answer:

    Say you have two complex numbers z1 and z2, what is arg(z1 z2) and arg(z1/z2) in terms of arg(z1) and arg(z2)?

    What happens to the angle of each factor in the denominator as s = jω, ω → 0?
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