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How does phase affect the Nyquist Diagram - imaginary axis, how should it look?

  1. Apr 18, 2012 #1
    Please consider


    I agree with all notes made on that slide, but I don't actually get how they constructed the diagram from that? I understand that they line represents frequency so going to 0 to infinity means the line would travel from -0.5 to 0, but HOW DO THEY KNOW what the size values the curve peak at on the imaginary axis???

    Further, it says phase decreases from -180 to -270, I agree from the transfer function, but how does this look on the Nyquist diagram? How does phase affect the Nyquist diagram?


    EDIT: Apologies if this may seem like a double post
  2. jcsd
  3. Apr 19, 2012 #2


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    Staff: Mentor

    Perhaps you should consider reducing the given transfer function into real and imaginary components. So then:
    [tex]\frac{1}{(jω + 1)(jω + 2)(jω - 1)} = \frac{-2}{(1 + ω^2)(4 + ω^2)} + j\frac{ω}{(1 + ω^2)(4 + ω^2)} [/tex]
    This should help pick out any particular relationships or extrema of the real and imaginary components, as well as phase relationships since ##\phi = tan^{-1}(Im/Re)##.
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