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

by thomas49th
Tags: affect, axis, diagram, imaginary, nyquist, phase
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thomas49th
#1
Apr18-12, 11:04 AM
P: 656
Please consider

http://gyazo.com/e5c5b4f7808a63e7e664440259ac3058

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?

Thanks

EDIT: Apologies if this may seem like a double post
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gneill
#2
Apr19-12, 07:56 AM
Mentor
P: 11,689
Quote Quote by thomas49th View Post
Please consider

http://gyazo.com/e5c5b4f7808a63e7e664440259ac3058

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?

Thanks

EDIT: Apologies if this may seem like a double post
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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