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I'd always understood that in a transfer function, every pole contributes -90 degrees in phase and every zero contributes +90 degrees in phase. Based on this assumption, I would expect the following transfer function to settle at a total phase of 6*90 - 90 = 450 degrees. However, wolfram alpha shows the phase at 360 degrees. Do you know where I've gone wrong with my logic?

enter image description here

enter image description here

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    \$\begingroup\$ Where did the 6 come from? Also I see third degree denominator polynomial; i.e. three poles. First degree numerator; i.e. one zero. More over that zero seems to be on the left hand plane. So it may also contribute -90 degrees instead of +90 degrees. So -360 looks like a reasonable number. \$\endgroup\$
    – AJN
    Nov 11, 2020 at 5:55
  • \$\begingroup\$ Please explain how you got the equation 6*90-90. Each number and sign in that equation. \$\endgroup\$
    – AJN
    Nov 11, 2020 at 5:57
  • \$\begingroup\$ I did 6*90 because I see 6 poles (one from s, two from s^2, and three from s^3). And then I subtracted 90 degrees to account for the single zero. It should have probably been written as 6*-90 + 90. Do you mean that just because I have a pole in the transfer function, the contributing phase is not necessarily -90? \$\endgroup\$
    – syz
    Nov 11, 2020 at 6:18
  • \$\begingroup\$ If you are familiar with complex numbers, try evaluating the transfer function at \$s \rightarrow j \cdot \infty\$. Find \$\angle\$ of the resulting complex number. This should give you the "final" phase you are looking for. Alternately, look up the definition of "pole". You need to factorise the polynomial into monomials. You can't just count the number of "s" in the polynomials. The rule of 90 is probably a thumb rule. The methods I suggested above are better IMHO. \$\endgroup\$
    – AJN
    Nov 11, 2020 at 6:25
  • \$\begingroup\$ 3 Poles are s=0, s=-(5+sqrt(13))12, s=-(5-sqrt(13))12 \$\endgroup\$
    – mhaselup
    Nov 11, 2020 at 7:29

1 Answer 1

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There is no error in the phase response given by Wolfram Alpha. If you look at your transfer function, you can see the combination of three components:

  1. a right-half-plane zero (RHPZ) lies in the numerator
  2. a pole at the origin for \$s=0\$
  3. two poles located at higher frequency

The first component will shift the phase down to -90°. This is because the zero is located in the right half-plane, indicated by the minus sign in the numerator. Should you have a plus sign instead, the zero would be in the left half-plane (a LHPZ) and the phase would lead up to 90°.

Then the pole at the origin (a 1st-order integrator) will permanently lag the phase by 90°. So together with the RHPZ, you already have -90-90=-180°. Then, the two poles are located in the left half-plane and contribute another 180° phase lag, totaling the phase response to -180-180=-360° or 0°.

A quick Mathcad sheet shows the response of your formula while going through different arrangements. As expected, at frequencies beyond 100 Hz, the phase is 0° as predicted:

enter image description here

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  • \$\begingroup\$ Gotcha, so the s^2 and s^3 components don't mean that they contribute -180 (for s^2) and -270 degrees (for s^3)? In other words, when the transfer function is factored as shown in your screenshot, does the (s/wo)^2 component not add -180 degrees phase? \$\endgroup\$
    – syz
    Nov 12, 2020 at 4:16
  • \$\begingroup\$ The transfer function remains the same in all the presented forms, 3 poles in the denominator and 1 RHP zero in the numerator. The factorization helps to see these contributors in a clearer way, that's all. \$\endgroup\$ Nov 12, 2020 at 8:00

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