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I am reviewing bode plots and s-plane (pole-zero plots). I seem to have some confusion about them and how they relate to each other. I believe one is for closed-loop systems, the other is for open-loop.

My question is : If I have a first order, single pole system and the pole lies on the imaginary axis. From what I know about the s-plane, this would make it an unstable system.

If I plot the bode plot, I will see an unbounded peak at the pole frequency in the magnitude plot.

However, the phase shift will be a maximum of -90 degrees - nowhere close to the -180 which is required for instability. How does this make sense? The bode plot is still indicating the system is stable???

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  • \$\begingroup\$ Could you add pictures of the plots you’re talking about to the question? Also, a system with a pole on the imaginary axis is marginally stable, not unstable. The step response of such a system is a sinusoid. \$\endgroup\$
    – Ryan
    Feb 1, 2022 at 2:06

2 Answers 2

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A single-pole system is first-order system which can only have a real pole. I suppose you mean a single pair of poles?

More than that - I am afraid you are mixing open-loop and closed-loop analyses.

For example, the mentioned magnitude peak can be observed for a closed-loop system at the stability limit. In addition, there will be an abrupt phase jump from +90deg to -90deg at the same frequency.

The mentioned 180deg criterion (part of Barkhausens oscillation condition) is applicable for open-loop conditions only.

In detail: At the stability limit (frequency fo) the loop gain function (transfer functon of the open loop) will exhibit a 360deg phase shift (180deg without consideration of the phase inversion at the summing node) and a magnitude of unity - whereas the closed-loop magnitude will exhibit an unbounded peak and a phase jump of 180deg.

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  • \$\begingroup\$ Thanks for the response. That made a lot of sense. Does that mean that it's not possible to have a magnitude peak in an open-loop bode plot? If I have an open loop system with a pair of poles on the imaginary axis, will the open loop bode plot show the 180 degree instability criterion being met? \$\endgroup\$
    – AlfroJang80
    Feb 1, 2022 at 12:11
  • \$\begingroup\$ Correction: If you have "a CLOSED loop system" with a pair of poles on the imag. axis, the corresponding open-loop Bode plot will show 180deg (or 360 deg, respectively) phase shift at this frequency. \$\endgroup\$
    – LvW
    Feb 1, 2022 at 13:25
  • \$\begingroup\$ @LxW The open-loop bode plot will show a 180 deg phase shift and a > 0dB gain at the frequency of the closed loop pole pair which is on the imaginary axis of the pole-zero plot? Is that right? \$\endgroup\$
    – AlfroJang80
    Feb 1, 2022 at 14:26
  • \$\begingroup\$ Not quite. When the magnitude of the loop gain is exact unity, the poles of the closed-loop will be placed on the imag. axis. For larger gain values (and 180 eg loop phase) the poles will have a pos. real part (right from the imag. axis). \$\endgroup\$
    – LvW
    Feb 1, 2022 at 16:54
  • \$\begingroup\$ @LxW Ah that makes sense. I believe I am struggling with the differences between closed-loop poles vs open-loop poles. I will read up on this. Do you have any good texts that go into detail from the ground up on this? \$\endgroup\$
    – AlfroJang80
    Feb 1, 2022 at 18:21
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You are confusing the open-loop Bode plot with the closed-loop Bode plot.

Putting aside the fact that a system with a single complex or imaginary pole is unrealizable, if you have a control loop whose open-loop gain is $$H_o(s) = \frac{k}{s + j\omega},$$ and then you close the loop, you will get a system whose closed-loop gain is $$H_c(s) = \frac{k}{s + k + j\omega}.$$

For all positive values of \$k\$, this system has a single complex pole with a positive real part, and is thus a stable system in closed loop.

Similar results can be had with \$H_o(s) = \frac{k}{s}\$, with the advantage that it's actually a physically realizable system.

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  • \$\begingroup\$ Thanks for the response. I'm still a bit unsure - if my open loop transfer function has a pair of poles on the imaginary axis, will my open loop bode plot still not show a magnitude peak at that pole frequency? \$\endgroup\$
    – AlfroJang80
    Feb 1, 2022 at 12:12
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    \$\begingroup\$ @TimWescott I must admit that I have problems to understand your formulas. First question: Is w a fixed frequency (such as wo)? In this case, k mus have the unit rad/s, Second question: How did you arrive at the function Hc ? Unity feedback? I cannot confirm this expression. How can you add k and ks ? Both are not unit-free! Third question: What is "a single complex pole"? If complex - we always have a pole pair. \$\endgroup\$
    – LvW
    Feb 1, 2022 at 12:13
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    \$\begingroup\$ @AlfroJango80 There must be still a misunderstanding on your side. When a system has a pair of poles on the imag. axis it cannot be open-loop. Such a pole pair belongs to an oscillatory system and, therefore, is always a closed-loop system (positive loop gain). \$\endgroup\$
    – LvW
    Feb 1, 2022 at 12:21
  • \$\begingroup\$ @LvW Ahhh. That makes sense now! Thank you! \$\endgroup\$
    – AlfroJang80
    Feb 1, 2022 at 12:32
  • \$\begingroup\$ @AlfroJang80 I left out a \frac in the closed-loop equation. And I wasn't so much assuming unity gain as asserting that \$H_o\$ was the open loop gain -- the thing and the whole of the thing. Open loop gain is typically defined as the gain around the entire loop. \$\endgroup\$
    – TimWescott
    Feb 1, 2022 at 20:32

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