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I have a system with the three poles 0+8i, 0-8i, and -0.5. When I input a sinusoidal signal, this is the output: enter image description here Based on the poles, -0.5 is in the left side of the plane so it should be stable. +8i and -8i are on the axis so the output will oscillate. I am confused on why the signal is gaining magnitude as time increases? Based on the poles I would not think this would be the behavior. I thought it would be a steady sinusoid. Where is the amplification coming from?

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2 Answers 2

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You have a pair of poles on the Y-axis. That is the abstract representation of an oscillator without damping. If your input is of finite duration, once the input goes to zero the oscillation will be stable at constant amplitude. However, the input you chose continues to feed energy in without limit, so the oscillator amplitude grows without limit. With no damping, the oscillator cannot dissipate the energy fed to it. This is despite the fact that your other degree of freedom is damped (pole on negative X axis).

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    \$\begingroup\$ This; another way to think of it is an integrator (pole at zero), transformed to a rotating reference frame. For a similarly rotating input (the input and output appear in phase here so I guess it's at exactly the right frequency), the constant amplitude input is simply the step response of that integrator. Which of course is a ramp output, hence the ramp output amplitude. \$\endgroup\$ Feb 27 at 2:55
  • \$\begingroup\$ Thanks! Do either of you know of a resource that shows what different arrangements of poles mean (like how John knew that that pair of poles meant it was an oscillator without damping?) Some sort of sheet, maybe? \$\endgroup\$
    – newbie
    Feb 27 at 6:52
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It has to be excited at the resonant frequency for the amplitude to blow up.

I'm using the following transfer function with a DC gain of 1 and poles as you have specified.

$$\frac{32}{(s-8 i) (s+8 i) \left(s+\frac{1}{2}\right)}$$

I can reproduce the resonant behavior you are observing.

enter image description here

This does not happen with a sinusoid input having a non-resonant frequency.

enter image description here

Or a step input.

enter image description here

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    \$\begingroup\$ Good point! +1. \$\endgroup\$
    – John Doty
    Feb 27 at 21:02
  • \$\begingroup\$ Ahhh ok so it does produce a steady sinusoid under this circumstance. Thanks! \$\endgroup\$
    – newbie
    Feb 28 at 3:00

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