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Transfer function pole on the Imaginary axis indicates that the system is marginally stable which in time domain can be represented as a sinusoidal motion with constant amplitude and frequency of the Imaginary axis pole. In some applications, oscillations with small amplitude might be acceptable.

  1. Is amplitude of oscillation a (only) function of initial condition of the system?
  2. How to intuitively quantify amplitude of the oscillations for marginally stable system?
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Is amplitude of oscillation a (only) function of initial condition of the system?

No. For a system with a transfer function of the form $H(s) = \frac{\cdots}{s^2 + \omega_0^2}$, the amplitude of oscillation will increase any time it is excited by a signal that has a component at $x(t) = \cos \omega_0 t + \phi$. Even if you're not exciting it intentionally, random noise will excite that pole. Basically, the output will be of the form $a(t) \cos \omega_0 t + b(t) \sin \omega_0 t$, where $a(t)$ and $b(t)$ will be Wiener (random-walk) processes. Such processes have variances that tend to infinity as $t \to \infty$.

How to intuitively quantify amplitude of the oscillations for marginally stable system?

"Big and growing, boss, and I don't think they'll get smaller!"

Possibly by how fast they grow. But grow they will.

In some applications, oscillations with small amplitude might be acceptable.

If you have a system that's exhibiting persistently small oscillations, then you're seeing the signs of a nonlinear phenomenon called a "limit cycle". Basically, there's some oscillator in there that's either inherently small-valued (like the oscillation you may see around the least significant bit of a DAC or ADC), or there's a big oscillation that's mostly not getting to the output. Either way, if it persists at some amplitude, there's some nonlinear process that's keeping it that way.

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  • $\begingroup$ I should have clarified that this example is referring to unexcited (inherent) characteristics of the system. Assuming no external force/noise, given eigen values of the system, how to quantify amplitude of oscillations? Do we need to analyze eigen vectors for this purpose ? $\endgroup$
    – Alborz
    Nov 22 '20 at 22:02
  • $\begingroup$ If the poles are truly on the stability margin (imaginary axis in the Laplace domain, unit circle in the $z$ domain), then the way to quantify the amplitude is "unbounded". There's only one unbounded -- it's the number that has no bound. It's big. And getting bigger. Without end. Big. Bigger. Bigger yet. If you're dealing with something that isn't described by "unbounded", then you're dealing with something that isn't a linear system with poles on the stability boundary. $\endgroup$
    – TimWescott
    Nov 22 '20 at 23:23
  • $\begingroup$ I'm not clear on "the way to quantify the amplitude is "unbounded". Could you please elaborate ? $\endgroup$
    – Alborz
    Nov 23 '20 at 3:17
  • $\begingroup$ I mean the antonym of "bounded" in the mathematical sense. A signal $x(t)$ is bounded if, for all possible values of $t$, there exists some finite $x_{max}$ such that $\left | x(t) \right | < x_{max}$. If there is no such $x_{max}$ (i.e., if $x_{max}$ is infinity), then $x(t)$ is unbounded. en.wikipedia.org/wiki/Bounded_function $\endgroup$
    – TimWescott
    Nov 23 '20 at 16:06
  • $\begingroup$ I understand what "bounded" means. But this is not answering my question. I theory, a marginally stable system manifests a finite amplitude oscillation. Again, how one can measure this amplitude? $\endgroup$
    – Alborz
    Nov 24 '20 at 17:43
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I have to disagree with Tim here. And it's because I once took that position and was shown wrong by James McCartney, the author of SuperCollider.

There's a difference between "marginally stable" and either "stable" or "unstable".

For the marginally-stable oscillator:

$$\begin{align} y[n] &= 2 \cos(\omega_0) y[n-1] - y[n-2] \\ \\ y[-1] &= A \cos(-\omega_0 + \phi) \\ y[-2] &= A \cos(-2\omega_0 + \phi) \\ \end{align}$$

will result in this output:

$$ y[n] = A \cos(\omega_0 n + \phi) \qquad \forall n \in \mathbb{Z} \ge 0 $$

and with double-precision floating point is stable to the extent that I have run it overnight and no change in amplitude.

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