For a system to be stable you must not reach a phase of 180 degrees before gain reaches 1 (for an ideal system). The layman explanation I was offered was that otherwise, your negative feedback becomes positive and you just built an oscillator.

What if you created a system (ie 4 poles H(s) = 10^100/(s-10)^4 that underwent a phase change of -360 and then limit inputs to the system to be >Fmin (which is the frequency after the phase changes). Would the system be stable for these inputs Fin > Fmin?

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  • \$\begingroup\$ Maybe because of the fact that the negative feedback provides 180 degrees of a phase shift from start. \$\endgroup\$
    – G36
    Commented Jun 24, 2018 at 7:16
  • \$\begingroup\$ Updated title as it wasn't well related to my question. \$\endgroup\$
    – EasyOhm
    Commented Jun 24, 2018 at 7:18

2 Answers 2


At first, some clarifications:

1.) It is the loop gain (open the loop and measure/simulate the gain of the complete loop) which is subject of this stabilization criterion you have mentioned (if the loop phase reaches -180deg the loop gain must be <1 already). If this criterion is fulfilled, the closed loop is stable.

2.) However, this criterion applies only to a system where the critical phase (-180 deg) is crossed only once! Otherwise (when the phase goes again above this critical line) you must apply the Nyquist criterion to check stability (which might be a "conditional" stability only).

3.) Regarding your last example: The loop gain criterion (first point) applies to a 4th-order system as well. No - it does not help at all to "limit" the frequency.....


If you can completely block out any frequency from the input that would make it oscillate, then the overall system will also not oscillate. However as LvW pointed out, stability of a system is not defined that way. Also, constructing such an input is impossible to do.

  • For example, any non-periodic signal is bound to have spectral content over higher frequencies (can be limited, but not eliminated). So just switching on the input will generate spectral content with higher frequencies causing the oscillations to start. Any component that changes even slightly (eg. temperature, vibration, etc.) will also result in higher frequency spectral content.
  • Then there is noise, which is also typically white (present over all frequencies). Even a simulator can introduce quantization noise, or computational errors, also leading to high-frequency content. Noise is usually inherently present and can never be eliminated.

So there is no real-life situation where band-limiting the input signal helps. However, in theory it certainly is possible. The negative feedback loop situates itself in an unstable singular state, where even the smallest perturbation will cause oscillation.

On a side note: What you're describing is sometimes the case in simulation. When designing oscillators (where there is no input, or the input is \$0\$ having no spectral content at all), the simulator often calculates such an unstable singular state with such high precision that it won't start oscillating (or that it starts very slowly). Ie. you then usually need to apply some perturbation to it yourself to make it start. Usually the quantization noise is enough to get it started (very slowly) though.

  • 2
    \$\begingroup\$ Sven - your first sentence is not correct. If a system with feedback is able to oscillate (or is unstable in a more general sense), it does not matter at all if the input is zero or has any arbitrary form. Unstability arises in the feedback loop only - independent on the input condition! \$\endgroup\$
    – LvW
    Commented Jun 24, 2018 at 10:38
  • 1
    \$\begingroup\$ I used stable just as "not exploding" which I assumed the op meant, but I don't mind using the "strict" definition of stability of a system. \$\endgroup\$
    – Sven B
    Commented Jun 24, 2018 at 14:27

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