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I was reading in some textbooks and places online that a single pole introduces a -90 degree phase shift and that in the phase plot, it never actually reaches -90, it goes asymtotic to it.

Similarly, for a two-pole system, it never reaches -180 degrees, just reaches very close to it.

Does that mean that single-pole and two-pole systems can never be unstable since there is no phase crossover frequency?

AlfroJang80
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    If you build what you think is a 2 pole system, you will often have additonal poles at high frequency due to finite opamp GBW products, strays etc, which at some frequency will tip you over into instability. Have you ever seen the humble emitter follower become an oscillator due to strays! – Neil_UK Jan 19 '20 at 07:34
  • What do you mean by system? – Andy aka Jan 19 '20 at 10:27
  • Are you talking about a two pole system in feedback or open loop? – Voltage Spike Dec 28 '21 at 15:34

2 Answers2

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Unless there is more to your question, here it goes,

  • If a system is unstable, with any pole on the open right plane, you will not be able to plot its bode plot (or at least that plot will not have any physical meaning, since for any sinusoidal/DC input the system will grow unbounded).
  • If a system is unstable, with no poles on the open right plane but some on the imaginary axis, you will be able to plot most of its bode plot, but for some frequencies it will grow unbounded. But in this case, the indication that the system is not stable will come from an infinite peak at some frequencies.
  • There are plenty two pole systems that are unstable, such as, $$ H(s) = \frac{1}{(s-1)(s+1)}, ~~ H(j\omega) = \frac{1}{(-1+j\omega)(1+j\omega)}.$$
jDAQ
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    There are plenty of unstable systems whose Bode plot can be made -- that is, in fact, a very good way to tell that a candidate system is unstable. – TimWescott Jan 19 '20 at 00:42
  • Do you mean a feedback system? Or just any system (some generic Transfer function)? – jDAQ Jan 19 '20 at 00:44
  • What do you mean if a system is unstable, you cannot plot a bode plot? Then, a bode plot can't be used to check stablility at all? What's the point of it then? – AlfroJang80 Jan 19 '20 at 00:49
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    I usually see people analyzing the bode plot of an open loop system to evaluate how the closed loop system will behave, in that sense, looking for the gain and phase margin would work. – jDAQ Jan 19 '20 at 00:52
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My understanding of this....

Considering a theoretical 2 pole amplifier. With a theoretical 2 pole amplifier it shouldn't be possible to actually reach the loop gain=1, loop phase=-360 degrees condition necessary for instability where the closed loop gain theoretically goes to infinity, obviously limited by the power rails.

But for low phase margins there will be some peaking in the closed loop response at the top end of the closed loop bandwidth. If the phase margin is low enough then this closed loop response peaking can be large enough to result in a small amplitude high frequency oscillation riding on the signal waveform or indeed as a small amplitude high frequency oscillation even if there is no input signal (apart from noise). As the phase margin approaches zero degrees the high end closed loop gain (peaking) becomes greater and even though it can't get to infinity, with only two poles, it can become large enough (close enough to the loop gain=1, loop phase = -360 condition) to theoretically be large enough that it will become limited by the power rails. I wouldn't actually class this as instability because the output isn't theoretically ever increasing but has a maximum limit although the power rails limit the amplitude to lower then its theoretical peak.