When we look for stability of a feedback system, why do we only concentrate where loop gain is unity and check the phase margin there. The phase of the closed loop system might be passing through 180 but it would still be stable if the phase near the unity loop gain frequency is less than 180. Like in the figure shown below (for negative feedback system), the phase reaches -180 at frequency A but since it is less than 180 at B (unity loop gain) system is stable. enter image description here

Why doesn't the system gets unstable for frequency A? enter image description here

Shouldn't the signal Vi (in the figure) add constructively to signal Vf (inverted twice once due to phase shift of 180 and then due to negative feedback) and grows with time giving unstable system, if operated at frequency A?
I know the negative feedback equations (in figure) and the transfer function getting infinite only for |GH| = -1? But still why doesn't the above reasoning holds valid for frequency A?


Some comments from my side:

1.) The stability check in the BODE diagram concerns the LOOP GAIN response only (because once you did mention "closed-loop system" in your text.)

2.) The shown system is "conditionally stable". That means: It is stable - regardless the properties at the frequency A. However, if you REDUCE the gain within the loop until the gain crosses the point A (the phase remains unchanged) the closed-loop system will be unstable.

Such conditional stable system should be avoided because a gain reduction can happen due to aging or other damping effects. Remember: Classical feedback systems with a continuos decreasing loop phase will become unstable (under closed-loop conditions) for rising gain values (beyond a certain limit) only.

As to your next question - the input signal Vi does not influence stability properties at all. Stability is determined by the loop components only. That is the reason, we investigate the loop gain only.

EDIT: Here is an explanation why the closed loop (your example) will be stable: If a closed-loop system is unstable, this point of instability also must be "stable". That means - either we will have "stable" and continuous oscillations or the output is latched at one of the supply voltage rails. In both cases, this point of instability is fixed.

Now - what happens at the point A in your example? Here we have a rising phase which is identical to a NEGATIVE group delay at this point (group delay is defined as the negative phase slope). This is an indication for the unability of the closed-loop system to let the amplitudes rise (oscillations or latching at the supply rail). Rather, the system returns to a stable operating point.

A final information: The stability check investigates either (a) the -180deg line or (b) the -360 deg line. This depends on what you are investigating: (a) Either the simple product GH or (b) the loop gain LG which is LG=-GH.

  • \$\begingroup\$ I understand we investigate loop gain for stability but why we only investigate loop gain near the frequency where its magnitude is unity. \$\endgroup\$ – sarthak Nov 23 '14 at 14:08
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    \$\begingroup\$ This is based on Nyquist `s stability criterion. In general, there are two cases: (a) If the loop gain is 0 dB and the phase, for example, -190 deg, the gain at -180deg will be >0 dB (closed-loop unstable) or (b) if the loop phase is -180deg and the gain already <0 dB (closed-loop stable). However, there are some exceptions. for example when the phase crosses the 180 deg twice (as in your case). \$\endgroup\$ – LvW Nov 23 '14 at 14:42
  • \$\begingroup\$ See some additional explanations in my answer (EDIT). \$\endgroup\$ – LvW Nov 23 '14 at 14:54
  • \$\begingroup\$ what if I apply a sine input at the frequency A? In this case the phase delay of the sine input when it reaches Vf (in figure) is 180 lagging. Thus while subtraction, the two voltages (input and Vf) add constructively. Shouldn't the system become unstable for this input? \$\endgroup\$ – sarthak Nov 26 '14 at 4:02
  • \$\begingroup\$ No - stability of a system with feedback is independent on the input signal. Stability properties depend on the loop gain only. Either a system is stable or it is not. In the latter case, it will oscillate or go into saturation - with and without an input signal. \$\endgroup\$ – LvW Nov 26 '14 at 8:49

The concept of checking the phase angle at unity gain only applies to simple systems in which the phase-vs-frequency plot is monotonic, where the assumption is that the phase angle is only increasing with frequency, and that as long as the phase margin is sufficient at unity gain, then it can only be better at lower frequencies.

This assumption does not hold in the example you give, since the phase plot is not monotonic. Therefore, you must do a more complete analysis. Any system that has greater than unity gain and a total of 360° of phase shift (including systems with inverting amplifiers and 180° phase shift) at some frequency will oscillate.

  • \$\begingroup\$ .... will oscillate. Yes - agreed, in most cases. However, this is a necessary oscillation condition only, and not a sufficient one. \$\endgroup\$ – LvW Nov 23 '14 at 15:48
  • \$\begingroup\$ @LvW: Why is it not sufficient? What else is required? \$\endgroup\$ – Dave Tweed Nov 23 '14 at 16:02
  • \$\begingroup\$ Dave, that´s hard to answer because - up to now - there is no oscillation condition that is sufficient (as far as I know). The classical Barkhausen condition is a necessary one only. \$\endgroup\$ – LvW Nov 23 '14 at 18:26
  • \$\begingroup\$ @LvW: So, you're splitting a hair that has no relevance to the original question. When we're talking about control system stability, it's a given that we're talking about its response to disturbances. If the system meets the Barkhousen condition, it will oscillate in response to any disturbance. \$\endgroup\$ – Dave Tweed Nov 23 '14 at 18:32
  • \$\begingroup\$ Dave,you have asked me "why" and I gave you an answer (I agree without too much relevance to the original question, but it was YOUR question). I repeat (just in the interests of accuracy) that a feedback circuit that fulfills Barkhausen does NOT necessarily oscillate. \$\endgroup\$ – LvW Nov 24 '14 at 7:55

Let's assume the phase continues to fall at frequencies greater than point B and ultimately crosses -180 degrees (again). This assumption makes Bode analysis valid (because a rising phase and falling magnitude is indicative of a RHP pole that calls for further analysis). Otherwise you need to use the Nyquist Stability Criterion which is more universally valid (e.g. right-half poles).

enter image description here

This type of a system is very common in DC-DC Buck converters. It's a Conditionally Stable system. This means that lowering or increasing the gain by a sufficient amount (the Gain Margin) will cause oscillation. A conditionally Stable system has both an Upper and a Lower Gain Margin.

Using Bode analysis, you're analysing the Characteristic Equation: L-1=0 where L is the Loop gain. The solution to the characteristic equation occurs at L=1 or L=0dB. That's why you analyse the phase shift only at 0dB.

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    \$\begingroup\$ when you are talking of conditional stability you are assuming the oscillations will only occur when magintude of loop gain is unity. My question is why this? Why phase lag of 180deg not enough (sufficient condition) to cause oscillations in the negative feedback system by the reasoning that the fed back voltage will be in phase with the input and hence never dies? Shouldn't the magnitude of the oscillations rise and make the system unstable? \$\endgroup\$ – sarthak Dec 5 '14 at 6:55

To cause instability (sustained self oscillation), the phase has to be exactly 0 degrees and gain has to be equal to or greater than 1. Given that there is 180 degrees phase shift due to the circuit being an inverting amplifier, if the amplifier itself doesn't produce an additional phase shift of at least 180 degrees, then it won't oscillate.

Having said that, the frequency response when the loop is closed will not be ideal and there will likely be severe "ringing" on some signals but, it won't reach "classic" instability.

  • \$\begingroup\$ Suppose I have an ideal system where phase shift due to loop gain is 180 at some frequency and this is not unity gain frequency. Will the system become unstable? \$\endgroup\$ – sarthak Nov 23 '14 at 14:10
  • \$\begingroup\$ In a "classical" system with a decreasing phase and if the loop gain >0 db - yes, the closed-loop system will be unstable. \$\endgroup\$ – LvW Nov 23 '14 at 14:37
  • \$\begingroup\$ In a system where the "phase shift due to loop gain is 180 at some frequency and this is not unity gain frequency" this is NOT necessarily an unstable system. If the crossing occurs when gain >1 then the system is called Conditionally Stable because it has an upper and a low Gain Margin. \$\endgroup\$ – akellyirl Nov 23 '14 at 16:17
  • \$\begingroup\$ @akellyirl. Please check what you wrote regards ">1". Also note that loop gain phase for oscillation is zero not 180 degrees. Were you aiming your comment at anyone in particular btw? \$\endgroup\$ – Andy aka Nov 23 '14 at 16:30
  • \$\begingroup\$ @akellyirl, If the gain>1 then shouldn't the input at the "G" block (in figure) grow with time? Also, gain margin seems to be negative with gain>1? Could you please explain conditional stability? \$\endgroup\$ – sarthak Nov 24 '14 at 16:25

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