Having trouble understanding the stability of following system

I ran your numbers with

$$H(s) = \frac{2.123\times 10^6 \times (0.15s +1)}{s^3+12566.37s^2}$$ and unity negative feedback.

A system is stable if its parts are stable and all of the poles in its transfer function have negative real parts.

If a system has simple, unity negative feedback and the loop transfer function is \$P(s)/Q(s)\$, then the closed loop transfer function is

$$F(s) = \frac{\frac{P(s)}{Q(s)}}{1+\frac{P(s)}{Q(s)}} = \frac{P(s)}{P(s)+Q(s)}$$

The poles of \$F(s)\$ are then just the zeros of \$P(s)+Q(s)\$.

Your loop transfer function, with \$K_{vp}=0.15\$, and \$K_{vi}=1\$

is

$$\frac{P(s)}{Q(s)}=\frac{2.123\times 10^6 \times (0.15s +1)}{s^3+12566.37s^2}$$

If I have done my math correctly, we thus want to find the roots of

$$P(s)+Q(s) = (2.123\times 10^6 \times (0.15s +1))+(s^3+12566.37s^2)$$ $$P(s)+Q(s) = s^3 + 12566.37s^2 + 318450s + 2123000$$

The zeros of the above expression turn out to be

1. r = -12541
2. r = -12.6896 - 2.87376 j
3. r = -12.6896 + 2.87376 j

Since the all of the roots have real parts which are negative, the closed loop system is stable. This is the definitive test for stability when either the loop gain, or closed loop transfer function is known in the form of a rational function. The Nyquist method works, but to be accurate, one still needs to know the number of right-half plane zeros, and the number of encirclements of -1 in each direction. The Bode plot method works with simple systems, but the gain margin test requires looking at all frequencies where the phase is \$180^{\circ}\$ (or \$0^{\circ}\$, depending). So, to get a definitive answer, when you know the transfer function, calculate as I have done, and rely on that, rather than on gain margin.

The key word in the text is "may"

In summary, if a negative feedback circuit has a loop gain that satisfies two condictions $$|H(j\omega_0)| \ge 1$$ $$\angle H(j\omega_0) = 180^{\circ}$$ then the circuit may oscillate at \$\omega_0\$.

(emphasis added).

Note: Roots of a polynomial may be found using online calculators such as Wolfram Alpha. In the days of slide rules, finding roots of higher order polynomials was not trivial. This fact is a contributing factor to the spread of graphical methods for testing stability. Another obvious factor is that we often don't know an accurate representation of the transfer function as a rational function, but can obtain Bode plots through simulation and/or Vector Network Analyzers.

Just a short and general comment to the subject under discussion in this thread:

• Question: How is negative feedback defined (in contrast to positive feedback) ?

• Answer: NEGATIVE LOOP GAIN.

• Therefore, it seems to be logical (and to avoid misunderstandings/misinterpretations) to include all sign inversions within the complete feedback loop in the definition for the gain of the loop ("loop gain"). More than that, all classical methods for loop gain simulation (using programs for circuit analyses) consider, automatically, all sign inversions within the loop. Unfortunately, some authors (as Razavi) use another definition - and, as we can see in this thread, this can lead to severe misunderstandings and problems.

After discussion, interpreting your question as:

How can a negative feedback loop be stable if the loop TF has 180deg phase and gain>1 for some frequency?

(i.e., a version of the Barkhausen criterion - which is intended for oscillation rather than stability)

In short, it can happen - and it doesn't necessarily mean instability. The concepts of Gain Margin and Phase Margin, associated with Bode plots, are sufficient to assess stability for minimum phase systems.

Counterexample to this interpretation of the Barkhausen criterion can be found here: https://web.mit.edu/klund/www/weblatex/node4.html

The example you chose has the nuance of poles at the origin. This makes it, strictly speaking, non-minimum-phase. But in systems of this pattern (ie double integrator plus other poles and zeros), the Bode Plot based GM/PM tests can be used without problems.

the loop gain is still greater than 1 at lower frequencies where phase is -180. Shouldn't this lead to growing oscillations and instability?

If this output was used for the feed back, you have an inverted signal i.e. nearly (+/-) 180° of phase shift so, it would be totally stable. This is classic negative feedback in action.

Everywhere I read, loop gain greater than or equal to 1 at phase 180 degree is considered very very bad.

Not if the signal is fed back without alteration (because it is naturally inverting). I mean, think about negative feedback of an op-amp in unity gain. The loop gain has a phase angle of (+/-) 180° at low frequencies but, does this cause the op-amp instability (no of course not).

Here's the problem with your edited image from the book: -

In the above you are assuming that negative feedback involves a junction which is actually a negator for the feedback signal when, in fact, if H(s) is naturally negative at DC (low frequencies) then this is fine for your transfer function: -

As I said above (and in comments) you do appear to think that the summing node where feedback is added should always have a negative sign on the feedback port <-- this is absolutely not the case, if the plant naturally inverts (and is stable at higher frequencies, which it is) then you don't need a negate sign on the summing node where feedback is inputted.

This reiterates what I said in comments below this answer: -

what extra 180 degrees are you talking about in the latter part of your comment?