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#### VeryItchy

Example 8.6 in the oscillator chapterof

*RF Microelectronics 2nd edition*by Razavi asks to solve for the frequency and amplitude of the oscillation of an ideal double integrator oscillator.

He approached the problem by converting the transfer function into a differential equation and then solved it.

Upto this point I can still understand him, but then he wrote this:

Why does the circuit not oscillate at frequencies below ω1 = K? It appears that the loop has enough gain and a phase shift of 180 degrees at these frequencies. As mentioned earlier, oscillation build-up occurs with a loop gain of greator than unity only if the closed-loop system contains poles in the right-half plane. This is not the case for the two-integrator loop: Y/X can have only poles on the imaginary axis, failing to produce oscillation if s = jω != jK. (K being the constant nominator of the integrator's transfer function K/s)

To make the question clearer and to avoid copyright issues, I have redrawn the circuit and its loop gain myself. Example 8.6 is on page 506, if you happens to have the book.

Why won't the circuit oscillate at frequencies below ω1? They have more than enough loop gain and a phase shift of exactly 180 degrees. I think all the frequencies below ω1 should be able to oscillate, it is just that their amplitudes will be growing with time indefinitely rather than being stable like that of the frequency ω1. I wanted to verify my idea by assuming the solution to y(t) to take the form of a more general complex expression e^(st) and solve for the same diffential equation. I found that to satisfy the equation, the complex number s has to be purely imaginary, which left me more confused. I failed to see why the lack of right hand plane poles prevents frequency components below ω1 from oscillating despite the fact that they satisfy the Barkhausen's criteria.

Can you really find a negative feedback system that has a loop gain of greater than unity while having a phase shift of 180 degrees at that same frequency which also has a closed-loop transfer function with only left hand plane poles? The two conditions seem contradictory. A negative feedback system that has negative phase margin should be unstable, meanwhile a closed-loop transfer function with only LHP poles implies the system to be stable. Am I right? If I am right, is there a more formal proof to that statement?