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I am trying to learn about RF oscillators.

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.

enter image description here

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?

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First, indeed why the frequencies below ω1 won't oscillate?

In the context of linear control systems, oscillation occurs due to poles on the jω (imaginary) axis of the s-plane. If the system has poles in the right-half plane (RHP), then the system response grows exponentially with time, which can be interpreted as instability. The statement "oscillation build-up occurs with a loop gain of greater than unity only if the closed-loop system contains poles in the right-half plane" hints at the exponential growth associated with RHP poles. So, if a system has enough loop gain and 180 degrees phase shift but doesn't have RHP poles, it won't exhibit the exponential buildup of oscillation; instead, it may show some transient oscillation which will die out after some time.

My next question is can you really find an negative feedback system that has a loop gain of greater than unity (while having a phase shift of 180 degrees at that same frequency) also has a closed-loop transfer function with obly LHP poles? Because the two condition seems 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? And if I am right, is there a more formal proof to that statement?

You're right; there is a seeming contradiction here. Here's the resolution: The Barkhausen criteria for sustained oscillations demand: A loop gain magnitude of 1 (0 dB). A total phase shift of 0 or 360 degrees (which includes the 180 degrees of the oscillator itself and the other 180 degrees which can come from the feedback network). If both these conditions are satisfied for a particular frequency, then that frequency can oscillate. On the other hand, in feedback control theory, the Phase Margin is a measure of the system's distance to instability. A negative phase margin indicates a system that would oscillate if perturbed. These two concepts are seemingly at odds, but the difference lies in the initial conditions and the type of analysis. The Barkhausen criteria assume a small perturbation and then check if it grows or dies out. The phase margin concept from feedback control assumes some input and checks the system's behavior to that input. Theoretically, if you have a system with a negative phase margin but no poles in the RHP, then if you introduce a signal at the frequency corresponding to the negative phase margin, the system can oscillate, but these oscillations won't grow exponentially (no RHP poles). In real systems, however, noise and non-linearities often come into play, meaning even if you've designed an oscillator with poles on the imaginary axis, any small perturbation can move those poles into the RHP and create sustained oscillations.

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  • \$\begingroup\$ Hi Jerzy Przezdziecki, thank you for answering my question. However, I still have some problem regarding this issue. You wrote "if a system has enough loop gain and 180 degrees phase shift but doesn't have RHP poles, it won't exhibit the exponential buildup of oscillation." But if that is the case, wouldn't that mean the Barkhausen's criteria is useless. Since Barkhausen's criteria alone can't guarantee oscillation, designers still need to find at least one RHP pole. But for an causal system, which is always the case for real life circuits, the existance of RHP pole alone promises instability. \$\endgroup\$
    – VeryItchy
    Oct 5, 2023 at 15:33
  • \$\begingroup\$ I wanted to thank you for your help by upvoting your answer, but I am new to this forum, so I don't have enough reputation to do that. So, I thought I can express my gratitudes by actually typing them down. \$\endgroup\$
    – VeryItchy
    Oct 5, 2023 at 15:37
  • \$\begingroup\$ The Barkhausen criterion isn't "useless"; it's an essential starting point for oscillator design. But as with many engineering principles, the real-world application often requires understanding beyond the basic criterion and considering the full non-linear, dynamic behavior of the circuit. \$\endgroup\$ Oct 5, 2023 at 15:46
  • \$\begingroup\$ Also, most textbooks explain why signal amplifies itself when Barkhausen's criteria is "over-met" (180 degrees phase shift and greater than 1 loop gain magnitude) by tracking the signal through the loop multiple times. The explaination is that every time a signal propagates through the loop it gets amplified by the magnitude of the loop gain therefore exhibits exponential growth. And I don't see the connection of pole's location here. In other words, how would the absence of the RHP poles makes this explaination invalid. \$\endgroup\$
    – VeryItchy
    Oct 5, 2023 at 15:53

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