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I am studying analog design, and my instructor told me that you could determine whether or an amplifier is stable without knowing the transfer function, or the circuit diagram, just by examining the Bode plot. Below is an example of an unstable system. He claimed that it is unstable because there is non zero gain at the 180 degree phase shift point. My understand is that this only unstable if there is feedback, which we don't necessarily know.

My question is this: How do we know if the system is stable from just the Bode plot?

enter image description here

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The open loop Bode plot provides relative stability information on the closed loop system.

If an open loop can provide unity gain (or 0dB gain) in response to a sinusoid at a particular frequency where it also provides a -180deg phase shift then it can use the output sinusoid to replace the input sinusoid (the negative f/b provides the necessary additional -180deg phase shift). This is critical stability. Oscillation is self-sustaining. The loop does not see any difference between the externally applied sinusoid, and the one that it has created itself at the output terminal.

An open loop gain greater than 0dB at the frequency where the phase is -180deg will, clearly, give an unstable closed loop. And an open loop gain less than 0dB at that particular frequency will mean that the sinusoid gradually reduces in amplitude as it travels around the closed loop. That's a stable system.

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I think he means 0 dB or unity gain.

The negative feedback ends at 180 degrees and becomes positive feedback. If the gain is less than one, this is still stable.

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    \$\begingroup\$ Many analysis of stability are done by examining the open loop gain of a system. Then the system is connected with a feedback in closed loop. The analysis is done in open loop, the conclusions about stability apply to the closed loop system. \$\endgroup\$ – Claudio Avi Chami Mar 16 '17 at 20:06
  • \$\begingroup\$ @ClaudioAviChami Can any conclusions be made if we don't have a closed loop system, but rather the function is simply open loop with no feedback? \$\endgroup\$ – Dace Mar 16 '17 at 21:29
  • \$\begingroup\$ An open loop system can't really be unstable -- the whole point of stability theory is to see whether a small excitation will get repeatedly amplified and cause exponential divergence. That can't happen without feedback. You analyze stability of a system with feedback by breaking the loop, and then measuring the transfer function of the resulting 2-port network. \$\endgroup\$ – Evan Mar 16 '17 at 21:47
  • \$\begingroup\$ A system can be stable with >0 gain and 180 degrees of phase shift. It's called conditional stability. As long as the phase shift at 0 dB of gain shows sufficient phase margin the system will be stable. \$\endgroup\$ – John D Mar 16 '17 at 22:09
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Please check out the first slides of http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202009.pdf to understand how an oscillator theoretically works. With your amplifier, as you do not want to make it an oscillator, you have to stay away from conditions where it can oscillate or become unstable. These conditions are a gain of 1 (0 dB) and a phase lag of 180° or 360° depending where you observe the return path. Because the amplifier is naturally compensated to decrease its gain as frequency increases, there is a frequency point at which the magnitude plot crosses the 0-dB axis: this is the crossover frequency \$f_c\$. At this frequency, you read the phase curve and the distance to the -180° or -360° (0° then) line is your phase margin. The more phase margin you have, the more damped your transient response will be. As phase margin diminishes, the system becomes less damped (faster response with overshoot) and ringing grows in amplitude. Even if the phase margin disappears before crossover, as long as you still have gain (magnitude greater than 1), the system will be stable if the phase margin is sufficient at \$f_c\$. This is called conditional stability: should the magnitude curve go up or down and the gain is now 1 at the point where the phase margin was 0° then you have oscillations at this point. Very often Bode plots can be misleading and a lot of designers prefer Nyquist, especially for non-minimum phase transfer functions (unstable poles or zeros or pure delay). You can read more about phase margin in a classic document from TI http://www.ti.com/lit/an/slyt087/slyt087.pdf. Below Figure 2, the author discusses phase margin from the open-loop plot.

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This "system" oscillates. Does this fit your definition of "system"?

With a varactor at end of the cable, you can tune the oscillator to 10MHz, with little phase noise.

schematic

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