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If your system/transfer function has greater than zero gain for the frequency associated with -180 degrees phase, the system is said to be unstable. What does this mean for other input frequencies?

Does this mean that any input frequency would result in an unstably growing output? Or just a certain range of frequencies?

Bode plot for use as reference:

Bode Plot

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In a real system there will always be some component of noise or input signal at the frequency, so the output will grow until it hits some limit (saturates or whatever).

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  • \$\begingroup\$ Cheers! So have I got this right? We'd expect some non-zero noise at the frequencies that result in unstable feedback and then the system sees them growing from negligible to saturated? \$\endgroup\$ – tpixel Jun 11 '14 at 16:11
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    \$\begingroup\$ precisely. Every signal carries a bit of every frequency because every physic signal is time limited, so its fourier transform can't be frequency limited. \$\endgroup\$ – Vladimir Cravero Jun 11 '14 at 16:14
  • \$\begingroup\$ In reality, it is not the noise that brings the amplifier into saturation but the supply voltage switch-on transients (because these effects dominate over the noise). In short: If a circuit is unstable it will oscillate or go into saturation - indepent on any input signal. \$\endgroup\$ – LvW Jun 11 '14 at 16:52
  • \$\begingroup\$ @LvW True, but my point was even that if you allowed the system to reach equilibrium then close the loop, in a real system it will oscillate (it may not in simulation). It's like it was balanced on a knife edge, anything will set it off. \$\endgroup\$ – Spehro Pefhany Jun 11 '14 at 17:10

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