How do I know the value at which the Nyquist plot crosses -180 degrees from looking at the Phase plot from the Bode?
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The Nyquist plot is at -90 degrees when the phase lag is -90 degrees. This occurs over the frequency range during which the slope of the magnitude curve on the bode plot is truly -20dB/decade.
The -20dB/ decade slope will start to become steeper than -20dB/decade from about a decade below that double pole.
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\$\begingroup\$ But how do I know the value at which it crosses -180 degrees is -10? \$\endgroup\$– HectorOct 24, 2019 at 15:10
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1\$\begingroup\$ You don't, at -180 degrees the gain is 10. You've marked the gain as 20dB (=10) on the bode plot at -180 degrees which is why the system is unstable. ie. The gain is greater than 1 when the phase is -180 degrees. \$\endgroup\$– user173271Oct 24, 2019 at 15:23
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\$\begingroup\$ The minus in -10 on the Nyquist plot represents 180 degrees lag. So the gain is 10 on the bode plot but is shown as -10 on the Nyquist plot. \$\endgroup\$– user173271Oct 24, 2019 at 15:32
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\$\begingroup\$ Thanks, how do I know that at a frequency of 10^5, the dB gain is 20 dB? \$\endgroup\$– HectorOct 24, 2019 at 16:01
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1\$\begingroup\$ If the 1Hz gain is 10^6 (120dB) then dropping at 20dB/decade leaves 20dB at 10^5Hz. But I don't understand how you've arrived at your bode plot, either the low frequency pole frequency or the double pole frequency of 10^5. I'd be pleased to try and understand if you're willing to explain. I assume that A & B represent the open loop transfer function and the feedback transfer function which when multiplied together give the loop transfer function, Aol*beta otherwise known as G(s)H(s). \$\endgroup\$– user173271Oct 24, 2019 at 18:00