Correct Answer - Option 4 : Negative real axis in Nyquist plot
Concept:
In G(s) plane, the Nyquist plot of G(S) passes through the negative real axis at the point (-a, j0)
a = magnitude of G(s) at ω = ωpc
ωpc is phase cross over frequency.
at \({\rm{\omega}} = {{\rm{\omega }}_{{\rm{pc}}}},\angle G\left( {j\omega } \right) = - 180^\circ\)
Application:
The -180° phase line of Bode diagram indicates the phase angle of the function is -180°.
At this phase, the Nyquist plot of G(S) passes through the negative real axis.
The slope of magnitude plot of a Bode plot at any frequency is given by
Slope = (Z – P) × 20 dB/decade
Where, Z is the number of zeros at that frequency
P is the number of poles at that frequency
If a closed-loop pole of the system is on the jω axis, then the magnitude equals 1 and the phase is 180 degrees at that frequency. If that occurs, the Nyquist plot will pass through the -1 point at that frequency. It indicates that there are equal number of poles and zeros.
Bode plot is logarithmic scale plot, so we take log value there.
Therefore, the unit circle of the Nyquist plot transforms into 0 dB line of the amplitude plot of the Bode diagram at any frequency
