bode plot
Why 180 degree is taken as phase cross over frequency in bode plot?
In bode plot, grid impedance(Zg) and inverter impedance(Zi) is plotted for both magnitude and phase. The frequency at which Zg and Zi crosses in magnitude plot,
see the phase in phase plot of bode plot at the same frequency.if the phase is above 180 degree,the system is unstable and stable when phase is below 180 degree.
here my question is , why we choose 180 degree as the point to be a system stable or unstable
Any complex number \$Z\$ can be represented in rectangular coordinates or polar coordinates $$Z = x + jy = A\angle\phi$$ where, \$A = \sqrt{x^2+y^2}\$ and \$\phi = atan2(y,x)\$ where, \$atan2(y,x)\$ is defined as
As you can see your conversion formula is valid only if \$x>0\$.
Hence \$\phi\$ for \$a+jb\$ and \$\phi\$ for \$-a-jb\$ will be different.
I think this solves your problem.
My answer applies to higher-than-1st-order systems.
There will always be a resonant point even if you can't see it. You need to understand how "poles" work. Take a look at this: -
Even if there doesn't appear to be a resonance in the bode plot there will be a "pole" that is present and this pole represents the resonant frequency even though the "dampening" is causing it not to appear in the bode plot. Here is what a 2nd order low pass filter looks like with varying degrees of Q (where Q = \$\dfrac{1}{2\zeta}\$): -
If you could determine the phase angle where the output shifted by 90 degrees from the input you would find the resonant frequency even if it doesn't appear to have a "peak" in the bode plot.
Best Answer
Without seeing the diagrams your are talking about is a little difficult, but let me guess. At 180 phase delay, any controller loop is at critical point. Why? Because the negative feedback loop becomes a positive feedback loop. The feedback signal is not subtracted from setpoint value, but rather added. So what you get is an oscillator.