If there exists ω=φ such that |KG(jφ)| = 1 and ∠(KG(jφ)) = -180°, then we know that s=jφ must be a pole.

This is incorrect, and is leading to a misunderstanding.

The poles of the system are the roots of the characteristic equation of the **Open Loop** transfer function $$KG(s)=0$$

the roots are of the form $$s=\sigma+j\omega$$

These are the poles and zeros that are analysed in the Bode Plot.

Once the loop is closed the poles move location to be the roots of the characteristic equation $$1+KG(s)=0$$ You can view loop compensation as a method of moving the open loop poles into more suitable (stable) locations in the closed loop. This can even by performed directly using **pole placement**.

However, Bode stability analysis is based on the Nyquist Stability Criterion. So the condition for oscillation in a negative feedback system is unity gain and 180 degree phase shift :$$KG(s) = -1$$
and therefore the Bode plot illustrates the "stability condition" by rearranging $$KG(s)+1=0$$

This happens to be the same equation as the characteristic equation of the **Closed Loop**. But it is a misunderstanding to see this as relating to Bode stability plots, which are an Open Loop analysis.

The Nyquist Stability Criterion also tells us that in general (but there are exceptions), a closed loop system is stable if the unity gain crossing of the magnitude plot occurs at a lower frequency than the -180 degree crossing of the phase plot.

When evaluating the stability of a control system, it's most insightful to plot the *loop gain*, either as a Bode plot or as a Nyquist plot. They do not differ in this respect.

The Nyquist plot is useful for employing the *Nyquist stability criterion*. In summary, loop gain encirclements of the point \$(-1, 0)\$ on the Nyquist plot indicate instability. Unless the system was already unstable (i.e., has RHP poles), in which case a *counter-clockwise* encirclement must be made for each RHP pole.

On a Bode plot, the usual technique for evaluating stability is to investigate the gain margin and phase margin of the loop gain. If both of these values are greater than zero, then the system is stable (as long as it doesn't have RHP poles). This technique isn't quite as general as the Nyquist criterion, but for the vast majority of control systems it's good enough. It's possible to evaluate the Nyquist criterion by looking at a Bode plot, but it's more difficult.

So, why would you evaluate stability with a Bode plot when the Nyquist criterion is more general? Because the Bode plot gives you a lot of insight that the Nyquist plot doesn't. The Bode plot shows gain and phase versus frequency, helping you identify what frequencies to place compensating poles and zeroes, as well as lending insights into closed-loop response that are impossible to see on a Nyquist plot (such as closed-loop bandwidth).

Finally, once you've determined your system is stable, you can re-use the Bode plot for a meaningful demonstration of *closed-loop response* as well. Plotting closed-loop gain on a Nyquist plot isn't nearly as meaningful.

## Best Answer

You can easily find the phase margin by simply solving for the open loop gain from the closed loop gain. If \$F(s)\$ is the closed loop transfer function, and \$G(s)\$ is the open loop, then:

$$ F(s)=\frac{G(s)}{1+G(s)}$$ so $$ G(s)=\frac{F(s)}{1-F(s)}$$

You can work out the closed loop gain that corresponds to the open loop unity gain with a particular phase margin.

and just look at the closed loop gain bode plot to see where your loop is. The one you posted seems to have a phase margin of about 70 degrees.

The bode plot is not the easiest way to do this, if you plotted the closed loop gain on a polar plot, then the table above would be a curve and you could just find the point of intersection.