# Why can’t one see for 2nd order system that it is at its stability limit neither in the Nyquist plot nor Bode plot?

Consider $$\hat{G}(s) = \frac{1}{s^2+s}$$ than the Nyquist plot is

and the Bode plot is

In both plots it seems that the closed-loop system is stable even when the eigenvalues are {0, -1}. For a higher order closed-loop system one can see that a system like

$$\hat{G}(s) = \frac{1}{s^3+s^2+s}$$

is at its stability limit since ne Nquist plots hits the real axis at -1 and in the Bode plot there is no phase reserve when magnitude hits 0. Why dose the Nyquist and the Bode plot fail for a closed-loop system of 2nd order?

## 1 Answer

I think you are mixing up the closed- and open-loop systems.

The closed-loop system is $$\frac{\frac{1}{s^2+s}}{1+\frac{1}{s^2+s}}=\frac{1}{s^2+s+1}$$

This has poles at $-0.5\pm 0.866025 i$ which is stable.

• Am i only allowed to view the plots for closed-loops? Nov 29, 2016 at 18:15
• The Nyquist stability criterion and things like gain and phase margin that are seen on the Bode plot are for the closed-loop system. The utility of these plots is that you can get insights into the closed-loop behavior using the open-loop transfer function. Nov 29, 2016 at 18:17
• I don't get your point, so let the open-loop system be $\frac{1}{s^2+s-1}$ so the closed on is $\frac{1}{s^2+s}$ now the eigenvalues of the closed one are still {0, -1} but my bode plot still shows a phase reserve of 51.8 deg when the magnitude hits 0 Nov 29, 2016 at 18:58
• For $\frac{1}{s^2+s-1}$, when the magnitude is 0 db, the phase is -180 degrees. The phase margin is 0. Nov 29, 2016 at 19:29
• but not in the matlab plot above Nov 29, 2016 at 19:40