# Difference between revisions of "FAQ: What happens when the Nyquist plot goes exactly through -1?"

From FBSwiki

m (spelling) |
m |
||

Line 5: | Line 5: | ||

[[Category: Frequently Asked Questions]] | [[Category: Frequently Asked Questions]] | ||

− | [[Category: | + | [[Category: Loop Analysis FAQ]] |

[[Category: Nyquist FAQ]] | [[Category: Nyquist FAQ]] |

## Revision as of 16:21, 17 May 2008

*Posted by Tim Chung, 11 November 2002*

If the Nyquist plot passes through the critical point, s=-1+0j, then this means that the closed-loop poles, i.e. the zeros of the closed-loop characteristic equation, lie on the jw-axis. Hence, the system cannot be asymptotically stable. Whether it is stable or unstable depends on the multiplicity of the poles at the origin.

From a practical point-of-view, purely imaginary poles in the closed-loop system (as described above) are not usually desirable, in that this means the system will have oscillatory behavior. Thus, a well-designed closed-loop system should avoid such poles.