WebNov 18, 2015 · Poles on the imaginary axis, i.e. poles with \$\text{Re}(s_{\infty})=0\$ do not satisfy (1), and, consequently, systems with such poles are not stable in the BIBO sense. In some contexts, systems with poles on the imaginary axis are called marginally stable, but such systems will generally produce unbounded outputs for bounded input signals.

While the question is long, please answer all parts Chegg.com

WebMar 29, 2024 · As a rule of thumb, if you have a transfer function with repeated poles on the imaginary axis, then the system is unstable. This may not be the case in the state-space … Web9. The characteristic equation of a control system is given by s 6 +2s 5 +8s 4 +12s 3 +20s 2 +16s+16=0 . The number of the roots of the equation which lie on the imaginary axis of s-plane: a) 0 b) 2 c) 4 d) 6 View Answer the community hospice of albany

Concept of Stability in Control Systems - Inst Tools

WebRoots on the imaginary axis makes the system : a) Stable b) Unstable c) Marginally stable d) Linear View Answer. Answer: c Explanation: Roots on the imaginary axis makes the system marginally stable. 12. If the roots of the have negative real parts then the response is … WebThis happens when there exist conjugate poles on the imaginary axis. In this case, we must use the auxiliary polynomial, which is built from the coefficients of the last non-zero row, and then differentiate it. The details can be found, for example, in here. WebApr 6, 2024 · As seen, the oscillation of the response keeps growing in amplitude and hence the system response becomes unbound and unstable. Case 6 – Poles on the imaginary axis. The poles lying on the imaginary axis are purely imaginary. Consider the poles to be at s = 3j and at s = -3j. The corresponding transfer function would be, the community home inc