without using the routh hurwitz criterion, determing if the followign systems are asymptotically stable, marginally stable, or unstable. In each case, the closed-loop system transfer function is givenMs = [10 (s+2) ] / (s^3 + 3s^2 + 5s)

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Engineering

Golden

16 March, 18:37

without using the routh hurwitz criterion, determing if the followign systems are asymptotically stable, marginally stable, or unstable. In each case, the closed-loop system transfer function is givenMs = [10 (s+2) ] / (s^3 + 3s^2 + 5s)

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Maia Glover · Answer 1

The system is marginally stable.

Explanation:

Transfer function, M (s) = [10 (s+2) ] / (s³ + 3s² + 5s)

In control the stability properties of a system can be obtained from just the characteristic equation of its closed loop transfer function.

- The condition for stability is that all the roots of the characteristic equation be negative and real.

- The condition for asymptotic stability is that all the real parts of the roots must all be negative, since there'll be complex roots.

- The condition for marginal stability is that the real part of all the complex roots are negative, the roots without real parts must have distinct imaginary parts.

- The condition for instability is for at least one of the roots to be positive. Or if there are complex roots, the real part of the roots being positive indicates instability.

The characteristic equation for this transfer function is (s³ + 3s² + 5s)

Solving this polynomial

s = 0

s = [-3 - √ (11i) ]/2

s = [-3 + √ (11i) ]/2

These roots have all their real parts to be negative, and the zero root has a distinct imaginary part, hence the system is marginally stable