compensator19

# compensator19 - 2 1 180 0 1 2 a i i n m θ = = ± ±-K...

This preview shows pages 1–2. Sign up to view the full content.

Compensators Q UESTION 19 For the closed–loop system in Figure 19 with ( 29 1 C G s = and ( 29 2 1 G s s = , sketch the Root Locus Diagram. Qualitatively describe the closed–loop system stability and transient characteristics for 0 < K < ∞. Turn in your Matlab code. G(s) KG C (s) U(s) R(s) + - Y(s) E(s) Figure 19 There are m = 0 finite open–loop zeros There are n = 2 finite open–loop poles located at 0 and 0 There are n = 2 branches and n m = 2 branches go to infinity The Root Locus is on the real axis only at 0 The asymptote real–axis intercept is [ ] [ ] 0 0 0 0 a n m σ + - = = - The asymptote angles are

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: [ ] 2 1 180 0, 1, 2, a i i n m θ + = = ± ±-K There are n – m = 2 distinct asymptotes with angles 90 , 90 a =-The closed–loop characteristic equation is ( 29 ( 29 2 1 C KG s G s s K + = → + = Solving, s K j = ± . Therefore, the closed–loop system response is always undamped and the closed–loop system is always marginally stable. Compensators The Root Locus Diagram is given below.-0.2-0.15-0.1-0.05 0.05 0.1 0.15-1.5-1-0.5 0.5 1 1.5 Real Axis Imaginary Axis 2...
View Full Document

## This note was uploaded on 02/01/2012 for the course MECH ENG 279 taught by Professor Landers during the Spring '11 term at Missouri S&T.

### Page1 / 2

compensator19 - 2 1 180 0 1 2 a i i n m θ = = ± ±-K...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online