compensator22

# compensator22 - K s z is Therefore K = 2.8 and z = 5.71 for...

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Compensators Q UESTION 22 For the closed–loop system in Figure 22 with and , complete the following: a. Design a PD compensator such that the closed–loop system has a damping ratio of 0.35 and a natural frequency of 4 rad/s. b. For the zero location computed in part a, sketch the Root Locus Diagram. Qualitatively describe the closed–loop system stability and transient characteristics for 0 < K < ∞. c. For the PD compensator designed in part a, calculate the steady–state error for r ( t ) = 5, r ( t ) = 10 t , and r ( t ) = 8 t 2 . d. For the PD compensator designed in part a, simulate the closed–loop system for r ( t ) = 5, r ( t ) = 10 t , and r ( t ) = 8 t 2 . For each case, plot y ( t ) and e ( t ). Turn in your Matlab code. G(s) KG C (s) U(s) R(s) + - Y(s) E(s) Figure 22 There is m = 1 finite open–loop zero located at –4 There are n = 2 finite open–loop poles located at 0 and 0 There are n = 2 branches and n m = 1 branch goes to infinity The closed–loop characteristic equation for a general PD compensator [

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Unformatted text preview: K ( s + z )] is . Therefore, K = 2.8 and z = 5.71 for Î¶ = 0.35 and Ï‰ n = 4 rad/s. The closedâ€“loop characteristic equation is Since the characteristic polynomial is second order, the necessary and sufficient condition for stability is that all of the coefficients must have the same sign. Since only positive values for K are considered, the closedâ€“loop system is stable for K > 0. Compensators The closedâ€“loop system is underdamped for small values of K and becomes overdamped when the two branches break into the real axis. The 2% settling time decreases as K increases and approaches 4/5.71 s as K approaches âˆž. For r ( t ) = 5 For r ( t ) = 10 t For r ( t ) = 8 t 2 , The Root Locus Diagram and simulation plots are given below.-25-20-15-10-5 5-6-4-2 2 4 6 Real Axis Imaginary Axis Compensators 2 4 6 5 10 time (s) r y 2 4 6-5 5 time (s) e(t) 2 4 6 50 100 time (s) r y 2 4 6-2 2...
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## 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.

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compensator22 - K s z is Therefore K = 2.8 and z = 5.71 for...

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