compensator11

# compensator11 - Substituting s = j into the closedloop...

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Compensators Q UESTION 11 For the closed–loop system in Figure 11 with and , complete the following: a. Sketch the Root Locus Diagram. Qualitatively describe the closed–loop system stability and transient characteristics for 0 < K < ∞. b. For K = 12 and r ( t ) = 0.8 t , calculate the steady–state error. c. For K = 12 and r ( t ) = 0.8 t , simulate the closed–loop system and plot y ( t ), e ( t ), and u ( t ). Turn in your Matlab code. G(s) KG C (s) U(s) R(s) + - Y(s) E(s) Figure 11 There are m = 2 finite open–loop zeros at –10 and –10 There are n = 4 finite open–loop poles at 0, –3, –5, and –15 There are n = 4 branches and n m = 2 branches go to infinity The Root Locus is on the real axis between 0 and –3 and between –5 and –15 The asymptote real–axis intercept is The asymptote angles are There are n m = 2 distinct asymptotes with angles The closed–loop characteristic equation is

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Unformatted text preview: Substituting s = j into the closedloop characteristic equation and rearranging Setting the real and imaginary parts equal to zero and solving simultaneously, 1500 K 2 + 26875 K + 1500 = 0, which has solutions K = 5.59910 2 and 17.86. Since only positive values for K are considered, the closedloop system is stable for K > 0. Compensators The closedloop system is overdamped for small values of K and underdamped after the two imaginary branches break out of the real axis. As K approaches , approaches 0. The 2% settling time decreases as K increases and approaches 4/1.5 s as K approaches . For r ( t ) = 0.8 t , The Root Locus Diagram and simulation plots are given below. 5 2 4 time (s) r y 5-0.2 0.2 time (s) e(t) 5-0.5 0.5 1 time (s) u(t)-20-10 10-20 20...
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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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compensator11 - Substituting s = j into the closedloop...

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