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compensator10 - Setting the real and imaginary parts equal...

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Compensators Q UESTION 10 For the closed–loop system in Figure 10 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 ). d. For K = 12 and r ( t ) = 0.6, 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 10 There are m = 2 finite open–loop zeros at –7 and –10 There are n = 3 finite open–loop poles at 0, –3, and –5 There are n = 3 branches and n m = 1 branch go to infinity The Root Locus is on the real axis between 0 and –3, between –5 and –7, and between –10 and –∞ The closed–loop characteristic equation is Substituting s = into the closed–loop characteristic equation and rearranging
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Unformatted text preview: Setting the real and imaginary parts equal to zero and solving simultaneously, 425 K 2 + 405 K + 120 = 0, which has solutions K = –0.477 ± 0.235 j . Since only real, positive values for K are considered, the closed–loop system is stable for K > 0. The closed–loop system is overdamped for small values of K , underdamped for intermediate values of K after the two branches break out of the real axis, and becomes overdamped again Compensators when the two branches break back into the real axis. The 2% settling time decreases as K increases and approaches 4/7 s as K approaches ∞. For r ( t ) = 0.8 t , The Root Locus Diagram and simulation plots are given below.-40-30-20-10-15-10-5 5 10 15 Real Axis Imaginary Axis 5 0.5 1 time (s) r y 5 2 4 time (s) r y 5-0.5 0.5 1 time (s) e(t) 5 0.005 0.01 0.015...
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