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compensator14

# compensator14 - K 2 61 K 96 = 0 which has solutions K =...

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Compensators Q UESTION 14 For the closed–loop system in Figure 14 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 = 5 and r ( t ) = 5, calculate the steady–state error. c. For K = 5 and r ( t ) = 5, simulate the closed–loop system and plot y ( t ), e ( t ), and u ( t ). d. Graphically determine the percent overshoot and 2% settling time. Turn in your Matlab code. G(s) KG C (s) U(s) R(s) + - Y(s) E(s) Figure 14 There are m = 2 finite open–loop zeros located at –5 and –4 There are n = 3 finite open–loop poles located at 0, –2, and –6 There are n = 3 branches and n m = 1 branch goes to infinity The Root Locus is on the real axis from 0 to –2, –4 to –5, and –6 to –∞ The closed–loop characteristic equation is Substituting s = into the closed–loop characteristic equation and rearranging Setting the real and imaginary parts equal to zero and solving simultaneously, 9

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Unformatted text preview: K 2 + 61 K + 96 = 0, which has solutions K = –2.485 and –4.293. Since only 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 4/4 s as K approaches ∞. For r ( t ) = 5, The steady state output is 5. The 2% bounds for the output are 4.9 and 5.1. The output reaches 5.1 and stays within the bounds at 1.23 s. Therefore, the graphically determined 2% settling time is 1.23 s. The percent overshoot is . The Root Locus Diagram and simulation plots are given below. 2 4 6 5 10 time (s) r y 2 4 6-5 5 time (s) e(t) 2 4 6 10 15 20 25 time (s) u(t)-8-6-4-2-2 2...
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