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compensator16

# compensator16 - stability is that all of the coefficients...

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Compensators Q UESTION 16 For the closed–loop system in Figure 16 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 16 There are m = 2 finite open–loop zeros located at –5 and –4 There are n = 2 finite open–loop poles located at–2 and –6 There are n = 2 branches and n m = 0 branches go to infinity The Root Locus is on the real axis from –2 to –4 and –5 to –6 The closed–loop characteristic equation is Since the characteristic polynomial is second order, the necessary and sufficient condition for

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Unformatted text preview: stability is that all of the coefficients must have the same sign. Since only positive values for K are considered, the closed–loop system is stable for K > 0. The closed–loop system is overdamped for all K . The 2% settling time decreases as K increases and approaches 4/4 s as K approaches ∞. For r ( t ) = 5, Compensators The steady state output is 4.46. The 2% bounds for the output are 4.37 and 4.55. The output reaches 4.37 and stays within the bounds at 0.412 s. Therefore, the graphically determined 2% settling time is 0.412 s. The percent overshoot is 0% since the response is overdamped. The Root Locus Diagram and simulation plots are given below. 5 3.5 4 4.5 5 5.5 time (s) r y 5 0.4 0.6 0.8 1 time (s) e(t)-8-6-4-2-1 1 Real Axis Imaginary Axis...
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compensator16 - stability is that all of the coefficients...

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