compensator13 - stability is that all of the coefficients...

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Compensators Q UESTION 13 For the closed–loop system in Figure 13 with ( 29 1 C G s = and ( 29 ( 29 ( 29 5 2 6 s G s s s + = + + , 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 13 There is m = 1 finite open–loop zero located at –5 There are n = 2 finite open–loop poles located at –2 and –6 There are n = 2 branches and n m = 1 branch goes to infinity The Root Locus is on the real axis from –2 to –5 and –6 to –∞ The closed–loop characteristic equation is ( 29 ( 29 ( 29 ( 29 2 1 0 8 5 12 0 C KG s G s s K s K + = + + + + =
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Compensators 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 closedloop system is stable for K &gt; 0. The closedloop system is always overdamped. The 2% settling time approaches 4/5 s as K approaches . For r ( t ) = 5, ( 29 ( 29 ( 29 ( 29 1 5 lim 1.62 5 5 1 2 6 s e s s s s s = = + + + + The steady state output is 3.38. The 2% bounds for the output are 3.31 and 3.45. The output reaches 3.31 and stays within the bounds at 0.669 s. Therefore, the graphically determined 2% settling time is 0.669 s. The percent overshoot is 0% since the response is overdamped. The Root Locus Diagram and simulation plots are given below. 2 4 6 2 4 6 time (s) r y 2 4 6 2 4 6 time (s) e(t) 2 4 6 10 20 30 time (s) u(t)-8-6-4-2-1 1 2...
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compensator13 - stability is that all of the coefficients...

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