compensator2 - n – m = 2 distinct asymptotes with angles...

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Compensators Q UESTION 2 For the closed–loop system in Figure 2 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. Calculate K such that the steady–state error for a unit ramp input is 0.15. c. For the value of K calculated in part b, simulate the closed–loop system for r ( t ) = t and plot y ( t ), e ( t ), and u ( t ). d. For the value of K calculated in part b, calculate the closed–loop pole locations. For each complex conjugate pair, determine its natural frequency and damping ratio. For each real pole, determine its time constant. Turn in your Matlab code. G(s) KG C (s) U(s) R(s) + - Y(s) E(s) Figure 2 There is m = 1 finite open–loop zero at –1 There are n = 3 finite open–loop poles located at 0 and There are n = 3 branches and n m = 2 branches go to infinity The Root Locus is on the real axis between 0 and –1 The asymptote real–axis intercept is The asymptote angles are There are
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Unformatted text preview: n – m = 2 distinct asymptotes with angles The closed–loop characteristic equation is Substituting s = jω into the closed–loop characteristic equation and rearranging Compensators Setting the real and imaginary parts equal to zero and solving simultaneously, K = –4/3. Since only positive values for K are considered, the closed–loop system is stable for K > 0. The closed–loop system is always underdamped. As K approaches ∞, ζ approaches 0. The 2% settling time decreases as K increases and approaches 4/1 s as K approaches ∞. For r ( t ) = t , For K = 6.67, the closed–loop poles are located at The damping ratio and natural frequency, respectively, of the poles located at –1.53±5.756 j are and . The time constant of the pole located at –0.9401 is . The Root Locus Diagram and simulation plots are given below. 5 10 5 10 time (s) r y 5 10 0.1 0.2 time (s) e(t) 5 10 5 10 15 time (s) u(t)-3-2-1 1-10 10...
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