(b) A control system in Figure Q.1b has an open-loop transfer function of 1 (s+3)(s+5)(s+8) 100, the closed-loop poles a

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answerhappygod
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(b) A control system in Figure Q.1b has an open-loop transfer function of 1 (s+3)(s+5)(s+8) 100, the closed-loop poles a

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B A Control System In Figure Q 1b Has An Open Loop Transfer Function Of 1 S 3 S 5 S 8 100 The Closed Loop Poles A 1
B A Control System In Figure Q 1b Has An Open Loop Transfer Function Of 1 S 3 S 5 S 8 100 The Closed Loop Poles A 1 (27.87 KiB) Viewed 22 times
(b) A control system in Figure Q.1b has an open-loop transfer function of 1 (s+3)(s+5)(s+8) 100, the closed-loop poles are at $1,2 = -2.77 ± j3.65 and s3 = -10.46. Simulation result of the system when applied with a unit step input is shown in Figure Q.1d. G(s): = R(s) + . The system root locus is given in Figure Q.1c. At K = K G(s) C(s) Figure Q.1b By adjusting the value of the proportional controller K, the system could give a close approximation to a second order system. Design the proportional controller K such that the system can be approximated as a second order system. Justify the validation of second order approximation. (12 marks)
Imaginary 10 9 8 7 160 5 -1 N w + n 16 ņ -8 -9 -10 288 Root Locus ===============I J MON Hom ===== -16 -15 -14 -13 -12 -11 -10 -9 BONDA -8 -7 -6 -5 4 -3 -2 -1 0 1 2 Real Figure Q.1c
Magnitude 0.6 0.5 0.4 33 0.2 0.1 0 0 0.5 1 Step Response 1.5 2 Time (seconds) Figure Q.1d 2.5 3 3.5
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