Question

Problem Consider the system shown in Figure 5–74(a). The damping ratio of this system is 0.158 and the undamped natural frequency is 3.16 rad/sec. To improve the relative stability, we employ tachometer feedback. Figure 5–74(b) shows such a tachometer-feedback system. Determine the value of Kn so that the damping ratio of the system is 0.5. Draw unit-step response curves of both the original and tachometer-feedback systems. Also draw the error-versus-time curves for the unit-ramp response of both systems. R(3) C(s) 10 s(s+1) R(5) C(s) 10 S+1 KA Figure 5-74 (a) Control system; (b) control system with tachometer feedback.

Answer 1

Transducer Feedback System: - RCS) ((S) 10 S+ in kn 10 RCS) cis) -In S+1+10km R(3) + 10 (0) S(S+1+10kn) 10 CCS) R(S) s(s+HOK 10 10 it s(s+itlokn) +10 s(s+Itlokin) 10 C(s) RLS) $+5 (1+10kn)+10

Pg_2 Given E=0.5 Wn=510 = 3.16 2 & Wn= I+10Kn 9X0.5 x 3.16 = 1+10kn 3.16 = 1+10kn =) lokn = 2.16 kn = 0.216 Error due to Ramp Ilp Css = ter Error Kva Lt SGCS) HCS) S->0 10 3.16 It s S=0 SCs +1+lokn) 10 -3.16 I+1000.216) Time Time Domain specifications: - Eq=0.5 wn = 3.16 rad/sec TI - Wd $ = cos 0.5 Rise Time to = = 60 TO = 0.79 sec. 3,16 91-0.g?

Peak Time to وفا tp = 17 TT =1147 sec. 3.16 JT-0.52 &TT e Ji- 0.52 * 100, Peak overshoot Mp = 1 -0.5 * TT = e 1-0.52 e * 100 Mp = = 16.03 ) Exoor due to step Input - 11 It kp Kp Lt G(S) HCS) = Lt * 50 570 S(5+1 +lok) ess Кр Time Response CCT) Ito 1 1.6 Peakovershoot A Peak Time (1.47 Sec) Rise Time (0.19 Sec)

Pg:-@ Original System' Res) - 10 CCS) SCS+1) Wn a Undamped Natural frequency = 3.16 rad/sec sampling ratio -0.158 Error due to Ramp Input A ess = I ku a given is unit Ramp ile ku Where Kv It S. G(s)H(S) S-> 10 *1 It s. 2 SCS+) S->0 error 10 sk 1 Ot1 - 10 1:01 ess: 10 Time error for Ramp Input. 10 CCS) SCS+1) RCS) It 10 SCS +1) 10 SCSH) s(s+1) + sest

19.0 10 CCS) R(5) s²15+10 Wn = 3.16rad/sec 3.16rad/sec & = 0.158 Time Domain Specifications: coste T-0 Ø = cos Rise Time tr= π-p - los '10.158) White ป4 - 80.9 13 TT -80.9* I 180 3-16, 1-(0-153) = 0.554 sec TT T Peak time, to Wd Writer = 1006 sec 11 tp 3.16 ST - (0.1583? & TIL Sree *100 Maximum Peak over shoot Mp4.ce 0 158 *T ST-(0.15012 * 100 e (1 = 60.4901

Pg!- © Error due to step Input les = A, 1+KP 17 It kp Kp It GCS) HCS) It S>0 SCS+1) S-> ess =O kp=00 too Unit Steppesponce 1.6 i Peakconstant Am ma Time to-Rise Time (0:554 sec)