Question

a) Consider the control system in Figure 2(a). Determine the transient response characteristics (rise time, peak time, maximum overshoot and settling time) and the steady state error for the system. (2 marks) b) To improve the relative stability, the tachometer feedback are employed as shown in Figure 2(b). i. Determine the value Kn so that the damping ratio of the system is 0.5. (1 % marks) ii. From the result obtained in (), evaluate the transient response characteristics (rise time, peak time, maximum overshoot and settling time) and steady state error. (1 % marks) iii. Sketch the output response for the both systems in Figure (a) and (b). (2 marks) iv. State what type of the controller that can replace the tachometer feedback which can give the similar output response. Discuss your answer. (3 marks) [10 MARKS R(s) C(s) 10 s(s+1) R(s) C(s) 10 5+1 Figure 2 (a) Control system; (b) Control system with tachometer feedback

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Answer 1

9 (ii) Wn= TO 2 & wn= 3. 1622 &wna 1-5811 0.5 an 2 War wn. 11-42 = Sioxy - Tox 2-4386 Mad I see & = Tant 11-21 Wds 2 = Tant 1-14 11 60 x 즈 وا & Rise time (tr) 7-O 180°- 60 2.7386 = 0.7647 sec1 t peak time (tp) = a T 2.7386 = 1.1471 sec. wd -&T 11 MP e 0.5 x e LI Mp = 0-16303 3 Setting time (ts - 54. Tollerance) = 7.531 {wn 1.5811 for 2-1. Tollerance 1.8974sec 4 {wn 1-5811 2.52 98501 Scanned with CamScanner

open Loop transfer function tillo 10 ve Gle) HOB) s(s+ itlokn) S($+Sio Vio s(S+ UTO) = STO Sio Vio ku kra It S. Gus) hes) Sto it s 10 LO -1918 10 10 ess = 0.3162 Ces) RES) 10 Stst10 1 Response for steep input R(S) = 5 10 => lis) Sts+10 Sto: 5+105539 S B A S + + S+05+30.5 339 Sto.5+j0.5 539 11 = A s'ts to S=0 B= (s)(5 +0.5-10.5.839) 15 = -05 5-j0.5.539 ls BE CE (-0.5-30-3539)x (-339) B= US89) (0.5410-5 150) (5) CS+ (5) Sto.5+jog J39) 57) \s (1 557 -05 +505 839). -0.5 +/0.5 539 CS Scanned with CamScanner

(05-30:5 (39) Clt) = t ce = {A.* B. where 1 UiO.S+10 - (0.5+30.5 3977 .# B. e A = 1 B = (1 599) (05+1 0.5 537) (139) (-05+jo: 5.439) for tiglb); gesponse for Strep input Res)=21s closopen loop transfor function ordon ces s?+ JTÓ. S+101 ces) 10 RCS) 10 (sa+TO S+10) s(5+To Sto) 10 A=1 10 Ris) + IS A TO (مژاند (s+ vb – jv30) s(s+ Sip.ira )s - Tho-iso B (5330) Ciot j 530) Grad Celio + 110) CS Scanned with CamScanner

(99 +jJ30) (+502 +2 (LE) = SATB toe Ult) A= BE (30) (10 41J30) (judo) 6- so ti s3o). civ) Generally tachometer will be employed to improve the speed of response le transient performance is improved so to improve the transiat performance to Derivative controller can also be 01 used. SO we can replace tachometer with derivative controll 091 1 CS Scanned with CamScanner

a closed loop transfer function for the figca) is given as 10 COS) SC+1) Res) 1+ 10 SCS+1) 10 SCS+1) +10 $*+5+10 By comparing the above transfer function with Standard second order troms for function cls). 10 wno 5+2&wn S + wn RES) $4+s+10 wn= To 2&wn = 1 & wn= 1 / 2 = 0.1581 & = 2 wn 2UTO wa= wn 11-42 ton 510 x 539 - 39 20 J40 40 Wd = 3,122 rad /sec &= Taulti- I(JT- & ? ) = Tanisl to Tan 1539 2 To 2500 2.Vio Tan' (539) = 80.90259 weer 1990-90.9025 rise time (to) = -0% wd 3:1222 0.5539 sec 1.0062 See peak time (tp ) = wd 3.122 è &ALUTE2 X100 %. Maximum over shoot (1.MP) CS Scanned with CamScanner

-0,1581x T MT-0.15817 X100 e to Mp = 60.4706 %. 33 = 6 sec.- 51. Toloyance 42 Settiling time (ts) &wn for at. Tolerance band, 4 = 8 sec. ts= 4 12 qwn Steady state Error it S. ECS) $> ولية Gis) = 10 S(S+1) As it is type 1 system, ct s lo = 10 kv i S. Ges) 5-10 SCS+1) S>0 Kv 10 A ess == KW kv 2 to 30.1 tise time (tor) = 0.5539 see. peak time ( tp) = 1.0062 sec. Maximum over shoot (mp) = 0.604706 Settling time (ts - for 51. Tollerance) = 6 sec (ts- for 27. Toll rance) = & Sec Steady state Error (ess) CS Scanned with CamScanner

S(St 1+10kn) 6) 10 ces) RES) ş S+itlokh 10 cos) Res) 011 10 1+ s(s+i+10kn) cool 10 s(s+1+10. kn) + 10 10 CCS) R(S) $? + s (l+lokh) + 10 Itlokh. Wn= To 24 wn Given = 4:05 2x + x STO = 1+ lokh : ) rool lookh kh= 0.2162 .: Ces) = 10 RCS) s+SC3-1622) +0. CS Scanned with CamScanner

9 (ii) Wn= TO 2 & wn= 3. 1622 &wna 1-5811 0.5 an 2 War wn. 11-42 = Sioxy - Tox 2-4386 Mad I see & = Tant 11-21 Wds 2 = Tant 1-14 11 60 x 즈 وا & Rise time (tr) 7-O 180°- 60 2.7386 = 0.7647 sec1 t peak time (tp) = a T 2.7386 = 1.1471 sec. wd -&T 11 MP e 0.5 x e LI Mp = 0-16303 3 Setting time (ts - 54. Tollerance) = 7.531 {wn 1.5811 for 2-1. Tollerance 1.8974sec 4 {wn 1-5811 2.52 98501 Scanned with CamScanner

open Loop transfer function tillo 10 ve Gle) HOB) s(s+ itlokn) S($+Sio Vio s(S+ UTO) = STO Sio Vio ku kra It S. Gus) hes) Sto it s 10 LO -1918 10 10 ess = 0.3162 Ces) RES) 10 Stst10 1 Response for steep input R(S) = 5 10 => lis) Sts+10 Sto: 5+105539 S B A S + + S+05+30.5 339 Sto.5+j0.5 539 11 = A s'ts to S=0 B= (s)(5 +0.5-10.5.839) 15 = -05 5-j0.5.539 ls BE CE (-0.5-30-3539)x (-339) B= US89) (0.5410-5 150) (5) CS+ (5) Sto.5+jog J39) 57) \s (1 557 -05 +505 839). -0.5 +/0.5 539 CS Scanned with CamScanner