Question

6.24 The transfer function for a second-order system is 35+5 32+20s + 500 (a) Determine the impulsive response of the system. (b) Deter- mine the step response of the system. (e) Determine the 2 per- cent settling time. (d) Determine the 10-90 percent rise time. (e) Determine the percent overshoot of the step response. (f) Determine the peak time of the step response.

Answer 1

Pages 35+ 5 = AS²+ 20 AS + 500 A + BS²+cs 35+ 5 = A²+Br² + 20Astcs +500A (A+B)5² + (20A+C) s + 5000 Equating the we eget 3515 A+B=0 Soo A 5 A = 5 500 A 100 200tc - 3 20x(1x) to +0=3 2 tca 3 10 1/3 + C=3 (= 3-1 5 c=15-1 5 C-14 5

Page 6 B = 100 Putting the All constant value in equation (i) -1005 + 1 1100 t Y(s) = s s²+20st 500 Y(S) = (700) + (-1/2008 + 19/6) s ²+ 205 + 500 (-Yoost + 4(5) - (410) 14/5) 3²+ zos+10² +500 (10) (-100 14/5) (5+10)2 +(500-100) st (%100+ Y(s) = (-1005 + 19/5) уроо Y(62= (400) + (st10) + 400 s + 1415 (s+10) ²400 (200) - Yoo (S+10) + 400 Y(S) = 1415 + 1/100 Y(S) = Vol) Yoo (5+10-10) (s+10)² + 400 (s+10) ²+ 400

(page y(s) = + 145 (4000) - You(s+10) +( 1/100) (sto) ²+ (2072 (st103+ 400 (5+10)² + 400 29/10 + Y(S) = (400) 1/100 (st10) (s+10)² + (2072 (5+10)² + (20) 2 Inverse Laplace pansform on both side thes weget Taking yle) = 4* { teoreet 100 elot x I sin sot -lot e coszot +29 20 1 y(t) = 100 loo ** e lot cossot & 29 e lod sinzot 200 100 y(t) = 100 Determine the (2%) setting time 35+5 Cy(s) = s²+2051500 comparing 5 + 2ęwnston =5242057500 2

Page 8 wy² = 500 whe √500 rad/see un - 22.36 2o 2{wn { n = 10 lo the wn 10 ou 22.36 da - 0.44 72 For (2% esos band) 4 (ts) settling time = १७) Its). 0.4472x 22.36 4 Its) 10 (ts) = 0.4 sec

Page 9 Determine the percentage overshoot of the step respouse. -918 Mp = 1-2 Xloo Com 3.1416x 0.44 72 I-(0.44722 X100 » Mp = e Mp = el ( 1.404923 VI-0.199987 XIoo Mp = - е - 1.404923 0.89 4433 X100 > Mp = e ㅋ Mp - (€1570*3)* 100 Mp = 0.207893 X100 IN Mp = 20-7893% (f Determine the peak time of the step response we know that, IT Peak time EL وحين (t) = (tp) = wndrez

3.1416 page (tp) - ( 22:36) J7-10-4402) 3.1416 (+p) = (22.36) - -0.199987 3.1416 (to) 19.99954 Appottimate (te) = 0.157083 see Final Auswer

we know that the laplace pansform of the ratio Page (6.24 The Transfer function for a second Order system is- 35+5 Gils) = s²+206 +500 solution: - @ Determine the impulsive response of the system of output and input which initial condition is equal to zero is know as transferoferaction y(s) X(S) 35+5 s242051500 impulsive response z(t) = 8(t) Taking Laplace foems forom ons both side then we get L6071 La(t) = (st) x(s) = 1 Y(S) - 35+5 5+205+500 35+5 y(s) - 5242051500 2os = 2 ab las 208=286

Page 20 cb 2 29 1o-b 35+5 Y(S) - 32+205+(10)2 +500 (10) 35+5 Y(S) = (3 +10)2 + (500-100) 2 (a+b) = a + b² + 2ab Y(S) = 35+5 (S40)2 + 400 35 + y (6)= (5+10)² + 400 (5+10) 24400 + Y(S) = 3 (5+10-10) (5+102 +400 S+10)2 + 400 + 3o Y(S) = 3 (s+lo) (5+10)+400 (SH100%+400 (+0)?+400 25 y (s) - 315+10) (5+10)+400 (8+10)²+400

page Taking then we get Inverse Laplace transform on both side I ys) [{ 3 (stio) (5+10² + 400 & - [{ 240 '{412023 şi face I ' 25 e eloti y (t) = 3 elot we know that, ILT cosat [ { 2 } [ IL 1 Sinat q 57928 plot 25 e lot cos aot - 2 x 1 Sinzot 20 y(t) = 3 wszot e lot sinzot y(t) = 3 e lot 20 -lot y(t) = 3 e or Coszot - 1.25 e sinzot -107

Determine the Step response of the systeros 4 Page b Y(S) D(0) 35+5 54205 +500 unit step response alt) = u(t) Taking Laplace tooms foom on 20th side then we Weget * * La(t) = Lucto Lut) = / u(s) = y (6) You 35+5 $2+206 +500 (/) 35+5 (s’+205+500) S Y(s) = 35+5 Y(s) = s(s? +205+500) (partial Fraction) д BSTC y (5) s²4 2084500 + S 35+5 Als+205+500) +(85+C)s 31542057500) str'+205+500) = A(3²+ 2054500) + BSc)s 3515