Solutions to Skill-Assessment Exercises

Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright © 2004 by John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any from or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750- 8400, fax (978) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, e-mail: . To order books please call 1 (800) 225-5945. ISBN John Wiley & Sons, Inc. 111 River Street Hoboken, NJ 07030 USA Solutions to Skill-Assessment Exercises Chapter 2 2.1. The Laplace transform of t is 1 s 2 using Table 2.1, Item 3. Using Table 2.2, Item 4, F(s) = 1 (s + 5)2 . 2.2. Expanding F(s) by partial fractions yields: F(s) = A s + B s + 2 + C (s + 3)2 + D (s + 3) where, A s s S = + + = → 10 2 3 5 9 2 0 ( )( ) B = 10 s(s + 3)2 S→−2 = −5 C = 10 s(s + 2) S→−3 = 10 3 , and D = (s + 3)2 dF(s) ds s→−3 = 40 9 Taking the inverse Laplace transform yields, f (t) = 5 9 − 5e −2t + 10 3 te−3t + 40 9 e −3t 2.3. Taking the Laplace transform of the differential equation assuming zero initial conditions yields: s 3C(s) + 3s2C(s) + 7sC(s) + 5C(s) = s2R(s) + 4sR(s) + 3R(s) Collecting terms, (s 3 + 3s 2 + 7s + 5)C(s) = (s 2 + 4s + 3)R(s) Thus, 2 Solutions to Skill-Assessment Exercises C(s) R(s) = s 2 + 4s + 3 s 3 + 3s 2 + 7s + 5 2.4. G(s) = C(s) R(s) = 2s +1 s 2 + 6s + 2 Cross multiplying yields, d 2 c dt 2 + 6 dc dt + 2c = 2 dr dt + r 2.5. C(s) = R(s)G(s) = 1 s 2 * s (s + 4)(s + 8) = 1 s(s + 4)(s + 8) = A s + B (s + 4) + C (s + 8) where A = 1 (s + 4)(s + 8) S→0 = 1 32 B = 1 s(s + 8) S→−4 = − 1 16 , and C = 1 s(s + 4) S→−8 = 1 32 Thus, c(t) = 1 32 − 1 16 e −4t + 1 32 e −8t 2.6. Mesh Analysis Transforming the network yields, Now, writing the mesh equations, Chapter 2 3 (s +1)I 1 (s) − sI2 (s) − I 3 (s) = V(s) −sI1 (s) + (2s +1)I 2 (s) − I 3 (s) = 0 −I 1 (s) − I 2 (s) + (s + 2)I 3 (s) = 0 Solving the mesh equations for I2 (s), I 2 (s) = (s +1) V(s) −1 −s 0 −1 −1 0(s + 2) (s +1) −s −1 −s (2s +1) −1 −1 −1 (s + 2) = (s 2 + 2s +1)V(s) s(s 2 + 5s + 2) But, VL (s) = sI2 (s) Hence, VL (s) = (s 2 + 2s +1)V(s) (s 2 + 5s + 2) or VL (s) V(s) = s 2 + 2s +1 s 2 + 5s + 2 Nodal Analysis Writing the nodal equations, ( 1 s + 2)V1 (s) − VL (s) = V(s) −V1 (s) + ( 2 s +1)VL (s) = 1 s V(s) Solving for VL (s), VL (s) = ( 1 s + 2) V(s) −1 1 s V(s) ( 1 s + 2) −1 −1 (2 s +1) = (s 2 + 2s +1)V(s) (s 2 + 5s + 2) or VL (s) V(s) = s 2 + 2s +1 s 2 + 5s + 2 4 Solutions to Skill-Assessment Exercises 2.7. Inverting G(s) = − Z2 (s) Z1 (s) = −100000 (105 / s) = −s Noninverting G(s) = [Z1 (s) + Z2 (s)] Z1 (s) = ( 105 s +105 ) ( 105 s ) = s +1 2.8. Writing the equations of motion, (s 2 + 3s +1)X1 (s) − (3s +1)X2 (s) = F(s) −(3s +1)X1 (s) + (s 2 + 4s +1)X2 (s) = 0 Solving for X2 (s), X2 (s) = (s 2 + 3s +1) F(s) −(3s +1) 0 (s 2 + 3s +1) −(3s +1) −(3s +1) (s 2 + 4s +1) = (3s +1)F(s) s(s 3 + 7s 2 + 5s +1