Solution of skill Assessment Control Systems Engineering By Norman S.Nise 6th edition

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Solution of Skill Assessment Control Systems Engineering By Norman S.Nise 6th edition International Student Version

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<ul><li>1.E1SM 11/11/2010 9:29:8 Page 1 Solutions to Skill-Assessment Exercises CHAPTER 2 2.1 The Laplace transform of t is 1 s2 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 10 s 2 s 3 2S!0 5 9 B 10 s s 3 2S!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 5e2t 10 3 te3t 40 9 e3t 2.3 Taking the Laplace transform of the differential equation assuming zero initial conditions yields: s3 C s 3s2 C s 7sC s 5C s s2 R s 4sR s 3R s Collecting terms, s3 3s2 7s 5 C s s2 4s 3 R s Thus, C s R s s2 4s 3 s3 3s2 7s 5 1 </li></ul><p>2. E1SM 11/11/2010 9:29:8 Page 2 2.4 G s C s R s 2s 1 s2 6s 2 Cross multiplying yields, d2 c dt2 6 dc dt 2c 2 dr dt r 2.5 C s R s G s 1 s2 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 e4t 1 32 e8t 2.6 Mesh Analysis Transforming the network yields, +V(s) I1(s) I2(s) V1(s) V2(s) I9(s) + 1 1 s s s _ _ Now, writing the mesh equations, s 1 I1 s sI2 s I3 s V s sI1 s 2s 1 I2 s I3 s 0 I1 s I2 s s 2 I3 s 0 Solving the mesh equations for I2(s), I2 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 s2 2s 1 V s s s2 5s 2 2 Solutions to Skill-Assessment Exercises 3. E1SM 11/11/2010 9:29:8 Page 3 But, VL s sI2 s Hence, VL s s2 2s 1 V s s2 5s 2 or VL s V s s2 2s 1 s2 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 s2 2s 1 V s s2 5s 2 or VL s V s s2 2s 1 s2 5s 2 2.7 Inverting G s Z2 s Z1 s 100000 105 =s s Noninverting G s Z1 s Z s Z1 s 105 s 105 ! 105 s ! s 1 2.8 Writing the equations of motion, s2 3s 1 X1 s 3s 1 X2 s F s 3s 1 X1 s s2 4s 1 X2 s 0 Chapter 2 Solutions to Skill-Assessment Exercises 3 4. E1SM 11/11/2010 9:29:9 Page 4 Solving for X2(s), X2 s s2 3s 1 F s 3s 1 0 s2 3s 1 3s 1 3s 1 s2 4s 1 3s 1 F s s s3 7s2 5s 1 Hence, X2 s F s 3s 1 s s3 7s2 5s 1 2.9 Writing the equations of motion, s2 s 1 u1 s s 1 u2 s T s s 1 u1 s 2s 2 u2 s 0 where u1 s is the angular displacement of the inertia. Solving for u2 s , u2 s s2 s 1 T s s 1 0 s2 s 1 s 1 s 1 2s 2 s 1 F s 2s3 3s2 2s 1 From which, after simplification, u2 s 1 2s2 s 1 2.10 Transforming the network to one without gears by reflecting the 4 N-m/rad spring to the left and multiplying by (25/50)2 , we obtain, Writing the equations of motion, s2 s u1 s sua s T s su1 s s 1 ua s 0 where u1 s is the angular displacement of the 1-kg inertia. Solving for ua s , ua s s2 s T s s 0 s2 s s s s 1 sT s s3 s2 s 4 Solutions to Skill-Assessment Exercises 5. E1SM 11/11/2010 9:29:9 Page 5 From which, ua s T s 1 s2 s 1 But, u2 s 1 2 ua s : Thus, u2 s T s 1=2 s2 s 1 2.11 First find the mechanical constants. Jm Ja JL 1 5 1 42 1 400 1 400 2 Dm Da DL 1 5 1 42 5 800 1 400 7 Now find the electrical constants. From the torque-speed equation, set vm 0 to find stall torque and set Tm 0 to find no-load speed. Hence, Tstall 200 vnoload 25 which, Kt Ra Tstall Ea 200 100 2 Kb Ea vnoload 100 25 4 Substituting all values into the motor transfer function, um s Ea s KT RaJm s s 1 Jm Dm KTKb Ra 1 s s 15 2 where um s is the angular displacement of the armature. Now uL s 1 20 um s . Thus, uL s Ea s 1=20 s s 15 2 2.12 Letting u1 s v1 s =s u2 s v2 s =s Chapter 2 Solutions to Skill-Assessment Exercises 5 6. E1SM 11/11/2010 9:29:9 Page 6 in Eqs. 2.127, we obtain J1s D1 K s v1 s K s v2 s T s K s v1 s J2s D2 K s v2 s From these equations we can draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively. + J1 J1 J2 J2 D1 D1 D2 D2 T(t) T(t) w1(t) w1(t) w2(t) w2(t) 1 1 K 1 1 Series analog Parallel analog K 2.13 Writing the nodal equation, C dv dt ir 2 i t But, C 1 v vo dv ir evr ev evodv Substituting these relationships into the differential equation, d vo dv dt evodv 2 i t 1 We now linearize ev . The general form is f v f vo % df dv vo dv Substituting the function, f v ev , with v vo dv yields, evodv evo % dev dv vo dv 6 Solutions to Skill-Assessment Exercises 7. E1SM 11/11/2010 9:29:10 Page 7 Solving for evodv , evodv evo dev dv vo dv evo evo dv Substituting into Eq. (1) ddv dt evo evo dv 2 i t 2 Setting i t 0 and letting the circuit reach steady state, the capacitor acts like an open circuit. Thus, vo vr with ir 2. But, ir evr or vr lnir. Hence, vo ln 2 0:693. Substituting this value of vo into Eq. (2) yields ddv dt 2dv i t Taking the Laplace transform, s 2 dv s I s Solving for the transfer function, we obtain dv s I s 1 s 2 or V s I s 1 s 2 about equilibrium: CHAPTER 3 3.1 Identifying appropriate variables on the circuit yields + + C1 iL iC2 iC1 iR C2 R L vo(t)v1(t) Writing the derivative relations C1 dvC1 dt iC1 L diL dt vL C2 dvC2 dt iC2 1 Chapter 3 Solutions to Skill-Assessment Exercises 7 8. E1SM 11/11/2010 9:29:10 Page 8 Using Kirchhoffs current and voltage laws, iC1 iL iR iL 1 R vL vC2 vL vC1 vi iC2 iR 1 R vL vC2 Substituting these relationships into Eqs. (1) and simplifying yields the state equations as dvC1 dt 1 RC1 vC1 1 C1 iL 1 RC1 vC2 1 RC1 vi diL dt 1 L vC1 1 L vi dvC2 dt 1 RC2 vC1 1 RC2 vC2 1 RC2 vi where the output equation is vo vC2 Putting the equations in vector-matrix form, _x 1 RC1 1 C1 1 RC1 1 L 0 0 1 RC2 0 1 RC2 2 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 5 x 1 RC1 1 L 1 RC2 2 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 5 vi t y 0 0 1 x 3.2 Writing the equations of motion s2 s 1 X1 s sX2 s F s sX1 s s2 s 1 X2 s X3 s 0 X2 s s2 s 1 X3 s 0 Taking the inverse Laplace transform and simplifying, x1 _x1 x1 _x2 f x2 _x1 _x2 x2 x3 x3 _x3 x3 x2 Defining state variables, zi, z1 x1; z2 _x1; z3 x2; z4 _x2; z5 x3; z6 _x3 8 Solutions to Skill-Assessment Exercises 9. E1SM 11/11/2010 9:29:10 Page 9 Writing the state equations using the definition of the state variables and the inverse transform of the differential equation, _z1 z2 _z2 x1 _x1 x1 _x2 f z2 z1 z4 f _z3 _x2 z4 _z4 x2 _x1 _x2 x2 x3 z2 z4 z3 z5 _z5 _x3 z6 _z6 x3 _x3 x3 x2 z6 z5 z3 The output is z5. Hence, y z5. In vector-matrix form, _z 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 0 1 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 1 2 6 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 7 5 z 0 1 0 0 0 0 2 6 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 7 5 f t ; y 0 0 0 0 1 0 z 3.3 First derive the state equations for the transfer function without zeros. X s R s 1 s2 7s 9 Cross multiplying yields s2 7s 9 X s R s Taking the inverse Laplace transform assuming zero initial conditions, we get x 7_x 9x r Defining the state variables as, x1 x x2 _x Hence, _x1 x2 _x2 x 7_x 9x r 9x1 7x2 r Using the zeros of the transfer function, we find the output equation to be, c 2_x x x1 2x2 Putting all equation in vector-matrix form yields, _x 0 1 9 7# x 0 1# r c 1 2 x Chapter 3 Solutions to Skill-Assessment Exercises 9 10. E1SM 11/11/2010 9:29:11 Page 10 3.4 The state equation is converted to a transfer function using G s C sI A 1 B 1 where A 4 1:5 4 0 ! ; B 2 0 ! ; and C 1:5 0:625 : Evaluating sI A yields sI A s 4 1:5 4 s ! Taking the inverse we obtain sI A 1 1 s2 4s 6 s 1:5 4 s 4 ! Substituting all expressions into Eq. (1) yields G s 3s 5 s2 4s 6 3.5 Writing the differential equation we obtain d2 x dt2 2x2 10 df t 1 Letting x xo dx and substituting into Eq. (1) yields d2 xo dx dt2 2 xo dx 2 10 df t 2 Now, linearize x2 . xo dx 2 x2 o d x2 dx xo dx 2xodx from which xo dx 2 x2 o 2xodx 3 Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation gives us the linearized intermediate differential equation, d2 dx dt2 4xodx 2x2 o 10 df t 4 The force of the spring at equilibrium is 10 N. Thus, since F 2x2 ; 10 2x2 o from which xo 5 p 10 Solutions to Skill-Assessment Exercises 11. E1SM 11/11/2010 9:29:11 Page 11 Substituting this value of xo into Eq. (4) gives us the final linearized differential equation. d2 dx dt2 4 5 p dx df t Selecting the state variables, x1 dx x2 _dx Writing the state and output equations _x1 x2 _x2 dx 4 5 p x1 df t y x1 Converting to vector-matrix form yields the final result as _x 0 1 4 5 p 0 ! x 0 1 ! df t y 1 0 x CHAPTER 4 4.1 For a step input C s 10 s 4 s 6 s s 1 s 7 s 8 s 10 A s B s 1 C s 7 D s 8 E s 10 Taking the inverse Laplace transform, c t A Bet Ce7t De8t Ee10t 4.2 Since a 50; Tc 1 a 1 50 0:02s; Ts 4 a 4 50 0:08 s; and Tr 2:2 a 2:2 50 0:044 s. 4.3 a. Since poles are at 6 j 19:08; c t A Be6t cos 19:08t f . b. Since poles are at 78:54 and 11:46; c t A Be78:54t Ce11:4t . c. Since poles are double on the real axis at 15 c t A Be15t Cte15t : d. Since poles are at j 25; c t A B cos 25t f . 4.4 a. vn 400 p 20 and 2zvn 12; ;z 0:3 and system is underdamped. b. vn 900 p 30 and 2zvn 90; ;z 1:5 and system is overdamped. c. vn 225 p 15 and 2zvn 30; ;z 1 and system is critically damped. d. vn 625 p 25 and 2zvn 0; ;z 0 and system is undamped. Chapter 4 Solutions to Skill-Assessment Exercises 11 12. E1SM 11/11/2010 9:29:13 Page 12 4.5 vn 361 p 19 and 2zvn 16; ;z 0:421: Now, Ts 4 zvn 0:5 s and Tp p vn 1 z2 p 0:182 s. From Figure 4.16, vnTr 1:4998. Therefore, Tr 0:079 s. Finally, %os e zp 1 p z2 100 23:3% 4.6 a. The second-order approximation is valid, since the dominant poles have a real part of 2 and the higher-order pole is at 15, i.e. more than five-times further. b. The second-order approximation is not valid, since the dominant poles have a real part of 1 and the higher-order pole is at 4, i.e. not more than five-times further. 4.7 a. Expanding G(s) by partial fractions yields G s 1 s 0:8942 s 20 1:5918 s 10 0:3023 s 6:5 . But 0:3023 is not an order of magnitude less than residues of second-order terms (term 2 and 3). Therefore, a second-order approximation is not valid. b. Expanding G(s) by partial fractions yields G s 1 s 0:9782 s 20 1:9078 s 10 0:0704 s 6:5 . But 0.0704 is an order of magnitude less than residues of second-order terms (term 2 and 3). Therefore, a second-order approximation is valid. 4.8 SeeFigure4.31inthetextbookfortheSimulinkblockdiagramandtheoutputresponses. 4.9 a. Since sI A s 2 3 s 5 ! ; sI A 1 1 s2 5s 6 s 5 2 3 s ! : Also, BU s 0 1= s 1 # . The state vector is X s sI A 1 x 0 BU s 1 s 1 s 2 s 3 2 s2 7s 7 s2 4s 6# . The output is Y s 1 3 X s 5s2 2s 4 s 1 s 2 s 3 0:5 s 1 12 s 2 17:5 s 3 . Taking the inverse Laplace transform yields y t 0:5et 12e2t 17:5e3t . b. The eigenvalues are given by the roots of jsI Aj s2 5s 6, or 2 and 3. 4.10 a. Since sI A s 2 2 s 5 ! ; sI A 1 1 s2 5s 4 s 5 2 2 s ! . Taking the Laplace transform of each term, the state transition matrix is given by F t 4 3 et 1 3 e4t 2 3 et 2 3 e4t 2 3 et 2 3 e4t 1 3 et 4 3 e4t 2 6 6 4 3 7 7 5: 12 Solutions to Skill-Assessment Exercises 13. E1SM 11/11/2010 9:29:13 Page 13 b. Since F t t 4 3 e tt 1 3 e4 tt 2 3 e tt 2 3 e4 tt 2 3 e tt 2 3 e4 tt 1 3 e tt 4 3 e4 tt 2 6 6 4 3 7 7 5 and Bu t 0 e2t ! ; F t t Bu t 2 3 et et 2 3 e2t e4t 1 3 et et 4 3 e2t e4t 2 6 6 4 3 7 7 5: Thus, x t F t x 0 Rt 0 F t t BU t dt 10 3 et e2t 4 3 e4t 5 3 et e2t 8 3 e4t 2 6 6 4 3 7 7 5 c. y t 2 1 x 5et e2t CHAPTER 5 5.1 Combine the parallel blocks in the forward path. Then, push 1 s to the left past the pickoff point. 1 s s s s s2 + 1 + R(s) C(s) Combine the parallel feedback paths and get 2s. Then, apply the feedback formula, simplify, and get, T s s3 1 2s4 s2 2s . 5.2 Find the closed-loop transfer function, T s G s 1 G s H s 16 s2 as 16 , where and G s 16 s s a and H s 1. Thus, vn 4 and 2zvn a, from which z a 8 . But, for 5% overshoot, z ln % 100 p2 ln2 % 100s 0:69. Since, z a 8 ; a 5:52. Chapter 5 Solutions to Skill-Assessment Exercises 13 14. E1SM 11/11/2010 9:29:13 Page 14 5.3 Label nodes. ++ + R(s) s s N1 (s) N2 (s) N3 (s) N4 (s) N6 (s)N5 (s) N7 (s) s C(s) 1 s 1 s Draw nodes. R(s) N1(s) N2(s) N5(s) N7(s) N6(s) N3(s) N4(s) C (s) Connect nodes and label subsystems. R(s) C(s)1 1 s s 1 ss 1 1 1 1 1 sN1 (s) N2 (s) N5 (s) N6 (s) N7 (s) N3 (s) N4 (s) Eliminate unnecessary nodes. R(s) C(s)1 ss 1 s 1 s s 1 14 Solutions to Skill-Assessment Exercises 15. E1SM 11/11/2010 9:29:14 Page 15 5.4 Forward-path gains are G1G2G3 and G1G3. Loop gains are G1G2H1; G2H2; and G3H3. Nontouching loops are G1G2H1 G3H3 G1G2G3H1H3 and G2H2 G3H3 G2G3H2H3. Also, D 1 G1G2H1 G2H2 G3H3 G1G2G3H1H3 G2G3H2H3: Finally, D1 1 and D2 1. Substituting these values into T s C s R s P k TkDk D yields T s G1 s G3 s 1 G2 s 1 G2 s H2 s G1 s G2 s H1 s 1 G3 s H3 s 5.5 The state equations are, _x1 2x1 x2 _x2 3x2 x3 _x3 3x1 4x2 5x3 r y x2 Drawing the signal-flow diagram from the state equations yields 1 s 1 s 1 s11 1 1 5 4 23 3 r x1x2x3 y 5.6 From G s 100 s 5 s2 5s 6 we draw the signal-flow graph in controller canonical form and add the feedback. 1 5 6 100 500 1 y r 1 s 1 sx1 x2 Chapter 5 Solutions to Skill-Assessment Exercises 15 16. E1SM 11/11/2010 9:29:14 Page 16 Writing the state equations from the signal-flow diagram, we obtain x 105 506 1 0# x 1 0# r y 100 500 x 5.7 From the transformation equations, P1 3 2 1 4 ! Taking the inverse, P 0:4 0:2 0:1 0:3 ! Now, P1 AP 3 2 1 4# 1 3 4 6# 0:4 0:2 0:1 0:3# 6:5 8:5 9:5 11:5# P1 B 3 2 1 4# 1 3# 3 11# CP 1 4 0:4 0:2 0:1 0:3# 0:8 1:4 Therefore, _z 6:5 8:5 9:5 11:5# z 3 11# u y 0:8 1:4 z 5.8 First find the eigenvalues. jlI Aj l 0 0 l ! 1 3 4 6 ! l 1 3 4 l 6 l2 5l 6 From which the eigenvalues are 2 and 3. Now use Axi lxi for each eigenvalue, l. Thus, 1 3 4 6 ! x1 x2 ! l x1 x2 ! For l 2, 3x1 3x2 0 4x1 4x2 0 16 Solutions to Skill-Assessment Exercises 17. E1SM 11/11/2010 9:29:15 Page 17 Thus x1 x2 For l 3 4x1 3x2 0 4x1 3x2 0 Thus x1 x2 and x1 0:75x2; from which we let P 0:707 0:6 0:707 0:8 ! Taking the inverse yields P1 5:6577 4:2433 5 5 ! Hence, D P1 AP 5:6577 4:2433 5 5# 1 3 4 6# 0:707 0:6 0:707 0:8# 2 0 0 3# P1 B 5:6577 4:2433 5 5# 1 3# 18:39 20# CP 1 4 0:707 0:6 0:707 0:8# 2:121 2:6 Finally, _z 2 0 0 3# z 18:39 20# u y 2:121 2:6 z CHAPTER 6 6.1 Make a Routh table. s7 3 6 7 2 s6 9 4 8 6 s5 4.666666667 4.333333333 0 0 s4 4.35714286 8 6 0 s3 12.90163934 6.426229508 0 0 s2 10.17026684 6 0 0 s1 1.18515742 0 0 0 s0 6 0 0 0 Since there ar...</p>

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