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

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Solutions to Skill-Assessment Exercises 2.. Chapter 2 The Laplace transform of t is using Table 2., Item 3. Using Table 2.2, Item 4, 2 s F(s) 2.2. (s + 5) 2. Expanding F(s) by partial fractions yields: F(s) A s + where, 0 A 2 ( s+ 2)( s+ 3) S D (s + 3) 2 df(s) ds B s + 2 + C (s + 3) + D 2 (s + 3) 0 s 3 5 9 40 9 B 0 5C s(s + 3) 2 S 2 Taking the inverse Laplace transform yields, f (t) 5 9 5e 2t + 0 3 te 3t + 40 9 e 3t 2.3. 0 s(s + 2) S 3 0 3, and Taking the Laplace transform of the differential equation assuming zero initial conditions yields: s 3 C(s) + 3s 2 C(s) + 7sC(s) + 5C(s) s 2 R(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 + s 2 + 6s + 2 Cross multiplying yields, d 2 c dt 2 2.5. + 6 dc dt + 2c 2 dr dt + r C(s) R(s)G(s) s 2 * where A Thus, s (s + 4)(s + 8) s(s + 4)(s + 8) A s + B (s + 4) + C (s + 8) (s + 4)(s + 8) S 0 32 B s(s + 8) S 4 6, and C s(s + 4) S 8 32 c(t) 32 6 e 4t + 32 e 8t 2.6. Mesh Analysis Transforming the network yields, Now, writing the mesh equations,

Chapter 2 3 (s + )I (s) si 2 (s) I 3 (s) V(s) si (s) + (2s + )I 2 (s) I 3 (s) 0 I (s) I 2 (s) + (s + 2)I 3 (s) 0 Solving the mesh equations for I 2 (s), I 2 (s) (s + ) V(s) s 0 0 (s + 2) (s + ) s s (2s + ) (s + 2) (s2 + 2s + )V(s) s(s 2 + 5s + 2) But, V L (s) si 2 (s) Hence, V L (s) (s2 + 2s + )V(s) (s 2 + 5s + 2) or V L (s) V(s) s2 + 2s + s 2 + 5s + 2 Nodal Analysis Writing the nodal equations, ( s + 2)V (s) V L (s) V(s) V (s) + ( 2 s + )V L (s) s V(s) Solving for V L (s), V L (s) ( + 2) V(s) s s V(s) ( + 2) s ( 2 s + ) (s2 + 2s + )V(s) (s 2 + 5s + 2) or V L (s) V(s) s2 + 2s + s 2 + 5s + 2

4 Solutions to Skill-Assessment Exercises 2.7. Inverting G(s) Z (s) 2 Z (s) 00000 (0 5 / s) s Noninverting G(s) [Z (s) + Z 2 (s)] Z (s) 2.8. ( 05 s + 05 ) ( 05 s ) Writing the equations of motion, (s 2 + 3s + )X (s) (3s + )X 2 (s) F(s) (3s + )X (s) + (s 2 + 4s + )X 2 (s) 0 Solving for X 2 (s), (s 2 + 3s + ) F(s) (3s + ) 0 X 2 (s) (s 2 + 3s + ) (3s + ) (3s + ) (s 2 + 4s + ) Hence, X 2 (s) F(s) (3s + ) s(s 3 + 7s 2 + 5s + ) 2.9. Writing the equations of motion, (s 2 + s + )θ (s) (s + )θ 2 (s) T(s) (s + )θ (s) + (2s + 2)θ 2 (s) 0 s + (3s + )F(s) s(s 3 + 7s 2 + 5s + ) where θ (s) is the angular displacement of the inertia. Solving for θ 2 (s), (s 2 + s + ) T(s) (s + ) 0 θ 2 (s) (s 2 + s + ) (s + ) (s + ) (2s + 2) From which, after simplification, (s + )F(s) 2s 3 + 3s 2 + 2s +

Chapter 2 5 θ 2 (s) 2.0. 2s 2 + s + 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, T(t) θ (t) kg N-m-s/rad θ a (t) N-m/rad Writing the equations of motion, (s 2 + s)θ (s) sθ a (s) T(s) sθ (s) + (s + )θ a (s) 0 where θ (s) is the angular displacement of the -kg inertia. Solving for θ a (s), (s 2 + s) T(s) s 0 θ a (s) (s 2 + s) s s (s + ) From which, θ a (s) T(s) s 2 + s + st(s) s 3 + s 2 + s But, θ 2 (s) 2 θ a (s). Thus, θ 2 (s) T(s) /2 s 2 + s + 2.. First find the mechanical constants. J m J a + J L ( 5 * 4 )2 + 400( 400 ) 2 D m D a + D L ( 5 * 4 )2 5 + 800( 400 ) 7

6 Solutions to Skill-Assessment Exercises Now find the electrical constants. From the torque-speed equation, set ω m 0 to find stall torque and set T m 0 to find no-load speed. Hence, T stall 200 ω no load 25 which, K t R a T stall E a 200 00 2 K b E a 00 ω no load 25 4 Substituting all values into the motor transfer function, θ m (s) E a (s) K T R a J m s(s + J m (D m + K T K b R a ) s(s + 5 2 ) where θ m (s) is the angular displacement of the armature. Now θ L (s) 20 θ m (s). Thus, θ L (s) E a (s) /20 s(s + 5 ) 2 ) 2.2. Letting θ (s) ω (s)/s θ 2 (s) ω 2 (s)/s in Eqs. 2.27, we obtain (J s + D + K s )ω (s) K s ω 2 (s) T(s) K s ω (s) + (J 2 s + D 2 + K s )ω 2 (s) From these equations we can draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively.

Chapter 2 7 Series analog 2.3. Writing the nodal equation, C dv dt + i 2 i(t) r But, C v v o + δv i r e v r e v e v o +δv Parallel analog Substituting these relationships into the differential equation, d(v o + δv) + e v o +δv 2 i(t) () dt We now linearize e v. The general form is f (v) f (v o ) df dv vo δv Substituting the function, f (v) e v, with v v o + δv yields, e v o +δv e v o dev dv vo δv Solving fore v o +δv,

8 Solutions to Skill-Assessment Exercises e v o +δv e v o + dev dv vo δv e v o + ev o δv Substituting into Eq. () dδv dt + e v o + ev oδv 2 i(t) (2) Setting i(t) 0 and letting the circuit reach steady state, the capacitor acts like an open circuit. Thus, v o v r with i r 2. But, i r e v r or v r lni r. Hence, v o ln 2 0.693. Substituting this value of v o into Eq. (2) yields dδv dt + 2δv i(t) Taking the Laplace transform, (s + 2)δv(s) I(s) Solving for the transfer function, we obtain δv(s) I(s) s + 2 or V(s) I(s) s + 2 about equilibrium.

9 3.. Chapter 3 Identifying appropriate variables on the circuit yields Writing the derivative relations C dv C dt i C L di L dt v L () C 2 dv C2 dt i C2 Using Kirchhoff s current and voltage laws, i C i L + i R i L + R (v L v C 2 ) v L v C + v i i C2 i R R (v L v C 2 ) Substituting these relationships into Eqs. () and simplifying yields the state equations as dv C dt di L dt dv C2 dt RC v C + C i L RC v C2 + RC v i L v C + L v i RC 2 v C RC 2 v C2 RC 2 v i where the output equation is v o v C2 Putting the equations in vector-matrix form,

0 Solutions to Skill-Assessment Exercises RC C RC RC x 0 0 x + v i (t) L 0 L RC 2 RC 2 RC 2 y [ 0 0 ]x 3.2. Writing the equations of motion (s 2 + s + )X (s) sx 2 (s) F(s) sx (s) + (s 2 + s + )X 2 (s) X 3 (s) 0 X 2 (s) + (s 2 + s + )X 3 (s) 0 Taking the inverse Laplace transform and simplifying, x x x + x 2 + f x 2 x x2 x2 + x 3 x 3 x3 x3 + x 2 Defining state variables, z i, z x ; z 2 x ; z3 x 2 ; z 4 x 2; z5 x 3 ; z 6 x 3 Writing the state equations using the definition of the state variables and the inverse transform of the differential equation, z z2 z 2 x x z 3 x2 z 4 x + x 2 + f z2 z + z 4 + f z 4 x2 x x2 x2 + x 3 z 2 z 4 z 3 + z 5 z 5 x3 z 6 z 6 x3 x3 x3 + x 2 z 6 z 5 + z 3 The output is z 5. Hence, y z 5. In vector-matrix form,

Chapter 3 0 0 0 0 0 0 0 0 0 z 0 0 0 0 0 0 0 0 z + 0 0 0 0 0 0 0 0 0 0 0 3.3. f (t); y [ 0 0 0 0 0]z First derive the state equations for the transfer function without zeros. X(s) R(s) s 2 + 7s + 9 Cross multiplying yields (s 2 + 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, x x x 2 x Hence, x x2 x 2 x 7 x 9x + r 9x 7x 2 + r Using the zeros of the transfer function, we find the output equation to be, c 2 x + x x + 2x 2 Putting all equation in vector-matrix form yields, x 0 9 7 x + 0 r c [ 2]x 3.4. The state equation is converted to a transfer function using G(s) C(sI A) B () where

2 Solutions to Skill-Assessment Exercises 4.5 A 4 0, B 2 0, and C [.5 0.625]. Evaluating (si A) yields (si A) s + 4.5 4 s Taking the inverse we obtain (si A) s 2 + 4s + 6 s.5 4 s + 4 Substituting all expressions into Eq. () yields G(s) 3s + 5 s 2 + 4s + 6 3.5. Writing the differential equation we obtain d 2 x dt + 2 2x2 0 + δf (t) () Letting x x o + δx and substituting into Eq. () yields d 2 (x o + δx) dt 2 + 2(x o + δx) 2 0 + δf (t) (2) Now, linearize x 2. (x o + δx) 2 x o 2 d(x2 ) dx from which (x o + δx) 2 x o 2 + 2x o δx (3) x o δx 2x o δx Substituting Eq. (3) into Eq. () and performing the indicated differentiation gives us the linearized intermediate differential equation, d 2 δx dt 2 + 4x o δx 2x 2 o + 0 + δf (t) (4) The force of the spring at equilibrium is 0 N. Thus, since F 2x 2, 2 0 2x o from which x o 5

Chapter 3 3 Substituting this value of x o into Eq. (4) gives us the final linearized differential equation. d 2 δx dt 2 + 4 5δx δf (t) Selecting the state variables, x δx x 2 δx Writing the state and output equations x x2 x 2 δx y x 4 5x + δf (t) Converting to vector-matrix form yields the final result as x 0 4 5 0 x + 0 δf (t) y [ 0]x

4 Chapter 4 4.. For a step input C(s) 0(s ) 4)(s ) 6) s(s ) )(s ) 7)(s ) 8)(s ) 0) A s + Taking the inverse Laplace transform, c(t) A + Be t + Ce 7t + De 8t + Ee 0t 4.2. B s + + C s + 7 + D s + 8 + E s + 0 Since a 50, T c a 50 0.02s; T 4 s a 4 50 0.08 s; and T 2.2 r a 2.2 0.044 s. 50 4.3. a. Since poles are at 6 ± j9.08, c(t) A + Be 6t cos(9.08t + φ). b. Since poles are at 78.54 and.46, c(t) A + Be 78.54t + Ce.4t. c. Since poles are double on the real axis at 5 c(t) A + Be 5t + Cte 5t. d. Since poles are at ±j25, c(t) A + Bcos(25t + φ). 4.4. a. ω n 400 20 and 2ζω n 2; ζ 0.3 and system is underdamped. b. ω n 900 30 and 2ζω n 90; ζ.5 and system is overdamped. c. ω n 225 5 and 2ζω n 30; ζ and system is critically damped. d. ω n 625 25 and 2ζω n 0; ζ 0 and system is undamped. 4.5. ω n 36 9 and 2ζω n 6; ζ 0.42. Now, T s 4 ζω n 0.5 s and T p π 0.82 s. 2 ω n ζ From Figure 4.6, ω n T r.4998. Therefore, T r 0.079 s. -ζπ ζ Finally, %os e 2 *00 23.3%

Chapter 4 5 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 5, i.e. more than five-times further. b. The second-order approximation is not valid, since the dominant poles have a real part of 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) s + 0.8942 s + 20.598 s + 0 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) s + 0.9782 s + 20.9078 s + 0 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. See Figure 4.3 in the textbook for the Simulink block diagram and the output responses. 4.9. a. Since si A s 2 3 s + 5, (si s + 5 2 A) s 2 + 5s + 6 3 s. Also, 0 BU(s) /(s + ). 2(s 2 + 7s + 7) The state vector is X(s) (si A) [x(0) + BU(s)] (s + )(s + 2)(s + 3) s 2 4s 6. 5s The output is Y(s) [ 2 + 2s 4 0.5 3]X(s) (s + )(s + 2)(s + 3) s + 2 s + 2 + 7.5 s + 3. Taking the inverse Laplace transform yields y(t) 0.5e t 2e 2t + 7.5e 3t. b. The eigenvalues are given by the roots of si A s 2 + 5s + 6, or 2 and 3.

6 Solutions to Skill-Assessment Exercises 4.0. a. Since (si A) s 2 2 s + 5, (si A) s 2 + 5s + 4 transform of each term, the state transition matrix is given by 4 Φ(t) 3 e t 2 3 e 4t 3 e t 2 3 e 4t 2 3 e t + 2 3 e 4t 3 e t + 4. 3 e 4t s + 5 2 2 s. Taking the Laplace 4 b. Since Φ(t τ) 3 e (t τ ) 3 e 4(t τ ) 2 3 e (t τ ) 2 3 e 4(t τ ) 0 2 3 e (t τ ) + 2 3 e 4(t τ ) 3 e (t τ ) + 4 and Bu(τ) τ ) e 4(t 3 2 Φ(t τ)bu(τ) 3 e τ e t 2 3 e2τ e 4t 3 e τ e t + 4. 3 e2τ e 4t 0 t Thus, x(t) Φ(t)x(0)+ Φ(t τ)bu(τ) dτ 3 e t e 2t 4 3 e 4t 0 5 3 e t + e 2t + 8. 3 e 4t c. y(t) [ 2 ]x 5e t e 2t e 2τ,

7 5.. Chapter 5 Combine the parallel blocks in the forward path. Then, push s pickoff point. to the left past the s R(s) + - - s 2 + s s C(s) s Combine the parallel feedback paths and get 2s. Then, apply the feedback 3 s + formula, simplify, and get, Ts () 4 2 2s + s + 2 s. 5.2. G(s) Find the closed-loop transfer function, T(s) + G(s)H(s) 6 s 2 + as + 6, where G(s) 6 s(s + a) and H(s). Thus, ω n 4 and 2ζω n a, from which ζ a 8. But, for 5% overshoot, ζ ln( % 00 ) a 5.52. 5.3. Label nodes. π 2 + ln 2 ( % 00 ) 0.69. Since, ζ a 8,

8 Solutions to Skill-Assessment Exercises N (s) N 2 ( s) N 3 (s ) N 4 ( s) N 5 (s) N 6 (s) N 7 (s) Draw nodes. R( s ) N (s) N 2 (s) N 3 ( s ) N 4 ( s) C (s) N 5 ( s) N 6 ( s) N 7 ( s) Connect nodes and label subsystems. R(s ) N ( s) N 2 ( s) N 3 ( s) N 4 ( s) s s s C ( s) N 5 (s) N 6 ( s) s N 7 ( s) s Eliminate unnecessary nodes. - R(s) s s s C(s) s -s 5.4. Forward-path gains are G G 2 G 3 and G G 3.

Chapter 5 9 Loop gains are G G 2 H, G 2 H 2, and G 3 H 3. Nontouching loops are [ G G 2 H ][ G 3 H 3 ] G G 2 G 3 H H 3 and [ G 2 H 2 ][ G 3 H 3 ] G 2 G 3 H 2 H 3. Also, + G G 2 H + G 2 H 2 + G 3 H 3 + G G 2 G 3 H H 3 + G 2 G 3 H 2 H 3. Finally, and 2. Substituting these values into T(s) C(s) R(s) T(s) 5.5. k T k k yields G (s)g 3 (s)[ + G 2 (s)] [ + G 2 (s)h 2 (s) + G (s)g 2 (s)h (s)][ + G 3 (s)h 3 (s)] The state equations are, x 2x + x 2 x 2 x 3 3x 2 + x 3 3x 4x 2 5x 3 + r y x 2 Drawing the signal-flow diagram from the state equations yields r s x 3 s x 2 s x y -5-3 -2-4 -3 5.6. 00(s + 5) From G(s) we draw the signal-flow graph in controller canonical s 2 + 5s + 6 form and add the feedback.

20 Solutions to Skill-Assessment Exercises 00 r 500 y -5-6 - Writing the state equations from the signal-flow diagram, we obtain. x 05 506 x+ 0 0 r y 00 500 x [ ] 5.7. From the transformation equations, P 3 2 4 Taking the inverse, 0.4 0.2 P 0. 0.3 Now, P AP 3 2 3 0.4 0.2 6.5 8.5 4 4 6 0. 0.3 9.5.5 P B 3 2 4 3 3 0.4 0.2 CP [ 4] 0. 0.3 Therefore, z 6.5 8.5 9.5.5 z + 3 u y [ 0.8. 4]z [ 0.8. 4]

Chapter 5 2 5.8. First find the eigenvalues. λi A λ 0 0 λ 3 4 6 λ 3 4 λ + 6 λ2 + 5λ + 6 From which the eigenvalues are 2 and 3. Now use Ax i λx i for each eigenvalue, λ. Thus, 3 4 6 For λ 2, x x 2 3x + 3x 2 0 4x 4x 2 0 Thus x x 2 For λ 3 4x + 3x 2 0 4x 3x 2 0 λ x x 2 Thus x x 2 and x 0.75x 2 ; from which we let 0.707 0.6 P 0.707 0.8 Taking the inverse yields 5.6577 4.2433 P 5 5 Hence, 5.6577 4.2433 3 0.707 0.6 D P AP 5 5 4 6 0.707 0.8 2 0 0 3 5.6577 4.2433 P B 5 5 3 8.39 20 0.707 0.6 CP [ 4] 0.707 0.8 [ 2.2 2.6]

22 Solutions to Skill-Assessment Exercises Finally, z 2 0 0 3 z + 8.39 20 u y [ 2.2 2.6]z

23 6.. Make a Routh table. Chapter 6 s 7 3 6 7 2 s 6 9 4 8 6 s 5 4.666666667 4.333333333 0 0 s 4-4.3574286 8 6 0 s 3 2.9063934 6.426229508 0 0 s 2 0.7026684 6 0 0 s -.855742 0 0 0 s 0 6 0 0 0 Since there are four sign changes and no complete row of zeros, there are four right half-plane poles and three left half-plane poles. 6.2. Make a Routh table. We encounter a row of zeros on the s 3 row. The even polynomial is contained in the previous row as 6s 4 + 0s 2 + 6. Taking the derivative yields 24s 3 + 0s. Replacing the row of zeros with the coefficients of the derivative yields the s 3 row. We also encounter a zero in the first column at the s 2 row. We replace the zero with ε and continue the table. The final result is shown now as s 6-6 - 6 s 5 0-0 s 4-6 0 6 0 s 3-24 0 0 0 ROZ s 2 ε 6 0 0 s 44/ε 0 0 0 s 0 6 0 0 0 There is one sign change below the even polynomial. Thus the even polynomial (4 th order) has one right half-plane pole, one left half-plane pole, and 2 imaginary axis poles. From the top of the table down to the even polynomial yields one sign change. Thus, the rest of the polynomial has one right half-plane root, and one left

24 Solutions to Skill-Assessment Exercises half-plane root. The total for the system is two right half-plane poles, two left half-plane poles, and 2 imaginary poles. 6.3. K(s + 20) Since G(s) s(s + 2)(s + 3), T(s) G(s) + G(s) K(s + 20) s 3 + 5s 2 + (6 + K)s + 20K Form the Routh table. s 3 (6 + K) s 2 5 20K s s 0 30 5K 5 20K From the s row, K < 2. From the s 0 row, K > 0. Thus, for stability, 0 < K < 2. 6.4. First find s 0 0 2 (s 2) si A 0 s 0 7 (s 7) s 3 4s 2 33s + 5 0 0 s 3 4 5 3 4 (s + 5) Now form the Routh table. s 3-33 s 2-4 5 S -20.25 S 0 5 There are two sign changes. Thus, there are two rhp poles and one lhp pole.

25 7.. a. First check stability. T(s) Chapter 7 G(s) + G(s) 0s 2 + 500s + 6000 s 3 + 70s 2 + 375s + 6000 0(s + 30)(s + 20) (s + 26.03)(s + 37.89)(s + 6.085) Poles are in the lhp. Therefore, the system is stable. Stability also could be checked via Routh-Hurwitz using the denominator of T(s). Thus, 5 5 5u(t): e step ( ) + limg(s) + 0 s 0 5 5tu(t): e ramp ( ) lim sg(s) 5 2.875 0 * 20 * 30 s 0 25* 35 5 5t 2 u(t): e parabola ( ) lim s 0 s2 G(s) 30 0, since L [5t 2 ] 30 s 3 b. First check stability. T(s) G(s) + G(s) 0s 2 + 500s + 6000 s 5 + 0s 4 + 3875s 3 + 4.37e04s 2 + 500s + 6000 0(s + 30)(s + 20) (s + 50.0)(s + 35)(s + 25)(s 2 7.89e 04s + 0.372) From the second-order term in the denominator, we see that the system is unstable. Instability could also be determined using the Routh-Hurwitz criteria on the denominator of T(s). Since the system is unstable, calculations about steadystate error cannot be made. 7.2. a. The system is stable, since T(s) G(s) + G(s) 000(s + 8) (s + 9)(s + 7) + 000(s + 8) 000(s + 8) and is of s 2 + 06s + 8063 Type 0. Therefore, K p limg(s) 000 * 8 s 0 7*9 b. e step ( ) 27; K v lim s 0 + limg(s) 7.8e 03 + 27 s 0 sg(s) 0; and K a lim s 2 G(s) 0 s 0

26 Solutions to Skill-Assessment Exercises e ramp ( ) lim sg(s) 0 s 0 e parabola ( ) lim s 0 s2 G(s) 0 7.3. System is stable for positive K. System is Type 0. Therefore, for a step input e step ( ) 0.. Solving for K p yields K p 9 limg(s) 2K + K s 0 p 4 *8 ; from which we obtain K 89. 7.4. System is stable. Since G (s) 000, and G 2 (s) e D ( ) 7.5. lim s 0 G 2 (s) + limg (s) s 0 (s + 2) (s + 4), 9.98e 04 2 + 000 System is stable. Create a unity-feedback system, where H e (s) s + s s +. The system is as follows: R(s) + E a (s) 00 C(s) s + 4 - - s s + Thus, G e (s) G(s) + G(S)H e (s) 00 (s + 4) 00s (s + )(s + 4) 00(s + ) S 2 95s + 4 Hence, the system is Type 0. Evaluating K p yields

Chapter 7 27 K p 00 4 25 The steady-state error is given by e step ( ) 3.846e 02 + K P + 25 7.6. K(s + 7) Since G(s) s 2 + 2s + 0, e( ) + K p + 7K 0 Calculating the sensitivity, we get S e:k K e 7.7. Given e K K 0 0 + 7K ( 0)7 (0 + 7K) 7K 2 0 + 7K A 0 3 6 ; B 0 ; C [ ]; R(s) s. Using the final value theorem, e step ( ) lim s 0 sr(s)[ C(sI A) B] lim s 0 [ [ ] lim s 0 [ Using input substitution, s + 6 3 s 0 s 2 + 6s + 3 ] lim s 0 ( ) + step CA B 3 [ ] + 6 [ ] 0 0 0 0 + 7K. [ ] s 3 s + 6 s 2 + 5s + 2 s 2 + 6s + 3 2 3 6 3 0 0 3 + [ ] - 3 0 2 3 0 ]

28 Chapter 8 8.. a. F( 7 + j9) ( 7 + j9 + 2)( 7 + j9 + 4)0.0339 ( 7 + j9)( 7 + j9 + 3)( 7 + j9 + 6) ( 5 + j9)( 3 + j9) ( 7 + j9)( 4 + j9)( + j9) ( 66 j72) 0.0339 j0.0899 0.096 < 0.7o (944 j378) b. The arrangement of vectors is shown as follows: jω (-7+j9) s-plane M M2 M 3 M 4 M 5-7 X X -6-5 -4-3 -2 - X 0 σ From the diagram, F( 7 + j9) M 2M 4 ( 3 + j9)( 5 + j9) M M 3 M 5 ( + j9)( 4 + j9)( 7 + j9) 8.2. a. First draw the vectors. ( 66 j72) 0.0339 j0.0899 0.096 < 0.7o (944 j378)

Chapter 8 29 jω X j3 j2 s-plane j -3-2 - 0 σ -j -j2 X -j3 From the diagram, 3 angles 80 o tan 3 tan 80o 08.43 o + 08.43 o 80 o. b. Since the angle is 80 0, the point is on the root locus. ( )( 2 + 3 ) 2 Π pole lengths c. K Π zero lengths 2 + 3 2 8.3. First, find the asymptotes. σ a poles - zeros # poles-# zeros ( 2 4 6) (0) 3 0 0 4 (2k + )π θ a π 3 3, π, 5π 3 Next draw root locus following the rules for sketching.

30 Solutions to Skill-Assessment Exercises 5 4 3 2 Imag Axis 0 - -2-3 -4-5 -8-7 -6-5 -4-3 -2-0 2 3 Real Axis 8.4. a. jω j3 X s-plane O -2 0 2 σ -j3 X b. Using the Routh-Hurwitz criteria, we first find the closed-loop transfer function. T(s) G(s) + G(s) K(s + 2) s 2 + (K 4)s + (2K + 3) Using the denominator of T(s), make a Routh table.

Chapter 8 3 s 2 2K+3 s K-4 0 s 0 2K+3 0 We get a row of zeros for K 4. From the s 2 row with K 4, s 2 + 2 0. From which we evaluate the imaginary axis crossing at 2. c. From part (b), K 4. d. Searching for the minimum gain to the left of 2 on the real axis yields 7 at a gain of 8. Thus the break-in point is at 7. e. First, draw vectors to a point ε close to the complex pole. At the point ε close to the complex pole, the angles must add up to zero. Hence, angle from zero angle from pole in 4 th quadrant angle from pole in st quadrant 3 80 0, or tan 4 90o θ 80 o. Solving for the angle of departure, θ - 233..

32 Solutions to Skill-Assessment Exercises 8.5. a. ζ 0.5 jω X j4 s-plane -3 0 o 2 o 4 σ X -j4 b. Search along the imaginary axis and find the 80 0 point at s ±j4.06. c. For the result in part (b), K. d. Searching between 2 and 4 on the real axis for the minimum gain yields the break-in at s 2.89. e. Searching along ζ 0.5 for the 80 0 point we find s 2.42 + j4.8. f. For the result in part (e), K 0.08. g. Using the result from part (c) and the root locus, K <. 8.6. a. ζ 0.59 jω s-plane X -6 X -4 X -2 0 σ

Chapter 8 33 b. Searching along the ζ 0.59 (0% overshoot) line for the 80 0 point yields - 2.028+j2.768 with K 45.55. c. T s 4 Re 4 2.028.97 s; T p π Im π.3 s; 2.768 ω n T r.8346 from the rise-time chart and graph in Chapter 4. Since ω n is the radial distance to the pole, ω n 2.028 2 + 2.768 2 3.43. Thus, T r 0.53 s; since the system is Type 0, K p e step ( ) + K p 0.5. K 2*4*6 45.55 0.949. Thus, 48 d. Searching the real axis to the left of 6 for the point whose gain is 45.55, we find 7.94. Comparing this value to the real part of the dominant pole, -2.028, we find that it is not five times further. The second-order approximation is not valid. 8.7. Find the closed-loop transfer function and put it the form that yields p i as the root locus variable. Thus, 00 T(s) G(s) + G(s) 00 s 2 + p i s + 00 00 (s 2 + 00) + p i s s 2 + 00 p + i s s 2 + 00 p Hence, KG(s)H(s) i s. The following shows the root locus. s 2 + 00

34 Solutions to Skill-Assessment Exercises jω s-plane X j0 O 0 σ X-j0 8.8. Following the rules for plotting the root locus of positive-feedback systems, we obtain the following root locus: jω s-plane o -4 X -3 X -2 X - 0 σ

Chapter 8 35 8.9. The closed-loop transfer function is T(s) denominator with respect to K yields 2s s s s + (K + 2) + (s + ) (2s + K + 2) + (s + ) 0 K K K Solving for s s, we get K K (s + ) (2s + K + 2). Thus, S s:k K s Substituting K 20 yields S s:k 0(s + ) s(s + ). K(s + ). Differentiating the s 2 + (K + 2)s + K s K K(s + ) s(2s + K + 2). Now find the closed-loop poles when K 20. From the denominator of T(s), s,2-2.05, - 0.95, -when K 20. For the pole at 2.05, s s(s s:k ) K K 2.05 0( 2.05 + ) 0.05 0.9975. 2.05( 2.05 + ) For the pole at 0.95, s s(s s:k ) K K 0.95 0( 0.95 + ) 0.05 0.0025. 0.95( 0.95 + )

36 Chapter 9 9.. a. Searching along the 5% overshoot line, we find the point on the root locus at 3.5 + j5.8 at a gain of K 45.84. Thus, for the uncompensated system, K v lim sg(s) K /7 45.84 / 7 6.55. s 0 Hence, e ramp_uncompensated ( ) / K v 0.527. b. Compensator zero should be 20x further to the left than the compensator pole. (s + 0.2) Arbitrarily select G c (s) (s + 0.0). c. Insert compensator and search along the 5% overshoot line and find the root locus at 3.4 + j5.63 with a gain, K 44.64. Thus, for the compensated system, K v 44.64(0.2) 27.5 and e ramp_compensated ( ) 0.0078. (7)(0.0) K v d. e ramp_uncompensated e ramp_compensated 0.527 0.0078 9.58 9.2. a. Searching along the 5% overshoot line, we find the point on the root locus at 3.5 + j5.8 at a gain of K 45.84. Thus, for the uncompensated system, T s 4 Re 4.43 s. 3.5 b. The real part of the design point must be three times larger than the uncompensated pole s real part. Thus the design point is 3(-3.5) + j 3(5.8) -0.5 + j7.4. The angular contribution of the plant s poles and compensator zero at the design point is 30.8 0. Thus, the compensator pole must contribute 80 0 30.8 0 49.2 0. Using the following diagram,

Chapter 9 37 jω j7.4 s-plane 49.2 0 σ -p c -0.5 we find 7.4 p c 0.5 tan 49.2o, from which, p c 25.52. Adding this pole, we find the gain at the design point to be K 476.3. A higher-order closed-loop pole is found to be at.54. This pole may not be close enough to the closed-loop zero at 0. Thus, we should simulate the system to be sure the design requirements have been met. 9.3. a. Searching along the 20% overshoot line, we find the point on the root locus at 3.5 + 6.83 at a gain of K 58.9. Thus, for the uncompensated system, T s 4 Re 4.43 s. 3.5 b. For the uncompensated system, K v lim sg(s) K /7 58.9 / 7 8.4. Hence, s 0 e ramp_uncompensated ( ) / K v 0.89. c. In order to decrease the settling time by a factor of 2, the design point is twice the uncompensated value, or 7 + j3.66. Adding the angles from the plant s poles and the compensator s zero at 3 to the design point, we obtain 00.8 0. Thus, the compensator pole must contribute 80 0 00.8 0 79.2 0. Using the following diagram,

38 Solutions to Skill-Assessment Exercises jω j3.66 s-plane 79.2 0 σ -p c -7 we find 3.66 p c 7 tan 79.2o, from which, p c 9.6. Adding this pole, we find the gain at the design point to be K 204.9. Evaluating K v for the lead-compensated system: K v lim s 0 sg(s)g lead K(3) / [(7)(9.6)] (204.9)(3) / [(7)(9.6)] 9.38. K v for the uncompensated system was 8.4. For a 0x improvement in steadystate error, K v must be (8.4)(0) 84.. Since lead compensation gave us K v 9.38, we need an improvement of 84./9.38 9.2. Thus, the lag compensator zero should be 9.2x further to the left than the (s + 0.092) compensator pole. Arbitrarily select G c (s) (s + 0.0). Using all plant and compensator poles, we find the gain at the design point to be K 205.4. Summarizing the forward path with plant, compensator, and gain yields G e (s) 205.4(s + 3)(s + 0.092) s(s + 7)(9.6)(s + 0.0). Higher-order poles are found at 0.928 and 2.6. It would be advisable to simulate the system to see if there is indeed pole-zero cancellation. 9.4. The configuration for the system is shown in the figure below.

Chapter 9 39 R(s) + - K + - s(s + 7)(s +0) C(s) K f s Minor-Loop Design: K For the minor loop, G(s)H(s) f. Using the following diagram, we (s + 7)(s + 0) find that the minor-loop root locus intersects the 0.7 damping ratio line at 8.5 + j8.67. The imaginary part was found as follows: θ cos - ζ 45.57 0. Hence, Im 8.5 tan 45.570, from which Im 8.67. ζ 0.7 (-8.5 + j8.67) jω Im s-plane X 0 8.5 X -7 θ σ The gain, K f, is found from the vector lengths as K f.5 2 + 8.67 2.5 2 + 8.67 2 77.42 Major-Loop Design: Using the closed-loop poles of the minor loop, we have an equivalent forwardpath transfer function of G e (s) K s(s + 8.5 + j8.67)(s + 8.5 j8.67) K s(s 2 + 7s + 47.4).

40 Solutions to Skill-Assessment Exercises Using the three poles of G e (s) as open-loop poles to plot a root locus, we search along ζ 0.5 and find that the root locus intersects this damping ratio line at 4.34 + j7.5 at a gain, K 626.3. 9.5. a. An active PID controller must be used. We use the circuit shown in the following figure: where the impedances are shown below as follows: Matching the given transfer function with the transfer function of the PID controller yields G c (s) (s + 0.)(s + 5) s Equating coefficients R C 2 0.5 () R 2 C (2) s2 + 5.s + 0.5 s s + 5. + 0.5 s R 2 + C + R 2 C s + R C 2 R C 2 s R 2 + C 5. (3) R C 2 In Eq. (2) we arbitrarily let C 0 5. Thus, R 2 0 5. Using these values along with Eqs. () and (3) we find C 2 00 µf and R 20 kω.

Chapter 9 4 b. The lag-lead compensator can be implemented with the following passive network, since the ratio of the lead pole-to-zero is the inverse of the ratio of the lag pole-to-zero: Matching the given transfer function with the transfer function of the passive laglead compensator yields G c (s) (s + 0.)(s + 2) (s + 0.)(s + 2) (s + 0.0)(s + 20) s 2 + 20.0s + 0.2 Equating coefficients R C 0. () R 2 C 2 2 (2) + + 20.0 (3) R C R 2 C 2 R 2 C Substituting Eqs. () and (2) in Eq. (3) yields R 2 C 7.9 (4) Arbitrarily letting C 00 µf in Eq. () yields R 00 kω. Substituting C 00 µf into Eq. (4) yields R 2 558 kω. Substituting R 2 558 kω into Eq. (2) yields C 2 900 µf. s + s + R C R 2 C 2 s 2 + + + s + R C R 2 C 2 R 2 C R R 2 C C 2

42 0.. a. G(s) Chapter 0 (s + 2)(s + 4) ; G(jω) (8 - ω 2 ) + j6ω M(ω) (8 - ω 2 ) 2 + (6ω) 2 6ω For ω < 8, φ(ω) -tan - 8-ω 2. 6ω For ω > 8, φ(ω)- π + tan -. 8-ω 2 b. Bode Diagrams 0-20 -40 Phase (deg); Magnitude (db) -60-80 -00 0-50 -00-50 -200 0-0 0 0 0 2 Frequency (rad/sec)

Chapter 0 43 c. Nyquist Diagrams 0.08 0.06 0.04 Imaginary Axis 0.02 0-0.02-0.04-0.06-0.08-0.05 0 0.05 0. 0.5 0.2 Real Axis 0.2. Asymptotic 20 log M -40-60 -80 Actual -20 db/dec -40 db/dec -20 db/dec -40 db/dec -00-20 0. 0 00 000 Frequency (rad/s) -45 o /dec Phase (degrees) -50-00 -50-90 o /dec -45 o /dec -90 o /dec -45 o /dec Asymptotic -200 0. 0 00 000 Frequency (rad/s) Actual -45 o /dec

44 Solutions to Skill-Assessment Exercises 0.3. The frequency response is /8 at an angle of zero degrees at ω 0. Each pole rotates 90 0 in going from ω 0 to ω. Thus, the resultant rotates 80 0 while its magnitude goes to zero. The result is shown below. Im ω ω 0 0 8 Re 0.4. a. The frequency response is /48 at an angle of zero degrees at ω 0. Each pole rotates 90 0 in going from ω 0 to ω. Thus, the resultant rotates 270 0 while its magnitude goes to zero. The result is shown below. Im ω 6.63-480 ω 0 ω 0 48 Re b. Substituting jω into G(s) (s + 2)(s + 4)(s + 6) s 3 + 2s 2 + 44s + 48 and simplifying, we obtain G( jω) (48 2ω 2 ) j(44ω ω 3 ). The Nyquist ω 6 + 56ω 4 + 784ω 2 + 2304

Chapter 0 45 diagram crosses the real axis when the imaginary part of G( jω) is zero. Thus, the Nyquist diagram crosses the real axis at ω 2 44, or ω 44 6.63 rad/s. At this frequency G( jω). Thus, the system is stable for K < 480. 480 0.5. If K 00, the Nyquist diagram will intersect the real axis at 00/480. Thus, G M 20log 480 3.62 db. From Skill-Assessment Exercise Solution 0.4, the 00 80 0 frequency is 6.63 rad/s. 0.6. a. -60-80 -00 20 log M -20-40 -60-80 0 00 000 Frequency (rad/s) 0-50 Phase (degrees) -00-50 -200-250 -300 0 00 000 Frequency (rad/s) b. The phase angle is 80 0 at a frequency of 36.74 rad/s. At this frequency the gain is 99.67 db. Therefore, 20log K 99.67, or K 96,270. We conclude that the system is stable for K < 96,270. c. For K 0,000, the magnitude plot is moved up 20log0,000 80 db. Therefore, the gain margin is 99.67-80 9.67 db. The 80 0 frequency is 36.7

46 Solutions to Skill-Assessment Exercises rad/s. The gain curve crosses 0 db at ω 7.74 rad/s, where the phase is 87. 0. We calculate the phase margin to be 80 0 87. 0 92.9 0. 0.7. Using ζ -ln(% / 00), we find ζ 0.456, which corresponds to 20% π 2 + ln 2 (% / 00) overshoot. Using T s 2, ω BW 4 T s ζ ( 2ζ 2 ) + 4ζ 4 4ζ 2 + 2 5.79rad/s. 0.8. For both parts find that G( jω) 60 27 * (6750000 0250ω 2 ) + j350(ω 2 350)ω. For a range of ω 6 + 2925ω 4 + 072500ω 2 + 25000000 values for ω, superimpose G( jω) on the a. M and N circles, and on the b. Nichols chart. a. 3 Im G-plane F 20 o 2 M.3.4.5.6.8 2.0 25 M.0 30 o 40 o 50 o 70 o o 0.4 0.5 0.6 M 0.7 Re - -2-70 o -50 o -40 o -30 o -25 o -20 o -3-4 -3-2 - 2

Chapter 0 47 b. Nichols Charts 0 0.5 0.25 db db db 3 db 6 db 0 db - db -3 db -6 db -2 db -20 db Open-Loop Gain (db) -50-00 -50-40 db -60 db -80 db -00 db -20 db -40 db -60 db -200-80 db -200 db -220 db -240 db -350-300 -250-200 -50-00 -50 0 Open-Loop Phase (deg) Plotting the closed-loop frequency response from a. or b. yields the following plot:

48 Solutions to Skill-Assessment Exercises 0-20 20 log M -40-60 -80-00 -20 0 00 000 Frequency (rad/s) 0-50 Phase (degrees) -00-50 -200-250 -300 0 00 000 Frequency (rad/s) 0.9. The open-loop frequency response is shown in the following figure:

Chapter 0 49 Bode Diagrams 40 20 0 Phase (deg); Magnitude (db) -20-40 -00-20 -40-60 0-0 0 0 0 2 Frequency (rad/sec) The open-loop frequency response is 7 at ω 4.5 rad/s. Thus, the estimated bandwidth is ω WB 4.5 rad/s. The open-loop frequency response plot goes through zero db at a frequency of 9.4 rad/s, where the phase is 5.98 0. Hence, the phase margin is 80 0 5.98 0 28.02 0. This phase margin corresponds to 2 ( ζπ / ζ ) ζ 0.25. Therefore, %OS e x00 44.4%, T s T p 0.0. 4 ω BW ζ ( 2ζ 2 ) + 4ζ 4 4ζ 2 + 2.64 s and π ω BW ζ 2 ( 2ζ 2 ) + 4ζ 4 4ζ 2 + 2 0.33 s The initial slope is 40 db/dec. Therefore, the system is Type 2. The initial slope intersects 0 db at ω 9.5 rad/s. Thus, K a 9.5 2 90.25 and K p K v.

50 Solutions to Skill-Assessment Exercises 0.. 0 a. Without delay, G( jω) jω( jω + ) 0, from which the zero db ω( ω + j) 0 frequency is found as follows: M. Solving for ω, ω ω 2 + ω ω 2 + 0, or after squaring both sides and rearranging, ω 4 + ω 2 00 0. Solving for the roots, ω 2 0.5, 9.5. Taking the square root of the positive root, we find the 0 db frequency to be 3.08 rad/s. At this frequency, the phase angle, φ - ( ω + j) - ( 3.08 + j) 62 o. Therefore the phase margin is 80 0 62 0 8 0. b. With a delay of 3 s, φ - ( ω + j) ωt - ( 3.08 + j) (3.08)(3) 62 o 9.24 o 7.24 o. Therefore the phase margin is 80 0 7.24 0 8.76 0. c. With a delay of 7 s, φ - ( ω + j) ωt - ( 3.08 + j) (3.08)(7) 62 o 2.56 o 83.56 o. Therefore the phase margin is 80 0 83.56 0-3.56 0. Thus, the system is unstable. 0.2. Drawing judicially selected slopes on the magnitude and phase plot as shown below yields a first estimate.

Chapter 0 5 20 Experimental 0 Phase(deg) Gain(dB) 0-0 -20-30 -40-45 -50-55 -60-65 -70-75 -80-85 -90-95 2 3 4 5 6 7 8 0 20 30 40 50 70 00 200 300 500 000 Frequency (rad/sec) We see an initial slope on the magnitude plot of 20 db/dec. We also see a final 20 db/dec slope with a break frequency around 2 rad/s. Thus, an initial estimate is G (s) s(s + 2). Subtracting G (s)from the original frequency response yields the frequency response shown below.

52 Solutions to Skill-Assessment Exercises 90 Experimental Minus /s(s+2) 80 Gain(dB) 70 60 50 40 00 80 Phase(deg) 60 40 20 0 2 3 4 5 6 7 8 0 20 30 40 50 70 00 200 300 500 000 Frequency (rad/sec) Drawing judicially selected slopes on the magnitude and phase plot as shown yields a final estimate. We see first-order zero behavior on the magnitude and phase plots with a break frequency of about 5.7 rad/s and a dc gain of about 44 db 20log(5.7K), or K 27.8. Thus, we estimate G 2 (s) 27.8(s + 7). Thus, 27.8(s + 5.7) G(s) G (s)g 2 (s). It is interesting to note that the original s(s + 2) 30(s + 5) problem was developed from G(s) s(s + 20).

53 Chapter.. The Bode plot for K is shown below. Bode Diagrams -60-80 -00-20 Phase (deg); Magnitude (db) -40-60 -80-00 -50-200 -250 0-0 0 0 0 2 0 3 Frequency (rad/sec) A 20% overshoot requires ζ log % 00 0.456. This damping ratio % π 2 + log 2 00 implies a phase margin of 48.0, which is obtained when the _ -800 + 48.0 3.9 0. This phase angle occurs at ω 27.6rad/s. The magnitude at this frequency is 5.5 x 0-6. Since the magnitude must be unity K 94,200. 6 5.5x0

54 Solutions to Skill-Assessment Exercises.2. To meet the steady-state error requirement, K,942,000. The Bode plot for this gain is shown below. Bode Diagrams 60 40 20 Phase (deg); Magnitude (db) 0-20 -40-00 -50-200 -250 0-0 0 0 0 2 0 3 Frequency (rad/sec) log % A 20% overshoot requires ζ 00 0.456. This damping ratio % π 2 + log 2 00 implies a phase margin of 48. 0. Adding 0 0 to compensate for the phase angle contribution of the lag, we use 58. 0. Thus, we look for a phase angle of 80 0 + 58. 0-29.9 0. The frequency at which this phase occurs is 20.4 rad/s. At this frequency the magnitude plot must go through zero db. Presently, the magnitude plot is 23.2 db. Therefore draw the high frequency asymptote of the lag compensator at 23.2 db. Insert a break at 0.(20.4) 2.04 rad/s. At this frequency, draw 20 db/dec slope until it intersects 0 db. The frequency of intersection will be the low frequency break or 0.4 rad/s. Hence the

Chapter 55 (s + 2.04) compensator is G c (s) K c, where the gain is chosen to yield 0 db at (s + 0.4) low frequencies, or K c 0.4 / 2.04 0.069. In summary, (s + 2.04) G c (s) 0.069 (s + 0.4) and G(s),942,000 s(s + 50)(s + 20)..3. A 20% overshoot requires ζ log % 00 0.456. The required % π 2 + log 2 00 bandwidth is then calculated as ω BW 4 T s ζ ( 2ζ 2 ) + 4ζ 4 4ζ 2 + 2 57.9 rad/s. In order to meet the steady-state error requirement of K v 50 K (50)(20), we calculate K 300,000. The uncompensated Bode plot for this gain is shown below. Bode Plot for K 300000 40 20 0 Phase (deg); Magnitude (db) -20-40 -60-00 -50-200 -250 0-0 0 0 0 2 0 3 Frequency (rad/sec)

56 Solutions to Skill-Assessment Exercises The uncompensated system s phase margin measurement is taken where the magnitude plot crosses 0 db. We find that when the magnitude plot crosses 0 db, the phase angle is -44.8 0. Therefore, the uncompensated system s phase margin is -80 0 + 44.8 0 35.2 0. The required phase margin based on the required damping 2ζ ratio is Φ M tan 48. o. Adding a 0 0 correction factor, the 2ζ 2 + + 4ζ 4 required phase margin is 58. 0. Hence, the compensator must contribute φ max 58. 0-35.2 0 22.9 0. Using φ max sin β + β, β sin φ max + sin φ max 0.44. The compensator s peak magnitude is calculated as M max.5. Now find the β frequency at which the uncompensated system has a magnitude / M max, or 3.58 db. From the Bode plot, this magnitude occurs atω max 50 rad/s. The compensator s zero is at z c T. But, ω max T β. Therefore, z c 33.2. The compensator s pole is at p c βt z c β 75.4. The compensator gain is chosen to yield unity gain at dc. Hence, K c 75.4 / 33.2 2.27. Summarizing, (s + 33.2) G c (s) 2.27 (s + 75.4), and G(s) 300,000 s(s + 50)(s + 20)..4. A 0% overshoot requiresζ is then calculated as ω BW log % 00 0.59. The required bandwidth % π 2 + log 2 00 π T p ζ 2 ( 2ζ 2 ) + 4ζ 4 4ζ 2 + 2 7.53 rad/s. In order to meet the steady-state error requirement of K v 0 K (8)(30), we calculate K 2400. The uncompensated Bode plot for this gain is shown below.

Chapter 57 Bode Diagrams Phase (deg); Magnitude (db) 40 20 0-20 -40-60 -80-00 -00-50 -200-250 0-0 0 0 0 2 0 3 Frequency (rad/sec) Let us select a new phase-margin frequency at 0.8ω BW 6.02 rad/s. The required phase margin based on the required damping ratio 2ζ is Φ M tan 58.6 o. Adding a 5 0 correction factor, the 2ζ 2 + + 4ζ 4 required phase margin is 63.6 0. At 6.02 rad/s, the new phase-margin frequency, the phase angle is which represents a phase margin of 80 0 38.3 0 4.7 0. Thus, the lead compensator must contribute φ max 63.6 0 4.7 0 2.9 0. Using φ max sin β + β,β sin φ max + sin φ max 0.456. We now design the lag compensator by first choosing its higher break frequency one decade below the new phase-margin frequency, that is, z lag 0.602 rad/s. The lag compensator s pole is p lag βz lag 0.275. Finally, the lag compensator s gain is K lag β 0.456.

58 Solutions to Skill-Assessment Exercises Now we design the lead compensator. The lead zero is the product of the new phase margin frequency and β, or z lead 0.8ω BW β 4.07. Also, p lead z lead β 8.93. Finally, K lead 2.9. Summarizing, β (s + 0.602) G lag (s) 0.456 (s + 0.275) ; G (s + 4.07) lead (s) 2.9 ; and K 2400. (s + 8.93)

59 2.. Chapter 2 We first find the desired characteristic equation. A 5% overshoot requiresζ log % 00 π 0.69. Also, ω n 4.47 rad/s. Thus, the % π 2 + log 2 2 T p ζ 00 characteristic equation is s 2 + 2ζω n s + ω n 2 s 2 + 9.97s + 209.4. Adding a pole at 0 to cancel the zero at 0 yields the desired characteristic equation, (s 2 + 9.97s + 209.4)(s + 0) s 3 + 29.97s 2 + 409.s + 2094. The compensated system 0 0 matrix in phase-variable form is A BK 0 0. The (k ) (36 + k 2 ) (5 + k 3 ) characteristic equation for this system is si (A BK)) s 3 + (5 + k 3 )s 2 + (36 + k 2 )s + (k ). Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as K [ k k 2 k 3 ] [ 2094 373. 4.97]. 2.2. The controllability matrix is C M [ B AB A 2 B] 2 4 9. Since C M 80, 6 C M is full rank, that is, rank 3. We conclude that the system is controllable. 2.3. First check controllability. The controllability matrix is 0 0 C Mz [ B AB A 2 B ] 0 7. Since C Mz, C Mz is full rank, that is, rank 9 8 3. We conclude that the system is controllable. We now find the desired characteristic equation. A 20% overshoot

60 Solutions to Skill-Assessment Exercises requiresζ log % 00 % π 2 + log 2 00 0.456. Also, ω n 4 ζt s 4.386 rad/s. Thus, the characteristic equation is s 2 + 2ζω n s + ω n 2 s 2 + 4s + 9.24. Adding a pole at 6 to cancel the zero at 6 yields the resulting desired characteristic equation, (s 2 + 4s + 9.24)(s + 6) s 3 + 0s 2 + 43.24s + 5.45. Since G(s) (s + 6) (s + 7)(s + 8)(s + 9) s + 6, we can write the phase- s 3 + 24s 2 + 9s + 504 0 0 0 variable representation as A p 0 0 ; B p 0 ; C p [ 6 0]. 504 9 24 The compensated system matrix in phase-variable form is 0 0 A p B p K p 0 0. The characteristic equation for (504 + k ) (9 + k 2 ) (24 + k 3 ) this system is si (A p B p K p )) s 3 + (24 + k 3 )s 2 + (9 + k 2 )s + (504 + k ). Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as K p [ k k 2 k 3 ][ 388.55 47.76 4]. We now develop the transformation matrix to transform back to the z-system. 0 0 C Mz [ B z A z B z A 2 z B z ] 0 7 and 9 8 C Mp [ B p A p B p A 2 p B p ] 0 0 0 24. 24 385 0 0 9 24 0 0 Therefore, P C Mz C Mx 0 7 24 0 7 0 9 8 0 0 56 5 Hence,

Chapter 2 6 0 0 K z K p P [ 388.55 47.76 4] 7 0 49 5 2.4. [ 40.23 62.24 4]. (24 + l ) 0 For the given system e x (A LC)e x (9 + l 2 ) 0 e x. The characteristic (504 + l 3 ) 0 0 polynomial is given by [si (A LC) s 3 + (24 + l )s 2 + (9 + l 2 )s + (504 + l 3 ). Now we find the desired characteristic equation. The dominant poles from Skill-Assessment Exercise 2.3 come from (s 2 + 4s + 9.24). Factoring yields (-2 + j3.9) and (-2 - j3.9). Increasing these poles by a factor of 0 and adding a third pole 0 times the real part of the dominant second-order poles yields the desired characteristic polynomial, (s + 20 + j39)(s + 20 j39)(s + 200) s 3 + 240s 2 + 992s + 384200. Equating coefficients of the desired characteristic equation to the system s characteristic 26 equation yields L 9730. 383696 2.5. C 4 6 8 The observability matrix is O M CA 64 80 78, where CA 2 674 848 84 25 28 32 A 2 7 4. The matrix is of full rank, that is, rank 3, since O M 576. 77 95 94 Therefore the system is observable. 2.6. The system is represented in cascade form by the following state and output equations: 7 0 0 z 0 8 z + 0 u 0 0 9 y [ 0 0]z

62 Solutions to Skill-Assessment Exercises 0 0 The observability matrix is O Mz C z A z 7 0, where 2 C z A z 49 5 C z 49 5 A 2 z 0 64 7. Since G(s) (s + 7)(s + 8)(s + 9) s 3 + 24s 2 + 9s + 504, we 0 0 8 can write the observable canonical form as 24 0 0 x 9 0 x + 0 u 504 0 0 y [ 0 0]x 0 0 The observability matrix for this form is O Mx C x A x 24 0, where 2 C x A x 385 24 385 24 A 2 x 4080 9 0. 2096 504 0 We next find the desired characteristic equation. A 0% overshoot requiresζ log % 00 % π 2 + log 2 00 C x 0.59. Also, ω n 4 ζt s 67.66 rad/s. Thus, the characteristic equation is s 2 + 2ζω n s + ω n 2 s 2 + 80s + 4578.42. Adding a pole at 400, or 0 times the real part of the dominant second-order poles, yields the resulting desired characteristic equation, (s 2 + 80s + 4578.42)(s + 400) s 3 + 480s 2 + 36580s +.83x0 6. For the system represented in observable canonical form e x (24 + l ) 0 (A x L x C x )e x (9 + l 2 ) 0 e x. The characteristic polynomial is given (504 + l 3 ) 0 0 by [si (A x L x C x ) s 3 + (24 + l )s 2 + (9 + l 2 )s + (504 + l 3 ). Equating coefficients of the desired characteristic equation to the system s characteristic equation yields

Chapter 2 63 456 L x 36,389.,830, 496 Now, develop the transformation matrix between the observer canonical and cascade forms. 0 0 P O Mz O Mx 7 0 49 5 0 0 0 0 0 0 24 0 7 0 24 0 385 24 56 5 385 24 0 0 7 0. 8 9 0 0 456 456 456 Finally, L z PL x 7 0 36,389 28,637 28,640. 8 9,830, 496,539,93,540,000 2.7. We first find the desired characteristic equation. A 0% overshoot requires log % ζ 00 0.59 % π 2 + log 2 00. π Also, ω n.948 rad/s. Thus, the characteristic equation is 2 T p ζ s 2 + 2ζω n s + ω 2 n s 2 + 2.3s + 3.79. Adding a pole at 4, which corresponds to the original system s zero location, yields the resulting desired characteristic equation, (s 2 + 2.3s + 3.79)(s + 4) s 3 + 6.3s 2 + 3s + 5.6. Now, x x N (A BK) BK e x C 0 x N + 0 x r; and y [ C 0] x N, where A BK 0 7 9 0 k [ k 2 ] 0 7 9 0 0 k k 2 0 (7 + k ) (9 + k 2 ) [ ] C 4

64 Solutions to Skill-Assessment Exercises Bk e 0 k 0 e k e Thus, x x 2 x N 0 0 (7 + k ) (9 + k 2 ) k e 4 0 Finding the characteristic equation of this system yields x x 2 x N + 0 r; y 4 0 [ ] (A BK) si BK s 0 0 0 0 e C 0 0 s 0 (7 + k ) (9 + k 2 ) k e 0 0 s 4 0 s 0 (7 + k ) s + (9 + k 2 ) k e s 3 + (9 + k 2 )s 2 + (7 + k + k e )s + 4k e 4 s Equating this polynomial to the desired characteristic equation, s 3 + 6.3s 2 + 3s + 5.6 s 3 + (9 + k 2 )s 2 + (7 + k + k e )s + 4k e Solving for the k s, K [ 2.2 2.7] and k e 3.79. x x 2 x N.

65 3.. Chapter 3 f (t) sin(ωkt); f * (t) k 0 F * (s) sin(ωkt) e kts But, Thus, k 0 x k k 0 x sin(ωkt) δ(t kt); k 0 F * (s) 2 j e T (s jω ) e 3.2. F(z) F(z) z (e jωkt e jωkt )e kts T (s+ jω ) 2 j 2 j k 0 (e T (s jω ) ) k (e T (s+ jω ) k e Ts e jωt e Ts e jωt 2 j (e Ts e jωt e Ts e jωt ) + e 2Ts sin(ωt) e Ts e Ts 2cos(ωT) + e 2Ts z sin(ωt) 2z cos(ωt) + z 2 z(z + )(z + 2) (z 0.5)(z 0.7)(z 0.9) (z + )(z + 2) (z 0.5)(z 0.7)(z 0.9) 46.875 z 0.5 4.75 z 0.7 + 68.875 z 0.9 z F(z) 46.875 z 0.5 4.75 z z 0.7 + 68.875 z z 0.9, f (kt) 46.875(0.5) k 4.75(0.7) k + 68.875(0.9) k 3.3. 8 Since G(s) ( e Ts ) s(s + 4), G(z) ( z 8 )z s(s + 4) z z z A s + B s + 4 z z z 2 s + 2 s + 4. Let G 2 (s) 2 s + 2 s + 4. Therefore, g 2 (t) 2 2e 4t, or g 2 (kt) 2 2e 4kT. Hence, G 2 (z) 2z z 2z z e 2z( e 4T ) 4T (z )(z e 4T ).

66 Solutions to Skill-Assessment Exercises Therefore, G(z) z z G 2(z) 2( e 4T ) (z e 4T ). For T 4 s, G(z).264 z 0.3679. 3.4. Add phantom samplers to the input, feedback after H(s), and to the output. Push G (s)g 2 (s), along with its input sampler, to the right past the pickoff point and obtain the block diagram shown below. Hence, T(z) 3.5. G G 2 (z) + HG G 2 (z). Let G(s) 20 s + 5. Let G G(s) 2 (s) s 20 s(s + 5) 4 s 4. Taking the inverse s + 5 Laplace transform and letting t kt, g 2 (kt) 4 4e 5kT. Taking the z-transform yields G 2 (z) 4z z 4z z e 4z( e 5T ) 5T (z )(z e 5T ). Now, G(z) z z G 2(z) 4( e 5T ) G(z). Finally, T(z) (z e 5T ) + G(z) 4( e 5T ) z 5e 5T + 4. The pole of the closed-loop system is at 5e 5T 4. Substituting values of T, we find that the pole is greater than if T > 0.022 s. Hence, the system is stable for 0 < T < 0.022 s. 3.6. Substituting z s + s into D(z) z3 z 2 0.5z + 0.3, we obtain D(s) s 3 8s 2 27s 6. The Routh table for this polynomial is shown below.

Chapter 3 67 s 3-27 s 2-8 -6 s -27.75 0 s 0-6 0 Since there is one sign change, we conclude that the system has one pole outside the unit circle and two poles inside the unit circle. The table did not produce a row of zeros and thus, there are no jω poles. The system is unstable because of the pole outside the unit circle. 3.7. Defining G(s) as G (s) in cascade with a zero-order-hold, G(s) 20( e Ts (s + 3) ) s(s + 4)(s + 5) 20 e Ts Taking the z-transform yields ( ) G(z) 20 z (3 / 20)z z- ( ) 3/20 ( / 4)z (2 / 5)z + 4T z-e z-e 5T s + /4 (s + 4) 2/5 (s + 5) 5(z -) 8(z -) 3 + 4T z-e z-e. 5T Hence for T 0. second, K p limg(z) 3, K v z T lim (z -)G(z) 0, and z K a T lim (z -) 2 G(z) 0. Checking for stability, we find that the system is 2 z stable for T 0. second, since T(z) inside the unit circle at 0.957 and +0.735. G(z) + G(z).5z.09 z 2 + 0.222z 0.703. has poles Again, checking for stability, we find that the system is unstable for T 0.5 second, since T(z) G(z) + G(z) 3.02z 0.6383 has poles inside and outside z 2 + 2.802z 0.6272 the unit circle at +0.208 and 3.0, respectively. 3.8. Draw the root locus superimposed over the ζ 0.5 curve shown below. Searching along a 54.3 0 line, which intersects the root locus and the ζ 0.5 curve, we find the point 0.587 54.3 o (0.348+j0.468) and K 0.3.

68 Solutions to Skill-Assessment Exercises z-plane Root Locus.5 Imag Axis 0.5 0 54.3 0 (0.348+j0.468) K0.3-0.5 - -.5-3 -2.5-2 -.5 - -0.5 0 0.5.5 2 Real Axis 3.9. Let G e (s) G(s)G c (s) 00K 2.38(s + 25.3) s(s + 36)(s + 00) (s + 60.2) The following shows the frequency response of G e ( jω). 342720(s + 25.3) s(s + 36)(s + 00)(s + 60.2).

Chapter 3 69 Bode Diagrams 40 20 0 Phase (deg); Magnitude (db) -20-40 -60-00 -50-200 -250 0-0 0 0 0 2 0 3 Frequency (rad/sec) We find that the zero db frequency, ω Φ M, for G e ( jω)is 39 rad/s. Using Astrom s guideline the value of T should be in the range, 0.5 / ω Φ M 0.0038 second to 0.5 / ω Φ M 0.028 second. Let us use T 0.00 second. Now find the Tustin transformation for the compensator. Substituting s into G c (s) 2.38(s + 25.3) with T 0.00 second yields (s + 60.2) (z 0.975) G c (z) 2.34 (z 0.946). 3.0. G c (z) X(z) E(z) 899z2 376z + 86. Cross-multiply and obtain z 2.908z + 0.9075 2(z ) T(z + ) (z 2.908z + 0.9075)X(z) (899z 2 376z + 86)E(z). Solve for the highest power of z operating on the output, X(z), and obtain z 2 X(z) (899z 2 376z + 86)E(z) (.908z + 0.9075)X(z). Solving for

70 Solutions to Skill-Assessment Exercises X(z) on the left-hand side yields X(z) (899 376z - + 86z 2 )E(z) (.908z - + 0.9075z 2 )X(z). Finally, we implement this last equation with the following flow chart: