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F I V E Reduction of Multiple Subsystems SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Designing a Closed-Loop Response a. Drawing the block diagram of the system: Thus, T(s) = 3 2 76.39K s 151.32s 198s 76.39K b. Drawing the signal flow-diagram for each subsystem and then interconnecting them yields: 5-2 Chapter 5: Reduction of Multiple Subsystems 1x = x2 2x = - 1.32x2 + 0.8x3 3x = -150x3 +150K( i 1 10 (q 0.2x )) = -95.49Kx1 - 150x3 + 477.46Ki o = 0.2x1 In vector-matrix notation, c. 1 3 10 1 1 1 76.39 ( )(150) (0.8) (0.2)T K s s s s 1 2 3 3 150 1.32 1 1 1 10 76.39 ; ; ( )(150) (0.8) (0.2)L L L K G G G K s s s s s s Nontouching loops: GL1GL2 = 2 198 s = 1 - [GL1 + GL2 + GL3] + [GL1GL2] = 1 + 150 s + 1.32 s + 3 76.39K s + 2 198 s 1 = 1 T(s) = 1 1T = 3 2 76.39K s +151.32s +198s+76.39K d. The equivalent forward path transfer function is G(s) = 10 0.16K s(s+1.32) . Therefore, T(s) = 2 2.55 s +1.32s+2.55 The poles are located at -0.66 ± j1.454. n = 2.55 = 1.597 rad/s; 2n = 1.32, therefore, = 0.413. θ θ i o 0 1 0 0 0 -1.32 0.8 0 -95.49K 0 -150 477.46K 0.2 0 0 x x x Solutions to Case Studies Challenges 5-3 2/ 1 % 100 24%OS e x ; Ts = ζωn 4 = 4 0.66 = 6.06 seconds; Tp = π ω ζ2n 1- = π 1.454 = 2.16 seconds; Using Figure 4.16, the normalized rise time is 1.486. Dividing by the natural frequency, Tr = 1.486 2.55 = 0.93 seconds. e. 2 2 2 2 0.66 2.55 ( ) ( 1.32 2.55) 1 1 25 33 ( ) 25 1.32 2.55 1 1 25( 0.66) 11.347 2.1144 ( ) 25 ( 0.66) 2.1144 ( ) 1 (cos[1.454 ] 0.454sin[1.454 ])t C s s s s s C s s s s s C s s s c t e t t f. Since G(s) = 0.51K s(s+1.32) , T(s) = 2 0.51K s +1.32s+0.51K . Also, = π2 2 %OS - ln ( ) 100 %OS + ln ( ) 100 = 0.517 for 15% overshoot; n = 0.51K ; and 2n = 1.32. Therefore, n = ζ 1.32 2 = 1.32 2(0.5147) = 1.277 = 0.51K . Solving for K, K=3.2. UFSS Vehicle: Pitch-Angle Control Representation a. Use the observer canonical form for the vehicle dynamics so that the output yaw rate is a state variable. b. Using the signal flow graph to write the state equations: 5-4 Chapter 5: Reduction of Multiple Subsystems 1 2 2 2 3 4 3 2 4 4 1 2 4 1.483 0.125 0.24897 (0.125*0.437) 2 2 2 2 x x x x x x x x x x x x x u In vector-matrix form: 0 1 0 0 0 0 1.483 0 0.125 0 0 0.24897 0 0.054625 0 2 2 0 2 2 1 0 0 0]y x x x c. Program: numg1=-0.25*[1 0.437]; deng1=poly([-2 -1.29 -0.193 0]); 'G(s)' G=tf(numg1,deng1) numh1=[-1 0]; denh1=[0 1]; 'H(s)' H=tf(numh1,denh1) 'Ge(s)' Ge=feedback(G,H) 'T(s)' T=feedback(-1*Ge,1) [numt,dent]=tfdata(T,'V'); [Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent) Computer response: ans = G(s) Transfer function: -0.25 s - 0.1093 -------------------------------------- s^4 + 3.483 s^3 + 3.215 s^2 + 0.4979 s ans = H(s) Transfer function: -s Answers to Review Questions 5-5 ans = Ge(s) Transfer function: -0.25 s - 0.1093 -------------------------------------- s^4 + 3.483 s^3 + 3.465 s^2 + 0.6072 s ans = T(s) Transfer function: 0.25 s + 0.1093 ----------------------------------------------- s^4 + 3.483 s^3 + 3.465 s^2 + 0.8572 s + 0.1093 Acc = -3.4830 -3.4650 -0.8572 -0.1093 1.0000 0 0 0 0 1.0000 0 0 0 0 1.0000 0 Bcc = 1 0 0 0 Ccc = 0 0 0.2500 0.1093 Dcc = 0 ANSWERS TO REVIEW QUESTIONS 1. Signals, systems, summing junctions, pickoff points 2. Cascade, parallel, feedback 3. Product of individual transfer functions, sum of individual transfer functions, forward gain divided by one plus the product of the forward gain times the feedback gain 4. Equivalent forms for moving blocks across summing junctions and pickoff points 5. As K is varied from 0 to ∞, the system goes from overdamped to critically damped to underdamped. When the system is underdamped, the settling time remains constant. 5-6 Chapter 5: Reduction of Multiple Subsystems 6. Since the real part remains constant and the imaginary part increases, the radial distance from the origin is increasing. Thus the angle is increasing. Since = cos the damping ratio is decreasing. 7. Nodes (signals), branches (systems) 8. Signals flowing into a node are added together. Signals flowing out of a node are the sum of signals flowing into a node. 9. One 10. Phase-variable form, cascaded form, parallel form, Jordan canonical form, observer canonical form 11. The Jordan canonical form and the parallel form result from a partial fraction expansion. 12. Parallel form 13. The system poles, or eigenvalues 14. The system poles including all repetitions of the repeated roots 15. Solution of the state variables are achieved through decoupled equations. i.e. the equations are solvable individually and not simultaneously. 16. State variables can be identified with physical parameters; ease of solution of some representations 17. Systems with zeros 18. State-vector transformations are the transformation of the state vector from one basis system to another. i.e. the same vector represented in another basis. 19. A vector which under a matrix transformation is collinear with the original. In other words, the length of the vector has changed, but not its angle. 20. An eigenvalue is that multiple of the original vector that is the transformed vector. 21. Resulting system matrix is diagonal. SOLUTIONS TO PROBLEMS 1. a. Combine the inner feedback and the parallel pair. Multiply the blocks in the forward path and apply the feedback formula to get, T(s) = 3 2 50(s-2) s +s +150s-100 . b. Program: Solutions to Problems 5-7 'G1(s)' G1=tf(1,[1 0 0]) 'G2(s)' G2=tf(50,[1 1]) 'G3(s)' G3=tf(2,[1 0]) 'G4(s)' G4=tf([1 0],1) 'G5(s)' G5=2 'Ge1(s)=G2(s)/(1+G2(s)G3(s))' Ge1=G2/(1+G2*G3) 'Ge2(s)=G4(s)-G5(s)' Ge2=G4-G5 'Ge3(s)=G1(s)Ge1(s)Ge2(s)' Ge3=G1*Ge1*Ge2 'T(s)=Ge3(s)/(1+Ge3(s))' T=feedback(Ge3,1); T=minreal(T) Computer response: ans = G1(s) Transfer function: 1 --- s^2 ans = G2(s) Transfer function: 50 ----- s + 1 ans = G3(s) Transfer function: 2 - s ans = G4(s) Transfer function: s 5-8 Chapter 5: Reduction of Multiple Subsystems ans = G5(s) G5 = 2 ans = Ge1(s)=G2(s)/(1+G2(s)G3(s)) Transfer function: 50 s^2 + 50 s ------------------------- s^3 + 2 s^2 + 101 s + 100 ans = Ge2(s)=G4(s)-G5(s) Transfer function: s - 2 ans = Ge3(s)=G1(s)Ge1(s)Ge2(s) Transfer function: 50 s^3 - 50 s^2 - 100 s ------------------------------- s^5 + 2 s^4 + 101 s^3 + 100 s^2 ans = T(s)=Ge3(s)/(1+Ge3(s)) Transfer function: 50 s - 100 ----------------------- s^3 + s^2 + 150 s - 100 2. Push G1(s) to the left past the pickoff point. Solutions to Problems 5-9 Thus, 1 2 31 2 3 1 1 1 1 1 11 ( ) 1 1 G G GG T s G G G H G G H 3. a. Split G3 and combine with G2 and G4. Also use feedback formula on G6 loop. Push G2 +G3 to the left past the pickoff point. 5-10 Chapter 5: Reduction of Multiple Subsystems Using the feedback formula and combining parallel blocks, Multiplying the blocks of the forward path and applying the feedback formula, 6 4 6 3 6 5 3 6 5 2 6 3 1 2 1 7 6 4 7 6 3 7 6 5 3 7 6 5 2 6 3 1 6 2 1 ( ) 1 G G G G G G G G G G T s G G G G G G G G G G G G G G G G G G G G G G G G G 4. Push G2(s) to the left past the summing junction. Solutions to Problems 5-11 Collapse the summing junctions and add the parallel transfer functions. Push G1(s)G2(s) + G5(s) to the right past the summing junction. Collapse summing junctions and add feedback paths. 5-12 Chapter 5: Reduction of Multiple Subsystems Applying the feedback formula, 3 1 2 2 4 3 1 2 3 1 2 3 1 2 3 1 2 2 4 ( ) ( ) ( ) ( ) ( ) ( ) 1 [ ( ) ( ) ( )] ( ) ( ) ( ) ( ) ( ) ( ) 1 [ ( ) ( ) ( )] ( ) ( ) G s G s G s T s G s G s G s G s G s H G s G s G s G s G s G s H G s G s G s G s G s 5. a. Push G7 to the left past the pickoff point. Add the parallel blocks, G3+G4. Push G3+G4 to the right past the summing junction. Solutions to Problems 5-13 Collapse the minor loop feedback. Push 7 3 4 6 7 G (G +G ) 1+G G to the left past the pickoff point. 5-14 Chapter 5: Reduction of Multiple Subsystems Push G1 to the right past the summing junction. Add the parallel feedback paths to get the single negative feedback, H(s) = 5 7 G G + 2 6 7 7 3 4 G (1+G G ) G (G +G ) - 8 1 G G . Thus, 7 1 4 3 7 6 2 1 4 3 5 1 8 7 7 6 ( ) T(s) = 1 ([ 1] [ ][ ]) ( 1) G G G GG GH G G G G G G G G G G G G Solutions to Problems 5-15 b. Program: G1=tf([0 1],[1 7]); %G1=1/s+7 input transducer G2=tf([0 0 1],[1 2 3]); %G2=1/s^2+2s+3 G3=tf([0 1],[1 4]); %G3=1/s+4 G4=tf([0 1],[1 0]); %G4=1/s G5=tf([0 5],[1 7]); %G5=5/s+7 G6=tf([0 0 1],[1 5 10]); %G6=1/s^2+5s+10 G7=tf([0 3],[1 2]); %G7=3/s+2 G8=tf([0 1],[1 6]); %G8=1/s+6 G9=tf([1],[1]); %Add G9=1 transducer at the input T1=append(G1,G2,G3,G4,G5,G6,G7,G8,G9); Q=[1 -2 -5 9 2 1 8 0 3 1 8 0 4 1 8 0 5 3 4 -6 6 7 0 0 7 3 4 -6 8 7 0 0]; inputs=9; outputs=7; Ts=connect(T1,Q,inputs,outputs); T=tf(Ts) 5-16 Chapter 5: Reduction of Multiple Subsystems Computer response: Transfer function: 6 s^7 + 132 s^6 + 1176 s^5 + 5640 s^4 + 1.624e004 s^3 + 2.857e004 s^2 + 2.988e004 s + 1.512e004 ----------------------------------------------------------- s^10 + 33 s^9 + 466 s^8 + 3720 s^7 + 1.867e004 s^6 + 6.182e004 s^5 + 1.369e005 s^4 + 1.981e005 s^3 + 1.729e005 s^2 + 6.737e004 s - 1.044e004 6. Combine G6 and G7 yielding G6G7. Add G4 and obtain the following diagram: Next combine G3 and G4+G6G7. Push G5 to the left past the pickoff point. Solutions to Problems 5-17 Notice that the feedback is in parallel form. Thus the equivalent feedback, Heq(s) = 2 5 G G + G3(G4+G6G7) + G8. Since the forward path transfer function is G(s) = Geq(s) = G1G5, the closed- loop transfer function is T(s) = eq eq eq G (s) 1+G (s)H (s) . Hence, 5 1 1 8 5 7 6 5 3 5 4 3 2 ( ) 1 ( ) G G T s G G G G G G G G G G G 7. Push 2s to the right past the pickoff point. 5-18 Chapter 5: Reduction of Multiple Subsystems Combine summing junctions. Combine parallel 2s and s. Apply feedback formula to unity feedback with G(s) = s. Combine cascade pair and add feedback around 1/(s+1). Solutions to Problems 5-19 Combine parallel pair and feedback in forward path. Combine cascade pair and apply final feedback formula yielding 2 2 5 2 ( ) 6 9 6 s s T s s s . 8. Push G3 to the left past the pickoff point. Push G6 to the left past the pickoff point. 5-20 Chapter 5: Reduction of Multiple Subsystems Hence, Thus the transfer function is the product of the functions, or θ θ 22 11 (s) (s) = 1 2 4 5 6 7 4 5 4 5 6 1 2 3 1 2 3 4 5 1 2 3 4 5 6 G G G G G G 1 - G G + G G G + G G G - G G G G G + G G G G G G 9. Combine the feedback with G6 and combine the parallel G2 and G3. Move G2+G3 to the left past the pickoff point. Solutions to Problems 5-21 Combine feedback and parallel pair in the forward path yielding an equivalent forward-path transfer function of Ge(s) = 2 3 64 5 1 2 3 2 3 6 G +G GG G + 1+G (G +G ) G +G 1+G But, T(s) = e e 7 G (s) 1+G (s)G (s) . Thus, 6 4 5 3 5 2 6 7 4 7 5 3 7 5 2 3 1 2 1 1 3 2 ( ) ( ) ( 1) ( ) 1 G G G G G G T s G G G G G G G G G G G G G G G G 10. Push G3(s) to the left past the pickoff point. 5-22 Chapter 5: Reduction of Multiple Subsystems Push G2(s)G3(s) to the left past the pickoff point. Push G1(s) to the right past the summing junction. Solutions to Problems 5-23 Collapsing the summing junctions and adding the feedback transfer functions, 1 2 3 1 2 3 ( ) ( ) ( ) ( ) 1 ( ) ( ) ( ) ( )eq G s G s G s T s G s G s G s H s where 3 1 2 4 3 2 3 1 3 1 ( ) ( ) ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ( ) ( ) eq H s H s H s H s H s G s G s G s G s G s G s 11. 2 225 ( ) 15 225 T s s s . Therefore, 2n = 15, and n = 15. Hence, = 0.5. 2/ 1 % 100 16.3%OS e x ; Ts = ζωn 4 =0.533; Tp = π ω ζ2n 1- =0.242. 12. 2 2 4 1 4( 3) 4 ( (3 4) 1.5 1.75 1 ( 3) ) A Bs Cs s C s s s s s s s s s 1A ; 24 (1 ) (3 ) 4B s C s ; 1B ; 3C . 2 2 2( ) ( ) 1 3 1 1.5 1.5 1.75 ( ) 1.5 1.75 1.5 1.75 1.5 1.751.75 ( ) s s C s s ss s s 1.5( ) 1 cos1.323 1.134sin1.( 323 )tc t e t t 13. The forward paths are 1 5T and 2 4T s . The loops are 1 5L , 2 20L s , 3 3 L s and 4 2L s . There are no non-touching loops. So 3 3 1 5 20 2 6 22s s s s s . When either of the forward paths is removed there are no loops left, so Δ Δ 1 2 1 . The closed loop transfer function is Δ Δ Δ 1 1 2 2 2 5 4 (4 5) ( ) 3 22 6 3 6 22 T T s s s T s s s s s Solving for the roots of the denominator one gets -0.136j0.343. 5-24 Chapter 5: Reduction of Multiple Subsystems 14. Since ( ) ( 30) K G s s s , 2 ( ) ( ) 1 ( ) 30 G s K T s G s s s K . Therefore, 2n = 30. Thus, = 15/n = 0.5912 (i.e. 10% overshoot). Hence, n = 25.37 = K . Therefore K = 643.6. 15. 2 ( ) K T s s s K ; 2 2 2 2 % ln( ) ln0.2100 0.456 % ln 0.2 ln ( ) 100 OS OS ; 4 0.12s n T . Therefore, n = 73.1; K = n 2 = 5343.5; = 2n = 66.67. 16. We first find , n necessary for the specifications. We have 4 2s n T and 2 1 1 p n T . Eliminating n from both equations we get 2 2 1 . Cross-multiplying, squaring both sides and solving, we get 2 4 0.537 4 . 3.724n . The closed loop transfer function of the system is: 1 1 2 1 2 21 50 50( 1) ( ) 50 50 (50 1) 50 1 ( 1) ( 1) K Ks s T s K K s s K s K s s s s From which we get that 2150 nK or 1 0.277K and 250 1 2 4nK or 2 0.06K . 17. a. For the inner loop, Ge(s) = 20 s(s 12) , and He(s) = 0.2s. Therefore, Te(s) = e e e G (s) 1 + G (s)H (s) = 20 s(s+16) . Combining with the equivalent transfer function of the parallel pair, Gp(s) = 20, the system is reduced to an equivalent unity feedback system with G(s) = Gp(s) Te(s) = 400 s(s+16) . Hence, T(s) = G(s) 1+G(s) = 2 400 s +16s+400 . Solutions to Problems 5-25 b. n 2 = 400; 2n = 16. Therefore, n = 20, and 2/ 1 % 100 25.38OS e x ; Ts = ζωn 4 =0.5; Tp = π ω ζ 2n 1- =0.171. From Figure 4.16, nTr = 1.463. Hence, Tr = 0.0732. d = Im = n ζ 21 - = 18.33. 18. The closed loop transfer function of the system is 2 7350 7350( 100) ( ) 7350 100 7350 1 ( 100) s s T s s s s s From which we get 85.73n and 100 0.583 2 n . The %OS= 21 100e =10.5%. 2 0.045 1 p n T sec. 4 0.08s n T sec. From Figure 4.16 we get that 1.85n rT from which we get 0.022rT sec. 19. For the generator, Eg(s) = Kf If (s). But, If (s)= i f fs E (s) R +L . Therefore, g i E (s) E (s) = 2 s+1 . For the motor, consider Ra = 2 the sum of both resistors. Also, Je = Ja+JL( 1 2 )2 = 0.75+1x 1 4 = 1; D e = D L ( 1 2 )2 = 1. Therefore, θ m g (s) E (s) = 1 ( ( )) t a e t b e e a K R J K K s s D J R = 1 ( 3)s s . But, θ θ o m (s) (s) = 1 2 . Thus, θ o g (s) E (s) = 0.5 ( 3)s s . Finally, 5-26 Chapter 5: Reduction of Multiple Subsystems θ o i (s) E (s) = θg o i g E (s) (s) E (s) E (s) = 1 ( 1)( 3)s s s . 20. For the mechanical system, J( 2 1 N N ) 2 s22(s) = T( 2 1 N N ) . For the potentiometer, Ei (s)= 10 θ π 2 (s) 2 , or 2(s) = π 5 Ei(s). For the network, Eo(s) = Ei(s) R 1 R+ Cs = Ei(s) s 1 s+ RC , or Ei(s) = Eo(s) 1 s+ RC . s Therefore, 2 o 1 s RC(s) E (s) 5 s . Substitute into mechanical equation and obtain, 1 o 2 5N E (s) J N 1T(s) s s RC . 21. The equivalent mechanical system is found by reflecting all mechanical impedances to the spring. Writing the equations of motion: 2 1 2 2 1 2 4 2 5 ( ) 5 ( ) 4 ( ) 5 ( ) 2 5 ( ) 0 s s s s T s s s s Solving for 2 (s), 2 2 4 3 22 2 4 2 5 4 ( ) 5 0 20 ( ) ( ) 8 4 30 104 2 5 5 5 2 5 s s T s T s s s s s ss s s Solutions to Problems 5-27 The angular rotation of the pot is 0.25 that of 2 , or 3 2 ( ) 2.5 ( ) 4 2 15 5 p s T s s s s s For the pot: ( ) 25 2.5 ( ) 5(2 ) p p E s s For the electrical network: Using voltage division, 5 ( ) 200,000 1 1( ) 200,000 10 2 o p E s s E s s s Substituting the previously obtained values, 3 2 6.25 ( ) ( )( ) ( ) 1( ) ( ) ( ) ( ) 4 2 15 5 2 p po o p p ss E sE s E s T s T s s E s s s s s s 22. a. 5-28 Chapter 5: Reduction of Multiple Subsystems b. Solutions to Problems 5-29 c. 23. a. 5-30 Chapter 5: Reduction of Multiple Subsystems 1 2 2 3 3 1 2 3 1 2 2 4 6 x x x x x x x x r y x x b. 1 2 2 2 3 3 1 2 3 1 2 3 3 4 5 2 x x x x x r x x x x r y x x c. Solutions to Problems 5-31 1 1 2 2 1 2 3 3 1 3 1 2 3 7 3 2 2 2 3 2 x x x r x x x x r x x x r y x x x 24. a. The transfer function is 3 2 ( ) 200 ( ) 60 1100 6000 C s R s s s s The corresponding differential equation is: 60 1100 6000 200c c c c r The state variables are defined as: 1 c x ; 2 c x and 3 c x . From which we get 1 2 x x 2 3 x x 1 1 2 3 6000 1100 60 200x x x x r 1 y x The corresponding signal flow diagram: 5-32 Chapter 5: Reduction of Multiple Subsystems b. The transfer function can be expressed as 1 1 1 ( ) 200 10 20 30 G s s s s from which the signal flow diagram is obtained. The state space representation is 1 1 2 30x x x 2 2 3 20x x x 1 3 10 200x x r 1 y x 25. a. Since G(s) = 4 3 2 20 15 66 80s s s s = C(s) R(s) , 15 66 80 20c c c c r Let, Solutions to Problems 5-33 1 2 3 4 c x c x c x c x Therefore, 1 2 2 3 3 4 4 2 3 4 1 80 66 15 20 x x x x x x x x x x r y x b. G(s) = ( 20 s ) ( 1 2s ) ( 1 5s ) ( 1 8s ). Hence, From which, 1 1 2 2 2 3 3 3 4 4 1 8 5 2 20 x x x x x x x x x x r y x 5-34 Chapter 5: Reduction of Multiple Subsystems 26. = 1 + [G2G3G4 + G3G4 + G4 + 1] + [G3G4 + G4]; T1 = G1G2G3G4; 1 = 1. Therefore, T(s) = 1 1T = 1 2 3 4 2 3 4 3 4 4 G G G G 2 + G G G + 2G G + 2G 27. Closed-loop gains: G2G4G6G7H3; G2G5G6G7H3; G3G4G6G7H3; G3G5G6G7H3; G6H1; G7H2 Forward-path gains: T1 = G1G2G4G6G7; T2 = G1G2G5G6G7; T3 = G1G3G4G6G7; T4 = G1G3G5G6G7 Nontouching loops 2 at a time: G6H1G7H2 = 1 - [H3G6G7(G2G4 + G2G5 + G3G4 + G3G5) + G6H1 + G7H2] + [G6H1G7H2] 1 = 2 = 3 = 4 = 1 T(s) = Δ Δ Δ Δ Δ 1 1 2 2 3 3 4 4 T + T + T + T = 1 2 4 6 7 1 2 5 6 7 1 3 4 6 7 1 3 5 6 7 3 6 7 2 4 2 5 3 4 3 5 6 1 7 2 6 1 7 2 G G G G G + G G G G G + G G G G G + G G G G G 1 - H G G (G G + G G + G G + G G ) - G H - G H + G H G H 28. Closed-loop gains: -s2; - 1 s ; - 1 s ; -s2 Forward-path gains: T1 = s; T2 = 2 1 s Nontouching loops: None = 1 - (-s2 - 1 s - 1 s - s2) 1 = 2 = 1 G(s) = 1 1 2 2T + T = 2 2 2 1 s + s 1 1 1 + (s + + + s ) s s = 3 4 2 s +1 2s +s +2s 29. T(s) = 2 3 4 5 1 2 1 4 2 2 3 4 5 6 7 8 2 1 4 2 7 4 G G G G G (1-G H )(1-G H ) G G G G G G G 1 - (1-G H )(1-G H )(1-G H ) = Solutions to Problems 5-35 1 2 3 4 5 7 4 2 1 4 2 2 4 1 2 7 4 2 7 1 4 4 7 2 4 2 4 7 1 2 4 2 3 4 5 6 7 8 G G G G G (1-G H ) 1-G H -G H +G G H H -G H +G G H H +G G H H -G G G H H H -G G G G G G G 30. a. 2 2 ( 1)( 2) 2 5 6 ( ) ( 3) ( 4) ( 3) 3 4 s s G s s s s s s Writing the state and output equations, 1x = -3x1 + x2 2x = -3x2 + r 3x = -4x3 + r y = 2x1 - 5x2 + 6x3 In vector-matrix form, 3 1 0 0 0 3 0 1 0 0 4 1 2 5 6 r y x x b. G(s) = 2 2 2 2 ( 2) 3 / 4 1 5 / 4 1 ( ) ( 5) ( 7) ( 5) 5 ( 7) 7 s G s s s s s s s 5-36 Chapter 5: Reduction of Multiple Subsystems Writing the state and output equations, 1x = -5x1 + x2 2x = -5x2 + r 3x = -7x3 + x4 4x = -7x4 + r y = - 3 4 x1 + x2 - 5 4 x3 - x4 In vector matrix form, -5 1 0 0 0 0 -5 0 0 1 = 0 0 -7 1 0 0 0 0 -7 1 3 5 = - 1 - -1 4 4 x x r y x c. 2 2 4 1 1 1 1 1 1 1 1 ( ) ( 2) ( 5)( 6) 8 6 9 5 72 2 6 ( 2) s G s s s s s s s s Solutions to Problems 5-37 Writing the state and output equations, 1x = - 2x1 + x2 2x =- 2x2 + r 3x = - 5x3 + r 4x = - 6x4 + r y = 1 6 x1 - 1 72 x2 - 1 9 x3 + 1 8 x4 In vector-matrix form, -2 1 0 0 0 0 2 0 0 1 x x 0 0 5 0 1 0 0 0 6 1 r 1 1 1 1 6 72 9 8 y x 31. a. 5-38 Chapter 5: Reduction of Multiple Subsystems Writing the state equations, 1x = x2 2x = - 7x1 - 2x2 + r y = 3x1 + x2 In vector matrix form, 0 1 0 r 7 2 1 y 3 1 x x x b. Solutions to Problems 5-39 Writing the state equations, 1 2 2 3 3 1 2 3 1 2 3 2 5 6 2 x x x x x x x x r y x x x In vector matrix form, 0 1 0 0 0 0 1 0 1 2 5 1 6 2 1 r y x X x c. 5-40 Chapter 5: Reduction of Multiple Subsystems 1x = x2 2x = x3 3x = x4 4x = - 4x1 - 6x2 - 5x3 - 3x4 + r y = x1 + 7x2 + 2x3 + x4 In vector matrix form, 0 1 0 0 0 0 0 1 0 0 r 0 0 0 1 0 4 6 5 3 1 y 1 7 2 1 x x x 32. a. Controller canonical form: From the phase-variable form in Problem 5.31(a), reverse the order of the state variables and obtain, 2x = x1 1x = - 7x2 - 2x1 + r y = 3x2 + x1 Putting the equations in order, 1x = - 2x1 - 7x2 + r 2x = x1 y = x1 + 3x2 In vector-matrix form, 2 7 1 1 0 0 1 3 r y x x x Observer canonical form: G(s) = 2 s+3 s +2s+7 . Divide each term by 2 1 s and get Solutions to Problems 5-41 2 2 1 3 + C(s)s sG(s)= = 2 7 R(s) 1+ + s s Cross multiplying, ( 1 s + 2 3 s ) R(s) = (1 + 2 s + 2 7 s ) C(s) Thus, 1 (R(s) - 2C(s)) s + 2 1 (3R(s) - 7C(s)) s = C(s) Drawing the signal-flow graph, Writing the state and output equations, 1x = - 2x1 + x2 + r 2x = - 7x1 + 3r y = x1 In vector matrix form, 2 1 1 7 0 3 1 0 r y x x x b. Controller canonical form: From the phase-variable form in Problem 5.31(b), reverse the order of the state variables and obtain, 5-42 Chapter 5: Reduction of Multiple Subsystems 3 2 2 1 1 3 2 1 3 2 1 2 5 6 2 x x x x x x x x y x x x Putting the equations in order, 1 1 2 3 2 1 3 2 1 2 3 5 2 2 6 x x x x x x x x y x x x In vector-matrix form, 5 2 1 1 1 0 0 0 0 1 0 0 1 2 6 r y x x x Observer canonical form: 2 3 2 s 2s 6 G(s) s 5s 2s 1 . Divide each term by 3 1 s and get 2 3 2 3 1 2 6 C(s)s s sG(s) 5 2 1 R(s) 1 s s s Cross-multiplying, 2 3 2 3 1 2 6 5 2 1 R(s)= 1 C(s) s s s s s s Thus, 2 3 1 1 1 (R(s) 5c(s)) (2R(s) 2C(s)) (6R(s) C(s)) C(s) s s s Drawing the signal-flow graph, Solutions to Problems 5-43 Writing the state and output equations, 1 1 2 2 1 3 3 1 5 2 2 6 1 0 0 x x x r x x x r x x r y x In vector-matrix form, 5 1 0 1 2 0 1 2 1 0 0 6 1 0 0 r y x x x c. Controller canonical form: From the phase-variable form in Problem 5.31(c), reverse the order of the state variables and obtain, 4x = x3 3x = x2 2x = x1 1x = - 4x4 - 6x3 - 5x2 - 3x1 + r y = x4 + 7x3 + 2x2 + x1 Putting the equations in order, 1x = - 3x1 - 5x2 - 6x3 - 4x4 + r 2x = x1 5-44 Chapter 5: Reduction of Multiple Subsystems 3x = x2 4x = x3 y = x1 + 2x2 +7x3 + x4 In vector-matrix form, 3 5 6 4 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 2 7 1 r y x X x Observer canonical form: G(s) = 3 2 4 3 2 s +2s +7s+1 s +3s +5s +6s+4 . Divide each term by 2 1 s and get 2 3 4 2 3 4 1 2 7 1 + + + C(s)s s s sG(s)= = 3 5 6 4 R(s) 1+ + + + s s s s Cross multiplying, ( 1 s + 2 2 s + 3 7 s + 4 1 s ) R(s) = (1 + 3 s + 2 5 s + 3 6 s + 4 4 s ) C(s) Thus, 1 (R(s) - 3C(s)) s + 2 1 (2R(s) - 5C(s)) s + 3 1 (7R(s) - 6C(s)) s + 4 1 (R(s) - 4C(s)) s = C(s) Drawing the signal-flow graph, Solutions to Problems 5-45 Writing the state and output equations, 1x = - 3x1 + x2 + r 2x = - 5x1 + x3 + 2r 3x = - 6x1 + x4 +7r 4x = - 4x1 + r y = x1 In vector matrix form, 3 1 0 0 1 5 0 1 0 2 r 6 0 0 1 7 4 0 0 0 1 y 1 0 0 0 x x x 5-46 Chapter 5: Reduction of Multiple Subsystems 33. a. Writing the state equations, 1 1 2 2 2 3 3 1 3 1 2 8 50 9 50 x x x x x x x x x r y x In vector-matrix form, 2 1 0 0 0 8 1 0 50 0 9 50 1 0 0 r y x x x b. Writing the state equations, Solutions to Problems 5-47 1 2 2 1 2 3 3 1 1 24 6 10 x x x x x x x x r y x In vector-matrix form, 0 1 0 0 24 6 10 0 1 0 0 1 1 0 0 r y x x x c. 1x = x2 2x = -x2 - x2 + 160(r-x1) = -160x1 -2x2 +160r y = x1 In vector-matrix form, 0 1 0 160 2 160 rx x y 1 0 x 5-48 Chapter 5: Reduction of Multiple Subsystems d. Since 2 1 (s+1) = 2 1 s +2s+1 , we draw the signal-flow as follows: Writing the state equations, 1x = x2 2x = -x1 - 2x2 + 16(r-c) = -x1 - 2x2 + 16(r - (2x1+x2) = -33x1 - 18x2 + 16r y = 2x1 + x2 In vector-matrix form, 0 1 0 33 18 16 rx x . y 2 1 x 34. a. Phase Variable form 3 2 30 ( ) 8 15 30 T s s s s 21 x x 32 x x 1 33 2 30 15 8 30xx x x r 1 y x Solutions to Problems 5-49 Or in matrix form 0 1 0 0 0 0 1 0 30 15 8 30 u x x 1 0 0y x b. Parallel form 30 2 5 3 ( ) ( 3)( 5) 3 5 G s s s s s s s The state equations are: 3 31 1 2 1 2 2( ) 2 2 2 2u x x x x x ux x 1 2 3 2 12 2 3 5( ) 3 5 2 5 5u x x x x x ux x x 3 33 1 2 3 1 2 3( ) 5 3 3 8 3u x x x x x x ux x 1 2 3 y x x x In matrix form: 2 2 2 2 5 2 5 5 3 3 8 3 u x x 1 1 1y x 5-50 Chapter 5: Reduction of Multiple Subsystems 35. a. 2 4 3 2 10( 5 6) ( ) 16 99 244 180 s s T s s s s s Drawing the signal-flow diagram, Writing the state and output equations, 1 2 2 3 3 4 1 1 2 3 4 1 2 3 180 244 99 16 10 6 5 x x x x x x x x x x x r y x x x In vector-matrix form, 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 180 244 99 16 10 6 5 1 0 r y x x x b. 10( 2)( 3) 1 / 3 10 / 3 15 12 ( ) ( 1)( 4)( 5)( 6) 1 4 5 6 s s G s s s s s s s s s Drawing the signal-flow diagram and including the unity-feedback path, Solutions to Problems 5-51 Writing the state and output equations, 1 1 2 3 4 1 2 1 2 3 4 2 3 1 2 3 4 3 4 1 2 3 4 4 1 23 4 1 ( ) 3 10 ( ) 4 3 15( ) 5 12( ) 12 x u x x x x x x u x x x x x x u x x x x x x u x x x x x y x x x x In vector-matrix form, 4 1 1 1 1 3 3 3 3 3 10 2 10 10 10 3 3 3 3 3 15 15 20 15 15 12 12 12 0 12 1 1 1 1 u y x x x 36. Program: '(a)' 'G(s)' 5-52 Chapter 5: Reduction of Multiple Subsystems G=zpk([-2 -3],[-1 -4 -5 -6],10) 'T(s)' T=feedback(G,1,-1) [numt,dent]=tfdata(T,'v'); 'Find controller canonical form' [Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent) A1=flipud(Acc); 'Transform to phase-variable form' Apv=fliplr(A1) Bpv=flipud(Bcc) Cpv=fliplr(Ccc) '(b)' 'G(s)' G=zpk([-2 -3],[-1 -4 -5 -6],10) 'T(s)' T=feedback(G,1,-1) [numt,dent]=tfdata(T,'v'); 'Find controller canonical form' [Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent) 'Transform to modal form' [A,B,C,D]=canon(Acc,Bcc,Ccc,Dcc,'modal') Computer response: ans = (a) ans = G(s) Zero/pole/gain: 10 (s+2) (s+3) ----------------------- (s+1) (s+4) (s+5) (s+6) ans = T(s) Zero/pole/gain: 10 (s+2) (s+3) ------------------------------------------ (s+1.264) (s+3.412) (s^2 + 11.32s + 41.73) ans = Find controller canonical form Acc = -16.0000 -99.0000 -244.0000 -180.0000 1.0000 0 0 0 Solutions to Problems 5-53 0 1.0000 0 0 0 0 1.0000 0 Bcc = 1 0 0 0 Ccc = 0 10.0000 50.0000 60.0000 Dcc = 0 ans = Transform to phase-variable form Apv = 0 1.0000 0 0 0 0 1.0000 0 0 0 0 1.0000 -180.0000 -244.0000 -99.0000 -16.0000 Bpv = 0 0 0 1 Cpv = 60.0000 50.0000 10.0000 0 ans = (b) ans = G(s) Zero/pole/gain: 10 (s+2) (s+3) ----------------------- (s+1) (s+4) (s+5) (s+6) ans = 5-54 Chapter 5: Reduction of Multiple Subsystems T(s) Zero/pole/gain: 10 (s+2) (s+3) ------------------------------------------ (s+1.264) (s+3.412) (s^2 + 11.32s + 41.73) ans = Find controller canonical form Acc = -16.0000 -99.0000 -244.0000 -180.0000 1.0000 0 0 0 0 1.0000 0 0 0 0 1.0000 0 Bcc = 1 0 0 0 Ccc = 0 10.0000 50.0000 60.0000 Dcc = 0 ans = Transform to modal form A = -5.6618 3.1109 0 0 -3.1109 -5.6618 0 0 0 0 -3.4124 0 0 0 0 -1.2639 B = -4.1108 1.0468 1.3125 0.0487 C = Solutions to Problems 5-55 0.1827 0.6973 -0.1401 4.2067 D = 0 37. 1 2 x x 2 1 3 2x x x 3 4 x x 4 1 2 5x x x r 1 2 5y x x Or in Matrix form 0 1 0 0 0 2 0 1 0 0 0 0 0 1 0 5 1 0 0 1 r x x 5 1 0 0y c x 38. a. θ 1 + 5θ1 + 61 - 3θ2 - 42 = 0 -3θ 1 - 41 + θ2 + 5θ2 + 52 = T or θ 1 = - 5θ 1 - 61 + 3θ2 + 42 5-56 Chapter 5: Reduction of Multiple Subsystems θ 2 = 3θ1 + 41 - 5θ2 - 52 + T Letting, 1 = x1 ; θ1 = x2 ; 2 = x3 ; θ2 = x4 , where x = . b. Using the signal-flow diagram, 1 x = x2 2 x = -6x1 - 5x2 + 4x3 + 3x4 3 x = x4 4 x = 4x1 + 3x2 - 5x3 - 5x4 + T y = x3 In vector-matrix form, 0 1 0 0 0 -6 -5 4 3 0 = + T 0 0 0 1 0 4 3 -5 -5 1 y 0 0 1 0 x x x 39. Program: numg=8; deng=poly([0 -8 -10]); Solutions to Problems 5-57 G=tf(numg,deng); T=feedback(G,1); [numt,dent]=tfdata(T,'v'); [A,B,C,D]=tf2ss(numt,dent); %Obtain controller canonical form '(a)'; %Display label A=rot90(A,2); %Convert to phase-variable form A=fliplr(A); %Convert to phase-variable form B=rot90(B,2); %Convert to phase-variable form C=fliplr(C); %Convert to phase-variable form '(b)'; %Display label [a,b,c,d]=canon(A,B,C,D); %Convert to parallel form Computer response: T = 8 ----------------------- s^3 + 18 s^2 + 80 s + 8 Continuous-time transfer function. numt = 0 0 0 8 dent = 1 18 80 8 ans = (a) A = 0 1 0 1 0 0 -18 -80 -8 B = 0 0 1 C = 5-58 Chapter 5: Reduction of Multiple Subsystems 8 0 0 ans = (b) a = -8 0 0 0 1 0 0 0 -1 b = 0.0625 0 0 c = 0 0.0442 0.0442 d = 0 40. 1x = A1x1 + B1r (1) y1 = C1x1 (2) 2x = A2x2 + B2y1 (3) y2 = C2x2 (4) Substituting Eq. (2) into Eq. (3), 1x = A1x1 + B1r 2x = B2C1x1 + A2x2 y2 = C2x2 In vector-matrix notation, 1 1 2 1 r 1 1 2 2 2 x A O x B x B C A x O 1 2 2 2 x y O C x Solutions to Problems 5-59 41. 1 • x = A1x1 + B1r (1) y1 = C1x1 (2) 2 • x = A2x2 + B2r (3) y2 = C2x2 (4) In vector-matrix form, 1 1 r 1 1 2 2 2 2 x A O x B x O A x B 1 1 2 1 2 2 x y y y C C x 42. 1x = A1x1 + B1e (1) y = C1x1 (2) 2x = A2x2 + B2y (3) p = C2x2 (4) Substituting e = r - p into Eq. (1) and substituting Eq. (2) into (3), we obtain, 1x = A1x1 + B1(r - p) (5) y = C1x1 (6) 2x = A2x2 + B2C1x1 (7) p = C2x2 (8) Substituting Eq. (8) into Eq. (5), 1x = A1x1 - B1C2x2 + B1r 2x = B2C1x1 + A2x2 y = C1x1 In vector-matrix form, -1 1 1 2 1 1 2 2 1 2 2 x A B C x B r x B C A x 0 1 1y 2 x C 0 x 43. 5-60 Chapter 5: Reduction of Multiple Subsystems 1 1 u y z P APz P B CPz 1 4 9 3 -0.2085 -0.3029 -0.1661 0 4 7 ; = 0.0228 -0.1075 -0.0912 1 4 9 0.0130 0.0814 -0.0521 P P 1 -1 18.5961 25.4756 5.6156 58 -12.9023 -28.8893 -8.3909 ; 63 ; C = 1.5668 3.0423 2.7329 -0.5733 11.4169 5.2932 12 P AP P B P 44. 5 4 9 0.3469 -0.3878 0.2653 6 7 6 ; P= 0.3673 -0.4694 0.1633 6 5 3 0.0816 0.0068 -0.0748A= -28.2857 40.8095 -40.9048 -18.3061 28.2245 -37.4694 5.3878 -6.5510 -5.9388 u y -1 -1 -1 -1 z P APz P B CPz P P AP 61 ; 82 ; -2.0816 2.6599 -1.2585 74 -1 P B CP 45. Eigenvalues are -1, -2, and -3 since, | - A | = (+ 3) (+ 2) (+ 1) Solving for the eigenvectors, Ax = x or, 3 2 1 3 2 1 2 3 4 5 ( 5) 0 2 2 0 2 ( 1) 0 x x x x x x x x For = -1, x2 = 0, x1 = x3. For = -2, x1 = x2 = 3x 2 . For = -3, x1 = - 2x 2 , x2 = x3 . Thus, z = P-1APz + P-1Bu ; y = CPz, where Solutions to Problems 5-61 1 1 1 0 0 12 0 2 0 ; 8 ; (1,4,7) 0 0 3 3 P AP P B CP 46. Eigenvalues are 1, -2, and 3 since, |I - A | = ( - 3) ( + 2) ( - 1) Solving for the eigenvectors, Ax = x or, 1 3 2 1 2 3 1 2 3 ( 10) 7 3 0 73 25 47 0 4 4 4 29 9 23 0 4 4 4 x x x x x x x x x For = 1, x1 = x2 = 3x 2 . For = -2, x1 = 2x3, x2 = -3x3. For = 3, x1 = x3 , x2 = -2x3 . Thus, z = P-1APz + P-1Bu ; y = CPz, where 3 2 1 1 1 2 3 2 1 0 0 0 2 0 ; ; (7,12,9) 0 0 3 P AP P B CP 47. Program: A=[-10 -3 7;18.25 6.25 -11.75;-7.25 -2.25 5.75]; B=[1;3;2]; C=[1 -2 4]; [P,d]=eig(A); Ad=inv(P)*A*P Bd=inv(P)*B Cd=C*P Computer response: Ad = -2.0000 0.0000 0.0000 -0.0000 3.0000 -0.0000 0.0000 0.0000 1.0000 Bd = 1.8708 -3.6742 3.6742 5-62 Chapter 5: Reduction of Multiple Subsystems Cd = 3.2071 3.6742 2.8577 48. Push Pitch Gain to the right past the pickoff point. Collapse the summing junctions and add the feedback transfer functions. Apply the feedback formula and obtain, 4 3 2 ( ) 0.25( 0.435) ( ) 1 ( ) ( ) 3.4586 3.4569 0.9693 0.15032 G s s T s G s H s s s s s 49. Program: numg1=-0.125*[1 0.435] deng1=conv([1 1.23],[1 0.226 0.0169]) 'G1' G1=tf(numg1,deng1) 'G2' Solutions to Problems 5-63 G2=tf(2,[1 2]) G3=-1 'H1' H1=tf([-1 0],1) 'Inner Loop' Ge=feedback(G1*G2,H1) 'Closed-Loop' T=feedback(G3*Ge,1) Computer response: numg1 = -0.1250 -0.0544 deng1 = 1.0000 1.4560 0.2949 0.0208 ans = G1 Transfer function: -0.125 s - 0.05438 ------------------------------------ s^3 + 1.456 s^2 + 0.2949 s + 0.02079 ans = G2 Transfer function: 2 ----- s + 2 G3 = -1 ans = H1 Transfer function: -s 5-64 Chapter 5: Reduction of Multiple Subsystems ans = Inner Loop Transfer function: -0.25 s - 0.1088 ------------------------------------------------ s^4 + 3.456 s^3 + 3.457 s^2 + 0.7193 s + 0.04157 ans = Closed-Loop Transfer function: 0.25 s + 0.1088 ----------------------------------------------- s^4 + 3.456 s^3 + 3.457 s^2 + 0.9693 s + 0.1503 50. Solutions to Problems 5-65 5-66 Chapter 5: Reduction of Multiple Subsystems 51. a. Since VL(s) = Vg(s) – VR(s), the summing junction has Vg(s) as the positive input and VR(s) as the negative input, and VL(s) as the error. Since I(s) = VL(s) (1/(Ls)), G(s) = 1/(Ls). Also, since VR(s) = I(s)R, the feedback is H(s) = R. Summarizing, the circuit can be modeled as a negative feedback system, where G(s) = 1/(Ls), H(s) = R, input = Vg(s), output = I(s), and error = VL(s), where the negative input to the summing junction is VR(s). b. 1 ( ) ( ) 1 ( ) 1( ) 1 ( ) ( ) 1g I s G s LsT s V s G s H s Ls R R Ls . Hence, 1 ( ) ( ) g I s V s Ls R . c. Using circuit analysis, ( ) ( ) g V s I s Ls R . 52. a. The first equation follows from the schematic. The second equation is obtained by applying the voltage divider rule at the op-amp’s inverting terminal, noting that since the op-amp considered is ideal, there is no current demand there. b. 1 T A ; i i f R L A R R ; 1 i i f R A R R ; 1 1 1 1 1 o ii i f v T A Rv A R R c. 1 1 1 o f A i ii i i f i f v RA Lim R Rv R A R R R R 53. a. Adding currents at the op-amp’s inverting terminal, under ideal condition we get 11 oi i f v vv v R R which after some algebraic manipulations gives 1 f i i o f i f i R R v v v R R R R Also from the circuits diagram 1o v Av b. These equations can be represented by the following block diagram Solutions to Problems 5-67 We have that 1 f f i R T A R R ; i i f R L A R R ; 1 i i f R A R R ; 1 1 1 1 1 f o i f ii i f R A v T R R Rv A R R c. 1 f f o i f i f f A i ii i i f i f R R A v R R R R R Lim R Rv R A R R R R 54. a. The three equations follow by direct observation from the small signal circuit. b. The block diagram is given by c. From the block diagram we get ( || ) 1 ( || ) o m s o i i m s o i s v g R r R v g R r R R 55. a. Using Mason’s rule 2 0 1 2 2 2 2 0 0 1t US K T Ms s ; Loops 2 0 1 2 2 0 rL m ss and 2 rmL s , no non-touching loops. 1 1 2 2 0 0 2 2 2 2 3 0 01 1 2 2 0 2 22 2 00 11 ( )( ) r r r X s sT R m s m m s s s ss a. From part (a) 5-68 Chapter 5: Reduction of Multiple Subsystems 2 0 2 2 3 01 2 2 2 0 1 ( )( ) r X s sX m sR R s s s 56. a. >> A=[-100.2 -20.7 -30.7 200.3; 40 -20.22 49.95 526.1;... 0 10.22 -59.95 -526.1; 0 0 0 0]; >> B=[208; -208; -108.8; -1]; >> C = [0 1570 1570 59400]; >> D = -6240; >> [n,d]=ss2tf(A,B,C,D) n = 1.0e+009 * Columns 1 through 3 -0.00000624000000 -0.00168228480000 -0.14206098728000 Columns 4 through 5 -3.91955218234127 -9.08349454230472 d = Solutions to Problems 5-69 1.0e+005 * Columns 1 through 3 0.00001000000000 0.00180370000000 0.09562734000000 Columns 4 through 5 1.32499100000000 0 >> roots(n) ans = 1.0e+002 * -1.34317654991673 -0.78476212102923 -0.54257777928519 -0.02545278053809 >> roots(d) ans = 0 5-70 Chapter 5: Reduction of Multiple Subsystems -92.38329312886714 -66.38046756013043 -21.60623931100260 Note that (0) 68555.14 Y s U , follows that 6348.17( 2.5)( 54.3)( 78.5)( 134.3) ( ) ( 21.6)( 66.4)( 92.4) Y s s s s s U s s s s b. >> [r,p,k]=residue(n,d) r = 1.0e+005 * -0.73309459854184 -0.51344619392820 -3.63566779304453 -0.68555141448543 p = -92.38329312886714 -66.38046756013043 -21.60623931100260 0 Solutions to Problems 5-71 k = -6240 or 73309.46 51344.6 363566.8 68555.14 ( ) 6240 21.6 66.4 92.4 Y s U s s s s c. d. The corresponding state space representation is: 1 1 2 2 3 3 4 4 0 0 0 0 73309.5 0 21.6 0 0 51344.6 ( ) 0 0 66.4 0 363566.8 0 0 0 9.4 68555.14 x x x x u t x x x x 5-72 Chapter 5: Reduction of Multiple Subsystems 1 2 3 4 1 1 1 1 6240 ( ) x x y u t x x 57. a. >> A = [0 1 0; 0 -68.3 -7.2; 0 3.2 -0.7] A = 0 1.0000 0 0 -68.3000 -7.2000 0 3.2000 -0.7000 >> [V,D]=eig(A) V = 1.0000 0.0147 -0.1016 0 -0.9988 0.1059 0 0.0475 -0.9892 D = Solutions to Problems 5-73 00 0 0 -67.9574 0 0 0 -1.0426 Matrix V is the sought similarity transformation. b. >> Ad = inv(V)*A*V Ad = 0 -0.0000 -0.0000 0 -67.9574 0.0000 0 -0.0000 -1.0426 >> B = [0;425.4;0] B = 0 425.4000 0 >> Bd = inv(V)*B Bd = 5-74 Chapter 5: Reduction of Multiple Subsystems 4.2030 -428.1077 -20.5661 The diagonalized system is: 1 1 2 2 3 3 0 0 0 4.2 0 67.96 0 428.11 0 0 1.0426 20.57 m z z z z e z z 58. a. b. There is only one forward path 1 e h T H H There are three loops: 1 e h L H H ; 2 i h L H H and 3 i r act L H H H 1 L and 3 L are non-touching loops so 1 2 3 1 3 1 1 e h i h i r act e h i r act L L L L L H H H H H H H H H H H H When 1 T is eliminated only 3 L is left so 1 3 1 1 r i act L H H H Finally 1 1 1 ( ) 1 e h r i acth e h i h i r act e h i r act H H H H HF T s D H H H H H H H H H H H H 59. Solutions to Problems 5-75 Block Diagram: Case Structure Details: 5-76 Chapter 5: Reduction of Multiple Subsystems 60. a. There are two forwards paths: 1 2 1 = h hs sM K K C s and 2 1 = h es es sM K T K C s The loops are: 1 2 1 h hs s hsL K K C T s 2 1 h es es s hsL K T K C T s 3 j ej ejL C T K 4 h j hjL K C T There are no non-touching loops. Therefore 2 1 1 1 h hs s hs h es es s hs j ej ej h j hjK K C T K T K C T C T K K C T ss Also 1 2 1 2 1 1 2 2 2 1 1 1 1 1 h hs s h es es s s h h hs s hs h es es s hs j ej ej h j hj K K C K T K C Y M M ss U K K C T K T K C T C T K K C T ss b. Solutions to Problems 5-77 There is only one forward path 1 h jM K C The loops and are the same as in part a. Also 1 1 . It follows that 1 1 2 1 1 1 j h j h h hs s hs h es es s hs j ej ej h j hj Y K CM U K K C T K T K C T C T K K C T ss 61. a. Assuming 0hZ there are two forward paths, 11 mM Z and 12 6 mM C Z The loops are 1 s s L G C 1 2 m m L Z C 1 3 1 4m s L Z C G C 1 4 2m s s e L Z C G Z C There are two non-touching loops 1 L and 2 L . 1 1 1 1 1 4 2 1 s s m m m s m s s e s s m m G C Z C Z C G C Z C G Z C G C Z C 1 1 1 2 2 1 1 ( ) s s m m m s s e s s m m s s e G C Z C Z C G Z C G C Z C C G Z C We also have that by eliminating 1M or 2M 1 2 1 s sG C 1 1 1 6 21 1 2 2 1 1 2 2 ( )(1 ) (1 ) ( ) 1 ( ) 1 ( ) h m m s s m s s h s s m m e s s s s m m e s s X Z Z C G C Z C G CM M Y s F G C Z C C Z G C G C Z C C Z G C b. The system can be described by means of the following diagram: It follows that ( ) 1 ( ) h h h X Y s F Y s Z 62. a. There are three forward paths: 5-78 Chapter 5: Reduction of Multiple Subsystems 1 1 2 1 3 1 ˆ i p p f f f K M K K s L C C s 1 2 1 2 1i p f f K M K s L C s 3 2ˆ ˆ 1 f f f f M L s R L C s The loops are: 1 1 2 1 2 1i p p f f K L K K s L C s 1 2 1 1i p f K L K s L s 3 f f R L L s There are no non-touching loops. 1 1 2 1 12 1 1 1 fi i p p p f ff f RK K K K K s s L s L sL C s and 1 2 3 1 1 1 2 1 1 2 23 1 1 2 2 3 3 1 1 2 1 12 1 1 1ˆ ˆ ˆ 1 1 1 i i p p p f f f f f ff f fLoad fCf i i p p p f ff f K K K K K L s R s s L C s L C sL C C sV M M M RI K K K K K s s L s L sL C s b. There are three forward paths: 1 1 2 1 3 1 ˆ i p p f f f K M K K s L C C s 1 2 1 2 1i p f f K M K s L C s 3 2ˆ ˆ 1 f f f f M L s R L C s The loops are: 1 1 2 1 2 1i p p f f K L K K s L C s 1 2 1 1i p f K L K s L s 3 f f R L L s There are no non-touching loops. Solutions to Problems 5-79 1 1 2 1 12 1 1 1 fi i p p p f ff f RK K K K K s s L s L sL C s and 1 2 3 1 1 1 2 1 1 2 23 1 1 2 2 3 3 1 1 2 1 12 1 1 1ˆ ˆ ˆ 1 1 1 i i p p p f f f f f ff f fLoad fCf i i p p p f ff f K K K K K L s R s s L C s L C sL C C sV M M M RI K K K K K s s L s L sL C s The loops and are the same as in part a. There is only one forward path 1 1 f M C s and 1 1 1 1 1 1 2 1 12 1 1 = 1 - 1 fLoad fDist i i p p p f ff f C sV M RI K K K K K s s L s L sL C s 63. a. Substituting the values given above into the block diagram, when vp = 0, we have: The Mould Level Block Diagram for vp = 0 Thus, the closed-loop transfer function is: ( ) ( ) ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) m C V X m C V X m H (s) G s G s G s G s T s R(s) G s G s G s G s = = 2 2 2 2 1.6 1.25 0.25 100 0.63 1 0.926 0.510 100 1.6 1.25 0.25 100 0.63 1 1 0.5 0.926 0.510 100 s s s s ss s s s s s ss s = 5-80 Chapter 5: Reduction of Multiple Subsystems = 2 2 2 2 201.6 1.25 0.25 10 100 0.926 0.5 201.6 1.25 0.25 s s s s s s s s = = 2 5 4 3 2 201.6 1.25 0.25 10.93 109.26 193.4 126 25.2 s s s s s s s b. Simulink was used to simulate the system. The model of that system is shown in Figure P5.x-4*. The parameters of the PID controller were set to: Kp = 2, Kd = 1.6, and KI = 0.4. The reference step, r(t) = 5 u(t), and the casting speed step, vp (t) = 0.97 u(t) were set to start at t = 0. An adder was used to add the initial value, Hm (0 — ) = – 75 mm, at the output, to the change in mould level, Hm. The time and mould level (in array format) were output to “workspace ” sinks, each of which was given the respective variable name. After the simulation ended, Matlab plot commands were used to obtain and edit the graph of hm(t) from 0 to t = 80 seconds. Simulink Model of the Mould Level Control System Solutions to Problems 5-81 Response of the Mould Level to Simultaneous Step Changes in Reference Input, r(t) = 5 u(t), and Casting Speed, vp (t) = 0.97 u(t) at an Initial Level, Hm (0 — ) = – 75 64. a. Following the procedure described in Chapter 3 we define 1 2 1X aV gbh s h and 1( ) ( ) V s X s A . In time domain 1 1 g aV x x h bh . 1 1 V x x a and we also define 1 2 x x . These equations give 1 1 2 2 00 1 0 x x aVg x x bhh 1 2 1 xV xa 0 10 20 30 40 50 60 70 80 -78 -76 -74 -72 -70 -68 -66 -64 Mould Level, hm, in Response to a Reference Signal r(t) = 5 u(t) and a Casting Speed vp (t) = 0.97 u(t). At t = 0, hm (0) = - 75.0 mm Steady state (final) value (at t = 80 sec), hm = - 65.0 mm Time, sec M o u ld L ev el , h m , m m 5-82 Chapter 5: Reduction of Multiple Subsystems b. The eigenvalues can be obtained directly from the transfer function poles. Thus 12 g h Consider 1 g h , the first eigenvector is found from the solution of 1 1 1 Ax x or 1 1 2 2 0 1 0 x xg g x xh h . Thisresults in 2 1 g x x h . Arbitrarily let 1 1x so the first eigenvector is 1 g h 1v . Similarly for 2 g h ; 2 1 2Ax x or 1 1 2 2 0 1 0 x xg g x xh h resulting in 2 1 g x x h . Letting arbitrarily 1 1x the second eigenvector is 2 1 g h v . c. The similarity transformation matrix is 1 1 g g h h 1 2P v v 1 1 1 1 1 2 2 1 1 2 2 2 g h hg gh g h h g P The matrices for the diagonalized are calculated as follows: 1 1 1 1 00 1 2 2 01 1 0 2 2 h g g h g g g h g h h h g h 1 dA P AP 11 1 0 22 2 11 1 22 2 aVh g b gh aV aVh bh b ghg 1 dB P B 1 1 1 V V g V g g g a a h a h h h d C CP Solutions to Problems 5-83 The diagonalized representation is: 1 1 2 2 1 0 2 1 0 2 aVg b ghz zh z z aVg b ghh 1 2 zV g V g za h a h 65. It can be easily verified that the closed loop transfer function for this system is identical to the original. 66. The closed loop transfer function is ( ) 1 ( ) ( ) t ref V G s V G s H s where 2 20.3 1.6 0.4 10 1 1 75 400 100 ( ) 0.1 1 0.4 1 1 ( 1)( 2.5)( 10) s s s s G s s s s s s s s s and 1 100 ( ) 0.01 1 100 H s s s Substituting 2 2 3 2 5 4 3 2 75 400 100 ( 1)( 2.5)( 10) 7500 40000 10000 1 ( 1)( 2.5)( 10)( 100) 75 7900 40100 10000 113.5 1387.5 11275 42500 10000 t ref s s V s s s s V s s s s s s s s s s s s s s s 67. The MATLAB M-file is: 5-84 Chapter 5: Reduction of Multiple Subsystems num1=25; num2=[1 1.2 12500]; den1=[1 0]; den2=[1 5.6 62000]; num=conv(num1, num2); den=conv(den1, den2); G=tf(num, den); D=feedback(G,0.1); [numd,dend]=tfdata(D,'v'); numcm=[40 5]; dencm=[1 0]; numOL=conv(numd, numcm); denOL=conv(dend, dencm); Omega_OL=tf(numOL, denOL); Omega_CL=260*feedback(Omega_OL,1); step(Omega_CL, 0:0.002:0.2); grid; After the above file is run, MATLAB’s command window may be used to obtain the requested minor-loop transfer function: D = 25 s^2 + 30 s + 312500 ---------------------------------------- s^3 + 8.1 s^2 + 62003 s + 31250 Continuous-time transfer function. The MATLAB figure, shown below, illustrates the step-response obtained with all of the requested important characteristics noted on it. Solutions to Problems 5-85 68. a. Note that due to the topology, the loop on the top should have no influence whatsoever on output ssx . Applying Mason’s: There are two forward paths 2 1 ss ssT s m H and 2 ss ssT C H . There are two loops 1 ss ssL C H and 2 ls lsL C H ; both loops are non-touching. Thus 1 ss ss ls ls ss ss ls lsC H C H C H C H . Eliminating forward path 1: 1 1 ls lsC H . Eliminating forward path 2: 2 1 ls lsC H . The closed loop transfer function is: 2 2 1 1 2 2 2 ( ) ( )(1 ) ( ) 1 1 1 1 ss ss ss ss ss ls ls ss ss ss ss ss ss ss ss ls lsss ss ls ls ss ss ls ls ls ls ss ss ss ss ss ss x s s m H C H C H s m H C HT T C H C H C Hr s C H C H C H C H C H s m H C H C H b. In this case there are three forward paths 21 ( )ss ls lsT s m m H ; 2 2 ss ss ls lsT s m H C H ; 3 ss ss ls lsT C H C H . The loops and are just as in part a. Eliminating forward path 1 1 1 ss ssC H . Eliminating either paths 2 or 3 2 3 1. The closed loop transfer function is: 5-86 Chapter 5: Reduction of Multiple Subsystems 1 1 2 2 3 3 2 2 ( ) ( ) ( ) (1 ) 1 ls ss ls ls ss ss ss ss ls ls ss ss ls ls ss ss ls ls ss ss ls ls x s T T T r s s m m H C H s m H C H C H C H C H C H C H C H 69. There are three forward paths 1 f m pi maT k F T T ; 2 pa maT T T and 3 faT T . There are two loops: 1 m pi i cL F T R H and 2 m pi ma rL F T T k . There are no non-touching loops. So 1 m pi i c m pi ma rF T R H F T T k . Eliminating forward paths 1 or 2 leaves no loops so 1 2 1 . Eliminating forward path 3 leaves one loop so 3 1 m pi i cF T R H . The closed loop transfer function is: 0 1 1 2 2 3 3 (1 )( ) 1( ) f m pi ma pa ma fa m pi i c m pi i c m pi ma ri k F T T T T T F T R Hv s T T T F T R H F T T kv s 70 . a. Showing the equivalent circuit here for reference we have: 1 ( ) ( ) ( )c acF acRV s I s I s Cs (1) Substituting the equation given in the problem into (1), re-arranging, and simplifying gives: 2 1 ( ) ( ) ( ) 1c a acR LCV s V s Ls I s R s s L LC (2) b. Substituting circuit parameters into the equation given in the problem statement & (2), and assuming zero initial conditions, we have: 3 3 2 3 7 2 3 7 10 ( 50000) 10 ( ) ( ) ( ) ( 10 5 10 ) 10 5 10 acF acR a s s I s I s V s s s s s (3) Solutions to Problems 5-87 7 4 1 2 2 3 7 2 3 7 5 10 5 10 ( ) ( ) ( ) ( ) ( ) 10 5 10 10 5 10 c c c a acR s V s V s V s V s I s s s s s (4) The Simulink model is shown below. The current source, IacR, (a step input) was set to model iacR (t) = 10 u(t), amps, and the voltage ramp, va (t) = 20 t u(t), was created using a step of 20 volts and an integrator with the upper saturation limit set to 20 volts and the lower to 0. The result of the simulation was captured on a “scope” that was set to display the two input variables (using a mux) from t = 0 to 15 ms, with the y-axis range set from – 60 to +60. Simulink model of the system 5-88 Chapter 5: Reduction of Multiple Subsystems Inductor Current, iacF(t) in amps, and Capacitor Voltage, vC(t) in volts, as Displayed on the scope at the end of simulation (set to 15 ms). SOLUTIONS TO DESIGN PROBLEMS 71. Je = Ja+JL( 1 20 )2 = 2+2 = 4; De = Da+DL( 1 20 )2 = 2+DL( 1 20 )2. Therefore, the forward-path transfer function is, Solutions to Problems 5-89 G(s) = (1000) e 1 14 1 20 s(s+ (D +2)) 4 . Thus, T(s) = G 1+G = 2 e 25 2 1 25 s + (D +2)s+ 4 2 . Hence, = π2 2 %OS - ln ( ) 100 %OS + ln ( ) 100 = 0.456; n = 25 2 ; 2n = e D +2 4 . Therefore De = 10.9; from which DL = 3560. 72. a. T(s) = 2 25 s +s+25 ; from which, 2n = 1 and n = 5. Hence, = 0.1. Therefore, 2/ 1 %OS e 100 72.92%x ; Ts = ζω n 4 = 8. b. T(s) = 1 2 2 1 25K s +(1+25K )s+25K ; from which, 2n = 1+25K2 and n = 5 1K . Hence, 2 2 2 2 % ln( ) ln 0.2100 0.456 % ln 0.2 ln ( ) 100 OS OS Also, Ts = ζωn 4 = 0.25, Thus, n = 16; from which K2 = 32 1.28 25 and n = 35.09. Hence, K1 = 49.25. 73. The equivalent forward path transfer function is Ge(s) = 2 K s(1+(1+K )) . Thus, T(s) = e e G (s) 1+G (s) = 2 2 K s +(1+K )s+K . Prior to tachometer compensation (K2 = 0), T(s) = 2 K s +s+K . Therefore K = n 2 = 5-90 Chapter 5: Reduction of Multiple Subsystems 100. Thus, after tachometer compensation, T(s) = 2 2 100 s +(1+K )s+100 . Hence, n= 10; 2n = 1+K2. Therefore, K2 = 2n - 1 = 2(0.69)(10) - 1 = 12.8. 74. At the N2 shaft, with rotation, ( )L s 2 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) eq eq L eq v J s D s s F s r T s F s Ms f s X s Thus, 2 2( ) ( ) ( ) ( ) ( ) eq eq L v eq J s D s s Ms f s X s r T s But, ( ) ( ) L X s r s . Hence, 2 2 2( ) ( ) ( ) ( ) eq eq v L eq J Mr s D f r s s T s where 2 2 (2) 5 (2) 4 2 eq a eq a J J J D D D D r Thus, the total load inertia and load damping is 2 2 2 5 4 4 (1)(2) 8 L eq L eq v J J Mr M D D f r D D Reflecting JL and DL to the motor yields, (5 4 ) (8 ) ; 4 4 m m M D J D Thus, the motor transfer function is 1 ( ) 11( ) ( ( 1))( ( )) t m a m m t aa mm mm a K s R J J K KE s s s Ds s D JJ R The gears are (10/20)(1) = 1/2. Thus, the forward-path transfer function is 1 1 ( ) (500) 1 2 ( ( 1)) m e m m J G s s s D J Finding the closed-loop transfer function yields, Solutions to Design Problems 5-91 2 ( ) 250 / ( ) 1 2501 ( ) e m me m m G s J T s DG s s s J J For Ts = 2, 1 4m m D J . For 20% overshoot, = 0.456. Thus, 1 2 2(0.456) 4m n n m D J Or, 250 4.386 n m J ; from which 13 m J and hence, 51 m D . But, (5 4 ) (8 ) ; 4 4 m m M D J D . Thus, M = 11.75 and D = 196. 75. a. The leftmost op-amp equation can be obtained by superposition. Let 0 o v , then the circuit is an inverting amplifier thus 1 10 10 in in k v v v k . Now let 0 in v , the circuit is a non-inverting amplifier with an equal resistor voltage divider at its input, thus 1 10 10 (1 ) 10 10 10 o o k k v v v k k k . Adding both input components 1 o in v v v b. The two equations representing the system are: 1 o in v v v and 1 1 0.1 0.1 1 1 0.1 0.1 o in v s R v R s s R The block diagram is: 5-92 Chapter 5: Reduction of Multiple Subsystems c. From the figure 1 0.1 1 1 0.1 0.1 1 2 0.1 0.1 1 1 0.1 o in R s v R R v s R R s R d. The system is first order so 4 0.2 1 sec 2 0.1 s T R m R from which 1 5 0.2 m R k e. 2000 4000 o i v v s For a unit step input the output will look as follows 76. a. The transfer function derived for that system was: 1 2 ( ) 1 ( ) ( ) ( ) eq eq eq s G s T s J s D s K , where: Solutions to Design Problems 5-93 21 2 1 ( ) eq N J N ; 21 2 ( ) eq N D N ; 21 2 ( ) eq N K N . or 2 1 2 2 2 2 2 1 ( ) 1( ) ( ) 1 1 s nG s n nT s s s n n 2 2 2 1 n n n and 2 2 21 n n n . From Ts = 10 sec = 4/n, we find n = 2. b. In Figure P5.58, KG(s) = 2 0.8 0.8 0.8s s ; and T(s) = 2 0.8 0.8 1.6s s , where 2n = 0.8; 2 n = 1.6 This results in: n = 1.265 rad/sec & = 0.32; % O.S. = 33% and Ts = 10 sec. 77. a. The block diagram of this system is shown below. b. The transfer function of GP(s) of the circuit representing the process may be derived as shown in example 2.6. That yields (see equation 2.66): 6 2 6 2 ( ) 1 10 ( ) ( ) 10 101 o P pf p V s LC G s RV s s R s s s L LC Hence: 310n and for a critically damped process ( =1), we have: 32 10 2 10 /10 200n p pR R 5-94 Chapter 5: Reduction of Multiple Subsystems c. In the circuit representing the final control element: 8 4 7 8 4 10 10 5.556 101.810K / /0.022 ( ) 10 ( 5556) 10 1.8 f f sZ F Z s s s ; 4( ) 10 i Z s 3( ) ( ) 5.556 10 ( ) ( ) ( ) 5556 f f F c i V s Z s G s V s Z s s The overall system transfer function is given by: 9 2 6 9 ( ) ( ) ( ) 5.556 10( ) ( ) ( ) 1 ( ) ( ) ( ) ( 5556)( 2000 10 ) 5.556 10 C F P P C F P P G s G s G s KC s T s R s G s G s G s s s s K Given that 0.5 and 4 0.004 2000 s n n T , the dominant poles should be: 2 1,2 1 1000 1732 n n s j j , which are the roots of 2 62000 4 10s s . Thus, the third closed-loop pole may be found by dividing the characteristic polynomial by 2 62000 4 10s s , which gives 1389(1 ) P s K . This third pole is non-dominant for KP > 2.6. Given that its tolerance is ± 10%, I would set the controller potentiometer, RF, to: 10 k 1.1 28.6 k F p R K Solutions to Design Problems 5-95 78. a. The transfer function obtained in Problem 3.32 is 3 2 1 520 10.3844 2.6817 0.11 0.0126 Y s U s s s by inspection we write the phase-variable form 1 1 2 2 1 3 3 0 1 0 0 0 0 1 0 0.0126 0.11 2.6817 1 x x x x u x x 1 2 3 10.3844 520 0 x y x x b. We renumber the phase-variable form state variables in reverse order 3 3 2 2 1 1 1 0 1 0 0 0 0 1 0 0.0126 0.11 2.6817 1 x x x x u x x 3 2 1 10.3844 520 0 x y x x And we rearrange in ascending numerical order to obtain the controller canonical form: 1 1 2 2 1 3 3 2.6817 0.11 0.0126 1 1 0 0 0 0 0 1 0 x x x x u x x 1 2 3 0 520 10.3844 x y x x c. To obtain the observer canonical form we rewrite the system’s transfer function as: 5-96 Chapter 5: Reduction of Multiple Subsystems 2 3 1 2 3 520 10.3844 2.6817 0.11 0.0126 1 Y s s U s s s We cross-multiply to obtain 12 3 2 3 520 10.3844 2.6817 0.11 0.0126 1U Y s s s s s Combining terms with like powers of integration: 2 3 1 1 1 2.6817 520 0.11 10.3844 0.0126 1 1 1 2.6817 520 0.11 10.3844 0.0126 Y Y R Y R Y s s s Y R Y R Y s s s We draw the signal flow graph: The following equations follow: 1 1 2 2.6817x x x 2 1 3 0.11 520x x x r 3 1 0.0126 10.38x x r Solutions to Design Problems 5-97 1 y x Which lead to observer canonical form: 1 1 2 2 1 3 3 2.6817 1 0 1 0.11 0 1 520 0.0126 0 0 10.38 x x x x u x x ; 1 2 3 1 0 0 x y x x d. >> A=[-0.04167 0 -0.0058; 0.0217 -0.24 0.0058; 0 100 -2.4]; >> B=[5.2;-5.2;0]; >> C=[0 0 1]; >> [V,D]=eig(A); >> Bd=inv(V)*B Bd = 1.0e+002 * -0.9936 + 0.0371i -0.9936 - 0.0371i 1.9797 >> Cd = C*V Cd = 0.9963 0.9963 1.0000 5-98 Chapter 5: Reduction of Multiple Subsystems >> D D = -0.0192 + 0.0658i 0 0 0 -0.0192 - 0.0658i 0 0 0 -2.6433 So a diagonalized version of the system is 1 1 2 2 1 3 3 0.0192 0.0658 0 0 99.36 3.71 0 0.0192 0.0658 0 99.36 3.71 0 0 2.6433 197.97 x j x j x j x j u x x 1 2 3 0.9963 0.9963 1 x y x x 79. a. Substituting all values and transfer functions into the respective blocks of the system (Figure 4), we get:Solutions to Design Problems 5-99 Moving the last pick-off point to the left past the 0.3 0.06154 4.875 tot r i block and changing the position of the back-emf feedback pick-off point, so that it becomes an outer loop, we obtain the block-diagram shown below. In that diagram the 1 7.226s block (representing the total inertia) has two parallel feedback blocks. Reducing these two blocks into one, we have the following equivalent feedback transfer function: 1 ( ) 0.13847.226 ( ) 0.13787( ) 0.01908 1 7.226 eq s s G s T s s s 5-100 Chapter 5: Reduction of Multiple Subsystems Replacing that feedback loop with its equivalent transfer function, Geq(s), we have: Moving the armature current pick-off point to the right past the ( ) ( ) a T s I s and Geq(s) blocks, the above block-diagram becomes as shown below. Solutions to Design Problems 5-101 The latter, in turn, can be reduced to that shown next as the cascaded blocks in the feedback to the torque controller are replaced by the single block: ( ) 0.01908 ( ) 0.4982 CS a K I s s s and the inner feedback loop is replaced by its equivalent transfer function: 0.2491 ( ) 0.24910.01908 0.2491( ) 0.5173 1 2 0.01908 A s s U s s s 5-102 Chapter 5: Reduction of Multiple Subsystems Thus: 10 6 0.2491 ( ) 0.2491 (10 6)0.5173 10 6 0.2491 0.01908( ) ( 0.5173) 0.5 (10s 6) (s 0.01908) 1 0.5173 0.4982 C s s ss s s sU s s s s s Finally 100 40 0.2491(10 6) ( 0.5173) 0.5(10s 6)(s 0.01908)( ) ( ) 100 40 0.2491(10 6) 1 0.0443 ( 0.5173) 0.5(10s 6)(s 0.01908) v s s s s ss R s s s s s s or 2 2 2 249.1 0.4 ( 0.6)( ) ( ) 6 3.613 0.0572 11.035 0.24 249.1 0.4 ( 0.6) 36 14.644 11.09s 2.65 s ss R s s s s s s v s s s s Hence: 3 2 15.33 0.4 ( 0.6)( ) ( ) 0.06154 ( ) ( ) 6 14.644 11.09s 2.65v v s sV s s R s R s s s b. Simulink was used to model the HEV cascade control system. That model is shown below. The reference signal, rv (t), was set as a step input with a zero initial value, a step time = 0 seconds, and a final value equal to 4 volts [corresponding to the desired final car speed, v () = 60 km/h, e.g. a desired final value of the change in car speed, v () = 5.55 m/s]. The variables of interest [time, change in car speed, acceleration, and motor armature current] were output (in array format) to four “workspace” sinks, each of which was assigned the respective variable name. After the simulation ended, Matlab plot commands were utilized to obtain and edit the required three graphs. These graphs are shown below. The simulations show that in response to such a speed reference command, car acceleration would go initially to a maximum value of 10.22 m/s 2 and the motor armature current would reach a maximum value of 666.7 A. That would require an electric motor drive rated around 80 kW or using both the electric motor and gas or diesel engine, when fast acceleration is required. Most practical HEV control systems, however, use current-limiting and acceleration-limiting devices or software programs. Solutions to Design Problems 5-103 Model of the HEV Cascade Control System Change in car speed in response to a speed reference signal step of 4 volts 5-104 Chapter 5: Reduction of Multiple Subsystems Car acceleration reponse to a speed reference signal step of 4 volts Motor armature current reponse to a speed reference signal step of 4 volts Solutions to Design Problems 5-105 80. a. There is only 1 forward path. The gain of the forward path is 1 M GP . There are two loops 1 L GP and 2 L GP , and no non-touching loop; so 1 GP GP . After the forward loop is eliminated 1 1 . The closed-loop transfer function from command input to output is: 1 1( ) 1 ( ) MY GP s R G P P b. There is only 1 forward path. The gain of the forward path is 1 1M There are two loops 1 L GP and 2 L GP , and no non-touching loop; so 1 GP GP . After the forward loop is eliminated one loop remains so 1 1 GP . The closed-loop transfer function from command input to output is: 1 1 1( ) 1 ( ) MY GP s D G P P c. The total output is 1 1 ( ) 1 ( ) GP GP Y R D G P P G P P d. In Figure P5.54(b) let ( ) ( ) 1 ( ) ( ) C G s G s G s P s . The closed-loop transfer function from command input to system output is: ( ) ( ) ( ) ( )1 ( ) ( ) ( ) 1 ( ) ( ) ( )1 ( ) 1 ( ) ( ) G s P s Y G s P sG s P s G sR G s P s P sP s G s P s Similarly the transfer function from disturbance to system outputs is: 1 1 ( ) ( ) ( ) 1 ( ) ( ) ( )1 ( ) 1 ( ) ( ) Y G s P s G sD G s P s P sP s G s P s ONLINEFFIRS 11/25/2014 13:29:37 Page 1 Copyright 2015 John Wiley & Sons, Inc. All rights reserved. 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