Reduction of Multiple Subsystems

F I V E Reduction of Multiple Subytem SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Deigning a Cloed-Loop Repone a. Drawing the block diagram of the ytem: u i + - Pot 0 Π Pre amp K Power amp 50 +50 Motor, load and gear 0.6 (+.32) u o Thu, T() = 76.39K 3 +5.32 2 +98+76.39K b. Drawing the ignal flow-diagram for each ubytem and then interconnecting them yield: u i 0 Π K x 50 0.8 3 x 0.2 u o pot pre amp -50 -.32 gear power amp - 0 Π motor and load pot

Solution to Cae Studie Challenge 27 ẋ =. = -.32 + 0.8x 3 ẋ 3 = -50x 3 +50K( 0 π (q i 0.2x )) = -95.49Kx - 50x 3 + 477.46Kθ i θ o = 0.2x In vector-matrix notation, ẋ = 0 0 0 -.32 0.8-95.49K 0-50 x + 0 0 477.46K θ i θ o = 0.2 0 0 x c. T = 0 π (K)(50) (0.8) 76.39 (0.2) = 3 G L = 50 Nontouching loop: G L G L2 = 98 2 ; G L 2 =.32 ; G L 3 = (K)(50) (0.8) 0 (0.2) π = 76.39K 3 = - [G L + G L2 + G L3 ] + [G L G L2 ] = + 50 = +.32 + 76.39K 3 + 98 2 T() = T 76.39K = 3 +5.32 2 +98+76.39K 0 π 0.6K d. The equivalent forward path tranfer function i G() = (+.32) Therefore, T() = 2.55 2 +.32+2.55. The pole are located at -0.66 ± j.454. ω n = 2.55 =.597 rad/; 2ζω n =.32, therefore, ζ = 0.43.

28 Chapter 5: Reduction of Multiple Subytem %OS = e ζπ / ζ 2 4 4 x00 = 24%; T = = ζω n 0.66 = 6.06 econd; T π π p = ω n -ζ 2 =.454 = 2.6 econd; Uing Figure 4.6, the normalized rie time i.486. Dividing by the natural frequency, T r =.486 = 0.93 econd. 2.55 e. f. Since G() = %OS - ln ( 0.5K (+.32), T() = 0.5K 2 +.32+0.5K. Alo, ζ = 00 ) π 2 + ln 2 ( %OS = 0.57 for 5% 00 ) overhoot; ω n = 0.5K ; and 2ζω n =.32. Therefore, ω n =.32 2ζ =.32 2(0.547) Solving for K, K=3.2. =.277 = 0.5K. UFSS Vehicle: Pitch-Angle Control Repreentation a. Ue the oberver canonical form for the vehicle dynamic o that the output yaw rate i a tate variable. u - 2-0.25 0.437 x 4 x 3 x y -2 -.483-0.24897 - b. Uing the ignal flow graph to write the tate equation:

Solution to Cae Studie Challenge 29 x Ý = x Ý 2 =.483 + x 3 0.25x 4 x Ý 3 = 0.24897 (0.25* 0.437)x 4 x Ý 4 = 2x +2 2x 4 2u In vector-matrix form: 0 0 0 0 0.483 0.25 0 x Ý = x + u 0 0.24897 0 0.054625 0 2 2 0 2 2 y = [ 0 0 0]x c. Program: numg=-0.25*[ 0.437]; deng=poly([-2 -.29-0.93 0]); 'G()' G=tf(numg,deng) numh=[- 0]; denh=[0 ]; 'H()' H=tf(numh,denh) 'Ge()' Ge=feedback(G,H) 'T()' T=feedback(-*Ge,) [numt,dent]=tfdata(t,'v'); [Acc,Bcc,Ccc,Dcc]=tf2(numt,dent) Computer repone: an = G() Tranfer function: -0.25-0.093 -------------------------------------- ^4 + 3.483 ^3 + 3.25 ^2 + 0.4979 an = H() Tranfer function: - an = Ge() Tranfer function: -0.25-0.093 -------------------------------------- ^4 + 3.483 ^3 + 3.465 ^2 + 0.6072

30 Chapter 5: Reduction of Multiple Subytem an = T() Tranfer function: 0.25 + 0.093 ----------------------------------------------- ^4 + 3.483 ^3 + 3.465 ^2 + 0.8572 + 0.093 Acc = Bcc = Ccc = Dcc = -3.4830-3.4650-0.8572-0.093.0000 0 0 0 0.0000 0 0 0 0.0000 0 0 0 0 0 0 0 0.2500 0.093 ANSWERS TO REVIEW QUESTIONS. Signal, ytem, umming junction, pickoff point 2. Cacade, parallel, feedback 3. Product of individual tranfer function, um of individual tranfer function, forward gain divided by one plu the product of the forward gain time the feedback gain 4. Equivalent form for moving block acro umming junction and pickoff point 5. A K i varied from 0 to, the ytem goe from overdamped to critically damped to underdamped. When the ytem i underdamped, the ettling time remain contant. 6. Since the real part remain contant and the imaginary part increae, the radial ditance from the origin i increaing. Thu the angle θ i increaing. Since ζ= co θ the damping ratio i decreaing. 7. Node (ignal), branche (ytem) 8. Signal flowing into a node are added together. Signal flowing out of a node are the um of ignal flowing into a node. 9. One 0. Phae-variable form, cacaded form, parallel form, Jordan canonical form, oberver canonical form. The Jordan canonical form and the parallel form reult from a partial fraction expanion. 2. Parallel form

Anwer to Review Quetion 3 3. The ytem pole, or eigenvalue 4. The ytem pole including all repetition of the repeated root 5. Solution of the tate variable are achieved through decoupled equation. i.e. the equation are olvable individually and not imultaneouly. 6. State variable can be identified with phyical parameter; eae of olution of ome repreentation 7. Sytem with zero 8. State-vector tranformation are the tranformation of the tate vector from one bai ytem to another. i.e. the ame vector repreented in another bai. 9. A vector which under a matrix tranformation i collinear with the original. In other word, the length of the vector ha changed, but not it angle. 20. An eigenvalue i that multiple of the original vector that i the tranformed vector. 2. Reulting ytem matrix i diagonal. SOLUTIONS TO PROBLEMS. a. Combine the inner feedback and the parallel pair. Multiply the block in the forward path and apply the feedback formula to get, 50(-2) T() = 3 + 2 +50-00. b. Program: 'G()' G=tf(,[ 0 0]) 'G2()' G2=tf(50,[ ]) 'G3()' G3=tf(2,[ 0]) 'G4()' G4=tf([ 0],) 'G5()' G5=2 'Ge()=G2()/(+G2()G3())' Ge=G2/(+G2*G3) 'Ge2()=G4()-G5()' Ge2=G4-G5 'Ge3()=G()Ge()Ge2()' Ge3=G*Ge*Ge2

32 Chapter 5: Reduction of Multiple Subytem 'T()=Ge3()/(+Ge3())' T=feedback(Ge3,); T=minreal(T) Computer repone: an = G() Tranfer function: --- ^2 an = G2() Tranfer function: 50 ----- + an = G3() Tranfer function: 2 - an = G4() Tranfer function: an = G5() G5 = an = 2 Ge()=G2()/(+G2()G3()) Tranfer function: 50 ^2 + 50 ------------------------- ^3 + 2 ^2 + 0 + 00 an = Ge2()=G4()-G5() Tranfer function: - 2

Solution to Problem 33 an = Ge3()=G()Ge()Ge2() Tranfer function: 50 ^3-50 ^2-00 ------------------------------- ^5 + 2 ^4 + 0 ^3 + 00 ^2 an = T()=Ge3()/(+Ge3()) Tranfer function: 50-00 ----------------------- ^3 + ^2 + 50-00 2. Puh G () to the left pat the pickoff point. H + - G + G 2 G 3 G + G Thu, T() = + G H G + 2 G G = ( G G + 2 )G 3 3 ( + G H ) 3. a. Split G 3 and combine with G 2 and G 4. Alo ue feedback formula on G 6 loop.

34 Chapter 5: Reduction of Multiple Subytem Puh G 2 +G 3 to the left pat the pickoff point. Uing the feedback formula and combining parallel block, Multiplying the block of the forward path and applying the feedback formula,

Solution to Problem 35 4. Puh G 2 () to the left pat the umming junction. Collape the umming junction and add the parallel tranfer function. Puh G ()G 2 () + G 5 () to the right pat the umming junction.

36 Chapter 5: Reduction of Multiple Subytem Collape umming junction and add feedback path. Applying the feedback formula, G3() + G() G2() T() = G2() G4() + [ G3() + G() G2()] H + G 3() G () G 2() + G () + G () G () HG [ () G() G()] G() G() 3 2 = + 3 + 2 + 2 4 5. a. Puh G 7 to the left pat the pickoff point. Add the parallel block, G 3 +G 4.

Solution to Problem 37 Puh G 3 +G 4 to the right pat the umming junction. Collape the minor loop feedback.

38 Chapter 5: Reduction of Multiple Subytem Puh G 7(G 3 +G 4 ) +G 6 G 7 to the left pat the pickoff point. Puh G to the right pat the umming junction. Add the parallel feedback path to get the ingle negative feedback, H() = G 5 G 7 + G 2(+G 6 G 7 ) G 7 (G 3 +G 4 ) - G 8 G. Thu,

Solution to Problem 39 T() = b. Program: G=tf([0 ],[ 7]); %G=/+7 input tranducer G2=tf([0 0 ],[ 2 3]); %G2=/^2+2+3 G3=tf([0 ],[ 4]); %G3=/+4 G4=tf([0 ],[ 0]); %G4=/ G5=tf([0 5],[ 7]); %G5=5/+7 G6=tf([0 0 ],[ 5 0]); %G6=/^2+5+0 G7=tf([0 3],[ 2]); %G7=3/+2 G8=tf([0 ],[ 6]); %G8=/+6 G9=tf([],[]); %Add G9= tranducer at the input T=append(G,G2,G3,G4,G5,G6,G7,G8,G9); Q=[ -2-5 9 2 8 0 3 8 0 4 8 0 5 3 4-6 6 7 0 0 7 3 4-6 8 7 0 0]; input=9; output=7; T=connect(T,Q,input,output); T=tf(T)

40 Chapter 5: Reduction of Multiple Subytem Computer repone: Tranfer function: 6 ^7 + 32 ^6 + 76 ^5 + 5640 ^4 +.624e004 ^3 + 2.857e004 ^2 + 2.988e004 +.52e004 ----------------------------------------------------------- ^0 + 33 ^9 + 466 ^8 + 3720 ^7 +.867e004 ^6 + 6.82e004 ^5 +.369e005 ^4 +.98e005 ^3 +.729e005 ^2 + 6.737e004 -.044e004 6. Combine G 6 and G 7 yielding G 6 G 7. Add G 4 and obtain the following diagram: Next combine G 3 and G 4 +G 6 G 7. Puh G 5 to the left pat the pickoff point.

Solution to Problem 4 Notice that the feedback i in parallel form. Thu the equivalent feedback, H eq () = G 2 G 5 + G 3 (G 4 +G 6 G 7 ) + G 8. Since the forward path tranfer function i G() = G eq () = G G 5, the cloedloop tranfer function i Hence, G eq () T() = +G eq ()H eq (). 7. Puh 2 to the right pat the pickoff point.

42 Chapter 5: Reduction of Multiple Subytem Combine umming junction. Combine parallel 2 and. Apply feedback formula to unity feedback with G() =. Combine cacade pair and add feedback around /(+).

Solution to Problem 43 Combine parallel pair and feedback in forward path. Combine cacade pair and apply final feedback formula yielding T() = 52 + 2 6 2 + 9 + 6. 8. Puh G 3 to the left pat the pickoff point. Puh G 6 to the left pat the pickoff point.

44 Chapter 5: Reduction of Multiple Subytem Hence, Thu the tranfer function i the product of the function, or θ 22 () θ () G G 2 G 4 G 5 G 6 G 7 = - G 4 G 5 + G 4 G 5 G 6 + G G 2 G 3 - G G 2 G 3 G 4 G 5 + G G 2 G 3 G 4 G 5 G 6 9. Combine the feedback with G 6 and combine the parallel G 2 and G 3. Move G 2 +G 3 to the left pat the pickoff point.

Solution to Problem 45 Combine feedback and parallel pair in the forward path yielding an equivalent forward-path tranfer function of G e () = G 2 +G 3 +G (G 2 +G 3 ) G 5 + G 4 G 2 +G 3 G 6 +G 6 G e () But, T() = +G e ()G 7 (). Thu, 0. Puh G3() to the left pat the pickoff point.

46 Chapter 5: Reduction of Multiple Subytem Puh G2()G3() to the left pat the pickoff point. Puh G() to the right pat the umming junction.

Solution to Problem 47 Collaping the umming junction and adding the feedback tranfer function, where G T() = ()G 2 ()G 3 () + G ()G 2 ()G 3 ()H eq (). T() = H eq () = H 3 () G 3 () + H () G 2 ()G 3 () + H 2 () G ()G 3 () + H 4 () G () + 225 2 +2 + 225. Therefore, 2ζω n = 2, and ω n = 5. Hence, ζ = 0.4. %OS = e ζπ / ζ 2 x00 = 6.3%; T = 4 π =0.667; T ζω p = n ω n -ζ 2 =0.229. 2. 3. C() = A = 5 ( 2 + 3 + 5) = A + B + C 2 + 3 + 5 5 = 2 + 3 + 5 + B 2 + C B =, C = 3 C() = + 3 2 + 3 + 5 = + 3 ( +.5) 2 + 2.75 = ( +.5) +.5 ( +.5) 2 + 2.75 = ( +.5) +.5 2.75 2.75 ( +.5) 2 + 2.75 c(t) = e.5t (co 2.75t +.5 in 2.75t) 2.75 Puh 2 to the left pat the pickoff point and combine the parallel combination of 2 and /.

48 Chapter 5: Reduction of Multiple Subytem Puh (2+)/ to the right pat the umming junction and combine umming junction. 4. 2(2 +) Hence, T() = + 2(2 +)H eq (), where H eq() = + 2 + + 5 2. K SinceG() = ( + 30), T() = G() + G() = K 2 + 30 + K. Therefore, 2ζω n = 30. Thu, ζ = 5/ω n = 0.456 (i.e. 20% overhoot). Hence, ω n = 32.89 = K. Therefore K = 082. 5. T() = K 2 + α + K ; ζ = 55.89. K = ω n 2 = 324. α = 2ζω n = 40. %OS ln( 00 ) π 2 + ln 2 ( %OS 00 ) = 0.358 ;T = 4 ζω n = 0.2. Therefore, ω n = 6. 7. 0K The equivalent forward-path tranfer function i G() = [ + (0 K 2 + 2)]. Hence, T() = G() + G() = 0K 2. Since + (0 K 2 + 2) + 0K T = 4 Re = 2, Re = 2; and T p = π =, Im = π. The pole are thu at 2+jπ. Hence, Im ω n = 2 2 +π 2 = 0K. Thu, K =.387. Alo, (0K 2 + 2)/2 = Re = 2. Hence, K 2 = /5. a. For the inner loop, G e () = 20 ( +2), and H e() = 0.2. Therefore, T e () = G e () + G e ()H e () 20 (+6). Combining with the equivalent tranfer function of the parallel pair, G p() = 20, the ytem =

Solution to Problem 49 400 i reduced to an equivalent unity feedback ytem with G() = G p () T e () = (+6) G() +G() = 400 2 +6+400.. Hence, T() = b. ω 2 n = 400; 2ζω n = 6. Therefore, ω n = 20, and ζ = 0.4. %OS = e ζπ / ζ 2 x00 = 25.38; 4 π T = =0.5; T ζω p = n ω n -ζ 2 =0.7. From Figure 4.6, ω nt r =.463. Hence, T r = 0.0732. 8. 9. ω d = Im = ω n - ζ 2 = 8.33. 28900 T() = 2 + 200 + 28900 ; from which, 2ζω n = 200 and ω n = 28900 = 70. Hence, π ζ = 0.588. %OS = e -ζπ/ -ζ 2 x00 = 0.8%; T p = = 0.0229. 2 ω n ζ Alo,T = 4 = 0.04. ζω n For the generator, E g () = K f I f (). But, I f () = E i() R f +L f. Therefore, E g() E i () = 2 +. For the motor, conider R a = 2 Ω, the um of both reitor. Alo, J e = J a +J L ( 2 )2 = 0.75+ 4 = ; D e = D L ( 2 )2 =. Therefore, θ m () E g () = K t R a J e (+ J e (D e + K tk a R a )) 0.5 = (+.5). But, θ o() θ m () = 2. Thu, θ o () E g () = 0.25 (+.5). Finally, θ o () E i () = E g() E i () θ o () E g () = 0.5 (+)(+.5). 20. For the mechanical ytem, J( N 2 N ) 2 2 θ 2 () = T( N 2 N ). For the potentiometer, E i () = 0 θ 2() 2π θ 2 () = π 5 E R i(). For the network, E o () = E i () R+ C = E i () + RC + RC, or E i () = E o ()., or

50 Chapter 5: Reduction of Multiple Subytem 2. Therefore, θ 2 () = π + 5 E RC o() 5N E o () T() = JπN 2 +. RC. Subtitute into mechanical equation and obtain, The equivalent mechanical ytem i found by reflecting mechanical impedance to the pring. Writing the equation of motion: ( 4 2 + 2 + 5)θ () 5θ 2 () = 4T() 5θ () + ( 2 2 + 5)θ 2 () = 0 Solving for θ 2 (), θ 2 () = ( 4 2 + 2 + 5) 4T() 5 0 ( 4 2 + 2 + 5) 5 5 ( 2 2 + 5) = 20T() 8 4 + 4 3 + 30 2 +0 The angular rotation of the pot i 0.2 that of θ 2, or θ p () T() = 2 4 3 + 2 2 +5 + 5 ( ) For the pot: E p () θ p () = 50 5( 2π ) = 5 π For the electrical network: Uing voltage diviion,

Solution to Problem 5 Subtituting the previouly obtained value, E o () E p () = 200,000 = 0 5 + 200,000 + 2 E o () T() = θ () p E () p T() θ p () E () o E p () = 0 π + 4 3 + 2 2 +5 + 5 2 ( ) 22. a. r 50 x 4 x 2 3 + x 2 2

52 Chapter 5: Reduction of Multiple Subytem b. x 3 x x 3 x 2 x x 5 x 4 x 5 x 4 - G - r x 5 G 2 G x 5 x G 3 2 6 x G 3 x 4 G 4 - G 7

Solution to Problem 53 c. x 5 x 4 x 3 x G 8 - G 6 r x 5 G x x 4 G 3 x G x 3 2 7 - G 2 G 4 - G 5 23. a. x x 3 = = x 3 = 2x 4 6x 3 + r y = x +

54 Chapter 5: Reduction of Multiple Subytem r x 3 x y -6-4 -2 b. x x 3 = = 3 + x 3 + r = 3x 4 5x 3 + r y = x + 2 c. x x 3 = 7x + + r = 3x + 2 x 3 + 2r = x + 2x 3 + r y = x + 3 + 2x 3

Solution to Problem 55 2-3 24. a. Since G() = 0 3 + 24 2 +9 + 504 = C() R(), c + 24 c + 9c + 504c = 0r Let, c = x c = c = x3 Therefore, x x 3 = = x 3 = 504x 9 24x 3 +0r y = x r x 0 3 x y -24-9 -504

56 Chapter 5: Reduction of Multiple Subytem r 0 b. G() = ( + 7 ) ( +8 ) ( + 9 ) x x 0 3 2 x y -7-8 -9 Therefore, x x 3 = 9x + = 8 + x 3 = 7x 3 +0r y = x 25. a. Since G() = 20 4 +5 3 + 66 2 + 80 = C() R(), c +5 c + 66 c +80 c = 20r Let, c = x c = c = x3 c = x4 Therefore, x x 3 x 4 = = x 3 = x 4 = 80 66x 3 5x 4 + 20r y = x

Solution to Problem 57 r 20 x 4 x 3 x y -5-66 -80 b. G() = ( 20 ) ( + 2 ) ( + 5 ) ( ). Hence, +8 x r 20 x x x 4 3 2 y -2-5 -8 From which, x x 3 x 4 = 8x + = 5 + x 3 = 2x 3 + x 4 = 20r 26. y = x = + [G 2 G 3 G 4 + G 3 G 4 + G 4 + ] + [G 3 G 4 + G 4 ]; T = G G 2 G 3 G 4 ; =. Therefore, T() = T G G 2 G 3 G 4 = 2 + G 2 G 3 G 4 + 2G 3 G 4 + 2G 4 27. Cloed-loop gain: G 2 G 4 G 6 G 7 H 3 ; G 2 G 5 G 6 G 7 H 3; G 3 G 4 G 6 G 7 H 3; G 3 G 5 G 6 G 7 H 3; G 6 H ; G 7 H 2 Forward-path gain: T = G G 2 G 4 G 6 G 7 ; T 2 = G G 2 G 5 G 6 G 7; T 3 = G G 3 G 4 G 6 G 7; T 4 = G G 3 G 5 G 6 G 7 Nontouching loop 2 at a time: G 6 H G 7 H 2 = - [H 3 G 6 G 7 (G 2 G 4 + G 2 G 5 + G 3 G 4 + G 3 G 5 ) + G 6 H + G 7 H 2 ] + [G 6 H G 7 H 2 ] = 2 = 3 = 4 =

58 Chapter 5: Reduction of Multiple Subytem 28. T() = T + T 2 2 + T 3 3 + T 4 4 = G G 2 G 4 G 6 G 7 + G G 2 G 5 G 6 G 7 + G G 3 G 4 G 6 G 7 + G G 3 G 5 G 6 G 7 - H 3 G 6 G 7 (G 2 G 4 + G 2 G 5 + G 3 G 4 + G 3 G 5 ) - G 6 H - G 7 H 2 + G 6 H G 7 H 2 Cloed-loop gain: - 2 ; - ; - ; -2 Forward-path gain: T = ; T 2 = 2 Nontouching loop: None = - (- 2 - - - 2 ) = 2 = G() = T + T 2 2 + 2 = + ( 2 + + + 2 ) = 3 + 2 4 + 2 +2 29. T() = - G G 2 G 3 G 4 G 5 (-G 2 H )(-G 4 H 2 ) G 2 G 3 G 4 G 5 G 6 G 7 G 8 (-G 2 H )(-G 4 H 2 )(-G 7 H 4 ) = G G 2 G 3 G 4 G 5 (-G 7 H 4 ) -G 2 H -G 4 H 2 +G 2 G 4 H H 2 -G 7 H 4 +G 2 G 7 H H 4 +G 4 G 7 H 2 H 4 -G 2 G 4 G 7 H H 2 H 4 -G 2 G 3 G 4 G 5 G 6 G 7 G 8 30. a. G() = ( +)( + 2) ( + 3) 2 ( + 4) = 2 ( + 3) 2 5 + 3 + 6 + 4

Solution to Problem 59 Writing the tate and output equation, ẋ = -3x + ẋ 2 = -3 + r ẋ 3 = -4x 3 + r In vector-matrix form, y = 2x - 5 + 6x 3 b. G() = G() = 3 0 0 x = 0 3 0 x + r 0 0 4 [ ] y = 2 5 6 ( + 2) = 3/ 4 + ( + 5) 2 ( + 7) 2 ( + 5) 2 + 5 5 / 4 ( + 7) 2 + 7 x r -5-5 x 4 x 3-3 4-5 4 y -7-7 - Writing the tate and output equation, ẋ = -5x + ẋ 2 = -5 + r ẋ 3 = -7x 3 + x 4 ẋ 4 = -7x 4 + r In vector matrix form, y = - 3 4 x + - 5 4 x 3 - x 4

60 Chapter 5: Reduction of Multiple Subytem ẋ = -5 0 0 0-5 0 0 0 0-7 0 0 0-7 x + 0 0 r y = - 3 4-5 4 - x c. Writing the tate and output equation, In vector-matrix form, ẋ = - 2x + ẋ 2 = - 2 + r ẋ 3 = - 4x 3 + r ẋ 4 = - 5x 4 + r y = 6 x + 36-4 x 3 + 2 9 x 4

Solution to Problem 6 3. a. Writing the tate equation, ẋ = ẋ 2 = - 7x - 2 + r In vector matrix form, y = 3x + b.

62 Chapter 5: Reduction of Multiple Subytem Writing the tate equation, x x 3 = = x 3 = x 2 5x 3 + r y = 6x + 2 + x 3 In vector matrix form, c. 0 0 0 x = 0 0 X + 0 r 2 5 y = [ 6 2 ]x ẋ = ẋ 2 = x 3 ẋ 3 = x 4 ẋ 4 = - 4x - 6-5x 3-3x 4 + r y = x + 7 + 2x 3 + x 4 In vector matrix form,

Solution to Problem 63 32. a. Controller canonical form: From the phae-variable form in Problem 5.3(a), revere the order of the tate variable and obtain, ẋ 2 = x ẋ = - 7-2x + r y = 3 + x Putting the equation in order, ẋ = - 2x - 7 + r ẋ 2 = x y = x + 3 In vector-matrix form, x 2 7 = 0 x + 0 r y = [ 3]x Oberver canonical form: +3 G() = 2 +2+7. Divide each term by Cro multiplying, Thu, Drawing the ignal-flow graph, G( ) = 2 and get + 3 2 + 2 + 7 2 = C( ) R( ) ( + 3 2 ) R() = ( + 2 + 7 2 ) C() (R() - 2C()) + 2 (3R() - 7C()) = C()

64 Chapter 5: Reduction of Multiple Subytem R() 3-2 Writing the tate and output equation, -7 ẋ = - 2x + + r ẋ 2 = - 7x + 3r In vector matrix form, y = x x = 2 7 0 x + 3 r y = [ 0]x b. Controller canonical form: From the phae-variable form in Problem 5.3(b), revere the order of the tate variable and obtain, x 3 x = = x = x 3 2 5x Putting the equation in order, In vector-matrix form, y = 6x 3 + 2 + x x x 3 = 5x 2 x 3 = x = y = x + 2 + 6x 3

Solution to Problem 65 5 2 x = 0 0 x + 0 r 0 0 0 y = [ 2 6]x Oberver canonical form: G() = 2 + 2 + 6 3 + 5 2 + 2 +. Divide each term by 3 and get Cro-multiplying, Thu, G() = + 2 + 6 2 3 + 5 + 2 + = C() R() 2 3 + 2 + 6 2 3 R() = + 5 + 2 + 2 3 C() (R() 5c()) + 2 (2R() 2C()) + 3 (6R() C()) = C() Drawing the ignal-flow graph, 2 R() 6 X 3 () X 2 () X () C() -5-2 Writing the tate and output equation, x x 3 - = 5x + + r = 2x + x 3 + 2r = x + 6r y = [ 0 0]x

66 Chapter 5: Reduction of Multiple Subytem In vector-matrix form, 5 0 x = 2 0 x + 2 r 0 0 6 y = [ 0 0]x c. Controller canonical form: From the phae-variable form in Problem 5.3(c), revere the order of the tate variable and obtain, ẋ 4 = x 3 ẋ 3 = ẋ 2 = x ẋ = - 4x 4-6x 3-5 - 3x + r Putting the equation in order, y = x 4 + 7x 3 + 2 + x In vector-matrix form, Oberver canonical form: G() = ẋ = - 3x - 5-6x 3-4x 4 + r ẋ 2 = x ẋ 3 = ẋ 4 = x 3 y = x + 2 +7x 3 + x 4 3 5 6 4 x 0 0 0 0 = X + r 0 0 0 0 0 0 0 0 y = [ 2 7 ]x 3 +2 2 +7+ 4 +3 3 +5 2 +6+4. Divide each term by 2 and get G( ) = + 2 + 7 + 2 3 4 + 3 + 5 2 + 6 3 + 4 4 = C( ) R( )

Solution to Problem 67 Cro multiplying, ( + 2 2 + 7 3 + 4 ) R() = ( + 3 + 5 2 + 6 3 + 4 4 ) C() Thu, (R() - 3C()) + 2 (2R() - 5C()) + 3 (7R() - 6C()) + 4 (R() - 4C()) = C() Drawing the ignal-flow graph, R() 7 2-3 -5-6 -4 Writing the tate and output equation, ẋ = - 3x + + r ẋ 2 = - 5x + x 3 + 2r ẋ 3 = - 6x + x 4 +7r In vector matrix form, ẋ 4 = - 4x + r y = x

68 Chapter 5: Reduction of Multiple Subytem 33. a. r x x 50 3 2 x c= y -2-5 -7 Writing the tate equation, - x x 3 = 7x + = 5 + x 3 = 50x 2x 3 + 50r y = x In vector-matrix form, b. 7 0 0 x = 0 5 x + 0 r 50 0 2 50 y = [ 0 0]x

Solution to Problem 69 r x 3 0 x 2 x c= y -8-25 - Writing the tate equation, x x 3 = = 25x 8 +0x 3 = x + r y = x In vector-matrix form, 0 0 0 x = 25 8 0 x + 0 r 0 0 y = [ 0 0]x c. r 00 - x c= y - Tach feedback before integrator - ẋ = ẋ 2 = - - + 00(r-x ) = -00x -2 +00r y = x

70 Chapter 5: Reduction of Multiple Subytem In vector-matrix form, x = 0-00 -2 x + 0 00 r y = 0 x d. Since (+) 2 = 2 +2+, we draw the ignal-flow a follow: r 0 x x 2 2-2 - - c= y Writing the tate equation, ẋ = ẋ 2 = -x - 2 + 0(r-c) = -x - 2 + 0(r - (2x + ) = -2x - 2 + 0r y = 2x + In vector-matrix form, x = 0-2 -2 x + 0 0 r 34. a. Phae-variable form: y = 2 x 0 T() = 3 +3 2 +2+0 x x x 0 3 2 r= u c= y -3 Writing the tate equation, -2-0

Solution to Problem 7 ẋ = ẋ 2 = x 3 ẋ 3 = -0x -2-3x 3 + 0u In vector-matrix form, x y = x = 0 0 0 0-0 -2-3 x + 0 0 0 u b. Parallel form: y= 0 0 x G() = 5 + -0 + + 5 +2 5 x r=u r=u -0 x 2 c=y - 5 x 3-2 - Writing the tate equation, ẋ = 5(u - x - - x 3 ) = -5x -5-5x 3 +5u ẋ 2 = -0(u - x - - x 3 ) - = 0x + 9 + 0x 3-0u ẋ 3 = 5(u - x - - x 3 ) - 2x 3 = -5x -5-7x 3 +5u In vector-matrix form, y = x + + x 3

72 Chapter 5: Reduction of Multiple Subytem 5 5 5 5 x = 0 9 0 x + 0 u 5 5 7 5 y = [ ]x 35. a. T() = 0( 2 + 5 + 6) 4 +6 3 + 99 2 + 244 +80 Drawing the ignal-flow diagram, r 0 6 x 4 x 3 6 5 x y 99 244 80 Writing the tate and output equation, In vector-matrix form, x x 3 x = = x 3 = x 4 = 80x 244 99x 3 6x 4 + 0r y = 6x + 5 + x 3 0 0 0 0 x 0 0 0 0 = x + r 0 0 0 0 80 244 99 6 0 y = [ 6 5 0]x

Solution to Problem 73 b. G() = 0( + 2)( + 3) ( +)( + 4)( + 5)( + 6) = / 3 + 0 / 3 + 4 + 5 + 5 2 + 6 Drawing the ignal-flow diagram and including the unity-feedback path, r = u x - 3 0 3-4 x 2 y 5-2 -5-6 x x 4 3 - Writing the tate and output equation, x = 3 (u x x 3 x 4 ) x x 3 x 4 = 0 3 (u x x 3 x 4 ) 4 =5(u x x 3 x 4 ) 5x 3 = 2(u x x 3 x 4 ) 2x 4 In vector-matrix form, y = x + + x 3 + x 4

74 Chapter 5: Reduction of Multiple Subytem 36. 4 3 3 3 3 x 0 = 2 3 0 0 x + 0 u 3 3 3 3 5 5 20 5 3 5 2 2 2 0 2 y = [ ]x Program: '(a)' 'G()' G=zpk([-2-3],[- -4-5 -6],0) 'T()' T=feedback(G,,-) [numt,dent]=tfdata(t,'v'); 'Find controller canonical form' [Acc,Bcc,Ccc,Dcc]=tf2(numt,dent) A=flipud(Acc); 'Tranform to phae-variable form' Apv=fliplr(A) Bpv=flipud(Bcc) Cpv=fliplr(Ccc) '(b)' 'G()' G=zpk([-2-3],[- -4-5 -6],0) 'T()' T=feedback(G,,-) [numt,dent]=tfdata(t,'v'); 'Find controller canonical form' [Acc,Bcc,Ccc,Dcc]=tf2(numt,dent) 'Tranform to modal form' [A,B,C,D]=canon(Acc,Bcc,Ccc,Dcc,'modal') Computer repone: an = (a) an = G() Zero/pole/gain: 0 (+2) (+3) ----------------------- (+) (+4) (+5) (+6) an = T() Zero/pole/gain: 0 (+2) (+3) ------------------------------------------ (+.264) (+3.42) (^2 +.32 + 4.73) an =

Solution to Problem 75 Find controller canonical form Acc = -6.0000-99.0000-244.0000-80.0000.0000 0 0 0 0.0000 0 0 0 0.0000 0 Bcc = Ccc = Dcc = an = 0 0 0 0 0 0.0000 50.0000 60.0000 Tranform to phae-variable form Apv = 0.0000 0 0 0 0.0000 0 0 0 0.0000-80.0000-244.0000-99.0000-6.0000 Bpv = Cpv = an = (b) an = G() 0 0 0 60.0000 50.0000 0.0000 0 Zero/pole/gain: 0 (+2) (+3) ----------------------- (+) (+4) (+5) (+6) an = T() Zero/pole/gain:

76 Chapter 5: Reduction of Multiple Subytem 0 (+2) (+3) ------------------------------------------ (+.264) (+3.42) (^2 +.32 + 4.73) an = Find controller canonical form Acc = -6.0000-99.0000-244.0000-80.0000.0000 0 0 0 0.0000 0 0 0 0.0000 0 Bcc = Ccc = Dcc = an = 0 0 0 0 0 0.0000 50.0000 60.0000 Tranform to modal form A = B = -5.668 3.09 0 0-3.09-5.668 0 0 0 0-3.424 0 0 0 0 -.2639-4.08.0468.325 0.0487 C = D = 0.827 0.6973-0.40 4.2067 37. 0

Solution to Problem 77 r x 4 3 x x 2 x - c= y - - Writing the tate equation, ẋ = ẋ 2 = - x + x 3 ẋ 3 = x 4 ẋ 4 = x - + r In vector-matrix form, y = -x + x = 0 0 0-0 0 0 0 0-0 0 x + 0 0 0 r y = c = [- 0 0] x 38. a. θ.. + 5θ. + 6θ - 3θ. 2-4θ 2 = 0 or.... -3θ - 4θ + θ 2 + 5θ 2 + 5θ 2 = T... θ =. - 5θ - 6θ + 3θ 2 + 4θ 2 θ.. 2 = 3θ. + 4θ - 5θ. 2-5θ 2 + T Letting, θ = x ; θ. = ; θ 2 = x 3 ; θ. 2 = x 4,

78 Chapter 5: Reduction of Multiple Subytem x 2 x 4 3-5 -6 T 3-5 x 4 4 x 3-5 where x = θ. b. Uing the ignal-flow diagram, ẋ = ẋ 2 = -6x - 5 + 4x 3 + 3x 4 ẋ 3 = x 4 ẋ 4 = 4x + 3-5x 3-5x 4 + T In vector-matrix form, ẋ = y = x 3 0 0 0-6 -5 4 3 0 0 0 4 3-5 -5 x + 0 0 0 T y = 0 0 0 x 39. Program: numg=7; deng=poly([0-9 -2]); G=tf(numg,deng); T=feedback(G,) [numt,dent]=tfdata(t,'v') [A,B,C,D]=tf2(numt,dent); %Obtain controller canonical form '(a)' %Diplay label A=flipud(A); %Convert to phae-variable form A=fliplr(A) %Convert to phae-variable form B=flipud(B) %Convert to phae-variable form C=fliplr(C) %Convert to phae-variable form

Solution to Problem 79 '(b)' [a,b,c,d]=canon(a,b,c,d) %Diplay label %Convert to parallel form Computer repone: Tranfer function: 7 ------------------------ ^3 + 2 ^2 + 08 + 7 numt = dent = an = (a) A = B = C = an = 0 0 0 7 2 08 7 0 0 0 0-7 -08-2 0 0 7 0 0

80 Chapter 5: Reduction of Multiple Subytem (b) a = -0.0657 0 0 0-2.807 0 0 0-8.7537 b = -0.0095-3.5857 2.5906 c = -6.9849-0.0470-0.0908 d = 0 40. ẋ = A x + B r () y = C x (2) ẋ 2 = A 2 + B 2 y (3) y 2 = C 2 (4) Subtituting Eq. (2) into Eq. (3), ẋ = A x + B r ẋ 2 = B 2 C x + A 2 In vector-matrix notation, y 2 = C 2 x = A B 2 C - O A 2 x + B O r 4. y 2 = O C 2 x ẋ = A x + B r () y = C x (2)

Solution to Problem 8 ẋ 2 = A 2 + B 2 r (3) In vector-matrix form, y 2 = C 2 (4) x = A - O x + B r O A 2 B 2 y = y + y 2 = C C 2 x 42. ẋ = A x + B e () y = C x (2) ẋ 2 = A 2 + B 2 y (3) p = C 2 (4) Subtituting e = r - p into Eq. () and ubtituting Eq. (2) into (3), we obtain, Subtituting Eq. (8) into Eq. (5), ẋ = A x + B (r - p) (5) y = C x (6) ẋ 2 = A 2 + B 2 C x (7) p = C 2 (8) ẋ = A x - B C 2 + B r ẋ 2 = B 2 C x + A 2 In vector-matrix form, x = A y = C x - B C 2 x + B B 2 C A 2 0 r 43. y = C 0 z = P APz + P Bu y = CPz x

82 Chapter 5: Reduction of Multiple Subytem 2 4 0.0606 0.3939 0.22 P = 2 0 ; P = 0.0303-0.3030 0.0606 4 6 2-0.22 0.22 0.0758.6.23 3.8 3 P AP = 3.33.33 2.33 ; P - B = 4 ; CP = -0.544-0.0702 0.92.63.79.26 4 [ ] 44. 45. Eigenvalue are -, -2, and -3 ince, Solving for the eigenvector, Ax = λx or, λι - A = (λ + 3) (λ + 2) (λ + ) For λ = -, = 0, x = x 3. For λ = -2, x = = x 3 2. For λ = -3, x = - 2, = x 3. Thu, ż = P - APz + P - Bu ; y = CPz, where 46. Eigenvalue are, -2, and 3 ince,

Solution to Problem 83 Solving for the eigenvector, Ax = λx or, λi - A = (λ - 3) (λ + 2) (λ - ) For λ =, x = = x 3 2. For λ = -2, x = 2x 3, = -3x 3. For λ = 3, x = x 3, = -2x 3. Thu, ż = P - APz + P - Bu ; y = CPz, where 47. Program: A=[-0-3 7;8.25 6.25 -.75;-7.25-2.25 5.75]; B=[;3;2]; C=[ -2 4]; [P,d]=eig(A); Ad=inv(P)*A*P Bd=inv(P)*B Cd=C*P Computer repone: Ad = -2.0000 0.0000 0.0000-0.0000 3.0000-0.0000 0.0000 0.0000.0000 Bd =.8708-3.6742 3.6742 Cd = 48. 3.207 3.6742 2.8577 a. Combine G () and G 2 (). Then puh K to the right pat the umming junction:

84 Chapter 5: Reduction of Multiple Subytem R() + + - - K K 2 + - G () G () 2 C() 2 K Puh K K 2 to the right pat the umming junction: R() + + + - - - K K 2 G () G () 2 C() 2 K K 2 K Hence, T() = K K 2 G ()G 2 () + K + 2 K K 2 + K K 2 G ()G 2 () b. Rearranging the block diagram to how commanded pitch rate a the input and actual pitch rate a the output:

Solution to Problem 85 K Commanded pitch - rate + - K 2 + - G () G () 2 Actual pitch rate 2 Puhing K 2 to the right pat the umming junction; and puhing to the left pat the pick-off point yield, K Commanded pitch rate - + + - - K 2 G () G () 2 Actual pitch rate K 2 Finding the cloed-loop tranfer function: K 2 G ()G 2 () T() = + K + K 2 + K 2 G ()G 2 () K 2 G ()G 2 () = + G ()G 2 ()( 2 + K 2 + K K 2 )

86 Chapter 5: Reduction of Multiple Subytem c. Rearranging the block diagram to how commanded pitch acceleration a the input and actual pitch acceleration a the output: K K 2 Commanded pitch acceleration - + - - + G () K 2 G () 2 2 Actual pitch acceleration Puhing 2 to the left pat the pick-off point yield, K K 2 2 K 2 Commanded pitch acceleration - + - - + 2 G () G () 2 Actual pitch acceleration Finding the cloed-loop tranfer function: T() = 2 G ()G 2 () + 2 G ()G 2 () + K K 2 2 + K 2 2 G ()G 2 () = + G ()G 2 ()( 2 + K 2 + K K 2 ) 49. K Etablih a inuoidal model for the carrier: T() = 2 +a 2

Solution to Problem 87 r K x -a K 2 Etablih a inuoidal model for the meage: T() = 2 +b 2 r 2 K 2 x 4 x 3 -b Writing the tate equation, ẋ = ẋ 2 = - a 2 x + K r ẋ 3 = x 4 ẋ 4 = - b 2 x 3 + K 2 r 50. 5. y = x x 3 The equivalent forward tranfer function i G() = K K 2 (+a ) i H() = K 3 + K 4 +a 2. Hence, the cloed-loop tranfer function i G() T() = + G()H() a. The equivalent forward tranfer function i. The equivalent feedback tranfer function K K 2 (+a 2 ) = 3 + (a +a 2 ) 2 + (a a 2 +K K 2 K 3 +K K 2 K 4 ) + K K 2 K 3 a 2 5 G e = + 2 K = + 5 + 3 + 2 G T = e = 5 K + G e 4 + 5 3 + 2 + 5 + 5 K 5 K + 3 2 + 2 + 5

88 Chapter 5: Reduction of Multiple Subytem b. Draw the ignal-flow diagram: K 5 u x x x 4 3 2 x y -2-3 - Writing the tate and output equation from the ignal-flow diagram: In vector-matrix form:. x.. x 3. x 4 = = 3 + x 3 = x 4 - = 5Kx 5x 3 2x 4 +5Ku y = x 0 0 0 0 0 3 0 0 ẋ = x + u 0 0 0 0 5K 0 5 2 5K y = [ 0 0 0]x c. Program: for K=::5 numt=5*k; dent=[ 5 5 5*K]; T=tf(numt,dent); hold on; ubplot(2,3,k); tep(t,0:0.0:20) title(['k=',int2tr(k)]) end

Solution to Problem 89 Computer repone: 52. a. Draw the ignal-flow diagram: u 666.67 0.06-720 5x0 6 x 4 x 3-82 -4x0 6 x y -2x0 7 - Write tate and output equation from the ignal-flow diagram:

90 Chapter 5: Reduction of Multiple Subytem In vector-matrix form:. x.. x 3. x 4 = = x 3 = 2*0 7 x 4*0 6 82x 3 +5*0 6 x 4 = 00x 720x 4 +00u y = x 0 0 0 0 0 0 0 0 ẋ = x + u 2*0 7 4*0 6 82 5*0 6 0 00 0 0 720 00 y = [ 0 0 0]x b. Program: numg=666.67*0.06*5e6; deng=conv([ 720],[ 82 4e6 2e7]); 'G()' G=tf(numg,deng) 'T()' T=feedback(G,) tep(t) Computer repone: an = G() Tranfer function:.5e009 ---------------------------------------------------- ^4 + 802 ^3 + 4.059e006 ^2 + 2.9e009 +.44e00 an = T() Tranfer function:.5e009 ---------------------------------------------------- ^4 + 802 ^3 + 4.059e006 ^2 + 2.9e009 +.59e00

Solution to Problem 9 53. a. Phae-variable from: G() = -272( 2 +.9+84) 3 +7. 2 +34.58-23.48 Drawing the ignal-flow diagram:.9-272 u x 3-7. x -34.58 23.48 84 y Writing the tate and output equation:. x.. x 3 = = x 3 = 23.48x 34.58 7.x 3 272u y = 84x +. 9 + x 3 In vector-matrix form:

92 Chapter 5: Reduction of Multiple Subytem 0 0 0 x. = 0 0 x + 0 u 23.48 34.58 7. 272 y = [ 84.9 ]x -272( b. Controller canonical form: G() = 2 +.9+84) 3 +7. 2 +34.58-23.48 Drawing the ignal-flow diagram:.9 u -272-7. x x 2-34.58 x 3 84 y 23.48 Writing the tate and output equation:. x.. x 3 = 7.x 34.58x. 2 +23.48x 3 272u = x = y = x +. 9 + 84x 3 In vector-matrix form: 7. 34.58 23.48 272 x. = 0 0 x + 0 u 0 0 0 y = [.9 84]x c. Oberver canonical form: Divide by highet power of and obtain G() = -272-56.8 2-22848 3 + 7. + 34.58 2-23.48 3 Cro multiplying,

Solution to Problem 93 [ -272-56.8 2-22848 7. 3 ]R() = [ + + 34.58 2-23.48 3 ]C() Rearranging, C() = [ -272R() - 7.C()] + 2 [ -56.8R() - 34.58C()] + 3 [ -22848R() + 23.48C()] Drawing the ignal-flow diagram, where r = u and y = c: -272-56.8 u 22848 x x 3 2 x y -7. -34.58 23.48 d. Draw ignal-flow ignoring the polynomial in the numerator: u -272 x x 3 2 x -4.8-4.9 Write the tate equation:. x.. x 3 = 4.9x. + =.8 + x 3 = 4x 3 272u

94 Chapter 5: Reduction of Multiple Subytem The output equation i But, and.. x = 4.9 x... y = x +. 9 x +84x. +. x. = 4.9x + = 4. 9( 4.9x + ) +.8 + x 3 () (2) (3) Subtituting Eq. (2) and (3) into () yield, In vector-matrix form: y = 98.7x.2 + x 3 4. 9 0 0 x. = 0.8 x + 0 u 0 0 4 272 y = [ 98.7.2 ]x e. Expand a partial fraction: G = 479.38 232.94 + 440.32 + 4.8 + 4.9 Draw ignal-flow diagram: u -479.38-232.94 440.32-4.8 x x 2 y -4.9 x 3 Write tate and output equation:. x.. x 3 = 4x + 479.38u. =.8 232.94u = 4. 9x 3 + 440.32u y = x + + x 3 In vector-matrix form:

Solution to Problem 95 4 0 0 479.38 x. = 0.8 0 x + 232.94 u 0 0 4.9 440.32 y = [ ]x 54. Puh Pitch Gain to the right pat the pickoff point. Collape the umming junction and add the feedback tranfer function. Apply the feedback formula and obtain, 55. G() T() = + G()H() = 0.25( + 0.435) 4 + 3.4586 3 + 3.4569 2 + 0.9693 + 0.5032 Program: numg=-0.25*[ 0.435] deng=conv([.23],[ 0.226 0.069]) 'G' G=tf(numg,deng) 'G2' G2=tf(2,[ 2]) G3=- 'H' H=tf([- 0],) 'Inner Loop' Ge=feedback(G*G2,H) 'Cloed-Loop' T=feedback(G3*Ge,)

96 Chapter 5: Reduction of Multiple Subytem Computer repone: numg = -0.250-0.0544 deng =.0000.4560 0.2949 0.0208 an = G Tranfer function: -0.25-0.05438 ------------------------------------ ^3 +.456 ^2 + 0.2949 + 0.02079 an = G2 Tranfer function: 2 ----- + 2 G3 = - an = H Tranfer function: - an = Inner Loop Tranfer function: -0.25-0.088 ------------------------------------------------ ^4 + 3.456 ^3 + 3.457 ^2 + 0.793 + 0.0457 an = Cloed-Loop

Solution to Problem 97 Tranfer function: 0.25 + 0.088 ----------------------------------------------- ^4 + 3.456 ^3 + 3.457 ^2 + 0.9693 + 0.503 56.

98 Chapter 5: Reduction of Multiple Subytem Linear Deadzone Backlah Linear 57. a. Since V L () = V g () V R (), the umming junction ha V g () a the poitive input and V R () a the negative input, and V L () a the error. Since I() = V L () (/(L)), G() = /(L). Alo, ince V R () = I()R, the feedback i H() = R. Summarizing, the circuit can be modeled a a negative feedback ytem, where G() = /(L), H() = R, input = V g (), output = I(), and error = V L (), where the negative input to the umming junction i V R (). b. T() = I() V g () = G() + G()H() = L + = L R c. Uing circuit analyi, I() = V g () L + R. L + R. Hence, I() = V g () L + R. SOLUTIONS TO DESIGN PROBLEMS 58. J e = J a +J L ( 20 )2 = 2+2 = 4; D e = D a +D L ( 20 )2 = 2+D L ( 20 )2. Therefore, the forward-path tranfer function i,

Anwer to Deign Problem 99 G() = (000) Hence, ζ = which D L = 3560. 4 (+ 4 (D e+2)) - ln ( %OS 00 ) 20. Thu, T() = G +G = π 2 + ln 2 ( %OS 00 ) = 0.456; ω n = 25 2 2 + 4 (D e+2)+ 25 2 25 2 ; 2ζω n = D e+2 4.. Therefore D e = 0.9; from 59. 25 a. T() = 2 ++25 ; from which, 2ζω n = and ω n = 5. Hence, ζ = 0.. Therefore, %OS = e ζπ / ζ 2 x00 = 72.92% ; T = 4 = 8. ζω n 25K b. T() = 2 ; from which, 2ζω +(+25K 2 )+25K n = +25K 2 and ω n = 5 ζ = K. Hence, - ln ( %OS 00 ) 4 π 2 + ln 2 ( %OS = 0.404. Alo, T = = 0.2, Thu, ζω 00 ) ζω n = 20; from which K 2 = 39 n 25 and ω n = 49.5. Hence, K = 98.0. 60. K The equivalent forward path tranfer function i G e () = (+(+K 2 )). Thu, T() = G e () +G e () K 2 +(+K 2 )+K. Prior to tachometer compenation (K 2 = 0), T() = 00. Thu, after tachometer compenation, T() = K 2 ++K. Therefore K = ω n 2 = 00 2 +(+K 2 )+00. Hence, ω n = 0; 2ζω n = +K 2. = Therefore, K 2 = 2ζω n - = 2(0.5)(0) - = 9. 6. At the N 2 haft, with rotation,θ L () (J eq 2 + D eq )θ L () + F()r = T eq () F() = (M 2 + f v )X() Thu, (J eq 2 + D eq )θ L () + (M 2 + f v )X()r = T eq () But, X() = rθ L (). Hence, where [(J eq + Mr 2 ) 2 + (D eq + f v r 2 )]θ L () = T eq ()

200 Chapter 5: Reduction of Multiple Subytem J eq = J a (2) 2 + J = 5 D eq = D a (2) 2 + D = 4 + D r = 2 Thu, the total load inertia and load damping i Reflecting J L and D L to the motor yield, Thu, the motor tranfer function i θ m () E a () J L = J eq + Mr 2 = 5 + 4M D L = D eq + f v r 2 = 4 + D + ()(2) 2 = 8 + D J m = (5 + 4M) (8 + D) ; D 4 m = 4 K t R a J m ( + (D m + K tk a )) J m R a J m = ( + (D m +)) J m The gear are (0/20)() = /2. Thu, the forward-path tranfer function i J G e () = (500) m ( + (D m +)) 2 J m Finding the cloed-loop tranfer function yield, For T = 2, D m + J m Or, ω n = 4.386 = 250 J m J m = (5 + 4M) ; D 4 m = T() = G e () + G e () = 250/ J m 2 + D m + + 250 J m J m = 4. For 20% overhoot, ζ = 0.456. Thu, 2ζω n = 2(0.456)ω n = D m + J m = 4 ; from which J m =3 and hence, D m = 5. But, (8 + D). Thu, M =.75 and D = 96. 4

Solution to Deign Problem 20 62. Deired force a. 00 Input tranducer Input voltage+ - K 000 Fup 0.7883( + 53.85 ) ( 2 + 5.47 + 9283 )( 2 + 8.9 + 376.3) Controller Actuator Pantograph dynamic Y h -Y cat Spring diplacement 82300 Spring F out 00 Senor b. G() = Y h () Y cat () F up () = 0.7883( + 53.85) ( 2 + 5.47 + 9283)( 2 +8.9 + 376.3) 648.7709 (+53.85) G e () = (K/00)*(/000)*G()*82.3e3 = ( 2 + 8.9 + 376.3) (^2 + 5.47 + 9283) 648.7709 (+53.85) T() = Ge/(+Ge) = (^2 + 8.89 + 380.2) ( 2 + 5.4 + 9279) 648.8 + 3.494e04 = 4 + 23.59 3 + 9785 2 + 8.84e04 + 3.528e06 c. For G() = (yh-ycat)/fup Phae-variable form Ap = 0 0 0 0 0 0 0 0 0-3.493e6-890 - 9785-23.59 Bp = 0 0 0 Cp = 42.45 0.7883 0 0 Uing thi reult to draw the ignal-flow diagram,

202 Chapter 5: Reduction of Multiple Subytem 0.7883 K f deired v 0.0 up = 000-23.59 x x 4 x x 3 2 42.45 82300 f out -9785-890 -3.493x0 6 Writing the tate and output equation But, x x 3 x 4 = = x 3 = x 4-0.0 = 23.59x 4 9785x 3 890 3493000x + 0.0f deired 0.0 f out Subtituting f out into the tate equation yield x 4 f out = 42.45* 82300x + 0.7883 *82300 = 3527936.35x 8838.7709 9785x 3 23.59x 4 + 0.0f deired Putting the tate and output equation into vector-matrix form. 0 0 0 0 x 0 0 0 0 = + 0 0 0 0 3.528x0 6 8840 9785 23.59 0.0 y = f out = [ 3494000 64880 0 0]x f deired