Solution Manual - Control Systems by Gopal

CHAPTER 2 DYNAMIC MODELS AND DYNAMIC RESPONSE2.1 Revisiting Examples 2.4 and 2.5 will be helpful. (a) R(s) = A; Y s KA s ( ) + J 1 y(t) = KA e t J J ÷ / ; y ss = 0 (b) R(s) = A s ; Y(s) = KA s s ( ) τ +1 y(t) = KA( ) / 1− − e t τ ; y ss = KA (c) R(s) = A s 2 ; Y(s) = KA s s 2 1 ( ) τ + y(t) = KA( ) / t e t − + − τ τ τ ; y ss = KA( ) t ÷J (d) R(s) = A s M M 2 2 + ; Y(s) = KA s s ω τ ω ( )( ) + + 1 2 2 y(t) = KA e KA t t ωτ τ ω τ ω ω θ τ 2 2 2 2 1 1 + + + + − / sin( ) θ = tan ( ) − − 1 ωτ y t t ( ) ÷~ = KA t τ ω ω θ 2 2 1 + + sin( ) 2.2 (a) The following result follows from Eqns (2.57 - 2.58) y(t) =1 1 2 1 2 1 2 − − + − F H G I K J − − e t n t d ζω ζ ω ζ ζ sin tan ω d = ω ζ n 1 2 − The final value theorem is applicable: the function sY(s) does not have poles on the jM-axis and right half of s-plane. y ss = lim ( ) s sY s ÷ 0 1 (b) The following result follows from Review Examples 2.2. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 3 y(t) = t e t n t d d n − + + − F H G I K J − − 2 2 1 1 2 ζ ω ω ω ζ ζ ζω sin tan y ss = t n − 2ζ ω The final value theorem is not applicable; the function sY(s) has a pole on the jM-axis. 2.3 Revisiting Review Example 2.1 will be helpful. E s E s R RCs i 0 4 10 1 ( ) ( ) / + ; E s s i ( ) 10 2 e t R t e RC t 0 4 10 10 ( ) ( ); / × ÷ + ÷ J J J J For the output to track the input with a steady-state delay 100 × 10 –6 sec, it is necessary that R 10 1 4 , and J × ÷ RC 100 10 6 This gives R = 10 kO; C = 0.1 µF Steady-state error = 10 t - (10 t – 10J ) = 0.001 2.4 Revisiting Review Example 2.2 will be helpful. M M n n 100 3 2 6 100 ; ; / / sec = 60 msec Steady-state error = 25 25 25 2 t t n ÷ ÷ × F H G I K J M = 1.5 2.5 My t By t Ky t ( ) ( ) ( ) + + = F(t) = 1000 (t) Y(s) = 1 10 100 2 s s s ( ) + + ω n = 10, ζ = 0.5, u d = 5 3 Using Eqns (2.57)-(2.58), we obtain http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN y(t) = 0.01 1 2 3 5 3 3 5 1 ÷ + L N M O Q P ÷ ÷ e t t sin( tan 2.6 Y s R s G s s s s s ( ) ( ) ( ) ( )( ) + + + + 6 7 6 6 1 6 2 | ( )| G jω ω =2 = 3 5 2 ; ∠G(j2) = – 81.87º y ss = 3 5 2 2 81 87 sin( . º ) t − = 3 50 sin 2t – 21 50 cos 2t 2.7 Y s R s ( ) ( ) = G(s) = s s s + + + 3 7 10 2 = s s s + + + 3 2 5 ( )( ) (a) R(s) = 1 1 s + y(t) = 1 2 1 3 1 6 2 5 e e e t t t t − − − − − F H I K µ( ) (b) (s 2 + 7s + 10) Y(s) – sy(0) – y(0) – 7y(0) = (s + 3) R(s) – sr(0) Initial conditions before application of the input are y(0 – ) = 1, y(0 – ) = 1 2 , r(0 – ) = 0 Y(s) = s s s s s s s + + + + + + + + 15 2 2 5 3 2 5 1 / ( )( ) ( )( )( ) y(t) = 1 2 3 2 2 5 e e e t t t ÷ ÷ ÷ + ÷ 2.8 X = X x I I i + + ; M d X x dt 2 2 ( ) + = F X x I i Mg ( , ) + + − http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 5 Mx = F X I F X x F I i Mg X I X I ( ) , , + + ∂ ∂ F H G I K J + ∂ ∂ F H G I K J − F X I ( ) + = Mg = × × − 8 4 10 9 8 3 . . Newtons For this value of force, we get from Fig. 2.8b, X I = = 0 0 6 .27 ; . cm amps Again from Fig. P 2.8b, K F X X I 1 0 14 o o , . Newtons/cm K F I X I 2 0 4 o o , . Newtons/amp ; ( ) ( ) . . x K M x K M i X s I s s + ÷ 1 2 2 47 6 16 67 2.9 Mx Bx Kx K y x ( ) + + ÷ 1 Gravitational effect has been eliminated by appropriate choice of the zero position. X s Y s ( ) ( ) = G s s s ( ) . ( . )( . ) = + + 0 1667 0 0033 1 0 0217 1 u = 2 1163 F v / . rad /sec | ( | . G jω ω=11 63 = 0.1615 x(peak) = 7.5 × 0.1615 = 1.2113 cm 2.10 (a) Mx Bx Kx F t ( ) + + Gravitational effect has been eliminated by appropriate choice of the zero position. X s F s ( ) ( ) = 1 2 Ms Bs K + + Force transmitted to the ground = K X(s) + Bs X(s) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN = B K s F s M K s B K s + F H I K + + 1 1 2 ( ) F(t) = A sin Mt Peak amplitude of the force transmitted to the ground at steady-state = A B K M K B K 1 1 2 2 2 2 + F H I K − F H G I K J + F H I K ω ω ω (b) Mx B x y K x y ( ) ( ) + ÷ + ÷ 0 X s Y s ( ) ( ) = Bs K Ms Bs K + + + 2 Peak amplitude of machine vibration = A B K v Mv K Bv K 1 1 2 2 2 2 + F H I K ÷ F H G I K J + F H I K ` 2.11 M y K y y B y y K y y B y y 1 1 1 1 0 1 1 0 2 1 2 2 1 2 ( ) ( ) ( ) ( ) + − + − + − + − = 0 M y K y y B y y 2 2 2 2 1 2 2 1 ( ) ( ) + − + − = 0 Gravitational effect has been eliminated by appropriate choice of the zero position. 2.12 The following result follows from Section 11.2 (Eqn (11.9)). (a) E s E s i 0 ( ) ( ) = 1 2 1 2 2 2 1 2 2 1 2 2 1 2 2 + + + + + RC s R C C s R C C s R C C s ( ) (b) E s E s i 0 ( ) ( ) = 1 2 1 2 1 1 2 2 2 1 2 1 2 2 2 + + + + + R Cs R R C s R R Cs R R C s ( ) 2.13 Refer Example 11.6 2.14 Refer Example 11.7 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 7 2.15 The following result follows from Example 12.9. x x x z x z r F t 1 2 3 4 B B , , , , ( ) x = Ax + br A = 0 1 0 0 4 4537 0 0 0 0 0 0 1 0 0 0 0 . .5809 L N M M M M O Q P P P P ; b = 0 0 3947 0 0 − L N M M M M O Q P P P P . .9211 2.16 Mixing valve obeys the following equations: ( ) [ ( )] ( ) Q q c Q Q q c Q c Q i i H i i C i i + + ÷ + + H G H G H G q i = k v x The perturbation equation is K H C v ( ) G G ÷ x(t) = Q t i G ( ) or, x(t) = K t K Q K i H C G G G ( ); / [ ( )] ÷ v The tank obeys the equations V c d dt H G = Q c t t id id i D H G G G G J ( ); ( ) ( ) ÷ ÷ This gives G G ( ) ( ) s s i = e s s ÷ + 1 5 1 . 2.17 C d dt 1 1 θ = q t R R UA m ( ) ; G G ÷ ÷ 1 2 1 1 1 C d dt 2 2 G = G G G G 1 2 1 2 2 ÷ + ÷ R R i ; C V c R Q c 2 2 1 H H ; From these equations, we get θ 1 = − + + 1 1 4 46 1 2 .92 .92 . θ θ q m θ 2 = 0 078 0 2 0125 1 2 . . . G G G ÷ + i http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 2.18 At steady-state 0 = G G G G 1 2 1 2 2 ÷ + ÷ R R i ; this gives θ 1 120 = º C 0 = Q m λ − θ θ 1 2 1 − R ; this gives Q m 17 26 . / min kg 2.19 Energy balance on process fluid: V c d dt 2 2 2 2 2 ρ θ θ ( ) + = Q c UA i 2 2 2 2 2 2 2 2 1 1 ρ θ θ θ θ θ θ θ [ ( )] [ ] − + − + − − At steady-state 0 = Q c UA i 2 2 2 2 2 2 1 H G G G G ( ) ( ); ÷ ÷ ÷ this gives G 1 40 º C The perturbation equation is V c d dt Q c UA 2 2 2 2 2 2 2 2 2 1 H G H G G G ÷ ÷ ÷ ( ) This gives 544 5 0 2 2 1 . .432 + + d dt G G G Energy balance on the cooling water: V c d dt 1 1 1 1 1 H G G + d i = ( ) [ ( )] Q q c i 1 1 1 1 1 1 1 + ÷ + H G G G + UA[ ] G G G G 2 2 1 1 + ÷ ÷ At steady-state 0 = Q c UA i 1 1 1 1 1 2 1 H G G G G [ ] [ ]; ÷ + ÷ this gives Q 1 = 5.28 ×10 –3 m 3 /sec The perturbation equation is V c d dt 1 1 1 1 H G = ( ) ( ) θ θ ρ ρ θ θ θ i c q Q c UA 1 1 1 1 1 1 1 1 1 2 1 − − + − http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 9 This gives 184.55 d dt G G 1 1 + = 0.465 G 1 – 1318.2 q 1 Manipulation of the perturbation equation gives G G 2 1 ( ) ( ) s s = – 557 6 100 10 729 0 8 3 2 . .5 . × + + s s 2.20 Tank 1: C dp dt 1 1 = q 1 – q 10 – q 11 q 10 = flow through R 0 ; q 11 = flow through R 1 A g g dh dt 1 1 H H × = q gh R g h h R 1 1 0 1 2 1 ÷ ÷ ÷ H H ( ) or dh dt 1 = ÷ + F H G I K J + + H H g A R R h g A R h A q 1 0 1 1 1 1 2 1 1 1 1 1 = – 3h 1 + 2h 2 + r 1 Tank 2: C dp dt 2 2 = q 2 + q 11 – q 20 q 20 = flow through R 2 A g g dh dt 2 2 H H × = H H g h h R gh R q ( ) 1 2 1 2 2 2 ÷ ÷ + or dh dt 2 = H H g A R h g A R R h A q 2 1 1 2 1 2 2 2 2 1 1 1 ÷ + F H G I K J + = 4h 1 – 5h 2 + r 2 2.21 A d dt H h ( ) + = Q q Q q Q q 1 1 2 2 + + + ÷ ÷ At steady-state 0 = Q Q Q 1 2 + ÷ ; this gives Q = 30 litres/sec The perturbation equation is http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN A dh dt = q 1 + q 2 – q The turbulent flow is governed by the relation Q(t) = K H t ( ) = f(H) Linearizing about the operating point, we obtain Q(t) = f H f H H H H H H ( ) ( ) ( ) + o o F H G I K J ÷ = Q K H h t + 2 ( ) Therefore q(t) = K H h t 2 ( ) = K H H h t 2 ( ) = Q H h t 2 ( ) dh t dt ( ) = ÷ + + 1 1 1 1 2 A q t A q t A q t ( ) ( ) ( ) = – 0.01 h(t) + 0.133 q 1 (t) + 0.133 q 2 (t) Mass balance on salt in the tank: A d dt H h t C c t [( ( ) ( ( ))] + + = C Q q t C Q q t 1 1 1 2 2 2 [ ( )] [ ( )] + + + − + + [ ( )][ ( )] C c t Q q t At steady state, 0 = C Q C Q CQ 1 1 2 2 + ÷ ; this gives C = 15 The perturbation equation is AH dc t dt AC dh t dt ( ) ( ) + = C q t C q t Cq t Qc t 1 1 2 2 ( ) ( ) ( ) ( ) + ÷ ÷ This gives dc t dt ( ) = – 0.02 c(t) – 0.004 q 1 (t) + 0.002 q 2 (t) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 1 2.22 e D s −τ = 1 2 3! 4 5 2 2 3 3 4 4 5 5 ÷ + ÷ + ÷ + J J J J J D D D D D s s s s s ! ! ! ... It is easy to calculate with long division that 1 2 1 2 ÷ + J J D D s s / / = 1 2 4 2 2 3 3 ÷ + ÷ + J J J D D D s s s ... 1 2 12 1 2 12 2 2 2 2 ÷ + + + J J J J D D D D s s s s / / / / = 1 2 6 2 2 3 3 ÷ + ÷ J J J D D D s s s + + ÷ J J D D s s 4 4 5 5 12 144 ... http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 3 MODELS OF INDUSTRIAL CONTROL DEVICES AND SYSTEMS 3.1 R G 2 G 1 Y G +G 3 4 H 2 H G 1 2 G G + G 2 3 4 + + + – – – Y s R s ( ) ( ) = G G G G G G G G H G H G 1 2 3 4 2 3 4 1 2 1 1 2 1 ( ) ( )( ) 3.2 R Y G 1 G 2 G 4 + + + + – – H 2 G 3 (1– )H G 1 1 Y s R s ( ) ( ) = G G G G G G H G H G 4 1 2 3 2 3 2 2 1 1 1 1 ( ) 3.3 Y s R s ( ) ( ) = ( ) / P P P P 1 1 2 2 3 3 4 4 , , , , , P 1 = G 1 G 2 ; P 2 = G 1 G 3 ; P 3 = G 4 H 2 G 1 G 2 ; P 4 = G 4 H 2 G 1 G 3 ∆ = 1 – (– G 1 G 3 H 1 H 2 – G 1 G 2 H 1 H 2 ) ∆ 1 = ∆ 2 = ∆ 3 = ∆ 4 = 1 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 3 Y s R s ( ) ( ) = ( )( ) ( ) 1 1 4 2 2 3 1 1 1 2 2 3 G H G G G G H H G G R 1 1 Y G 1 G 2 G 4 –H 2 – H 1 G 3 3.4 R G 2 G 1 Y G 3 1+G H 3 2 H /G 3 3 H 1 + + – – Y s R s W ( ) ( ) 0 = M(s) = G G G G H G H G G G H 1 2 3 2 3 3 2 1 2 3 1 1+ + + W G 3 Y H G 2 3 –(G 1 1 2 2 H G +H ) + + + – Y s W s R ( ) ( ) 0 = M W (s) = G H G H G G G H G H 3 3 2 3 2 3 1 1 2 2 1 1 ( ) ( ) + + + + http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 3.5 Y 1 Y 2 G 1 G 2 G 3 G 4 G 5 G 6 1 –1 –1 1 1 1 1 R 1 R 2 H 2 H 1 Y R R 1 1 0 2 = P 1 1 , , = G G G G G G G G H G G H G G G 1 2 3 4 1 2 4 1 2 4 5 1 1 2 4 1 1 ( ) ( ) Y R R 1 2 0 1 = = G G H GG G 4 5 1 1 2 3 , Y R R 2 1 0 2 = = G G G G H 1 4 5 6 2 , ; Y R R 2 2 0 1 = G G G G G 4 5 6 1 2 1 ( ) , 3.6 + + – – e r e t K A K g i f i a R a K b Js K T 1 1 w K t M( ) ( ) s E s r = 50 10 375 s . For E r (s) = 100/s, http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 5 M(s) = 100 50 10 375 s s ( . ) M( ) t = 481.8 (1 – e –10.375 t ) 3.7 + – q R q M q L K P K A K T sL + R f f Js + Bs 2 1 1 1 50 G G L R s s ( ) ( ) = 1 01 1 0 2 1 1 s s s ( . )( . ) 3.8 T M = K 1 e c + K M 2 θ = J B eq M eq M G G K 2 = slope of the characteristic lines shown in Fig. P3.8b = 3 0 0 300 = – 0.01 K 1 = ∆ ∆ T e M c M θ =0 = 3 2 30 20 = 0.1 J eq = J M + N N N N J L 1 3 2 4 2 F H G I K J = 0.0032 B eq = N N N N B L 1 3 2 4 2 F H G I K J = 0.00001 G M c s E s ( ) ( ) = 10 0 32 1 s s ( . ) ; G L c s E s ( ) = 1 0 32 1 s s ( . ) 3.9 (i) J eq = J M + J L G G L M F H G I K J 2 = 0.6 B eq = B M + B L G G L M F H G I K J 2 = 0.015 K T = K b http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN G M a s E s ( ) ( ) = K s J s B sL R K K T eq eq a a T b [ )( ) ] = 16 7 100 19 2 2 . ( . ) s s s (ii) A sketch of multi-loop configuration may be made on the lines of Example 3.5. (iii) It will behave as a speed control system. (iv) Relative stability and speed of response may become unsatisfactory. 3.10 J = J M = n 2 J L = 1.5 × 10 –5 ; n = 1 B = B M + n 2 B L = 1 × 10 –5 θ θ M R s s ( ) ( ) = 13 33 3 13 33 2 . .5 . s s + + + + + – – – K A K 1 K 2 1 1 K s e c Js +B T M q R q M K t s q M 3.11 + + – – K A K g K b K T Js + B 1 R a 1 1 K s sL +R f f q R q M s q M θ θ M R s s ( ) ( ) = θ θ L R s s ( ) ( ) = 15 10 54 200 15 10 3 3 2 3 × + + + × s s s 3.12 (a) ω( ) ( ) s E s r = K s s n n 1 2 1 2 2 ω ζ ω + + ; K = 17.2, = 1.066, M n = 61.2 (b) Modification in the drive system may be made on the lines of Fig. 3.64. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 7 + + – – K T Js + B 1 sL +R a a e r w K A K r K b K t 3.13 + + – – K P K A K T G s ( ) N s ( ) q L T W T M q R T M = J M ( ) ( ) G G G G G M M L M L B K B K M L M L ( ) ( ) G G G G = J T L L W G (J M s 2 + Bs + K) θ M (s) – (Bs + K) θ L (s) = T M (s) (Bs + K) θ M (s) – (J L s 2 + Bs + K) θ L (s) = T W (s) G L s ( ) = G(s) T M (s) – N(s) T W (s) G G L R s s ( ) ( ) = K K K G s K K K G s P A T P A T ( ) ( ) 1 G L W s T s ( ) ( ) = N s K K K G s P A T ( ) ( ) 1 G(s) = Bs K s J J s J J Bs J J K M L M L M L 2 2 [ ( ) ( ) ] http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN N(s) = J s Bs K s J J s J J Bs J J K M M L M L M L 2 2 2 [ ( ) ( ) ] Revisiting Example 2.7 will be helpful. 3.14 The required block diagram easily follows from Fig. 3.49. K x Ay 1 = K p 2 , ( ) K p A K 2 2 , = My (r – y) K p K A K = x Y s R s ( ) ( ) = 2 0 02 2 2 s s . 3.15 Y s Y s r ( ) ( ) = K K K AK K Ms B A K s K K K K A K A r A v v 1 2 2 2 2 1 2 / ( / ) / s y y Ms + B + + + – – – K r e r e 0 e a x y r K A K s K 1 K 2 F w A A 1 s 1 K v 3.16 Revisiting Example 3.7 will be helpful. From geometry of the linkage, we get E(s) = b a b X s a a b Y s ( ) ( ) Y s Y s r ( ) ( ) = bK AK a b K Ms B A K s K aAK a b K r 1 2 2 2 2 1 2 / ( ) ( / ) / ( ) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 9 y Ms + Bs + K 2 + + – – K r y r x e K 1 K 2 a + b a a + b A b sA 1 3.17 Let z(t) = displacement of the power piston Y(s) = e s D J Z(s) ; J D = d v sec Y s E s r ( ) ( ) = G s G s H s ( ) ( ) ( ) 1 G(s) = ( / ) ( / ) K K K A K e s Ms B A K A s D v 1 2 2 2 J ; H(s) = K s z + + + – – – F w K 2 s Ms + B A A y e –st D 1 z x i 1 K 1 K s K A e r e 0 K v 3.18 A dh dt 1 1 = q q gh R w 1 1 1 H A dh dt 2 2 = H H gh R gh R 1 1 2 2 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 2 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN + + – – R / g 2 r q 1 h r h 2 Q s w ( ) H s 2 ( ) Q s 1 ( ) (t s + 1) 1 t 1 t 2 ; = = (t s + 1) 2 e r e 0 K s K a rg rg A R 1 1 A R 2 2 K e K v K s 3.19 C d dt t G = h R a t G G + + + – q a q q 1/R t R t h e r ts + 1 t = R C t t R f R 2 K t – R 1 K h h e A R –1 3.20 Heater: C d dt 1 1 G = h R G G 1 1 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 2 1 Air: V c d dt H G = G G H G G 1 1 R Q c i ( ) These equations give us J = R 1 C 1 , K g = 1 Q c H , M n Q R C V 1 1 2M n = R C Q c V c C VR C c 1 1 1 1 1 H H H – + + + q i q r q q h ts + 1 1 + 1 s 2z K g s 2 w n 2 w n K g K s K A e A K h h K s e r e 0 3.21 Process: V c d dt 2 2 2 2 2 H G G ( ) = Q c UA i i 2 2 2 2 2 2 2 1 1 H G G G G G G ( ) ( ) – Q c 2 2 2 2 2 H G G ( ) Linearized equation is V c d dt 2 2 2 2 H G = Q c UA Q c i 2 2 2 2 2 1 2 2 2 2 H G G G H G ( ) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 2 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN Rearranging this equation and taking Laplace transform yields ( ) ( ) J G 2 2 1 s s = ′ + ′ K s K s i 1 1 2 2 θ θ ( ) ( ) J 2 = 544 0 423 0 1 2 .5; . ; .577 ′ = ′ = K K Cooling water: V c d dt 1 1 1 1 1 H G G ( ) = ( ) ( ) ( Q q c UA i i 1 1 1 1 1 1 2 2 H G G G G G G H G G 1 1 1 1 1 1 1 1 ) ( ) ( ) Q q c Linearized equation is V c d dt 1 1 1 1 H G = Q c c q UA i i 1 1 1 1 1 1 1 1 2 1 H G G H G G ( ) − − Q c c q 1 1 1 1 1 1 1 1 ρ θ θ ρ From this equation, we obtain ( ) ( ) J G 1 1 1 s s = − ′ + ′ + ′ K Q s K s K s i 3 1 4 2 5 1 ( ) ( ) ( ) θ θ J 1 = 182 1000 4184 3550 5 4 1000 4184 1 . . Q ′ K 3 = ( ) . 27 1000 4184 3550 5 4 1000 4184 1 1 G Q ′ K 4 = 3550 5 4 3550 5 4 1000 4184 1 . . Q ′ K 5 = Q Q 1 1 1000 4184 3550 5 4 1000 4184 . G 1 and Q 1 are obtained from the following energy-balance equations at steady state: Q c UA i 1 1 1 1 1 2 1 H G G G G ( ) ( ) = 0 Q c UA i 2 2 2 2 2 2 1 H G G G G ( ) ( ) = 0 These equations yield G 1 = 40ºC; Q 1 = 5.28 × 10 –3 m 3 /sec http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 2 3 + + + + + + – – q i1 q 1 q r q 2 q 2 q i2 K ¢ 5 q 1 q 1 K ¢ 2 K ¢ 3 K ¢ 4 e 1 1 t 1 s + 1 t 2 s + 1 K ¢ 1 K 1 K 1 K 2 K 3 K 4 With these parameters, J 1 = 184.55 sec, K 3 = 1318.2, K 4 = 0.465, K 5 = 0.535 G G 2 ( ) ( ) s s r = G s G s ( ) ( ) 1 ; G(s) = ( . ) . . 5 55 10 7 26 10 119 10 3 1 2 3 4 2 3 5 K K K K s s Refer Fig. 3.57 for suitable hardware required to implement the proposed control scheme. 3.22 + + – – q 1 q r q 2 Controller I Set-point adjustment Controller II Sensor II Sensor I Process II Process I Refer Fig. 3.57 for suitable hardware required to implement the proposed control scheme. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 4 BASIC PRINCIPLES OF FEEDBACK CONTROL 4.1 S G M = 1 1 G s H s ( ) ( ) = s s s s ( ) ( ) 1 1 50 | ( )| S j G M M M1 = 0.0289 S H M = G s H s G s H s ( ) ( ) ( ) ( ) 1 = 50 1 50 s s ( ) | ( )| S j H M M M1 = 1.02 4.2 Y s R s ( ) ( ) = M(s) = P 1 1 , , P 1 = K K s s 3 1 2 1 ( ) ; , = 1 5 1 1 1 2 1 2 3 1 2 F H G I K J s s K K s K K s s ( ) ( ) ; , 1 = 1 M(s) = 5 1 5 5 5 1 2 1 1 K s s K K ( ) S K M 1 = M K K M 1 1 = s s K K s s s K K 2 1 1 2 2 1 1 1 5 5 5 1 5 5 5 ( ) ( ) | ( )| S j K M 1 0 M M = 5 5 5 1 K = 0.5 4.3 For G(s) = 20/(s + 1), and R(s) = 1/s, y t ( )| open-loop = 20 (1 – e –t ) y t ( )| closed-loop = 20 21 1 21 ( ) e t For G'(s) = 20/(s + 0.4), and R(s) = 1/s, y t ( )| open-loop = 50 (1 – e –0.4t ) y t ( )| closed-loop = 20 20 4 1 20 4 . ( ) . e t The transient response of the closed-loop system is less sensitive to variations in plant parameters http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 2 5 4.4 For G(s) = 10/( ), J s 1 and R(s) = 1/s, e ss | open-loop = 0 e ss | closed-loop = 1 1 10 K p = 0.0099 For G'(s) = 11/( ) J s 1 , and R(s) = 1/s, e ss | open-loop = – 0.1 e ss | closed-loop = 0.009 The steady-state response of the closed-loop system is less sensitive to variations in plant parameters. 4.5 J f | open-loop = L R = 2 50 = 0.04 sec For the feedback system, K e Ki A f f ( ) = L di dt R R i f s f ( ) This gives I s E s f f ( ) ( ) = K sL R R K K A s A J f | closed-loop = L R R K K s A = 2 51 90 K = 0.004 This gives K = 4.99 4.6 The given block diagram is equivalent to a single-loop block diagram given below, + – 0.4 10( + ) K s 1 W s ( ) Y s ( ) s + 1 s + 1 Y s W s ( ) ( ) = 0 4 11 10 1 . s K http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 2 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN For W(s) = 1/s, y ss = lim ( ) s sY s →0 = 0 4 10 1 . K = 0.01 This gives K = 3.9 4.7 Case I: Y s W s ( ) ( ) = s s s s K ( ) ( ) J J 1 1 for W(s) = 1/s, y ss = 0 Case II: Y s W s ( ) ( ) = K s s K ( ) J 1 For W(s) = 1/s, y ss = 1 The control scheme shown in the following figure will eliminate the error in Case II. + + – K c 1 1 + T s I W s ( ) Y s ( ) Plant 4.8 A unity-feedback configuration of the given system is shown below. + – q r q D s ( ) e 200 0.02 ¥ ( + 1) s ( + 2) s G s ( ) = E s s r ( ) ( ) G = 1 1 D s G s ( ) ( ) ;e ss = lim ( ) s sE s →0 ; G r s ( ) = 1 s (i) e ss = 1/3 (ii) e ss = 0 (iii) e ss = 1/3 The integral term improves the steady-state performance and the derivative term has no effect on steady-state error. 4.9 (i) s 2 + 1 = 0; oscillatory (ii) s 2 + 2s + 1 = 0; stable (iii) s 3 + s + 2 = 0; unstable http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 2 7 The derivative term improves the relative stability; and the integral term has the opposite effect. 4.10 (a) e r = 50 × 0.03 = 1.5 volts (b) V s E s r ( ) ( ) = M(s) = 0 65 1 5 1 0 0195 . ( )( ) . K s s K e e S K M e = M K K M e e = ( )( ) ( )( ) . s s s s 1 5 1 1 5 1 9 75 (c) V(s) = 325 1 5 1 9 75 19 5 1 1 5 1 9 75 ( )( ) . ( ) . ( ) ( )( ) . ( ) s s E s s s s W s r Suppose that the engine stalls when constant percent gradient is x%. Then for E r (s) = 1.5/s and W(s) = x/s, v ss = lim ( ) s sV s 0 = 325 1 5 10 75 19 5 10 75 . . . . x = 0 This gives x = 25 4.11 (a) The block diagram is shown in the figure below. + + – – K A K A K g I L R a e o e r 1 sL + R f f (b) E s E s r I L 0 0 ( ) ( ) = 10 2 1 10 s K For E r (s) = 50/s, e oss = lim ( ) s sE s 0 0 = 500 1 10 K = 250 This gives K = 0.1 With K = 0.1, we have http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 2 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN E o (s) = 10 2 2 2 1 2 2 s E s s s I s r L ( ) ( ) When E r (s) = 50/s and I L (s) = 30/s, we get e oss = 10 50 2 30 2 = 235 V Let x V be the reference voltage required to restore the terminal voltage of 250 V. 10 2 30 2 x = 250; this gives x = 53V (c) E o (s) = 10 2 1 s E s I s r L ( ) ( ) e r = 25 V gives e oss = 250 V when I L = 0 e r = 25 V gives e oss = 220 V when I L = 30 amps (d) The effect of load disturbance on terminal voltage has been reduced due to feedback action. 4.12 + + – – w r w 0 K g K T R a Js K A K t 1 1 K b K t (a) M M o r s s ( ) ( ) = M(s) = K K s K K A g A g 2 5 ( ) For M r (s) = 10/s, M o (t) = 125 13 1 130 ( ) e t ; M oss = 9.61 rad/sec (b) When the feedback loop is opened, M M o r s s ( ) ( ) = 50 2 5 K s A ( ) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 2 9 K A = 0.192 gives M oss = 9.61 for M r = 10. M o t ( ) = 9.61 (1 – e –5t ) Time-constant in the open-loop case is 1/5 sec, and in closed-loop case is 1/130 sec; system dynamics becomes faster with feedback action (c) S K M A = M K K M A A 2 10 2 10 s s K K A g Kg = Kω g , where K is a constant. S M K K M M K K M s s K K g M g g g g g g A g M M M 2 10 2 10 The open-loop transfer function of the system is G s K K s A g ( ) ( ) 2 5 Therefore S S K G G A g 1 M S S S S K M K G M G A A g g ; M M 4.13 + + + – – – – K A K P K P K g sK b K T T w R a J s 2 eq q L q R q M = q C 1 sL + R f f 1 1 1 2 (b) G G L R g g s s M s K s s s K ( ) ( ) ( ) ( ) ( ) 5 1 4 10 S M K K M s s s s s s K M g g g ( ) ( ) ( ) ( ) 1 4 1 4 10 100 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 3 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN S j K M g ( ) ~ . M M 0 1 4 4 10 (c) G L w s T s ( ) ( ) = ( ) / ( ) ( ) s s s s K g 4 2 1 4 10 Wind gust torque on the load shaft = 100 N-m. Therefore, on the motor shaft T w (s) = n × 100/s = 50/s θ Lss = lim ( ) . s L s s 0 0 1 G rad (d) Replace K A by D(s) = K T s c I 1 1 F H G I K J to reduce the steady-state error to zero value. 4.14 (a) G G ( ) ( ) s s r open loop = G(s) = K K K K s s s s P A e ( ) ( . . ) 25 1 0 25 0 02 1 2 G G ( ) ( ) s s r closed loop = M(s) = G s G s ( ) ( ) 1 S K G = 1; S K M = s s s s s s K K K K s P A e ( . . ) ( . . ) ( ) 0 25 0 02 1 0 25 0 02 1 25 1 2 2 S S K M K G (b) A PD control scheme with proportional controller in the forward path and derivative action realized by feeding back G, will increase aircraft damping. 4.15 (a) G( ) ( ) s T s w = 1 1 / B s J B s sK K BR K K BR b T a T A a F H I K For T w (s) = K w /s, θ ss = lim s 0 sθ (s) = K R K K w a T A http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 3 1 (b) G G ( ) ( ) s s r = M (s) = K K BR s J B s sK K BR K K BR T A a b T a T A a / F H I K 1 S J M = M J J M = F H I K J B s s J B s sK K BR K K BR b T a T A a 2 1 (c) Excessive large magnitudes of signals at various levels in a control system can drive the devices into nonlinear region of operation. The requirement of linear operation of devices under various operating conditions imposes a constraint on the use of large values of K A . 4.16 (a) dV dt = q i − q; V = C h(t), q = h(t)/R (b) + – h r q i q h R 1 R RCs + 1 K (c) H s H s r ( ) ( ) = M (s) = KR RCs KR 1 S R M = M R R M RCs KR 1 1 S R M s = 0 = 1 1 KR (d) For H r (s) = 1/s, h ss = KR KR e ss 1 ; = 1 1 KR Replacing K by K T s c l 1 1 F H G I K J reduces the steady-state error to zero; and the system becomes insensitive to changes in R under steady dc conditions. 4.17 V d dt C c Q q C c Q C c i i i i ( ) ( ) ( ) ( ) + = + + − + The perturbation dynamics is given by J dc t dt ( ) + c(t) = C Q q t Q Q c t V Q i i i i ( ) ( ); J Block diagram of the closed-loop system is shown below. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 3 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN + + – q i 1 c ts + 1 K C /Q i c i Q /Q i C s C s i ( ) ( ) = M s Q Q s KC Q i i ( ) / / = + + τ 1 For C i (s) = A/s, c ss = AQ Q KC i i + With integral control, c ss becomes zero. 4.18 (b) Open-loop case: (i) G( ) ( ) s E s r = G(s) = KR R Cs S K G 1 1 1 1 ; + K + + – q q i e r K 1/R 1 K t e r e t K R C 1 s + 1 R 1 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 3 3 (ii) G G ( ) ( ) s s i 1 1 1 R Cs s s i i ss . ( ) / , For G G (iii) τ = R 1 C Closed-loop case: (i) G( ) ( ) s E s r = M(s) = KR R Cs KK R t 1 1 1 1 ; S R Cs R Cs KK R K M t 1 1 1 1 1 (ii) G G ( ) ( ) s s i = 1 1 1 1 R Cs KK R t . For θ i (s) = 1/s, θ ss = 1 1 1 KK R t (iii) τ = R C KK R t 1 1 1 + (c) K(e r − e t ) = C d dt R R R UA R R C R R i a G G G G G J 1 2 2 1 2 1 2 1 ; , + K + + + + – q q a q i e r K R 1 K t e r K R 2 ( + ) ( R R 1 2 ts + 1) R R 1 2 1 1 G G ( ) ( ) s s a open loop = R R R s 1 1 2 1 ( ) ( ) J ; G ss open loop = R 1 (R 1 + R 2 ) θ θ ( ) ( ) s s a closed loop − = 1 1 2 1 2 ( ) ( 1) t R R R s KK R R J θ ss closed loop − = R R R KK R R t 1 1 2 1 2 + + 4.19 (a) Y s W s ( ) ( ) = 0 1 1 2 3 1125 2 . ( ) . . s s s For W(s) = 1/s, y ss = 0.089 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 3 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) Replace K c by K T s c l 1 1 F H G I K J (c) Y s W s s s s K K s f c ( ) ( ) . ( ) . L N M O Q P 0 1 1 2 3 1125 1 1 2 For W(s) = 1/s, y ss = 0.089 (1 − K f K c ) y ss = 0 when K f = 1/K c = 0.8 The PI control scheme increases the order of the system; this makes it less stable. The feedforward scheme does not affect the characteristic roots of the system. A difficulty with feedforward compensation is that it is an open- loop technique; it contains no self-correcting action. If the value of K c is not accurately known, the gain K f will not cancel the disturbance completely. The usefulness of a method must be determined in the light of the specific performance requirements. If the uncertainty is large, self-correcting action can be obtained by using PI control law in conjunction with the feedforward compensation. 4.20 + + – – e f i f w 1 Js B + sL + R + 1 f f K A K t r K t K T Time-constant (without current feedback) = L f /(R f + 1) I s E s f f ( ) ( ) = K sL R K A f f A 1 Time-constant (with current feedback) = L R K f f A 1 Position control system is shown below. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 3 5 + + + – – – w w r K t i f 1 s K t K P K P K A sL + R + 1 f f Js B + 4.21 (a) s 2 + K c T D s + K c = (s + 1 + j1.732) (s + 1 − j 1.732) This gives K c = 4, T D = 0.5 (b) Alternative control scheme is shown below. + + – – q q r q 1 1 s s K 1 K 2 G G ( ) ( ) ; , s s K s K s K K K r 1 2 2 1 1 2 4 2 4.22 Fig. P4.22a: G G ( ) ( ) ( ) s s M s K s s K r 25 5 25 2 S M K K M s s s s K M ( ) 5 5 25 2 S j K M ( ) . M M 5 1 41 Fig. P4.22b: G G ( ) ( ) s s r = M (s) = KK s KK KK s 2 2 1 2 1 ( ) = 25 5 25 2 K s s K This gives K 2 = 25 and K 1 = 4 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 3 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN M (s) = 25 1 4 25 2 K s K s K ( ) S K M = s s s s ( ) 1 5 25 2 | ( )| S j K M M M5 ≈ 1 This shows the superiority of the two-loop system over a single-loop system. —- http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 5 CONCEPTS OF STABILITY AND THE ROUTH STABILITY CRITERION 5.1 (a) All the elements in the first column of the Routh array are +ve. Therefore, all the roots are in the left-half plane. (b) Two sign changes are found in the first column of the Routh array. Therefore, two roots are in the right-half plane and the rest in the left- half plane. (c) s 3 -row of the Routh array has zero pivot element, but the entire row is not all zeros. We replace the pivot element by A and then proceed with the construction of the Routh array. As A 0, two sign changes are found in the first column of the Routh array. Therefore, two roots are in right-half plane and the rest in the left-half plane. (d) s 1 -row of the Routh array is an all-zero row. Auxiliary polynomial formed using the elements of s 2 -row is given by A(s) = s 2 + 1 We replace the elements of s 1 -row with the coefficients of dA s ds ( ) = 2s + 0 and proceed with the construction of the Routh array. There are no sign changes in the resulting Routh array; the characteristic polynomial does not have any root in the right-half plane. The roots of the 2nd-order auxiliary polynomial are therefore purely imaginary. The given characteristic equation has two roots on the imaginary axis and the rest in the left-half plane. (e) Since all the coefficients of the given characteristic polynomial are not of the same sign, the system is unstable. The Routh array formation is required only if the number of roots in the right-half plane are to be determined. Only one sign change (the s 0 -row has – ve pivot element) is found in the Routh array. Therefore one root is in the right-half plane and the rest in the left-half plane. (f) s 3 -row of the Routh array is an all-zero row. The auxillary polynomial is A(s) = 9s 4 + 0s 2 + 36 We replace elements of s 3 -row with the coefficients of dA s ds ( ) = 36s 3 + 0s The resulting Routh array has s 2 -row with zero pivot element but the entire row is not all zeros. We replace the pivot element with A and proceed with the construction of the Routh array. Two sign changes are found in the first column; the characteristic polynomial ,(s) has http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 3 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN two roots in the right-half plane. Three possibilities exist: (i) the 4th- order auxiliary polynomial has all the four purely imaginary roots and the two right-half plane roots are contributed by the factor ,(s)/A(s), (ii) the 4th-order auxiliary polynomial has complex roots with quadrantal symmetry (two roots in the right-half plane) and the factor ,(s)/A(s) has all the roots is the left-half plane, and (iii) the 4th-order auxiliary polynomial has two real-axis roots and two imaginary-axis roots with quadrantal symmetry and the factor ,(s)/A(s) has one root in the right-half plane. Examining A(s) we find that auxiliary- polynomial roots are s = – 1 ± j1, 1 ± j1. The given system, therefore, has two roots in the right-half plane and the rest in the left-hand plane. 5.2 (a) s 1 -row of the Routh array is an all-zero row. The auxiliary polynomial is A(s) = s 2 + 9 By long division ,(s)/A(s) = (s 4 + 2s 3 + 11s 2 + 18s + 18)/A(s) = s 2 + 2s + 2 Therefore ,(s) = (s 2 + 9) (s 2 + 2s + 2) = (s + j3) (s – j3) (s + 1 + j1) (s + 1 – j1) (b) s 3 -row of the Routh array is an all-zero row. The auxiliary polynomial is A(s) = s 4 + 24s 2 – 25 ,(s)/A(s) = (s 5 + 2s 4 + 24s 3 + 48s 2 – 25s – 50)/A(s) = s + 2 ,(s) = (s + 2) (s 4 + 24s 2 – 25) = (s + 2) (s 2 – 1) (s 2 + 25) = (s + 2) (s + 1) (s – 1) (s + j5) (s – j5) (c) s 3 -row of the Routh array is an all-zero row. The auxiliary polynomial is A(s) = s 4 + 3s 2 + 2 ,(s)/A(s) = (s 6 + 3s 5 + 5s 4 + 9s 3 + 8s 2 + 6s + 4)/A(s) = s 2 + 3s + 2 ,(s) = (s 2 + 3s + 2) (s 4 + 3s 2 + 2) = (s 2 + 3s + 2) (s 2 + 1) (s 2 + 2) = (s + 1) (s + 2) (s + j1) (s - j1) (s + j 2 ) (s – j 2 ) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 3 9 5.3 With s = s – 1, the characteristic equation becomes s 3 + s 2 + s + 1 = 0 s 1 -row in the Routh array is an all-zero row. The auxiliary polynomial is A( s ) = s 2 + 1 = ( s + j1) ( s – j1) We replace the elements of s 2 -row with the coefficients of dA s ds ( ) = 2 s + 0 and proceed with the construction of the Routh array. There are no sign changes in the resulting Routh array. The s-polynomial does not have roots to the right of the line at s = –1, and there are two roots at s = – 1 ± j1. The largest time-constant is therefore 1 sec. 5.4 (a) Unstable for all values of K (b) Unstable for all values of K (c) K < 14/9 (d) K > 0.528 5.5 With s = s – 1, the characteristic equation becomes s 3 + 3K s 2 + (K + 2) s + 4 = 0 From the Routh array, we find that for K > 0.528, all the s-plane roots lie to the left of the line at s = –1. 5.6 G(s) = 4 2 1 20 1 1 20 0 1 K s s s L N M O Q P / ( ) .2 / ( ) H(s) = 0 05 4 1 . s 1 + G(s) H(s) = 0 gives 8s 3 + 46s 2 + 31s + 5 + 4K = 0 From the Routh array, we find that the closed-loop system is stable for K < 43.3. 5.7 (i) s 3 + 5s 2 + 9s + K = 0; K < 45 (ii) s 3 + 5s 2 + (9 – K) s + K = 0; K < 7.5 (iii) s 4 + 7s 3 + 19s 2 + (18 – K) s + 2K = 0; K < 10.1 5.8 (a) ,(s) = s 3 + 10s 2 + (21 + K)s + 13K = 0 (i) K < 70 (ii) K = 70 (iii) for K = 70, A(s) = s 2 + 91 = 0 ,(s) = (s 2 + 91) (s + 10) = (s + j9.54) (s – j9.54) (s + 10) (b) ,(s) = s 3 + 5s 2 + (K – 6)s + K = 0 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 4 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (i) K > 7.5 (ii) K = 7.5 (iii) For K = 7.5, A(s) = s 2 + 1.5 = 0 ,(s) = (s 2 + 1.5) (s + 5) = (s + j1.223) (s – j1.223) (s + 5) (c) ,(s) = s 4 + 7s 3 + 15s 2 + (25 + K)s + 2K = 0 (i) K < 28.1 (ii) K = 28.1 (iii) For K = 28.1, A(s) = s 2 + 7.58 = 0 ,(s) = (s 2 + 7.58) (s 2 + 7s + 7.414) = (s + j2.75) (s – j2.75) (s + 5.7) (s + 1.3) 5.9 (a) ,(s) = s 4 + 12s 3 + 69s 2 + 198s + 200 + K = 0 From the Routh array we find that the system is stable for K < 666.25. For K = 666.25, A(s) = 52.5s 2 + 866.25 = 0 This gives s = ± j 4.06 Frequency of oscillations is 4.06 rad/sec when K = 666.25. (b) ,(s) = s 4 + 3s 3 + 12s 2 + (K – 16) s + K = 0 From the Routh array, we find that the system is stable for 23.3 < K < 35.7. For K = 23.3, A(s) = 9.57s 2 + 23.3 = 0; s = ± j1.56 For K = 35.7, A(s) = 5.43s 2 + 35.7 = 0; s = ± j2.56 Frequency of oscillation is 1.56 rad/sec when K = 23.3; and 2.56 rad/ sec when K = 35.7. 5.10 ,(s) = s 3 + as 2 + (2 + K) s + 1 + K = 0 From the Routh array we find that for the system to oscillate, (2 + K)a = 1 + K Oscillation frequency = 1 K a = 2 These equations give a = 0.75, K = 2 5.11 ,(s) = 0.02s 3 + 0.3s 2 + s + K = 0 (a) K < 15 (b) For K = 15, the auxiliary equation is A(s) = s 2 + 50 = 0. The oscillation frequency is 7.07 rad/sec at K = 15. (c) With K = 7.5 and s = s – 1, ,() s = 0.2 s 3 + 2.4 s 2 + 4.6 s + 67.8 = 0 Two sign changes are there in the first column of the Routh array; therefore two roots have time-constant larger than 1 sec. 5.12 (a) Unstable for all K c (b) For stability, T D > J 5.13 (a) ,(s) = s 3 + 15s 2 + 50s + 25K = 0; K < 30 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 4 1 (b) ,(s) = Js 3 + (1 + 5J)s 2 + 5s + 50 = 0; J < 0.2 (c) ,(s) = Js 3 + (1 + 5J)s 2 + 5s + 2.5K = 0; K < 2 10 J F H I K 5.14 ,(s) = s 4 + 5s 3 + 6s 2 + K c s + K c = = 0; = = 1 T I From the Routh array, we find that K < (30 – 25=) results in system stability. Larger the = (smaller the T I ), lower is the limit on gain for stability. 5.15 ,(s) = s 3 + 15s 2 + (50 + 100K t )s + 100K = 0 From the Routh array, we find that for stability K < (7.5 + 15K t ). Limit on K for stability increases with increasing value of K t . http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 6 THE PERFORMANCE OF FEEDBACK SYSTEMS 6.1 (a) Refer Section 6.3 for the derivation (b) The characteristic equation is s 2 + 10s + 100 = s 2 + 2M n s + M n 2 = 0 (i) M n = 10 rad/sec; = 0.5; M d = M n 1 2 = 8.66 rad/sec (ii) t r = F M cos 1 d = 0.242 sec; t p = F M d = 0.363 sec M p = e F / 1 2 = 16.32%; t s = 4 M n = 0.8 sec (iii) K p = lim s0 G(s) = ; K v = lim s0 sG(s) = 10; K a = lim s0 s 2 G(s) = 0 (iv) e ss = 1 1 K p = 0; e ss = 1 K v = 0.1; e ss = 1 K a = 6.2 Characteristic equation is s 2 + 10s + 10K A = s 2 + 2M n s + M n 2 = 0 (a) K A = 2.5 gives = 1 which meets the response requirements. (b) K p = , K v = 2.5 e ss = 5 1 1 K K p v = 0.4 6.3 (a) Using Routh criterion, it can be checked that the close-loop system is stable. (i) e ss = 10 K v = 10 = 0 ; (ii) e ss = 10 0 2 K K a v . = 10 0 0 1 .2 . = 2 (b) The closed-loop system is unstable. 6.4 The closed-loop system is stable. This can easily be verified by Routh criterion. G(s) = 5 0 1 1 1 0 1 ( . )( )( .2 ) s s s http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 4 3 e ss | r = 10 = 10 1 K p = 10 1 5 Y s W s ( ) ( ) = 0 1 1 1 0 1 0 1 1 5 . ( )( .2 )( . ) s s s s For W(s) = 4/s, y ss = 4/6 e ss | total = 7/3 6.5 (a) The characteristic equation of the system is 900s 3 + 420s 2 + 43s + 1 + 0.8 K A = 0 Using Routh criterion, we find that the system is stable for K A < 23.84. (b) Rearrangement of the block diagram of Fig. P6.5 is shown below. + – q r q 50 1 0.016 30 1 s + 10 1 s + 3 1 s + K A G(s) = 0 016 50 3 1 30 1 . ( )( ) K s s A H(s) = 1 10 1 s ; the dc gain of H(s) is unity. K p = lim s0 G(s) = 0.8 K A ; e ss = 10 1 K p = 1 This gives K A = 11.25 6.6 (a) The system is stable for K c < 9. K p = lim s0 D(s) G(s) = K c ; e ss = 1 1 K c e ss = 0.1 (10%) is the minimum possible value for steady-state error. Therefore e ss less than 2% is not possible with propor- tional compensator. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 4 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) Replace K c by D(s) = 3 + K s I . The closed-loop system is stable for 0 < K 1 < 3. Any value in this range satisfies the static accuracy requirements. 6.7 (a) K v = 1000 10 K c = 1000 = 10 1000 2 100 K D = 0.5 These equations give K c = 10 and K D = 0.09. (b) The closed-loop poles have real part = – M n = – 50. The zero is present at s = – K c /K D = – 111.11. The zero will result in pronounced early peak. is not an accurate estimate of M p . 6.8 (a) K v = K I = 10 (b) With K I = 10, the characteristic equation becomes ,(s) = s 3 + 10s 2 + 100(1 + K c )s + 1000 = 0 With s = s – 1, ,( s ) = s s 3 2 7 + + [100(1 + K c ) – 17] s + 1009 – 100 (1 + K c ) = 0 In Routh array formation, K c = 0.41 results in auxiliary equation with purely imaginary roots. Therefore K c = 0.41 results in dominant s-plane poles with real part = – 1. (c) Routh array formation for ,( s ) gives the following auxiliary equation. A( s ) = 7 s 2 + 868 = 0; s = ± j11.14 The complex-conjugate s-plane roots are – 1 ± j11.14. M d = M n 1 2 = 11.14; M n = 1 These equations give = 0.09. (d) ,( s ) = ( s + 7) ( s + j11.14) ( s – j11.14) = 0 The third s-plane closed-loop pole is at s = – 8. The dominance condition in terms of third closed-loop pole is reasonably satisfied. The zero is at s = – K I /K c = – 24.39. The zero will not give pronounced early peak. Therefore approximately represents M p . 6.9 (a) With T I = , D(s)G(s) = 80 1 8 80 2 ( ) T s s s D http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 4 5 The characteristic polynomial becomes ,(s) = s 2 + (8 + 80T D )s + 160 T D = 0.216 gives = 1 (b) The zero is at s = – 1/0.216 = – 4.63. Real part of the closed-loop pole = – M n = – 12.65. Therefore, small overshoot will be observed. (c) The characteristic equation becomes s 3 + 25.28s 2 + 160s + 80 T I = 0 T I > 0.0198 for stability. 6.10 (a) G(s) = K s s ( ) 1 ; K v = lim s0 sG(s) = K K = 10 will give steady-state unit-ramp following error of 0.1. ,(s) = s 2 + s + K = s 2 + 2M n s + M n 2 K = 10 gives = 0.158. (b) G(s) = 10 1 10 s s K t ( ) ,(s) = s 2 + (1 + 10K t )s + 10 = s 2 + 2M n s + M n 2 K t = 0.216 gives = 0.5. K v = lim s0 sG(s) = 10 1 10 K t = 10 316 . e ss = 0.316 (c) G(s) = 10 1 10 K s s K A t ( ) K v = 10 1 10 K K A t ,(s) = s 2 + (1 + 10K t )s + 10 K A = s 2 + 2M n s + M n 2 From these equations, we find that K A = 10 and K t = 0.9 give e ss = 0.1 and = 0.5. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 4 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 6.11 (a) 2 = ( ) ( ) In In M M p p 2 2 2 + π M p = 0.1 gives = 0.59 t s = 4 M n = 0.05. Therefore, M n = 135.59 G(s) = 5 0 1 100 1 2 K s s K ( .2 ) ,(s) = 0.2s 2 + s(1 + 100K 2 ) + 5K 1 = 0 Therefore s 2 + 5(1 + 100K 2 )s + 25K 1 = s 2 + 2M n s + M n 2 This equation gives K 1 = 735.39 and K 2 = 0.31 (b) K p = , K v = 5 1 100 1 2 K K = 114.9, K a = 0 6.12 G( ) ( ) s T s w = 4 1 10 10 s s K K s t ( ) For T w (s) = 1/s, G ss = 4 10K = 0.05 This gives K = 8 ,(s) = s 2 + (1 + 10K t )s + 10K = s 2 + 2M n s + M n 2 K t = 0.79 gives = 0.5 6.13 Y s F s ( ) ( ) = 1 2 Ms Bs K . For F(s) = 9/s, y ss = 9 K = 0.03 This gives K = 300 Newtons/m e F / 1 2 = 0 003 0 03 . . = 0.1 This gives = 0.59 t p = F M n 1 2 = F M 0 81 . n = 2 This gives M n = 1.94 rad/sec From the equation http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 4 7 s 2 + B M s K M = s 2 + 2M n s + M n 2 we obtain M = 79.71 kg, B = 182.47 Newtons per m/sec 6.14 G G ( ) ( ) s s R = K J s B J s K J T T / 2 + + = M M M n n n s s 2 2 2 2 + + – – q R q 1 Js + Bs 2 K T T w 2 = ( ) ( ) ln ln M M p p 2 2 2 + π For M p = 0.25, = 0.4 Steady-state error to unit-ramp input = 2 M n = 0.04 This gives M n = 20 G( ) ( ) s T s w = 1 2 Js Bs K T G ss = 10 K T = 0.01 From these equations, we obtain the following values of system parameters. K T = 1000 N-m/rad B = 40 N-m/(rad/sec) J = 2.5 kg-m 2 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 4 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 6.15 J M + BM = K T (M r – M) M M ( ) ( ) s s r = 45 100 50 s M r = 50 × 2 60 F rad/sec = step input M(t) = 1.5F (1 – e –0.5t ) M ss = 1.5F rad/sec = 45 rpm; e ss = 5 rpm A control scheme employing gain adjustment with integral error compensation will remove the offset. 6.16 J G G B = K T (G r – G) G G ( ) ( ) s s r = 2400 150 600 2400 2 s s G r (s) = F 3s ; G(s) = F 3 16 4 16 2 s s s ( ) L N M O Q P G(t) = F M M 3 1 1 2 1 L N M M O Q P P e t n t d sin cos ( ) = F F 3 1 11547 3 46 3 2 F H I K L N M O Q P . sin . e t t M p = e F / 1 2 = 16.3% The following control schemes have the potential of eliminating overshoot: (i) gain adjustment with derivative error compensation; and (ii) gain adjustment with derivative output compensation. 6.17 + + – – q R e c e q L q M K s K A K 1 sK 2 1 1 50 Js + Bs 2 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 4 9 (b) G(s) = 4 100 K s s A ( .5) K v = 4K A /100.5 For a ramp input G R (t) = F t, the steady-state error = F K v The specification is 5 deg or 5 180 F rad. Therefore 100 4 .5F K A = F 180 ; this gives K A = 904.5 The characteristic equation is s 2 + 100.5s + 4 × 904.5 = 0 M n = 60.15; = 0.835; M p = 0.85%; t s = 0.0796 sec (c) With PI control the system becomes type-2; and the steady-state error to ramp inputs is zero provided the closed-loop system is stable. G(s) = 4 1 1 100 K s s s A F H I K ( .5) ; K A = 904.5 The characteristic equation is s 3 + 100.5s 2 + 4 × 904.5s + 4 × 904.5 = 0 It can easily be verified that the closed-loop system is stable. (d) The characteristic equation can be expressed as (s + 1.0291) (s 2 + 99.4508s + 3514.6332) = 0 or (s + 1.0291) (s + 49.7254 + j32.2803) (s + 49.7254 – j32.2803) = 0 The dominance condition is satisfied because the real pole is very close to the zero. Therefore, the transient response resembles that of a second-order system with characteristic equation s 2 + 99.4508s + 3514.6332 = 0 M n = 59.28; = 0.84; M p = 0.772%; t s = 0.08 sec http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 5 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 6.18 q L q R e J L L f K A R f e t I a (const) Ke t + + – – q R q M q L K A sKK t K P K P e K T 1 J s eq 2 1 50 (c) The characteristic equation of the system is s 2 + 100 KK A s + 6K A = s 2 + 2M n s + M n 2 = 0 This gives K A = 2.67; K = 0.024 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 5 1 6.19 + + – – q R q L K A K P K A sK b K T R a 1 Js + Bs 2 1 10 (b) Open-loop transfer function G(s) = 20 4 202 K s s A ( ) K v = 20 202 K A = 1 0 01 . ; this gives K A = 1010. The characteristic equation becomes s 2 + 50.5s + 5050 = 0 = s 2 + 2M n s + M n 2 This gives = 0.355; M p = 30.33% (c) G(s) = 20 4 202 ( ) ( ) K sK s s A D K v = 20 202 K A = 100; no effect on e ss . Characteristic equation of the system is s 2 + (50.5 + 5K D )s + 5050 = 0 = s 2 + 2M n s + M n 2 This gives K D = 6.95. For = 0.6, M p = 9.5% (d) System zero: s = K K A D = – 145.32 Real part of closed-loop poles = – M n = – 42.636. The zero will affect the overshoot. The peak overshoot will be slightly more than 9.5%. 6.20 (a) G(s) = 3 0 3 1 1 2 K s s s A ( . )( ) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 5 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN K v = 3K A ; e ss = 0 03 3 . K A = 0 01 . K A (b) Y s F s w ( ) ( ) = 0 3 1 2 0 3 1 1 2 3 . ( ) ( . )( ) s s s s K A y ss = 0 1 . K A (c) Characteristic equation is 0.6s 3 + 2.3s 2 + s + 3K A = 0 For stability, K A < 1.28. Minimum value of error in part (a) = 7.81 × 10 –3 Minimum value of error in part (b) = 0.078 6.21 – Controller Power amp 1 e u i 100 x 47.6 s 2 – 16.67 (a) Characteristic equation of the system is s 2 – 16.67s + 4760K c = 0 Unstable or oscillatory for all K c > 0. (b) U(s) = – K c (1 + sT D ) E(s) Characteristic equation becomes s 2 + 4760 K c T D s + 4760K c – 16.67 = 0 For stability, K c > 3.5 × 10 –3 and T D > 0. (c) 2 = [ ( .2)] [ ( .2)] ln ln 0 0 2 2 2 + π ; = 0.456 t s = 4 M n = 0.4; M n = 21.93 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 5 3 s 2 + 4760 K c T D s + 4760 K c – 16.67 = s 2 + 2M n s + M n 2 This gives K c = 0.1045, T D = 0.04. 6.22 (a) System-type number = 2 (b) H s Q s w ( ) ( ) = 0 0 1 1 0 1 1 0 2 .5 ( . ) ( . ) .5( ) s s s s K sK c D + + + + h ss = 0 0 .5 .5K c ; K c > 10 The characteristic equation is 0.1s 3 + s 2 + 0.5K D s + 0.5K c = 0 For stability, K D > 0.1K c 6.23 – + q q r K 2 1 + K 1 s s + 1 Characteristic equation is s 2 + (1 + K 1 )s + K 2 = 0 = s 2 + 2M n s + M n 2 M p = 20% = 0.456; t s = 2 M n = 4.386 K 2 = 19.237; K 1 = 3 —— http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 7 COMPENSATOR DESIGN USING ROOT LOCUS PLOTS 7.1 (a) – o A = – 2 ; o A = 60º, 180º, 300º ; intersection with ju-axis at ± j5.042; angle of departure o p from – 1 + j4 = 40º. (b) –o A = – 1.25 ; o A = 45º, 135º, 225º, 315º ; intersection with ju-axis at ± j1. 1; angle of departure o p from – 1 + j1 = – 71.6º; multiple roots at s = – 2.3. (c) Angle of arrival o z at – 3 + j4 = 77.5º ; multiple roots at s = – 0.45. (d) – o A = –1.33 ; o A = 60º; 180º, 300º; intersection with ju-axis at ± j 5 ; angle of departure o p from – 2 + j1 = – 63.43º; multiple roots at s = – 1, – 1.67. (e) – o A = – 1.5 ; o A = 90º, 270º (f) – o A = – 5.5 ; o A = 90º, 270º; multiple roots at s = – 2.31, – 5.18. (g) – o A = 0.5 ; o A = 90º, 270º; intersection with ju-axis at ± j2.92. 7.2 (i) – o A = – 2; o A = 60º, 180º, 300º; intersection with ju-axis at ± j 10 ; angle of departure o p from – 3 + j1 = – 71.5º. Two root loci break away from s = – 1.1835 at ± 90º. Two root loci approach s = – 2.8165 at ± 90º jw j ÷10 s – 3 + 1 j – 2 (i) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 5 5 (ii) Intersection with ju-axis at ± j2 3 ; angle of departure o p from – 3 + j 3 = – 60º. Three root loci approach the point s = – 2, and then break away in directions 120º apart. jw jw j ÷3 s s – 2 (ii) (iii) – 9 – 4 – 3 + 60° j2÷3 (iii) – o A = – 4 ; o A = 90º, 270º The characteristic equation has three roots at s = – 3; the three root loci originating from open-loop poles approach this point and then breakaway. Tangents to the three loci breaking away from s = – 3 are 120º apart. (iv) – o A = – 1 ; o A = 45º, 135º, – 45º, – 135º; intersection with the ju axis at ± j 1.58; angle of departure o p from – 1 + j2 = – 90º There are two roots at s = – 1, two roots at s = – 1 + j1.22, and two roots at s = – 1 – j1.22. The break away directions are shown in the figure. (v) There are four roots at s = – 1. The break away directions are shown in the figure. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 5 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN jw jw s s – 1 + 1 j – 1 + 2 j – 2 – 2 (v) (iv) 45° 45° 7.3 – o A = – 1 ; o A = 60º, 180º, 300º; intersection with the ju-axis at ± j 2 ; multiple roots at s = – 0.42. The ¸ = 0.5 loci passes through the origin and makes an angle of 0 = cos –1 ¸ = 60º with the negative real axis. The point s d = – 0.33 + j0.58 on this line satisfies the angle criterion. By magnitude criterion, the value of K at this point is found to be 1.04. Using this value of K, the third pole is found at s = – 2.33. Therefore, M(s) = 1 04 0 33 0 0 33 0 2 33) . ( . .58)( . .58)( . s j s j s - - - ÷ - 7.4 – o A = – 3 ; o A = 45º, 135º, 225º, 315º; intersection with the ju-axis at ± j3.25; departure angle o p from – 4 + j4 = 225º; multiple roots at s = – 1.5. At the intersection of the ¸ = 0.707 line with the root locus, the value of K, by magnitude criterion is 130. The remaining pair of complex roots for K = 130 can be approximately located graphically. It turns out that real part of complex pair away from ju-axis is approximately four times as large as that of the pair near ju-axis. Therefore, the transient response term due to the pair away from the ju-axis will decay much more rapidly than the transient response term due to the pair near ju-axis. 7.5 – o A = – 2.67 ; o A = 60º, 180º, 300º ; intersection with ju-axis at ± j 32 ; departure angle o p from – 4 + j4 = – 45º. The point s d = – 2 + j3.4 on the ¸ = 0.5 line satisfies the angle criterion. By magnitude criterion, the value of K at this point is found to be 65. Using this value of K, the third pole is found at s = – 4; the dominance condition is not satisfied. 7.6 – o A = – 2; o A = 60º, 180º, 300º; intersection with the ju-axis at ± j 5 ; multiple roots at s = – 0.473. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 5 7 The least value of K to give an oscillatory response is the value of K at the multiple roots (K = 1.128). The greatest value of K that can be used before continuous oscillations occur is the value of K at the point of intersection with the ju-axis. From the Routh array formulation it is found that the greatest value of K is 30 and at this gain continuous oscillations of frequency 5 rad/sec occur. 7.7 The proof given in Example 7.1. 7.8 The proof follows from Example 7.1. Draw a line from the origin tangential to the circular part of the locus. This line corresponds to damping ratio for maximum oscillatory response. The tangential line corresponds to ¸ = 0.82 and the value of K at the point of tangency is 2.1. 7.9 The proof given in Example 7.2 7.10 s = o + ju tan –1 u o u o - - ÷ ÷ 1 1 1 tan = tan tan º ( ) ÷ ÷ - - ± - 1 1 2 180 2 1 u o u o q Taking tangents on both sides of this equation, and noting that tan tan º − + ± L N M O Q P 1 2 180 ω σ = u o - 2 , we obtain u o u o u o u o - - ÷ ÷ - F H I K ÷ F H I K 1 1 1 1 1 = u o u o u o u o - - ÷ F H I K - F H I K 2 1 2 Manipulation of this equation gives o u ÷ F H I K - ÷ L N M O Q P 1 2 5 4 2 2 = 0 There exists a circular root locus with centre at o = ½, u = 0 and the radius equal to 5 2. 7.11 Use the result in Problem 7.7 for plotting the root locus. Complex-root branches form a circle. The ¸ = 0.707 line intersects the root locus at two points; the values of K at these points are 1 and 5. The point that corresponds to K = 5 results in lower value of t s ; therefore we choose K = 5(s d = – 3 + j3). http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 5 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN For ¸ = 0.707, M p = e ÷ ÷ *¸ ¸ / 1 2 = 4.3% t s = 4 ¸u n = 4 3 sec G(s) = K s s s ( ) ( )( ) - - ÷ 4 2 1 E(s) = 1 1 - G s ( ) R(s) = ( )( ) ( )( ) ( ) ( ) s s s s K s R s - - - ÷ - - 2 1 2 1 4 For R(s) = 1/s (assuming selection of K that results in stable closed-loop poles) e ss = lim ( ) s sE s ÷0 = ÷ ÷ - 2 2 4K For K = 5, e ss = – 1/9 Therefore, steady-state error is 11.11%. 7.12 Breakaway points obtained from the solution of the equation dK/ds = 0, are s = – 0.634, – 2.366. Two root loci break away from the real axis at s = – 0.634, and break into the real axis at s = – 2.366. By determining a sufficient number of points that satisfy the angle criterion, it can be found that the root locus is a circle with centre at – 1.5 that passes through the breakaway points. Both the breakaway points correspond to ¸ = 1. For minimum steady- state error we should select larger value of K. s = – 0.634 corresponds to K = 0.0718 s = – 2.366 corresponds to K = 14 7.13 The characteristic equation is 1 + 0 8 10 1 30 1 3 1 . ( )( )( ) K s s s - - - = 0 or 1 + ´ - F H I K - F H I K - F H I K K s s s 1 10 1 30 1 3 = 0 ; K' = 0.000888 K – o A = – 0.155 ; o A = 60º, 180º, 300º; intersection with the ju-axis at ± j 0.22; multiple roots at s = – 0.063 The point on the ¸ = 0.707 line that satisfies the angle criterion is s d = – 0.06 ± j0.06 ; the value of ′ K at this point, obtained by magnitude criterion, is 0.00131. Therefore K = 1.475. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 5 9 7.14 (a) Ideal feedback sensor: 1 + K s s ( . ) 0 1 1 - = 1 + 10 10 K s s ( ) - = 0 K = 10 gives ¸ = 0.5. (b) Sensor with appreciable time-constant: 1 + K s s s ( . )( . ) 0 1 1 0 02 1 - - = 0 or 1 + ´ - - K s s s ( )( ) 10 20 = 0 ; ´ K = 500 K Locate a root locus point on the real axis that gives ´ K = 5000 (i.e., K = 10). Complex-conjugate roots can then be found by long division. The process gives ¸ = 0.4. 7.15 Characteristic equation is given by A(s) = s 3 + 6s 2 + 8s + 0.1K(s + 10) = 0 which can be arranged as 1 + ´ - - - K s s s s ( ) ( )( ) 10 2 4 = 0 ; ´ K = 0.1K – o A = 2 ; o A = 90º, – 90º; intersection with the ju-axis at ± j 20 ; multiple roots at s = – 0.8951. The point on ¸ = 0.5 line that satisfies the angle criterion is s d = – 0.75 + j1.3. The value of K´ at this point, obtained by magnitude criterion, is 1.0154. Hence K = 10.154. The third closed-loop pole is found at s = – 4.5 7.16 Characteristic equation of the system is s 3 + s + 10K t s + 10 = 0 which can be rearranged as 1 + Ks s s 2 10 + + = 0 ; K = 10K t Notice that a zero is located at the origin and open-loop poles are located at s = – 0.5 ± j3.1225. As per the result of Problem 7.9, a circular root locus exists with the centre at zero and radius equal to 10 . The point on the ¸ = 0.7 line that satisfies the angle criterion is s d = – 2.214 + j2.258. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 6 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN The gain K corresponding to this point is 3.427. Hence the desired value of velocity feedback gain K t is 0.3427. 7.17 Characteristic equation of the system is A(s) = s(s + 1) (s + 4) + 20K t s + 20 = 0 which may be rearranged as 1 + Ks s j s j s ( )( )( ) - ÷ - 2 2 5 = 0 ; K = 20K t – o A = – 2.5 ; o A = 90º, – 90º; departure angle o p from s = j2 is 158.2º. The following two points on ¸ = 0.4 line satisfy the angle criterion: s 1 = – 1.05 + j2.41, s 2 = – 2.16 + j4.97 The value of K at s 1 is 0.449, and at s 2 is 1.4130. The third pole corresponding to K = 0.449 is found at s = – 2.9, and for K = 1.413 at – 0.68. From the root locus plot, one may infer that the closed-loop pole at s = – 0.68 is close to system zero at the origin. This is not the case. In fact the closed-loop system does not have a zero at the origin: Y s R s ( ) ( ) = 20 1 4 20 1 s s s K s t ( )( ) ( ) - - - - In the root locus plot, the zero at the origin was introduced because of the process of modifying the characteristic equation so that the adjustable variable K = 20K t appears as a multiplying factor. K t = 0.449 satisfies the dominance condition to a reasonable extent. For K t = 1.413, complex-conjugate poles are not dominant. 7.18 The characteristic equation of the system is s(s + 1) (s + 3) + 2s + 2u = 0 which may be rearranged as 1 + K s s s ( ) 2 4 5 - - = 0 ; K = 2u – o A = – 1.3333 ; o A = 60º, – 60º, 180º ; intersection with the ju-axis at ± j 5 ; departure angle o p from complex pole in the upper half of s-plane = – 63.43º; multiple roots at s = – 1, – 1.666. The point on ¸ = 0.5 line that satisfies the angle criterion is s d = – 0.63 + j1.09. The value of K at this point, obtained by magnitude criterion, is 4.32. Therefore u = 2.16. The third pole for K = 4.32 is at s = – 2.75. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 6 1 7.19 The characteristic equation of the system is s 3 + 9s 2 + 18s + 10u = 0 which may be rearranged as 1 + K s s s ( )( ) - - 3 6 = 0 ; K = 10u – o A = – 3 ; o A = 60º, – 60º, 180º; intersection with the ju-axis at ± j3 2 ; multiple roots at s = – 1.268. The point on ¸ = 0.5 line that satisfies the angle criterion is s d = – 1 + j1.732. The value of K at this point is 28. Therefore u = 2.8. The third pole for K = 28 is at s = – 7. 7.20 (a) 1 + K s s ( ) ÷ 2 = 0 System is unstable for all K > 0. (b) In practice, the poles and zeros can never exactly cancel since they are determined by two independent pieces of hardware whose numerical values are neither precisely known nor precisely fixed. We should therefore never attempt to cancel poles in the right-half plane, since any inexact cancellation will result in an unstable closed-loop system. (c) 1 + K s s s s ( ) ( )( ) + − + 1 2 8 = 0 From the Routh array, we find that the loci cross the ju-axis when K = 19.2. For K > 19.2, the closed-loop poles are always in the left- half plane and the system is stable. E s R s ( ) ( ) = 1 1- G s ( ) = s s s s s s Ks K ( )( ) ( )( ) ÷ - ÷ - - - 2 8 2 8 For R(s) = 1/s and K > 19.2 (required for stability), e ss = lim s÷0 sE(s) = 0 For R(s) = 1/s 2 and K > 19.2, e ss = lim s÷0 sE(s) = ÷16 K 7.21 – o A = – 2 ; o A = 60º, 180º, 300º; intersection with the ju-axis at ± j 5 ; multiple roots at s = – 0.473. The value of K at the point of intersection of the root locus and ¸ = 0.3 line is 7.0. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 6 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN K v = lim s÷0 sG(s) = K/5 = 1.4 ; t s = 4/¸u n = 11.6 sec. With the compensator, the characteristic equation becomes 1 + K s s s s s ( ) ( )( )( ) 10 1 100 1 1 5 - - - - = 0 or 1 + ´ - - - - K s s s s s ( . ) ( . )( )( ) 0 1 0 01 1 5 = 0 ; K´ = 0.1K The value of K´ at the point of intersection of ¸ = 0.3 line and the root locus is 6.0. Therefore K = 60. K v = lim s÷0 sD(s) G(s) = K/5 = 12, ; t s = 4/¸u n = 12.7 sec. 7.22 1 + K s s s s ( .5) ( )( ) - - - 2 1 u = 0 Desired closed-loop pole; s d = – 1.6 + j4. Angle criterion at s d is satisfied when u = 5.3. The value of K at s d is 23.2. The third pole is at s = – 3.1. 7.23 (a) 1 + K s s ( ) - 2 = 0 ¸ = 0.707 ; t s = 4 ¸u n = 2. Therefore, ¸u n = 2 Proportional control does not meet this requirement. (b) 1 + K sK s s c D - - ( ) 2 = 1 + K s K K s s D c D ( / ) ( ) - - 2 At the points of intersection of ¸ = 0.707 line and ¸u n = 2 line, the angle criterion is satisfied when K c /K D = 4. The root locus plot of 1 + K s s s D ( ) ( ) - - 4 2 = 0 has circular complex-root branches (refer Problem 7.7). At the point of intersection of ¸ = 0.707 line and ¸u n = 2 line, we find by magnitude criterion that K D = 2. Therefore K c = 8. 7.24 G(s) = K s s s s ( . ) ( . ) - - - 0 1 0 8 4 2 ; K = 4000 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 6 3 D(s) = s s s s 2 0 8 4 0 384 10 42 - - - - . ( . )( . ) D(s) improves transient response considerably, and there is no effect on steady-state performance. 7.25 ¸ = 0.45 (specified). Let us select the real part of the desired roots as ¸u n = 4 for rapid settling. The zero of the compensator is placed at s = – z = – 4, directly below the desired root location. For the angle criterion to be satisfied at the desired root location, the pole is placed at s = – p = – 10.6. The gain of the compensated system, obtained by magnitude criterion, is 96.5. The compensated system is then D(s)G(s) = 96.5 4 2 10 6 ( ) ( )( . ) s s s s - - - K v = lim s÷0 sD(s) G(s) = 18.2. The K v of the compensated system is less than the desired value of 20. Therefore, we must repeat the design procedure for a second choice of the desired root. 7.26 Desired root location may be taken as s d = – 1.588 + j3.152 It meets the requirements on M p and settling time. With D(s) = s s - - 1 10 , angle criterion is met. Magnitude criterion gives K = 147. The compensated system has K v = 2.94. 7.27 ¸ = 0.707; 4/¸u n = 1.4 ÷ ¸u n = 2.85 u n = 4 ; u ¸ n 1 2 ÷ = 2.85 s d = – 2.85 + j2.85 With D(s) = ( .5) ( .5) s s - - 1 3 2 2 , angle criterion is satisfied at s d . The value of K, found by magnitude criterion, is 37. 7.28 From the root locus plot of the uncompensated system, we find that for gain of 1.06, the dominant closed-loop poles are at – 0.33 ± j 0.58. The value of ¸ is 0.5 and that of u n is 0.66. The velocity error constant is 0.53. The desired dominant closed-loop poles are – 0.33 ± j0.58 with a K v of 5. The lag compensator http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 6 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN D´(s) = s s - - 0 1 0 001 . . gives K v boost of the factor of 10. The angle contribution of this compensator at – 0.33 + j0.58 is about seven degrees. Since the angle contribution is not small, there will be a small change in the new root locus near the desired dominant closed-loop poles. If the damping of the new dominant poles is kept at ¸ = 0.5, then the dominant poles from the new root locus are found at – 0.28 ± j 0.51. The undamped natural frequency reduces to 0.56. This implies that the transient response of the compensated system is slower than the original. The gain at – 0.28 + j0.51 from the new root locus plot is 0.98. Therefore K = 0.98/1.06 = 0.925 and D(s) = 0 0 1 0 01 .925( . ) . s s - - 7.29 The dominant closed-loop poles of uncompensated system are located at s = –3.6 ± j4.8 with ¸ = 0.6. The value of K is found as 820. Therefore K v = lim s÷0 sG(s) = 820/200 = 4.1. Lag compensator D(s) = s s - - 0 0 025 .25 . gives a K v boost of the factor of 10. The angle contribution of this compensator at – 3.6 + j4.8 is – 1.8º, which is acceptable in the present problem. 7.30 (a) 1 + K s s ( )( .5 ) 2 1 0 1 - - = 1 + K s s ( .5)( ) - - 0 2 = 0 From the root locus we find that no value of K will yield ¸u n = 0.75. (b) D(s) = K T s c I 1 1 - F H G I K J = K c + K I /s = K c s K K s I c - F H I K / D(s)G(s) = K s K K s s s c I c ( / ) ( .5)( ) - - - 0 2 At the point corresponding to ¸ = 0.6 and ¸u n = 0.75, the angle criterion is satisfied when K I /K c = 0.75. The gain at the desired point is K c = 2.06. This yields K I = 1.545. 7.31 K v = 20 demands K = 2000. However, using Routh criterion, we find that when K = 2000, the roots of the characteristic equation are at ± j10. Clearly the roots of the system when K v requirement is satisfied are a long way from satisfying the ¸ requirement. It will be difficult to bring the dominant http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 6 5 roots from the ju-axis to the ¸ = 0.707 line by using a lead compensator. Therefore, we will attempt to satisfy the K v and ¸ requirements by a lag compensator. From the uncompensated root locus we find that corresponding to ¸ = 0.707, the dominant roots are at – 2.9 ± j2.9 and the value of K is 236. Therefore necessary ratio of zero to pole of the compensator is | | | | z p = 2000 236 = 8.5 We will choose ¸ = 0.1 and p = 0.1/9. For this choice, we find that the angle contribution of the compensator at – 2.9 + j.2.9 is negligible. 7.32 D(s)G(s) = K s s s s s s s ( / )( / ) ( )( )( / )( / ) - - - - - - 1 1 1 5 1 1 2 1 2 . . u. u . 1 1 . = 0.05 ; 1 2 . = 1 ; 1 1 u. = 0.005 ; u . 2 = 10 D(s)G(s) = K s s s s s ( . ) ( . )( )( ) - - - - 0 05 0 005 5 10 For ¸ = 0.45, we obtain s d = – 1.571 + j3.119, K = 146.3, K v = 29.3, u n = 3.49 7.33 D(s) = K c + K D s = K c 1- F H G I K J K K s D c (a) D(s)G(s) = K K K s s c D c 1 1 2 - F H G I K J - s 2 + 1 + K c + K D s = (s + 1 + j 3 ) (s + 1 – j 3 ) This equation gives K c = 3 and K D = 2 K p = lim s÷0 D(s)G(s) = 3. (b) With PID controller D(s) = K s s s ( .2)( . ) - - 0 1 3332 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 6 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN the angle criterion at s d = – 1 + j 3 is satisfied. The gain K is 2.143. 7.34 (a) Uncompensated system: M p = 20% ÷ ¸ = 0.45 At the point of intersection of ¸ = 0.45 line and the root locus, ¸u n = 1.1 and K = 56 (b) t s = 4/¸u n = 3.6 sec K v = lim s÷0 sG(s) = 2.07 (c) Compensated system: M p = 15% ÷ ¸ = 0.55 t s = 3 6 2 . .5 = 1.45 ; ¸u n = 4/1.45 = 2.75 u n = 2 75 0 . .55 = 5 ; s d = – 2.75 + j4.18 K v > 20 Lead-lag compensated system: D 1 (s) D 2 (s) G(s) = K s s s s s s s ( .9) ( . )( )( . ) - - - - - - 3)( 0 0 15 3)( 9 13 6 ; K = 506 7.35 G(s) = K s s KK t ( ) - - 2 E s R s ( ) ( ) = 1 1- G s ( ) = s s KK s s KK K t t ( ) ( ) - - - - - 2 2 For R(s) = 1/s 2 , e ss = 2 0 35 - s KK K t . ¸ > 0.707 ; t s s 3 sec. The characteristic equation of the system is s 2 + 2s + KK t s + K = 0 which may be rearranged as http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 6 7 1 + KK s s s K t 2 2 - - = 0 The locus of the roots as K varies (set KK t = 0) is determined from the following equation: 1 + K s s ( ) - 2 = 0 For K = 20, the roots are – 1 ± j4.36. Then the effect of varying KK t is determined from the locus equation 1 + KK s s j s j t ( . )( . ) - - - ÷ 1 4 36 1 4 36 = 0 Note that complex root branches follow a circular path. At the intersection of the root locus and ¸ = 0.707 line, we obtain KK t = 4.3. The real part of the point of intersection is o = 3.15, and therefore the settling time is 1.27 sec. 7.36 The characteristic equation of the system is 1 + KK s s s K t 2 2 - - = 0 or 1 + KK s s j K s j K t ( ( )( ( ) - - ÷ - ÷ ÷ 1 1 1 1 = 0 The root contour plotted for various values of K with K´ t = = KK t varying from 0 to ~ are shown in the figure below. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 6 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN jw s – 1 j – 2 j 1 j K = ¢ t 0 K = ¢ t • • ¨ K ¢ t K = ¢ t 0 K = ¢ t 0 K = 2 K = 5 K = ¢ t 0 j2 – ÷5 – ÷2 – 1 7.37 G(s) = AK s s s KK s t ( )( ) - - - 1 5 The characteristic equation is s(s + 1) (s + 5) + KK t (s + A/K t ) = 0 which may be rearranged as 1 + KK s A K s s s t t ( / ) ( )( ) - - - 1 5 = 0 The angle criterion at s = – 1 + j2 is satisfied when A/K t = 2.5. Let us take A = 2 and K t = 0.8. K v = lim s÷0 sG(s) = AK KK t 5 - With K = 10, K v = 1.54 The closed-loop transfer function of the system is Y s R s ( ) ( ) = 20 1 2 1 2 4 ( )( )( ) s j s j s - - - ÷ - 7.38 The characteristic equation of the system is s(s + 1) (s + 5) + KK t s + AK = 0 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 6 9 or s(s + 1) (s + 5) + [s + u = 0 ; [ = KK t ; u = AK The root locus equation as a function of u with [ = 0 is 1 + u s s s ( )( ) - - 1 5 = 0 Sketch a root locus plot for u varying from 0 to ~. The points on these loci become the open-loop poles for the root locus equation 1 + [ u s s s s ( )( ) - - - 1 5 = 0 For some selected values of u, the loci for [ are sketched (refer Fig. 7.44). 7.39 e –s = ÷ ÷ - ( ) s s 2 2 1 – K s s s ( ) ( )( ) ÷ - - 2 1 2 = 0 = 1 – F(s) F(s) = K j j j ( ) ( )( ) o u o u o u - ÷ - - - - 2 1 2 = K j j ( ) ( )( ) ( o u o o u u o ÷ - - - ÷ - - 2 1 2 2 3) 2 ZF(s) = tan –1 u o u o o o u ÷ ÷ - - - ÷ ÷ 2 2 3) 1 2 1 2 tan ( ( )( ) = tan –1 u o u o o o u u o u o o o u ÷ ÷ - - - ÷ - ÷ - - - ÷ L N M O Q P R S | | T | | U V | | W | | ( ) 2 2 3) 1 2 1 2 2 3) 1 2 2 2 ( ( ) ( ( )( ) ZF(s) = 0º if u o u o o o u ÷ ÷ - - - ÷ 2 2 3) 1 2 2 ( ( )( ) = 0 Manipulation of this equation gives (o – 2) 2 + u 2 = 12 Centre = (2, 0) and radius = 12 = 3.464 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 7 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN It can easily be verified that at every point on this circle, ZF(s) = 0º. Revisiting Example 7.14 will be helpful. 7.40 The proof runs parallel to that in Example 7.2, with a change in angle criterion as K s 0. Centre at zero at s = 0 Radius = u [ 2 2 - = ( .5) ( .5) 0 0 2 2 - = 0.707 The root locus plot is shown in the figure below. j0.5 – 0.5 KÆ – • KÆ – • K = 0 K = 0     7.41 G(s) = 20 7 3) 2 8 . ( ( )( ) s s s s - - - (i) u = u 0 + Au = 8 + Au The characteristic equation is s(s + 2) (s + 8) + Au s(s + 2) + 20.7 (s + 3) = 0 When Au = 0, the roots are (may be determined by root-locus method or the Newton-Raphson method) i 1,2 = – 2.36 ± j2.48, x 3 = – 5.27 The root locus for Au is determined using the root locus equation 1 + Au x x x s s s s s ( ) ( )( )( ) - ÷ ÷ ÷ 2 1 2 3 = 0 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 7 1 The angle of departure at s = x 1 , is – 80º. Near s = x 1 , the locus may be approximated by a line drawn from – 2.36 + j2.48 at an angle of – 80º. For a change of Ax 1 = 0.2 Z – 80º along the departure line, Au is determined by magnitude criterion: Au = 0.48 Therefore the sensitivity at i 1 is S u x - 1 = A A x u u 1 0 / = 0 80 0 48 8 .2 º . / Z ÷ = 3.34 Z – 80º (ii) Sensitivity of the root s = x 1 to a change in zero at s = – 3 is determined as follows. [ = [ 0 + A[ = 3 + A[ The characteristic equation is s(s + 2) (s + 8) + 20.7 (s + 3 + A[) = 0 or 1 + 20 7 1 2 3 . ( )( )( ) A[ x x x s s s ÷ ÷ ÷ = 0 The angle of departure of x 1 is 50º. For a change of Ax 1 = 0.2 Z + 50º, we obtain A[ = 0.21. Therefore S [ x 1 = A A x [ [ 1 0 / = 0 50 0 3 .2 º .21 / Z - = 2.84 Z + 50º The sensitivity of the system to the pole can be considered to be less than the sensitivity to the zero because of the direction of departure from the pole at s = x 1 . 7.42 (a) The characteristic equation is 1 + K s s s s ( ) ( )( - - - 5 2 3) = 0 It can easily be verified from the root locus plot that for K = 8 the closed-loop poles are x 1,2 = – 0.5 ± j 3.12, i 3 = – 4 (b) s(s + 2 + ù) (s + 3) + 8(s + 5) = 0 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 7 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN or 1 + ù s s s s s s ( ) ( )( ) ( ) - - - - - 3 2 3 8 5 = 0 or 1 + ù x x x s s s s s s ( ) ( )( )( ) - ÷ ÷ ÷ 3 1 2 3 = 0 Case I: ù > 0 From the root locus plot we find that the direction of departure from the pole at s = x 1 , is away from the ju-axis. Case II: ù < 0 From the root locus plot we find that the direction of departure from the pole at s = x 1 is towards the ju-axis. Therefore the variation ù < 0 is dangerous. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 8 THE NYQUIST STABILITY CRITERION AND STABILITY MARGINS 8.1 (a) Revisiting Example 8.1 will be helpful. (i) 1Z 0º (ii) 0Z – 180º (iv) J J J J 1 2 1 2 90 + Z÷ º (b) Revisiting Example 8.2 will be helpful. (i) – J – j~ (ii) 0Z – 180º (c) Revisiting Example 8.3 will be helpful. (i) ~Z – 180º (ii) 0Z – 270º (d) (i) – (J 1 + J 2 ) – j ~ (ii) 0 Z – 270º (iii) J J J J 1 2 1 2 180 + Z÷ º (e) (i) ~Z÷180º (ii) 0 Z 0º (iv) J J J J J J 1 2 1 2 1 2 90 ( ) º + Z÷ 8.2 (a) Number of poles of G(s) in right half s-plane, P = 1 Number of clockwise encirclements of the critical point, N = 1 Z = number of zeros of 1 + G(s) in right half s-plane = N + P = 2 The closed-loop system is unstable. (b) P = 2 Number of counterclockwise encirclements of the critical point = 2 Therefore N = number of clockwise encirclements = – 2 Z = N + P = – 2 + 2 = 0 The closed-loop system is stable. (c) P = 0 N = (Number of clockwise encirclements of the critical point – number of counterclockwise encirclements of the critical point) = 0 Z = N + P = 0; the closed-loop system is stable. 8.3 The polar plot of G(jM) for M = 0 + to M = + ~ is given in Fig. P8.3b. Plot of G(jM) for M = – ~ to M = 0 – is the reflection of the given polar plot with respect to the real axis. Since G j ( ) M M0 = ~Z – 180º, G(s) has double pole at the origin. The map of Nyquist contour semicircle s = He jB , H ÷ 0, B varying from – 90º at M = 0 – through 0º to + 90º at M = 0 + , into the Nyquist plot is given by ~Z – 2B (an infinite semicircle traversed clockwise). http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 74 CONTROL SYSTEMS: PRINCIPLES AND DESIGN With this information, Nyquist plot for the given system can easily be drawn. (i) P = 0, N = 0; Z = N + P = 0 The closed-loop system is stable (ii) P = 1, N = 0; Z = N + P = 1 The closed-loop system is unstable (iii) P = 0, N = 0; Z = N + P = 0 The closed-loop system is stable. 8.4 (a) The key points to the polar plot are: G j H j ( ) ( ) M M M~ = 18Z 0º G j H j ( ) ( ) M M M~ = 0Z – 270º The intersections of the polar plot with the axes of G(s)H(s)-plane can easily be ascertained by identifying the real and imaginary parts of G(jM)H(jM). When we set Im [G(jM)H(jM)] to zero, we get M = 4.123 and G j H j ( ) ( ) . M M M4 123 = – 1.428 Similarly, setting Re [G(jM)H(jM)] to zero, we get intersection with the imaginary axis. Based on this information, a rough sketch of the Nyquist plot can easily be made. From the Nyquist plot, we find that N = 2. Since P = 0, we have Z = N + P = 2, i.e., two closed-loop poles in right half s-plane. (b) The key points of the polar plot are (refer Problem 8.1d): G j H j ( ) ( ) M M M0 = – 3 – j ~ G j H j ( ) ( ) M M M~ = 0Z – 270º Intersection with the real axis at M = 1/ 2 ; G j H j ( ) ( ) M M M 1 2 = ÷ 4 3 The map of Nyquist contour semicircle s = He jB , H ÷ 0, B varying from – 90º at M = 0 – through 0º to + 90º at M = 0 + , into the Nyquist plot is given by ~Z – B (an infinite semicircle traversed clockwise). Based on this information, a rough sketch of Nyquist plot can easily be made. P = 0, N = 2. Therefore Z = N + P = 2 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 7 5 (c) G j H j ( ) ( ) M M M0 = 2Z – 180º G j H j ( ) ( ) M M M~ = 0Z – 90º No more intersections with real/imaginary axis. From the Nyquist plot, we find that N = – 1. Since P = 1, we have Z = N + P = 0; the closed- loop system is stable. (d) G j H j ( ) ( ) M M M0 = 0Z 90º ; G j H j ( ) ( ) M M M1 = 2.6Z161º ; G j H j ( ) ( ) M M M2 = 2.6 Z 198º ; G j H j ( ) ( ) M M M10 = 0.8Z – 107º G j H j ( ) ( ) M M M~ = 0Z – 90º The Nyquist plot is given in Fig. P8.2b. P = 2, N = – 2 ; Z = N + P = 0 (e) G j H j ( ) ( ) M M M0 = ~Z – 180º G j H j ( ) ( ) M M M~ = 0 Z – 90º No intersections with real/imaginary axis. The map of Nyquist contour semicircle s = He jB , H ÷ 0, B varying from – 90º at M = 0 – through 0º to + 90º at M = 0 + , into the Nyquist plot is given by ~Z – 2B (an infinite semicircle traversed clockwise). With this information, Nyquist plot for the given system can easily be drawn. P = 0, N = 0 ; Z = N + P = 0 (f) G j H j ( ) ( ) M M M0 = 1 100 0 Z º G j H j ( ) ( ) u u u ÷ 10 = ~ Z 0º Consider the Nyquist contour shown in figure below. For the semi- circle s = j10 + He jB , H ÷ 0, B varying from – 90º at M = 0 – through 0º to + 90º at M = 10 + , G(s)H(s) = lim H÷0 1 10 100 2 ( ) j e j + + H B = lim ( ) H B B H H ÷ + 0 1 20 e j e j j = lim H B H ÷ ÷ 0 1 e j It is an infinite semicircle from M = 10 – to M = 10 + traversed clockwise. G j H j ( ) ( ) M M M~ = 0Z – 180º http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 76 CONTROL SYSTEMS: PRINCIPLES AND DESIGN The Nyquist plot is shown in the figure below. The Nyquist plot passes through – 1 + j0 point. The Nyquist criterion is not applicable. – 1 + 0 j 10 j – 10 j s – j• j• Re Im w = • jw w = 0 + w = 10 + Nyquist plot Nyquist contour 100 1 (g) G j H j ( ) ( ) M M M0 = – 8 + j~ G j H j ( ) ( ) M M M~ = 0º Z – 90º G j H j ( ) ( ) M M M 3 = – 2 + j0 No intersection with the imaginary axis. The map of Nyquist contour semicircle s = He jB , H ÷ 0, B varying from – 90º at M = 0 – through 0°to + 90° M = 0 + , into the Nyquist plot is given by ~Z – B (an infinite semicircle traversed clockwise). With this information, a rough sketch of Nyquist plot can easily be made (refer Fig. P8.2a). P = 1, N = – 1. Therefore Z = N + P = 0. 8.5 G(jM)H(jM) = ÷ + + + ÷ + K j K ( ) ( ) ( ) ( ) 1 1 1 2 1 2 2 2 1 2 1 2 2 2 1 2 M J J M M J M J J M M J (i) For J 1 < J 2 , the polar plot of G(jM)H(jM) is entirely in the third quadrant as shown in figure below. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 7 7 P = 0, N = 0. Therefore Z = N + P = 0; the closed-loop system is stable. (ii) For J 1 > J 2 , the polar plot of G(jM)H(jM) is entirely in the second quadrant, as shown in figure below. P = 0, N = 2. Therefore Z = N + P = 2; the closed-loop system is unstable. Re (i) (ii) Im w = • w = • w = – • w = – • w = 0 – w = 0 – w = 0 + w = 0 + 8.6 (a) G j ( ) M M0 = 4Z0º ; G j ( ) M M~ = 0 Z – 270º Intersection with the real axis at – 0.8. The critical point –1 + j0 will be encircled by the Nyquist plot if K > 1/0.8. Since P = 0, we want net encirclements of the critical point to be zero for stability. Therefore K must be less than 5/4. (b) G(s) = 4 1 1 0 1 2 ( ) ( . ) + + s s s = ¢ ´ + + K s s s ( ) ( ) J J 2 2 1 1 1 ; J 1 < J 2 The Nyquist plot for this transfer function has already been given in the Solution of Problem 8.5. From this plot we see that the closed-loop system is table for all K. (c) G(s) = 4 1 0 1 1 2 ( . ) ( ) + + s s s = ´ + + K s s s ( ) ( ) J J 2 2 1 1 1 ; J 1 > J 2 The Nyquist plot for this transfer function has already been given in the Solution of Problem 8.5. From this plot we see that the closed-loop system is unstable for all K. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 78 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (d) G(jM) = e j j ÷ + 0 8 1 . M M = 1 1 2 +M [(cos 0.8M – M sin 0.8M) – j(sin 0.8M + M cos 0.8M)] The imaginary part is equal to zero if sin 0.8M + M cos 0.8M = 0 This gives M = – tan 0.8M Solving this equation for smallest positive value of M, we get M = 2.4482. G j ( ) M M0 = 1Z 0º ; G j ( ) . M M2 4482 = – 0.378 + j0 G j ( ) M M~ = 0 The polar plot will spiral into the M ÷~ point at the origin (refer Fig. 8.39). The critical value of K is obtained by letting G(j 2.4482) equal – 1. This gives K = 2.65. The closed-loop system is stable for K < 2.65. 8.7 The figure given below shows the Nyquist plot of G(s)H(s) for K = 1. As gain K is varied, we can visualize the Nyquist plot in this figure expanding (increased gain) or shrinking (decreased gain) like a balloon. Since P = 2, we require N = – 2 for stability, i.e., the critical point must be encircled ccw two times. This is true if – 1.33 K < – 1 or K > 0.75. Re – 1.33 – 1 – 1 Im w = • w = 0 + w = ÷11 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 7 9 8.8 1 0 K G j H j ( ) ( ) M M M = – 1 + j0 ; 1 K G j H j ( ) ( ) M M M~ = 0.05 + j0 1 0 62 K G j H j ( ) ( ) . M M M = – 0.167 + j0 The Nyquist plot is shown in the figure below. If the critical point lies inside the larger loop, N = 1. Since P = 1, we have Z = N + P = 2 and the closed-loop system is unstable. If the critical point lies inside the smaller loop, N = – 1, Z = N + P = 0 and the closed-loop system is stable. Therefore, for stability – 0.167K < – 1 or K > 6. Re – 0.167 – 1 Im w = • w = 0 + 8.9 Eliminating the minor-loop we obtain forward-path transfer function of an equivalent single-loop system. G(s) = K s s s ( .5) + + + 0 1 3 2 By Routh criterion, we find that the polynomial (s 3 + s 2 + 1) has two roots in the right half s-plane. Therefore P = 2. 1 0 K G j ( ) M M = 0.5Z0º ; 1 K G j ( ) M M~ = 0Z – 180º http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 80 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 1 1 4 K G j ( ) . M M = – 0.5 + j0 The polar plot intersects the positive imaginary axis; the point of intersection can be found by setting real part of 1 K G j ( ) M , equal to zero. With this information, Nyquist plot can easily be constructed. From the Nyquist plot we find that the critical point is encircled twice in ccw if – 0.5 K < – 1 or K > 2. For this range of K, the closed loop system is stable. The figure given below illustrates this result. Re – 0.5 0.5 0.5 j 1.5 j 1 j Im w = 1.4 w = 0 + w = • 1 K G j ( ) – plot w 8.10 Fig. P8.10a: G(s) = Ke s s s s ÷ + + 2 1 4 1 ( )( ) 1 0 K G j ( ) M M = ~Z – 90º; 1 0 26 K G j ( ) . M M = – 2.55 + j0 1 K G j ( ) M M ~ = 0 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 8 1 The polar plot will spiral into the M ÷ ~ point at the origin (refer Fig. 8.39). It is found that the maximum value of K for stability is 1/2.55 = 0.385 Fig. P8.10b: D(s)G(s) = Ke s s s ÷ + 2 2 4 1 ( ) 1 0 K D j G j ( ) ( ) M M M = ~Z – 180º ; 1 K D j G j ( ) ( ) M M M ~ = 0 The polar plot will spiral into the M ÷ ~ point at the origin. A rough sketch of the Nyquist plot is shown in figure below. It is observed that the system is unstable for all values of K. Re Im w = 0 + 8.11 Revisiting Example 8.15 will be helpful. 1 + G(s) = 0 may be manipulated as e s s s D ÷ + J ( ) 1 = – 1 or G 1 (s) = – e s D J ; G 1 (s) = 1 1 s s ( ) + The polar plot of G 1 (jM) intersects the polar plot of – e j D M J at a point corresponding to M = 0.75 at the G(jM)-locus. At this point, the phase of the polar plot of –e j D M J is found to be 52º. Therefore, http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 82 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 0.75J D = 52(F/180) This gives J D = 1.2 sec 8.12 G j ( ) M M0 = – 2.2. – j ~; G j ( ) M M~ = 0Z – 270º G j ( ) . M M15 9 = – 0.182 + j0; G j ( ) . M M6 2 = 1 Z – 148.3º GM = 1/0.182 = 5.5 ; .M = 31.7º M g = 6.2 rad/sec ; M B = 15.9 rad/sec. 8.13 (a) This result can easily proved using Routh criterion. (b) General shape of the Nyquist plot of given transfer function is as shown in Fig. P8.2c. When K = 7, the polar plot intersects the negative real axis at the point – 1.4 + j0. The resulting Nyquist plot encircles the critical point once in clockwise direction and once in counter- clockwise direction. Therefore N = 0. Since P = 0, the system is stable for K = 7. Reducing the gain by a factor of 1/1.4 will bring the system to the verge of instability. Therefore GM = 0.7. The phase margin is found as +10º. 8.14 (a) The system is type-0; so the low-frequency asymptote has a slope of 0 dB/decade, and is plotted at dB = 20 log 25 = 27.96. Asymptote slope changes to –20 dB/decade at the first corner frequency M c1 = 1; then to – 40 dB/decade at M c2 = 10, and to – 60 dB/decade at M c3 = 20. Asymptotic crossing of the 0 dB-axis is at M g = 16. (b) The system is type-1; so the low-frequency asymptote has a slope of –20 dB/decade and is drawn so that its extension would intersect the 0 dB-axis at M = K = 50. The asymptote slope changes to – 40 dB/decade at the first corner frequency at M c1 = 1; then to – 20 dB/decade at M c2 = 5, and back to – 40 dB/decade at M c3 = 50. The asymptotic gain crossover is at M g = 10.2. (c) The system is type-2; the low-frequency asymptote is drawn with a slope of – 40 dB/decade, and is located such that its extension would intersect the 0 dB-axis at M = K = 500 = 22.36. The asymptote slope changes to – 60 dB/decade at the first corner frequency M c1 = 1; then to – 40 dB/decade at M c2 = 5, to – 20 dB/decade at M c3 = 10 and to – 40 dB/decade at M c4 = 50. The asymptotic gain crossover is at M g = 10. (d) The system is type-1, and the low-frequency asymptote has a slope of – 20 dB/decade and is drawn so that its extension would cross the http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 8 3 0 dB-axis at M = K = 50. The asymptote slope changes to –40 dB/decade at M c1 = 10; then to –20 dB/decade at M c2 = 20; to –40 dB/decade at M c3 = 50; and to – 80 dB/decade at M c4 = 200 (note that the corner frequency for the complex poles is M c4 = M n = 200). The asymptotic gain crossover is at M g = 26. 8.15 (a) The low-frequency asymptote has a slope of – 20 dB/decade and passes through the point (M = 1, dB = 20). The asymptote slope changes to – 40 dB/decade at M c = 10. Compensate the asymptotic plot by – 3 dB at M = 10, by – 1dB at M = 5 and by – 1 dB at M = 20. The compensated magnitude plot crosses the 0 dB line at M g = 7.86. The phase is computed from ZG(jM) = – 90º – tan –1 0.1M The phase shift at M g = 7.86 is –128.2º, and the resulting phase margin is – 128.2º – (– 180º) = 51.8º. Because the phase curve never reaches – 180º line, the gain margin is infinity. (b) Low-frequency asymptote has a slope of – 20 dB/decade and passes through the point (M = 1, dB = 20) The asymptote slope changes to –60 dB/decade at the corner frequency M c = 10. Compensate the asymptotic plot by – 6 dB at M = 10, by – 2 dB at M = 5 and – 2 dB at M = 20. The compensated magnitude plot crosses the 0 dB line at M g = 6.8. The phase is computed from ZG(jM) = – 90º – 2 tan –1 0.1M The phase shift at M g = 6.8 is – 158.6º. Therefore, the phase margin is 180º – 158.6º = 21.4º. At a frequency of M B = 10, the phase shift is – 180º; the gain margin is 6 dB. (c) The low-frequency asymptote has a slope of –20 dB/decade and passes through the point (M = 0.1, dB = 20log 200 = 46). The slope changes to – 40 dB/decade at the first corner frequency M c1 = 2, and back to – 20 dB/decade at the second corner frequency M c2 = 5. Compensate the asymptotic plot by – 3 dB at M = 2, by – 1 dB at M = 1, by – 1 dB at M = 4, by + 3 dB at M = 5, by + 1 dB at M = 2.5 and by + 1 dB at M = 10. The compensated magnitude plot crosses the 0 dB line at M g = 9. The phase shift at this frequency is – 106.6º. Therefore the phase margin is 73.4º. Phase never reaches – 180º line; the gain margin is infinity. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 84 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (d) Reconsider the bode plot for system of Problem 8.15c. The time-delay factor e –0.1s will not change the magnitude plot; it will change only the phase characteristics. For the system with dead-time, M g = 9 rad/sec; .M = 22º; M B = 13.65 rad/sec; GM = 4.2 dB (e) G(s) = 40 2 4 5 ( )( )( ) s s s + + + = 1 1 2 1 1 4 1 1 5 1 s s s + F H I K + F H I K + F H I K The low-frequency asymptote coincides with 0 dB line; its slope changes to – 20 dB/decade at M c1 = 2, to – 40 dB/decade at M c2 = 4, and to – 60 dB/decade at M c3 = 5. Compensated magnitude plot can easily be obtained by applying corrections. The phase plot crosses the – 180º line at M B = 7 rad/sec, and the gain margin is 20 dB. Since the gain never reaches 0 dB (M g = 0), the phase margin is infinity. (f) From the Bode plots of G(jM) we find that gain crossover frequency M g = 3.16 rad/sec, and the phase margin is –33º. The phase plot is asymptotic to the –180º line in the low-frequency range; it never reaches – 180º Therefore the GM = – ~ 8.16 G(s) = K s s s / ( ) 5 1 1 5 1 + + F H I K From the Bode plot of the system sketched for K = 10, we find that .M = 21º, GM = 8 dB. Increasing the gain from K = 10 to K = 100 shifts the 0 dB axis down by 20 dB. The phase and gain margins for the system with K = 100 are .M = – 30º, GM = – 12 dB Thus the system is stable for K = 10 but unstable for K = 100. 8.17 (a) G(s) = K s s s / 40 1 2 1 1 4 1 1 5 1 + F H I K + F H I K + F H I K Since G(s) has all poles in left half s-plane, the open-loop system is stable. Hence the closed-loop system will be stable if the frequency response has a gain less than unity when the phase is 180º. When K = 40, the gain margin is 20 dB (refer Problem 8.15e). Therefore, an increase in gain of +20 dB (i.e., by a factor of 10) is possible before the system becomes unstable. Hence K < 400 for stability. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 8 5 (b) From the Bode plot of G(jM) with K = 1, we find that the phase crossover frequency M B = 15.88 rad/sec and gain margin = 34.82 dB. This means that the gain can be increased by 34.82 dB (i.e., by a factor of 55) before the system becomes unstable. Hence K < 55 for stability. (c) From the Bode plot of G(jM) with K = 1, we find that M B = 0.66 rad/sec and GM = 4.5 dB. Thus the critical value of K for stability is 1.67, i.e., K < 1.67 for the system stability. 8.18 From the Bode plot of G(jM), it can easily be determined that (a) when J D = 0, the GM = 12 dB, and .M = 33º; and (b) when J D = 0.04 sec, the GM = 2.5 dB and .M = 18º. The relative stability of the system reduces due to the presence of dead-time. (c) The value of J D for the system to be on verge of instability is obtained by setting the phase margin equal to zero, i.e., G j g 1 ( ) M M M – M J F g D ×180º = – 180º where G 1 (jM) is the system without dead-time, and M g is the gain crossover frequency. From the Bode plot of G 1 (jM), we get M g = 7.4 rad/sec and ZG 1 (jM g ) = – 147º. Therefore, – 147º – 7 4 180 . º J F D × = – 180º or J D = 0.078 sec 8.19 Since the systems have minimum-phase characteristics, zeros (if any) are in left half s-plane. (a) The low-frequency asymptote has a slope of – 20 dB/decade and its extension intersects the 0 dB-axis at M = 4. The system is therefore type-1; 4/s is a factor of the transfer function. The asymptote slope changes from – 20 dB/decade to 0 dB/decade at M = 2 (this corresponds to the factor (1 + s/2)), then to – 20 dB/decade at M = 10 (this corresponds to the factor (1 + s/10)). The transfer function G(s) = 4 1 1 2 1 1 10 + F H I K + F H I K s s s (b) The low-frequency asymptote has a slope of + 20 dB/decade and it http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 86 CONTROL SYSTEMS: PRINCIPLES AND DESIGN intersects 0 dB-axis at M = 0.2. The system is therefore type-0; Ks is a factor of the transfer function with 20 log 0.2K = 0 or K = 5. The low-frequency asymptote has a magnitude of 20 dB at a frequency M c1 , where 20 log (5M c1 ) = 20. This gives M c1 = 2. The transfer function G(s) = 5 1 1 2 1 1 10 1 1 30 s s s s + F H I K + F H I K + F H I K (c) The low-frequency asymptote has a slope of – 6 dB/octave with a magnitude of – 9 dB at M = 1. The system is therefore type-1; K/s is a factor of the transfer function with 20 log (K/M) = – 9 at M = 1. This gives K = 0.355. The transfer function G(s) = 0 355 1 1 1 20 1 1 40 . ( ) + + F H I K + F H I K s s s s (d) The low-frequency asymptote has a slope of – 20dB/decade with a magnitude of 40 dB at M = 2.5. The system is therefore type-1; K/s is a factor of the transfer function with 20 log (K/2.5) = 40. This gives K = 250. G(s) = 250 1 1 2 1 1 40 s s s + F H I K + F H I K .5 (e) The low-frequency asymptote has a slope of +12 dB/octave; the system is therefore type-0 and Ks 2 is a factor of the transfer function. The first corner frequency M c1 = 0.5. The slope of the asymptote changes to +6 dB/decade at this corner frequency. Therefore 1/(2s + 1) is a factor of the transfer function. The asymptotic plot of Ks 2 /(2s + 1) is shown in the figure below. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 8 7 w 20 log Kw 2 – 20 log 2w 20 10 –20 –10 0.1 1 0.5 0 32 dB From this figure, we find that (20 log KM 2 – 20 log 2M)| M = 1 = 32 This gives K = 79.6. The transfer function G(s) = 79 6 1 2 1 1 0 2 . ( )( )( .2 ) s s s s + + + 8.20 The low-frequency asymptote has a slope of 20 dB/decade. Ks is a factor of the transfer function with 20logK = 30. This gives K = 31.623. The transfer function G(s) = 31 623 1 1 1 5 1 1 20 . ( ) s s s s + + F H I K + F H I K (i) 20 log (31.623M g1 ) = 0; this gives M g1 = 0.0316 (ii) 20 log (31.623M g2 ) – 20 log M g2 – 20 log 1 5 2 M g F H I K – 20 log 1 20 2 M g F H I K = 0; this gives M g2 = 56.234 8.21 The low-frequency asymptote has a slope of – 6 dB/octave and has a magnitude of 11 dB at M = 3. K/s is therefore a factor of the transfer function with 20 log (K/3) = 11. This gives K = 10.64. The transfer function http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 88 CONTROL SYSTEMS: PRINCIPLES AND DESIGN G(s) = 10 64 1 3 1 1 8 1 . s s s + F H I K + F H I K ZG(jM) = – 90º – tan –1 1 3 M – tan –1 1 8 M ZG(jM) = – 180º at M = 4.9 The gain obtained from asymptotic magnitude plot at M = 4.9 is 20 log 10 64 4 20 4 3 . .9 log .9 ÷ = 2.5 dB Therefore, GM = – 2.5 dB 8.22 Revisit Review Example 8.3. G(s) = 5 1 1 10 1 1 2 1 0 6 50 1 2500 2 + F H I K + F H I K + + L N M O Q P s s s s s . 8.23 From the asymptotic magnitude plot we find that the system is type-0 with a double pole at s = – 3. The transfer function obtained from the magnitude plot is G(s) = 0 1 1 3 1 2 . s + F H I K ZG(jM) vs M may be compared with the given phase characteristics to check the accuracy of identification of corner frequency at M c = 3. —— http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 9 FEEDBACK SYSTEM PERFORMANCE BASED ON THE FREQUENCY RESPONSE 9.1 For derivation of the result, refer Section 9.3; Eqn. (9.9). 9.2 For derivation of the result, refer Section 9.3; Eqn. (9.18). 9.3 For derivation of the result, refer Section 9.3; Eqns (9.15) – (9.16). 9.4 The relations 4 – 2 + 1 4 2 M r = 0 ; M r = M n 1 2 2 ÷ yield = 0.6 and M n = 21.8 Note that M r – relation gives two values of for M r = 1.8; = 0.6 and = 0.8. We select = 0.6 as damping ratio larger than 0.707 yields no peak above zero frequency. The characteristic equation of the given system is s 2 + as + K = s 2 + 2M n s + M n 2 = 0 This gives K = 475 and a = 26.2 t s = 4 M n = 0.305 sec M b = M n 1 2 2 4 4 2 2 4 1 2 ÷ + ÷ + / = 25.1 rad/sec 9.5 From the response curve, we find M p = 0.135, t p = 0.185 sec 2 = ( ) ( ) ln ln M M p p 2 2 2 + F gives = 0.535 for M p = 0.135 F M n 1 2 ÷ = 0.185 gives M n = 20 The corresponding frequency response performance indices are: M r = 1 2 1 2 ÷ = 1.11 ; M r = M n 1 2 2 ÷ = 13.25 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 90 CONTROL SYSTEMS: PRINCIPLES AND DESIGN M b = M n 1 2 2 4 4 2 2 4 1 2 ÷ + ÷ + / = 24.6 9.6 - 0.01B m The characteristic equation of the system is J s 2 + (1 + KK t )s + K = 0 This gives M n = / K J ; = 1 2 + KK K t J = 0.01B m K = 1 2 10 2 10 10 4 2 2 4 2 K K K t m t m m t × ÷ ± × ÷ ÷ ÷ ÷ B J B J B J 9.7 The polar plot of G(jM) crosses the real axis at M = 1/ J J 1 2 . The magnitude |G(jM)| at this frequency is given by |G(jM)| = KJ J J J 1 2 1 2 + Therefore G m × KJ J J J 1 2 1 2 + = 1 This gives K = 1 1 1 1 2 G m J J + F H G I K J 9.8 (a) It can be determined from the Bode plot that M g = 1 rad/sec and .M = 59.2º (b) .M = tan –1 2 4 1 2 4 2 + ÷ .M = 59.2º ÷ = 0.6 M g = M n 4 1 2 4 2 + ÷ This gives M n = 1.4 Therefore, M p = e ÷ ÷ F / 1 2 = 9.48% http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 9 1 t s = 4 M n = 4.76 sec. 9.9 The low-frequency asymptote has a slope of – 20 dB/decade and intersects the 0 dB-axis at M = 8; 8/s is a factor of the transfer function. The forward- path transfer function can easily be identified as G(s) = 8 1 1 2 1 1 4 1 1 8 1 1 24 1 1 36 + F H I K + F H I K + F H I K + F H I K + F H I K s s s s s s The phase curve can now be constructed. From the phase curve and the asymptotic magnitude curve, we obtain .M = 50º ; GM = 24 dB Phase margin of 50º gives = 0.48 which corresponds to M p - 18%. 9.10 (a) From the Bode plot, we get .M = 12º (b) For a phase margin of 50º, we require that G(jM) H(jM) = 1 Z – 130º for some value of M. From the phase curve of G(jM)H(jM), we find that ZG(jM)H(jM) ~ ÷ –130º at M = 0.5. The magnitude of G(jM)H(jM) at this frequency is approximately 3.5. The gain must be reduced by a factor of 3.5 to achieve a phase margin of 50º. (c) .M of 50º gives = 0.48 which corresponds to M p ~ ÷ 18%. 9.11 G(jM) = K j j ( ) ( ) M M + 2 2 ZG(jM) = tan –1 M 2 – 180º = – 130º This equation gives M = 2.3835 The magnitude of G(jM) must be unity at M = 2.3835. K j j ( ) ( ) . M M M + 2 2 2 3835 = 1 This equation gives K = 1.826. Since the phase curve never reaches – 180º line, the gain margin = ~. 9.12 (a) From the open-loop frequency response table, we find that M B = 10 rad/sec and GM = 20 log 1 0 64 . = 3.88 dB; M g = 8 rad/sec and .M = 10º. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 92 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) For the desired gain margin of 20 dB, we must decrease the gain by (20 – 3.88) = 16.12 dB. It means that magnitude curve must be lowered by 16.12 dB, i.e., the gain must be changed by a factor of >, given by 20 log > = – 16.12. This gives > = 0.156. (c) To obtain a phase margin of 60º, we must first determine the frequency at which the phase angle of G(jM) is – 120º, and then adjust the gain so that |G(jM)| at this frequency is 0 dB. From the Bode plot, we find that lowering the magnitude plot by 16.65 dB gives the desired phase margin. It equivalently means that the gain should be changed by a factor of = where 20 log = = – 16.65. This gives = = 0.147. 9.13 A 9.48% overshoot implies = 0.6. For this damping, required phase margin is 59.2º. G(s) = 100 36 100 K s s s ( )( ) + + = K s s s / 36 1 36 1 1 100 1 + F H I K + F H I K Make a Bode plot for say K = 3.6. From the plot we find that at M = 14.8 rad/sec, ZG(jM) = – 120.8º. At M = 14.8, the gain is – 44.18 dB. The magnitude curve has to be raised to 0 dB at M = 14.8 to yield the required phase margin. The gain should be changed by a factor of = where 20 log = = 44.18. This gives = = 161.808. Therefore, K = 3.6 × 161.808 = 582.51 M g = 14.8 = M n 4 1 2 4 2 + ÷ This gives M n = 20.68 t p = F M n 1 2 ÷ = 0.19 sec 9.14 (a) From the Bode plot, it can easily be determined that GM = 5 dB and .M = 30º. The gain crossover frequency M g = 0.83 rad/sec. (b) The magnitude plot should be lowered by 5 dB to obtain a gain margin of 10 dB. This is achieved by reducing the gain by 5 dB. (c) The gain at M = 1.27 is –4.8 dB. Therefore, the gain should be increased by 4.8 dB to obtain a gain crossover frequency of 1.27 rad/sec. (d) Gain at the frequency which givesZ G(jM) = – 135º, should be 0 dB. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 9 3 From the Bode plot we find that this requires reducing the gain by 3.5 dB. 9.15 The requirements on .M and M g are satisfied if we increase the gain to the limit of zero GM with a phase margin ± 45º. From the Bode plot we find that K = 37.67 meets this requirement. 9.16 (a) From the Bode plot we find that 0 dB crossing occurs at a frequency of 0.47 rad/sec with a phase angle of – 145º. Therefore, the phase margin is 35º. Assuming a second-order approximation, .M = 35º ÷ = 0.33, M p = 33%. (b) The zero dB crossing occurs with a phase angle of – 118º. Therefore the phase margin is 62º. .M = 62º ÷ = 0.64, M P = 7.3%. 9.17 The 0 dB crossing occurs at M = 0.8, with a phase angle of – 140º when J D = 0, and – 183º when J D = 1. Therefore phase margin of the system without dead-time is 40º; and with dead-time added, the phase margin becomes – 3º. We find that with the dead-time added, the system becomes unstable. Therefore, the system gain must be reduced in order to provide a reasonable phase margin. We find from the Bode plot that in order to provide a phase margin of 30º, the gain would have to be decreased by 5 dB, i.e., by a factor of > where 20 log > = 5. This gives > = 1.78. We find that the dead-time necessitates the reduction in loop gain is order to obtain a stable response. The cost of stability is the resulting increase in the steady-state error of the system as the loop gain is reduced. 9.18 Make a polar plot of G(jM) = 50 18 1 3)(1 6 / ( / / ) j j j M M M + + G j ( ) M M0 = ÷ ÷ ~ 25 18 j ; G j ( ) M M~ = 0 Z – 270º The polar plot intersects the negative real axis at M = 4.24. The polar plot is tangential to the M = 1.8 circle (refer Eqn. (9.22)). Therefore, the resonance peak M r = 1.8. The bandwidth of a system is defined as the frequency at which the magnitude of the closed-loop frequency response is 0.707 of its magnitude at M = 0. For the system under consideration, closed-loop gain is unity at M = 0; therefore, bandwidth is given by the frequency at which the M = 0.707 circle intersects the polar plot. This frequency is found to be 3.61 rad/sec. Therefore M b = 3.61 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 94 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 4 – 2 + 1 4 2 M r = 0 ; M p = e ÷ ÷ × F / 1 2 100 M b = M n ( ) 1 2 4 2 2 2 4 2 ÷ + ÷ + t s = 4/M n The M r – relation gives = 0.29, 0.96 for M r = 1.8. We select = 0.29 because damping ratio larger than 0.707 yields no peak above zero frequency. = 0.29 corresponds to M p = 38.6%. For = 0.29 and M b = 3.61, we get M n = 2.4721. The setting time t s = 5.58 sec. 9.19 The polar plot is tangential to M = 1.4 circle (refer Eqn. (9.22)) at a frequency M = 4 rad/sec. Therefore M r = 1.4 and M r = 4. 4 – 2 + 1 4 2 M r = 0 M r = M n 1 2 2 ÷ M r = 1.4 ÷ = 0.39. The corresponding M p = 26.43%. M r = 4 and = 0.39 ÷ M n = 4.8. The settling time t s = 2.14 sec. The M r – relationship has been derived for standard second-order systems with zero-frequency closed-loop gain equal to unity. The answer is based on the assumption that zero-frequency closed-loop gain for the system under consideration is unity. 9.20 From the Bode plot, we find that M g = 8.3 rad/sec ; M B = 14.14 rad/sec; GM = 8.2 dB ; .M = 27.7º For each value of M, the magnitude and phase of G(jM) are transferred from the Bode plot to the Nichols chart. The resulting dB vs phase curve is tangential to M = 6.7 dB contour of the Nichols chart at M = 9 rad/sec. Therefore M r = 6.7 dB and M r = 9. The dB vs phase curve intersects the – 3 dB contour of the Nichols chart at M = 13.6. Therefore bandwidth M b = 13.6 rad/sec. The following points are worth noting in the use of Nichols chart. (i) The frequency parameters (M r or M b ) can be easily determined by transferring the dB and/or phase data from the dB vs phase curve to the Bode plot. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 9 5 (ii) The open-loop dB vs phase plot may not be tangential to one of the constant-M contours of the Nichols chart. One must do a little interpolation. In the present problem, M r = 6.7 dB has been obtained by interpolation. 9.21 Plot the given frequency response on a dB scale against phase angle. From the dB vs phase curve, we obtain, GM = 7 dB ; .M = 17º The dB vs phase curve when transferred on Nichols chart gives M r = 10 dB; M r = 2.75 rad/sec; M b = 4.2 rad/sec Note that the –3dB bandwidth definition is applicable to systems having unity closed-loop gain at M = 0. We have made this assumption for the system under consideration. 9.22 For parts (a) and (b) of this problem, refer Problem 9.10. (c) Since the data for –3dB contour of the Nichols chart has been provided, we don’t require the Nichols chart for bandwidth determination. On a linear scale graph sheet, dB vs phase cure of the open-loop frequency response is plotted with data coming from Bode plot. On the same graph sheet, the – 3dB contour using the given data is plotted. The intersection of the two curves occurs at M = 0.911 rad/sec. Therefore bandwidth M b = 0.911. (d) .M = tan –1 2 4 1 2 4 2 + ÷ M b = M n 1 2 2 4 4 2 2 4 ÷ + ÷ + .M = 50º ÷ = 0.48 ; M p - 18% M b = 0.911 and = 0.47 ÷ M n = 0.7025 ; t s = 11.93 sec. 9.23 The frequency-response data from the Bode plot of G(jM) when transferred to the Nichols chart, gives the following result. M r = 1.4 ; M r = 6.9 rad/sec From the relations M r = 1 2 1 2 ÷ ; M r = M n 1 2 ÷ 2 we obtain = 39 ; M n = 8.27 rad/sec http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 96 CONTROL SYSTEMS: PRINCIPLES AND DESIGN Note that the given type-0 system has zero-frequency closed-loop gain = 54/(25 + 54) = 0.6835. The approximation of the transient response of the system by = 0.39 and M n = 8.27 has large error since the correlations (M r , M r ) ÷ (, M n ) used above, have been derived from systems with unity zero-frequency closed-loop gain. 9.24 (a) Transfer the frequency-response data from the Bode plot of G(jM) to the Nichols chart. Shifting the dB vs phase curve down by 5.4 dB makes this curve tangent to the M = 20 log1.4contour. Therefore gain must be changed by a factor of >, where 20 log > = – 5.4. This gives > = 0.54. (b) The gain-compensated system intersects the –3 dB contour of the Nichols chart at M = 9.2. Therefore bandwidth M b = 9.2 rad/sec. Note that the – 3dB bandwidth definition given by Eqn. (9.2b) is applicable to systems having unity closed-loop gain at M = 0. The given system does not satisfy this requirement. The answer, therefore, is an approximation of the bandwidth. 9.25 (a) From the Bode plot we find the phase crossover frequency M B = 2.1 and GM = 14 dB. For the gain margin to be 20 dB, the magnitude plot is to be brought down by 6 dB. The value of K corresponding to this condition is given by the relation 20 log K = – 6 which gives K = 0.5 (b) From the Bode plot it is observed that .M = 60º will be obtained if gain crossover frequency M g is 0.4 rad/sec. This is achieved if magnitude plot is brought down by 7 dB. This condition corresponds to K = 0.446. (c) From the dB-phase plot on the Nichols chart, we find that for this plot to be tangent to M = 1 dB contour, the dB-phase curve must be brought down by 4 dB. This condition corresponds to K = 0.63. The corresponding value of M r (read off from Bode plot) is 0.5 rad/sec. The value of bandwidth (M b ) is the frequency at which the dB-phase plot intersects the – 3dB contour. It is found that M b = 1 rad/sec. (d) The dB vs phase plot on the Nichols chart when raised by 2.4 dB, intersects the –3 dB contour at M = 1.5 rad/sec. This condition corresponds to K = 1.35. 9.26 Plot the given frequency response on a dB scale against phase angle. From this dB vs phase curve, we obtain GM = 16.5 dB ; .M = 59º http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 9 7 Transfer the dB vs phase curve on the Nichols chart. We find that this curve when raised by about 5 dB, touches the M = 20 log1.4 contour of the Nichols chart. Therefore the gain should be increased by a factor of >, where 20 log > = 5. This gives > = 1.75. The gain-compensated system has GM = 11.6 dB and .M = 42.5º. 9.27 The dB vs phase plot becomes tangential to 20 log1.4 dB contour on Nichols chart if the plot is lowered by about 11 dB. This means that the gain K must be reduced by a factor of = where 20 log (1/=) = –11. This gives = = 3.5. Phase margin of the gain-compensated system is 42.5º M r = 1.4 ÷ = 0.387 ; .M = 42.5º ÷ = 0.394 The following comments may be carefully noted. (i) Zero-frequency closed-loop gain has been assumed to be unity. (ii) One is usually safe if the lower of the two values of is utilized for analysis and design purposes. 9.28 S G M (jM) = 1 1+ G j ( ) M = G j G j ÷ ÷ + 1 1 1 ( ) ( ) M M Magnitude of S G M (jM) can be obtained by plotting G –1 (jM) on the Nichols chart. The dB vs phase plot of G –1 (jM) is tangent to M = 20 log 2.18 dB contour of Nichols chart. Therefore | S G M (jM)| max = 2.18 The peak occurs at M = 7 rad/sec. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 10 COMPENSATOR DESIGN USING BODE PLOTS 10.1 (a) From the Bode plot of G(jM)H(jM), we find that GM = 6 dB and .M = 17º (b) Now the gain and phase of the compensator with K c = 1 are added to the Bode plot. From the new Bode plot we find that the gain must be raised by 1.5 dB to have a gain margin of 6 dB. 20 log K c = 1.5 gives K c = 1.2 10.2 (a) From the Bode plot of G(jM)H(jM) we find that M g = 4.08 rad/sec; .M = 3.9º; GM = 1.6 dB (b) Now the gain and phase of the compensator are added to the Bode plot. From the new Bode plot, we find that M g = 5 rad/sec; .M = 37.6º; GM = 18 dB The increase in phase and gain margins implies that lead compensation increases margin of stability. The increase in M g implies that lead compensation increases speed of response. 10.3 (a) K v =lim s 0 sD(s)G(s) = K = 12 (b) From the Bode plot and Nichols chart analysis of the system with K=12, we find that .M = 15º; M b = 5.5 rad/sec; M r = 12 dB; M r = 3.5 rad/sec (c) The phase margin of the uncompensated system is 15º. The phase lead required at the gain crossover frequency of the compensated magnitude curve = 40º – 15º + 5º = 30º = = 1 30 1 30 sin º sin º = 0.334 The frequency at which the uncompensated system has a magnitude of – 20 log (1/ = ) = – 4.8 dB is 4.6 rad/sec. Selecting this frequency as gain crossover frequency of the compensated magnitude curve, we set 1 =J = 4.6 The transfer function of the lead compensator becomes D(s) = 12 0 376 1 0 128 1 ( . ) . s s (d) .M = 42º; M b = 9 rad/sec; M r = 3 dB; M r = 4.6 rad/sec http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 9 9 10.4 Lead compensator D(s) = 10 0 5 1 01 1 ( . ) . s s meets the phase margin and steady- state accuracy requirements. These requirements are also met by the lag compensator D(s) = 10 10 1 100 1 ( ) s s In this particularly simple example, specifications could be met by either compensation. In more realistic situations, there are additional performance specifications such as bandwidth and there are constraints on loop gain. Had there been additional specifications and constraints, it would have influenced the choice of compensator (lead or lag). 105. The settling time and peak overshoot requirements on performance may be translated to the following equivalent specifications: = 0.45 ; M n = 2.22 is related to the phase margin by the relation .M 0 01 . = 45º M n is related to the bandwidth M b by the relation M b = M n 1 2 2 4 4 2 2 4 1 2 / For a closed-loop system with = 0.45, we estimate from this relation M b = 1.33 M n Therefore, we require a closed-loop bandwidth M b ~ 3. The gain K of the compensator D(s) = K s s ( ) J =J 1 1 may be set a value given by K = M n 2 . This gives K ~ 5. To provide a suitable margin for settling time, we select K = 10. The phase margin of the uncompensated system is 0º because the double integration results in a constant 180º phase lag. Therefore, we must add a 45º phase lead at the gain crossover frequency of the compensated magnitude curve. Evaluating the value of =, we have = = 1 45 1 45 sin º sin º = 0.172 To provide a margin of safety, we select = = 1/6. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 0 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN The frequency at which the uncompensated system has a magnitude of – 20 log (1/ = ) = – 7.78 dB is 4.95 rad/sec. Selecting this frequency as the gain crossover frequency of the compensated magnitude curve, we set 1 =J = 4.95 The closed-loop system with the compensator D(s) = 10 0 083 1 0 1 ( . ) .5 s s satisfies the performance specifications. 10.6 K v = A e ss = A A 0 05 . = 20 K = 20 realizes this value of K v . From the Bode plot of G(jM) = 20 0 1 j j M M ( .5 ) we find that the phase margin of the uncompensated system is 18º. The phase lead required at the gain crossover frequency of the compensated magnitude curve = 45º – 18º + 3º = 30º. = = 1 30 1 30 sin º sin º = 0.334 The frequency at which the uncompensated system has a magnitude of – 20 log (1/ = ) = – 4.8 dB is 8.4 rad/sec. Selecting this frequency as the gain crossover frequency of the compensated magnitude curve, we set 1 =J = 8.4 Therefore the compensator D(s) = 0 1 0 0668 1 .2 . s s The phase margin of the compensated system is found to be 43.7º. If we desire to have exactly a 45º phase margin, we should repeat the steps with a decreased value of =. From the Nichols chart analysis of the compensated and uncompensated systems, we find that the lead compensator has increased the bandwidth from 9.5 rad/sec to 12 rad/sec. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 0 1 10.7 From the Bode plot of G(jM) = 20 0 1 j j M M ( .5 ) we find that the phase margin of the uncompensated system is 18º. To realize a phase margin of 45º, the gain crossover frequency should be moved to M g where the phase angle of the uncompensated system is: – 180º + 45º + 5º = – 130º. From the Bode plot of uncompensated system, we find that M g = 1.5. The attenuation necessary to cause M g to be the new gain crossover frequency is 20 dB. The > parameter of the lag compensator can now be calculated. 20 log > = 20 This gives > = 10. Placing the upper corner frequency of the compensator a decade below M g , we have 1/J = 0.15. Therefore, the lag compensator D(s) = 6.66 1 66.6 1 s s As a final check, we numerically evaluate the phase margin and the bandwidth of the compensated system. It is found that that .M = 45º and M b = 2.5 rad/sec. 10.8 From the steady-state requirement, we set K = 100. From the Bode plot of G(jM) = 100 0 1 0 1 j j j M M M ( . )( .2 ) we find that the phase margin is – 40º, which means that the system is unstable. The rapid decrease of phase of G(jM) at the gain crossover frequency M g = 17 rad/sec, implies that single-stage lead compensation may be ineffective for this system. (The reader should, in fact, try a single-stage lead compensator). For the present system, in which the desired K v is 100, a phase lead of more than 85º is required. For phase leads greater than 60º, it is advisable to use two or more cascaded stages of lead compensation (refer Fig. 10.15). The design approach may be that of achieving a portion of the desired phase margin improvement by each compensator stage. We first add a single-stage lead compensator that will provide a phase lead of about 42.5º, i.e., the compensator will improve the phase margin to about 2.5º. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 0 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN Since the phase curve of the Bode plot of the uncompensated system at the gain crossover frequency has a large negative slope, the value of = = 0.19 given by = = 1 42 1 42 sin .5º sin .5º will not yield a phase margin of 2.5º. In fact, when we set = = 0.08 and J = 0.1, the single-stage compensator improves the phase margin to 3º. Therefore D 1 (s) = 0 1 1 0 008 1 . . s s Notice that the single-stage lead compensator not only improves the phase margin, but also reduces the slope of the phase curve of the gain crossover frequency. By adding another stage of the compensator with D 2 (s) = D 1 (s), we obtain D 2 (s)D 1 (s)G(s) = 100 0 1 1 0 1 0 008 1 2 ( . ) ( .2 )( . ) s s s s Note that pole and zero at s = – 10 cancel each other. The final compensated system has a phase margin of 45º. 10.9 G(s) = 2500 25 K s s ( ) K = 1 satisfies the steady-state performance requirement. The phase margin of the uncompensated system, read at the gain crossover frequency M g = 47 rad/sec, is 28º. The phase lead required at the gain crossover frequency of the compensated magnitude curve = 45º – 28º + 8º = 25º = = 1 25 1 25 sin º sin º = 0.405 The frequency at which the uncompensated system has a magnitude of – 20 log (1/ = ) = – 3.93 dB is 60 rad/sec. Selecting this frequency as gain crossover frequency of the compensated magnitude curve, we set 1 =J = 60 The transfer function of the lead compensator is D(s) = 0 026 1 0 01 1 . . s s http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 0 3 The phase margin of the compensated system is found to be 47.5º. By Nichols chart analysis, it is found that the lead compensator reduces the resonance peak from 2 to 1.25 and increases the bandwidth from 74 to 98 rad/sec. 10.10The phase margin of the uncompensated system G(s) = 2500 25 K s s ( ) ; K = 1 read at the gain crossover frequency M g = 47 rad/sec is 28º. To realize a phase margin of 45º, the gain crossover frequency should be moved to M g where the phase angle of the uncompensated system is: – 180º + 45º = – 135º. From the Bode plot of the uncompensated system we find that M g = 25. Since the lag compensator contributes a small negative phase when the upper corner frequency of the compensator is placed at 1/10 of the value of M g , it is a safe measure to choose M g at somewhat less than 25 rad/sec, say, 20 rad/sec. The attenuation necessary to cause M g = 20 to be the new gain crossover frequency is 14 dB. The > parameter of the lag compensator can now be calculated. 20 log > = 14. This gives > = 5 Placing the upper corner frequency of the compensator a decade below M g , we have 1/J = 2. Therefore the lag compensator D(s) = 0 1 2 1 .5 .5 s s From the Bode plot–Nichols chart analysis we find that the compensated system has .M = 51º; M r = 1.2 and M b = 27.5 rad/sec. 10.11K v = lim s0 sG(s) = 5 From the Bode plot we find that the uncompensated system has a phase margin of – 20º; the system is therefore unstable. We attempt a lag compensator. This choice is based on the observation made from Bode plot of uncompensated system that there is a rapid decrease in phase of G(jM) near M g . Lead compensator will not be effective for this system. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 0 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN To realize a phase margin of 40º, the gain crossover frequency should be moved to M g where the phase angle of the uncompensated system is: – 180º + 40º + 12º = – 128º From the Bode plot of the uncompensated system, it is found that M g = 0.5 rad/sec. The attenuation necessary to cause M g to be the new gain crossover frequency is 20 dB. The > parameter of the lag compensator can now be calculated. 20 log > = 20. This gives > = 10 Since we have taken a large safety margin of 12°, we can place the upper corner frequency of the compensator at 0.1 rad/sec, i.e., 1/J = 0.1. Thus the transfer function of the lag compensator becomes D(s) = 10 1 100 1 s s From the Bode plot of compensated system we find that the phase margin is about 40° and the gain margin is about 11 dB. 10.12It easily follows that K = 30 satisfies the specification on K v . From the Bode plot of G(jM) = 30 0 1 1 0 1 j j j M M M ( . )( .2 ) we find that gain crossover frequency is 11 rad/sec and phase margin is – 24º. Nichols chart analysis gives M b = 14 rad/sec. If lead compensation is employed, the system bandwidth will increase still further, resulting in an undesirable system which will be sensitive to noise. If lag compensation is attempted, the bandwidth will decrease sufficiently so as to fall short of the specified value of 12 rad/sec, resulting in a sluggish system. These facts can be verified by designing lead and lag compensa- tors. We thus find that there is need to go in for lag-lead compensation. Since the full lag compensator will reduce the system bandwidth excessively, the lag section of the lag-lead compensator must be designed so as to provide partial compensation only. The lag section, therefore, should move the gain crossover frequency to a value higher than the gain crossover frequency of the fully lag-compensated system. We make a choice of new gain crossover frequency as M g = 3.5 rad/sec. The attenuation necessary to cause M g to be the new gain crossover frequency is 18.5 dB. This gives the > parameter of the lag section as 20 log > = 18.5; > = 8.32, say, 10 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 0 5 Placing the upper corner frequency of the lag section at 1/J 1 = 1, we get the transfer function of the lag section as D 1 (s) = J >J 1 1 1 1 s s = s s 1 10 1 It is found that the lag-section compensated system has a phase margin of 24º. We now proceed to design the lead section. The implementation of the lag- lead compensator is simpler if = and > parameters of the lead and lag compensators, respectively, are related as = = 1/>. Let us first make this choice. If our attempt fails, we will relax this constraint on =. The frequency at which the lag-section compensated system has a magnitude of – 20 log (1/ = ) = – 10 dB is 7.5 rad/sec. Selecting this frequency as gain crossover frequency of the lag-lead compensated magnitude curve, we set 1 2 =J = 7.5 The transfer function of the lead-section becomes D 2 (s) = J =J 2 2 1 1 s s = 0 422 1 0 0422 1 . . s s The analysis of the lag-lead compensated system gives .M = 48º and M b = 13 rad/sec. 10.13 2 = ( ) ( ) ln ln M M p p 2 2 2 F M p = 0.2 = 0.456 t p = F M n 1 2 t p = 0.1 M n = 35.3 M b = M n [ ] / 1 2 2 4 4 2 2 4 1 2 = 46.576 .M = tan –1 2 4 1 2 4 2 ~ 48º http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 0 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN In order to meet the specification of K v = 40, K must be set at 1440. From the Bode plot of G(jM) = 144000 36 100 j j j M M M ( )( ) we find that the gain crossover frequency is 29.7 rad/sec and phase margin is 34º. The phase lead required at the gain crossover frequency of the compensated magnitude curve = 48º – 34º + 10º = 24º = = 1 24 1 24 sin º sin º = 0.42 The frequency at which the uncompensated system has a magnitude of – 20 log ( / ) 1 = = – 3.77 dB is 39 rad/sec. Selecting this frequency as gain crossover frequency of the compensated magnitude curve, we set 1 =J = 39 The transfer function of the lead compensator becomes D(s) = 1440 0 04 1 0 0168 1 ( . ) . s s It can easily be verified by Nichols chart analysis that the bandwidth of the compensated system exceeds the requirement. We assume the peak time specification is met. This conclusion about the peak time is based on a second-order approximation that should be checked via simulation. 10.14K = 2000 meets the requirement on K v . We estimate a phase margin of 65º to meet the requirement on . (a) From the Bode plot we find that the uncompensated system with K = 2000 has a phase margin of zero degrees. From the Bode plot we observe that there is a rapid decrease of phase at the gain crossover frequency. Since the requirement on phase lead is quite large, it is not advisable to compensate this system by a single-stage lead compensator (refer Fig. 10.15). (b) Allowing 10º for the lag compensator, we locate the frequency at which the phase angle of the uncompensated system is: – 180º + 65º + 10º = – 105º. This frequency is equal to 1.5 rad/sec. The gain crossover frequency M g should be moved to this value. The necessary attenuation is 23 dB. The > parameter of the lag com- pensator can now be calculated. 20 log > = 23. This gives > = 14.2 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 0 7 Placing the upper corner frequency of the compensator one decade below M g , we have 1/J = 0.15. Therefore the lag compensator D(s) = 6 66 1 94 66 1 . . s s The phase margin of the compensated system is found to be 67º. (c) The bandwidth of the compensated system is found to be M b = 2.08 rad/sec. 10.15Approximate relation between B m and is 0.01B m = 0.4 B m = 40º Let us try lag compensation. The realize a phase margin of 40º, the gain crossover frequency should be moved to M g where the phase angle of the uncompensated system is: – 180º + 40º + 10º = – 130º. From the Bode plot of the uncompensated system we find that, M g = 6. The attenuation necessary to cause M g to be the new gain crossover frequency is 9 dB. 20 log > = 9. This gives > ~ 3 Placing the upper corner frequency of the lag compensator two octaves below M g , we have 1/J = 1.5. The lag compensator D(s) = 0 67 1 2 1 . s s Nichols chart analysis of the compensated system gives M b = 11 rad/sec. Using second-order approximation M b = M n 1 2 2 4 4 2 2 4 1 2 / we get M n = 10 Therefore t s = 4/M n = 1 sec Note that the zero-frequency closed-loop gain of the system is nonunity. Therefore, the use of – 3 dB bandwidth definition, and the second-order system correlation between M b and & M n may result in considerable error. Final design must be checked by simulation. 10.16(a) From the Bode plot of uncompensated system we find that .M = 0.63º http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 0 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) Phase margin of the system with the second-order compensator is 9.47º. There is no effect of the compensator on steady-state performance of the system. (c) The lead compensator D(s) = 1 0 0378 1 0 0012 . . s s meets the requirements on relative stability. 10.17 K v = 4.8 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 11 HARDWARE AND SOFTWARE IMPLEMENTATION OF COMMON COMPENSATORS 11.1 (a) E s Y s ( ) ( ) = – R R C s D D 1 / = – T s T s D D = 1 ; = = R/R D ; T D = R D C D (b) E s Y s ( ) ( ) = – R R R sR C 1 2 2 1 / ( ) = – K T s T s c D D ( ) 1 1 = K c = R R R 1 2 ; = = R R R 1 1 2 ; T D = R 2 C 11.2 Z 1 (s) = R 2 + R 1 /(1 + R 1 Cs); Z 2 (s) = R 1 + R 2 E s E s 2 1 ( ) ( ) = – Z s Z s 2 1 1 ( ) ( ) [ ] = J =J s s 1 1 ; J = R 1 C, = = R R R 2 1 2 < 1 Refer Section 11.2 (Fig. 11.4) for the Bode plot and filtering properties of the lead compensator. 11.3 Z 1 (s) = R 1 + R 2 ; Z 2 (s) = R 2 + R 1 /(1 + R 1 Cs) E s E s 2 1 ( ) ( ) = – Z s Z s 2 1 ( ) ( ) = – J >J s s 1 1 ; J = R R R R C 1 2 1 2 , > = R R R 1 2 2 > 1 Add an inverting amplifier of gain unity. Refer Section 11.2 (Fig. 11.7) for the Bode plot and filtering properties of the lag compensator. 11.4 Z 1 (s) = (R 1 C 1 s + 1)R 3 /[(R 1 + R 3 )C 1 s + 1] Z 2 (s) = (R 2 C 2 s + 1)R 4 /[(R 2 + R 4 )C 2 s + 1] E s E s 2 1 ( ) ( ) = – Z s Z s R R 2 1 6 5 ( ) ( ) L N M O Q P = K s s s s c F H G I K J F H G I K J F H G I K J F H G I K J 1 1 1 1 1 2 1 2 J J =J >J J 1 = (R 1 + R 3 )C 1 ; J 2 = R 2 C 2 ; = = R R R 1 1 3 ; > = R R R 2 4 2 ; http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 1 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN K c = R R R R R R R R R R 2 4 6 1 3 5 1 3 2 4 F H G I K J 11.5 Z 1 (s) = R 1 /(1 + R 1 Cs); Z 2 (s) = R 2 + 1/C 2 s E s E s 2 1 ( ) ( ) = – Z s Z s R R 2 1 4 3 ( ) ( ) L N M O Q P = K c 1 1 1 L N M O Q P T s T s D K c = R R C R C R R C 4 1 1 2 2 3 1 2 ( ) = 39.42; T I = R 1 C 1 + R 2 C 2 = 3.077 T D = R R C C R C R C 1 2 1 2 1 1 2 2 = 0.7692 These equations give R 1 = R 2 = 153.85 k9; R 4 = 197.1 k9 11.6 Z 1 (s) = R 1 /(1 + R 1 C 1 s); Z 2 (s) = R 2 /(1 + R 2 C 2 s) E s E s 2 1 ( ) ( ) = L N M O Q P Z s Z s R R 2 1 4 3 ( ) ( ) = R R R R R C s R C s 4 2 3 1 1 1 2 2 1 1 F H G I K J = 2.51 0 345 1 0185 1 . . s s F H I K This equation gives R 1 = 34.5 k9; R 2 = 18.5 k9; R 4 = 45.8 k9 11.7 By Trapezoidal rule for integration: e t dt kT ( ) 0 z = T e e T e T e T e k T e kT ( ) ( ) ( ) ( ) ... ( ) ( ) 0 2 2 2 1 2 L N M O Q P = T e i T e iT T i k ( ) ( ) L N M M O Q P P 1 1 By backward-difference approximation for derivatives: de t dt t kT ( ) = e kT e k T T ( ) (( ) ) 1 A difference equation model of the PID controller is, therefore, given by http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 1 1 u(k) = K c e k T T e i e i T T e k e k I i k D ( ) ( ) ( ) [ ( ) ( )] R S T U V W 1 2 1 1 11.8 Taking sampling frequency twenty times the closed-loop natural frequency (which is a measure of bandwidth), we have M s = 20M n = 2F T ; this gives T = 0.5 sec Settling time t s = 4/M n = 13 1 10 th of t s is 1.3 sec; T = 0.5 sec is therefore a safe choice. U s E s ( ) ( ) = 2 2 0 1 0 01 . ( . ) . s s u(t) + 0.01 u(t) = 2.2 e(t) + 0.22 e(t) 1 T [u(k) – u(k – 1)] + 0.01 u(k) = 2 2 . T [e(k) – e(k – 1)] + 0.22 e(k) This gives u(k) = 2 2 01 . u(k – 1) + 4 62 2 01 . . e(k) – 4 4 2 01 . . e(k – 1) Configuration of the digital control scheme is shown in Fig. 11.19. 11.9 (a) u(t) + au(t) = K e(t) + Kb e(t) By backward-difference approximation: u k u k T au k ( ) ( ) ( ) 1 = K T [e(k) – e(k – 1)] + Kbe(k) By backward-rectangular rule for integration: u(t) = u(0) – a u d Ke t Ke Kb e d t t ( ) ( ) ( ) ( ) J J J J z z 0 0 0 u(k) = u(k – 1) – aT u(k) + K e(k) – K e(k – 1) + Kb T e(k) By both the approaches of discretization, we get the following computer algorithm. u(k) = 1 1 1 1 1 1 1 aT u k K bT aT e k K aT e k ( ) ( ) ( ) ( ) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 1 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) u(k) = u(k – 1) – aT u k u k ( ) ( ) L N M O Q P 1 2 + Ke(k) – Ke(k – 1) + KbT e k e k ( ) ( ) L N M O Q P 1 2 = 1 2 1 2 aT aT u(k – 1) + K bT aT 1 2 1 2 F H I K e(k) – K bT aT 1 2 1 2 F H I K e(k – 1) 11.10 For a choice of T = 0.015 sec, sampling frequency M s = 2F/T = 418.88 rad/sec which is about 11 times the closed-loop bandwidth. This is a safe choice. U s E s ( ) ( ) = s s s 2 2 2 36 112 36 1 740 1 ( ) . F H I K 1.826 × 10 –6 u(t) + 2.7 × 10 –3 u(t) + u(t) = 7.7 × 10 –4 e(t) + 0.03 e(t) + e(t) Using the results of Eqns (11.36) – (11.37), we get 1.826 × 10 –6 u k u k u k T ( ) ( ) ( ) L N M O Q P 2 1 2 2 + 2.7 × 10 –3 u k u k T ( ) ( ) L N M O Q P 1 + u(k) = 7.7 × 10 –4 e k e k e k T ( ) ( ) ( ) L N M O Q P 2 1 2 2 + 0.03 e k e k T ( ) ( ) L N M O Q P 1 + e(k) This gives 1.1881 u(k) – 0.1962 u(k – 1) + 0.0081 u(k – 2) = 6.42 e(k) – 8.84 e(k – 1) + 3.42 e(k – 2) By direct digital design, we will be able to achieve the desired closed-loop performance using longer value of sampling interval. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 1 3 11.11 B s Y s ( ) ( ) = K s s t J 1 J b(t) + b(t) = K t y(t) b(t) = b(0) – 1 0 0 J G G J b d K y t y t t ( ) [ ( ) ( )] z b(k) = b(k – 1) – T b k b k K y k y k t J J ( ) ( ) [ ( ) ( )] L N M O Q P 1 2 1 = 1 2 1 2 T T / / J J b(k – 1) + K T y k K T y k t t / / ( ) / / ( ) J J J J 1 2 1 2 1 11.12 The set-point control, proportional action and derivative action are all included in the op amp circuit of Fig. 11.43. Only the integral action is to be added. Connect the integral-action unit of Fig. 11.41 in cascade with the proportional action unit of Fig. 11.43; the output e 4 of the proportional- action unit of Fig. 11.43 becomes input to the integral-action unit (output e 3 ). The resulting op amp circuit is governed by the following relation (refer Examples 11.6 and 11.7); E 3 (s) = K c (1 + T D s) 1 1 F H G I K J T s I [Y(s) – R(s)] K c = R R 2 2 ; T D = R D C D ; T I = R I C I 11.13 E 2 (s) = Y(s) – R(s) ; E 3 (s) = – T D sY(s) ; T D = R D C D E 4 (s) = K c [E 3 (s) – E 2 (s)] ; K c = R R 2 2 = K c [R(s) – (1 + T D s) Y(s)] 11.14 K cu = 1.2, T u = 4.5 min From the tuning rules given in Table 11.1: K c = 0.45K cu = 0.54 or 1 0 54 . × 100 = 185% PB T I = T u /1.2 = 3.75 min or 1 3 75 . = 0.266 repeats/min http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 1 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) K c : 110% PB ; T I : 0.355 repeats/min ; T D : 0.45 min 11.15 From the process reaction curve (refer Section 11.9) we obtain J D = 6 sec, J = 12 sec, K = 1 From tuning rules given in Table 11.2; K c = 1.8, T I = 19.98 sec 11.16 The model obtained from process reaction curve (refer Section 11.9): K = 1 ; J = 1.875 ; J D = 1.375 From tuning rules given in Table 11.2: (a) K c = 1.36 (b) K c = 1.23 ; T I = 4.58 min 11.17 J ND = 0.2 for process A, 0.5 for process B, and 0.5 for Process C. Process A is more controllable than processes B and C, which are equally controllable. 11.18 (a) From the tuning rules given in Table 11.2: K c = 3 ; T I = 4 min ; T D = 1 min (b) J CD = J D + 1 2 T = 2 min + 1 2 × 8 sec = 2.067 min. Replacing J D by J CD in the tuning formulas, we obtain K c = 2.9 ; T I = 4.13 min ; T D = 1.03 min 11.19 Replace J D in tuning formulas of Table 11.2 by J CD = J D + 1 2 T. Increase in sampling time demands reduced K c and increased T I and T D . 11.20 From tuning rules given in Table 11.1: K c = 2.25 ; T I = 28.33 sec Correction to account for sampling is not required since the test has directly been conducted on the digital loop. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ CHAPTER 12 CONTROL SYSTEM ANALYSIS USING STATE VARIABLE METHODS 12.1 + – x 4 x 3 x 2 x 1 K g K b i a i f u K T 1 1 1 1 sL + R f f sJ + B e e s n s a g ( ) L + L + R +R a g q M q L q M J e = n 2 J = 0.4 ; B e = n 2 B = 0.01 x 1 = x 2 ; 0.4 x 2 + 0.01x 2 = 1.2x 3 0.1 x 3 + 19x 3 = 100 x 4 – 1.2x 2 ; 5 x 4 + 21x 4 = 4 or x = Ax + bu A = 0 1 0 0 0 0 025 3 0 0 12 190 1000 0 0 0 4 2 L N M M M M O Q P P P P . . ; b = 0 0 0 0 2 . L N M M M M O Q P P P P y = G L = nx 1 = 0.5x 1 12.2 x 3 x 2 x 1 i f u q q K A K T s f f L + R + 1 1 1 Js + B 1 s x 1 = x 2 ; 0.5 x 2 + 0.5x 2 = 10x 3 20 x 3 + 100x 3 = 50u ; y = x 1 A = 0 1 0 0 1 20 0 0 5 L N M M M O Q P P P ; b = 0 0 2 5 . L N M M M O Q P P P ; c = 1 0 0 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 1 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 12.3 x 1 = G M ; x 2 = G M , x 3 = i a ; y = G L x 1 = x 2 2 x 2 + x 2 = 38 x 3 2 x 3 + 21x 3 = e a – 0.5 x 2 e a = k 1 (G R – G L ) – k 2 G M = k 1 G R – k 1 20 x 1 – k 2 x 2 A = 0 1 0 0 0 5 19 40 0 5 2 21 2 1 2 L N M M M M M O Q P P P P P . ( . ) k k ; b = 0 0 2 1 k L N M M M M O Q P P P P ; c = 1 20 0 0 L N M O Q P 12.4 x 1 = M ; x 2 = i a J M + BM = K T i a R a i a + L a di dt a = e a – K b M e a = K c e c = K c [k 1 (e r – K t M) – k 2 i a } A = L N M M M M O Q P P P P B J K J k K K K L R k K L T t c b a a c a ( ) ( ) 1 2 ; b = 0 1 k K L c a L N M M O Q P P ; c = 1 0 12.5 A = P –1 AP = L N M O Q P 11 6 15 8 b = P –1 b = 1 2 L N M O Q P ; c = cP = 2 1 Y s U s ( ) ( ) = P 1 1 , , = s s s 2 1 2 1 3 2 ( ) = 1 3 2 2 s s http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 1 7 X 2 X = Y 1 s –1 –2 –3 1 1 U s –1 Y s U s ( ) ( ) = P P P P 1 1 2 2 3 3 4 4 , , , , , = 2 1 8 15 1 2 1 11 2 6 2 1 1 11 8 15 6 11 8 1 1 2 1 1 1 1 1 1 1 1 1 1 s s s s s s s s s s s s s ( ) ( ) ( )[ ] ( ) [ ] [ ( )( )] [( )( )] = 1 3 2 2 s s X 2 X 1 Y s –1 –1 6 2 –15 –11 2 1 8 U s –1 12.6 A = 0 1 0 0 L N M O Q P ; b = 0 1 L N M O Q P A= P –1 AP = 1 1 –1 –1 L N M O Q P ; b= P –1 b = 0 1 L N M O Q P |I – A| = |I – A| = 2 12.7 X(s) = (sI – A) –1 x 0 + (sI – A) –1 b U(s) = G(s)x 0 + H(s) U(s) G(s) = 1 , s s s s s s s s ( ) ( ) L N M M M O Q P P P 3 3 1 1 3 1 2 ; H(s) = 1 1 2 , s s L N M M M O Q P P P http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 1 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN , = s 3 + 3s 2 + 1 12.8 –3 –1 u x 3 0 x 2 0 x 1 0 x 1 x 2 x 3 + + + + + + + + + + 12.9 Taking outputs of integrators as state variables, we get (x 1 being the output of rightmost integrator), x 1 = x 2 x 2 = – 2x 2 + x 3 x 3 = – x 3 – x 2 – y + u y = 2x 1 – 2x 2 + x 3 A = 0 1 0 0 2 1 2 1 2 L N M M M O Q P P P ; b = 0 0 1 L N M M M O Q P P P ; c = 2 2 1 12.10 (a) G(s) = c (sI – A) –1 b = s s s 3 1 2 ( )( ) (b) G(s) = 1 1 2 ( )( ) s s 12.11 x 1 = – 3x 1 + 2x 2 + [– 2x 1 – 1.5x 2 – 3.5x 3 ] x 2 = 4x 1 – 5x 2 x 3 = x 2 – r http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 1 9 A = L N M M M O Q P P P 5 0 5 3 5 4 5 0 0 1 0 . . ; b = 0 0 1 L N M M M O Q P P P ; c = 0 1 0 G(s) = c(sI – A) –1 b = 14 1 2 7 ( )( )( ) s s s 12.12 (a) x 1 = output of lag 1/(s + 2) x 2 = output of lag 1/(s + 1) x 1 + 2x 1 = x 2 ; x 2 + x 2 = – x 1 + u y = x 2 + (– x 1 + u) A = L N M O Q P 2 1 1 1 ; b = 0 1 L N M O Q P ; c = 1 1 ; d = 1 (b) x 1 = output of lag 1/(s + 2) x 2 = output of lag 1/s x 3 = output of lag 1/(s + 1) x 1 + 2x 1 = y ; x 2 = – x 1 + u x 3 + x 3 = – x 1 + u ; y = x 2 + x 3 A = L N M M M O Q P P P 2 1 1 1 0 0 1 0 1 ; b = 0 1 1 L N M M M O Q P P P ; c = 0 1 1 12.13(i) Y s U s ( ) ( ) = s s s 3 1 2 ( )( ) = 2 1 1 2 s s – A = L N M O Q P 1 0 0 2 ; b = 1 1 L N M O Q P ; c = 2 1 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 2 0 CONTROL SYSTEMS: PRINCIPLES AND DESIGN –1 2 –2 u + + + + + + x 2 x 1 y –1 (ii) Y s U s ( ) ( ) = > = = = 3 s s s 3 1 2 2 3 = 5 4 5 2 3 2 s s s From Eqns (12.46), the second companion form of the state model is given below. A = 0 0 2 1 0 5 0 1 4 L N M M M O Q P P P ; b = 5 0 0 L N M M M O Q P P P ; c = 0 0 1 2 u + + + – – – x 2 x 1 x 3 = y 2 5 4 5 (iii) Y s U s ( ) ( ) = > > > > = = = 0 3 1 2 2 3 3 1 2 2 3 s s s s s s = s s s s s s 3 2 3 2 8 17 8 6 11 6 From Eqns (12.44), the state model in second companion form is given below. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 2 1 A = 0 1 0 0 0 1 6 11 6 L N M M M O Q P P P ; b = 0 0 1 L N M M M O Q P P P ; c = 2 6 2 6 8 y 8 1 6 17 11 u + + + + + + + + + + + – x 2 x 1 x 3 z z z 12.14 (i) Y s U s ( ) ( ) = s s s s 1 3 2 3 2 = > > = = = 2 3 3 1 2 2 3 s s s s From Eqns (12.46): A = 0 0 0 1 0 2 0 1 3 L N M M M O Q P P P ; b = 1 1 0 L N M M M O Q P P P ; c = 0 0 1 (ii) Y s U s ( ) ( ) = 1 6 11 6 3 2 s s s = > = = = 3 3 1 2 2 3 s s s From Eqns (12.44): A = 0 1 0 0 0 1 6 11 6 L N M M M O Q P P P ; b = 0 0 1 L N M M M O Q P P P ; c = 1 0 0 (iii) Y s U s ( ) ( ) = s s s s s s 3 2 3 2 8 17 8 6 11 6 = 1 1 1 2 2 1 3 s s s http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 2 2 CONTROL SYSTEMS: PRINCIPLES AND DESIGN A = L N M M M O Q P P P 1 0 0 0 2 0 0 0 3 b = 1 1 1 L N M M M O Q P P P c = 1 2 1 ; d = 1 12.15 (a) Y s U s ( ) ( ) = 1000 5000 52 100 3 2 s s s s = > > = = = 2 3 3 1 2 2 3 s s s s From Eqns (12.44): A = 0 1 0 0 0 1 0 100 52 L N M M M O Q P P P ; b = 0 0 1 L N M M M O Q P P P ; c = 5000 1000 0 (b) Y s U s ( ) ( ) = 1000 5000 52 100 3 2 s s s s = 50 3125 2 18 75 50 s s s . . A = 0 0 0 0 2 0 0 0 50 L N M M M O Q P P P ; b = 50 3125 18 75 L N M M M O Q P P P . . ; c = 1 1 1 12.16 Y s U s ( ) ( ) = 2 6 5 1 2 2 2 s s s s ( ) ( ) = 1 1 1 1 1 2 2 ( ) s s s From Eqns (12.55): A = L N M M M O Q P P P 1 1 0 0 1 0 0 0 2 ; b = 0 1 1 L N M M M O Q P P P ; c = 1 1 1 x(t) = 0 t z e A(t – J) bu(J)dJ e –At = –1 [(sI – A) –1 ] = e te e e t t t t L N M M M O Q P P P 0 0 0 0 0 2 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 2 3 e u d t t A b ( ) ( ) z J J J 0 = ( ) ( ) ( ) ( ) t e d e d e d t t t t t t L N M M M M M M M M M O Q P P P P P P P P P z z z J J J J J J J 0 0 2 0 = 1 1 1 2 1 2 L N M M M M O Q P P P P e te e e t t t t ( ) y = x 1 + x 2 + x 3 = 2.5 – 2e –t – te –t – 0.5 e –2t u y + + + + + + + – x 3 z z z 1 1 1 –1 –2 –1 1 x 2 x 1 12.17 F = e AT = 0 696 0 246 0123 0 572 . . . . L N M O Q P g = (e –AT – I) A –1 b = – . . . . 0 304 0 246 0123 0 428 L N M O Q P L N M M M O Q P P P L N M O Q P 3 4 1 2 1 4 1 2 1 5 = L N M O Q P 0 021 0 747 . . c = 2 4 ; d = 6 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 2 4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 12.18 Y s U s ( ) ( ) = e s s 0 4 1 . ( ) x t 1 = – x 1 (t) + u(t – 0.4) J D = 0.4 ; therefore, N = 0, , = 0.4, m = 0.6 F = e –1 = 0.3679 g 2 = e d z I I 0 0 6 . = 0.4512 ; g 1 = e –0.6 e d z G G 0 0 4 . = 0.1809 x 1 (k + 1) = 0.3679 x 1 (k) + 0.1809 u(k – 1) + 0.4512u(k) Introduce a new state x 2 (k) = u(k – 1) x(k + 1) = Fx(k) = gu(k) y(k) = cx(k) F = 0 3679 01809 0 0 . . L N M O Q P ; g = 0 4512 1 . L N M O Q P c = 1 0 12.19 s –1 s –1 s –1 s –1 –2 x 1 x 2 x 2 0 x 2 0 –2 1 1 1 1 u G(s) = G s G s G s G s 11 12 21 22 ( ) ( ) ( ) ( ) L N M O Q P ; H(s) = H s H s 1 2 ( ) ( ) L N M O Q P X s x 1 1 0 ( ) = G 11 (s) = s s s s s s s 1 1 1 1 2 1 1 1 2 1 2 2 2 2 ( ) ( ) ( )( ) http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 2 5 = s s 1 1 1 2 ( ) , = 1 2 1 1 2 3 / / s s X s x 1 2 0 ( ) = G 12 (s) = s 2 1 ( ) , = 1 2 1 1 2 3 / / s s X s x 2 1 0 ( ) = G 21 (s) = s 2 1 ( ) , = 1 2 1 1 2 3 / / s s X s x 2 2 0 ( ) = G 22 (s) = s s 1 1 1 2 ( ) , = 1 2 1 1 2 3 / / s s X s U s 1 ( ) ( ) = H 1 (s) = s s s 2 1 1 1 1 2 ( ) ( ) , = 1 1 s X s U s 2 ( ) ( ) = H 2 (s) = 1 1 s Zero-input response: x(t) = e At x 0 = -1 [G(s)x 0 ] = 1 2 3 3 3 3 1 0 2 0 e e e e e e e e x x t t t t t t t t L N M O Q P L N M O Q P Zero-state response: x(t) = e d t t A b ( ) ( ) z J u J J 0 = –1 [H(s)U(s)] = 1 1 L N M O Q P e e t t 12.20 A = 0 1 2 3 L N M O Q P ; eigenvalues are 1 = – 1, 2 = – 2 Y(s) = c(sI – A) –1 x 0 + c(sI – A) –1 b U(s) = s s s s s s 4 1 2 1 1 2 ( )( ) ( )( ) = 3 1 2 2 1 2 1 1 1 2 2 s s s s s / / y(t) = 1 2 2 3 2 2 e e t t http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 2 6 CONTROL SYSTEMS: PRINCIPLES AND DESIGN 12.21 x 1 = – 3x 1 + 2x 2 + [7r – 3x 1 – 1.5 x 2 ] x 2 = 4x 1 – 5x 2 A = L N M O Q P 6 0 5 4 5 . ; b = 7 0 L N M O Q P ; c = 0 1 e At = –1 [(sI – A) –1 ] = 1 3 2 3 0 5 3 0 5 3 4 3 4 3 2 3 1 3 4 7 4 7 4 7 4 7 e e e e e e e e t t t t t t t t L N M M M O Q P P P . . y(t) = x 2 (t) = 7 4 3 4 3 4 7 0 e e d t t t L N M O Q P z ( ) ( ) J J J = 28 3 1 4 1 1 7 1 4 7 ( ) ( ) L N M O Q P e e t t 12.22 e At = e –t L N M O Q P te te te e te t t t t t x(t) = e t t t A x ( ) ( ) 0 0 Given: x 1 (2) = 2 ; t 0 = 1, t = 2 Manipulation of the equation gives x 1 (2) = 2e –1 x 1 (1) + e –1 x 2 (1) = 2 If x 2 (1) = 2k, then x 1 (1) = e 1 – k Thus e k k 1 2 L N M O Q P is a possible set of states x x 1 2 1 1 ( ) ( ) L N M O Q P for any k 0 12.23 e e t t L N M O Q P 2 2 2 = a b c d L N M O Q P L N M O Q P 1 2 ; e e t t L N M O Q P = a b c d L N M O Q P L N M O Q P 1 1 This gives e At = a b c d L N M O Q P = 2 2 2 2 2 2 2 2 e e e e e e e e t t t t t t t t L N M O Q P http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 2 7 = –1 [(sI – A) –1 ] (sI – A) –1 = 2 1 1 2 1 1 1 2 2 2 2 1 2 2 1 1 s s s s s s s s L N M M M O Q P P P (sI – A) = s s L N M O Q P 1 2 3 A = 0 1 2 3 L N M O Q P 12.24 V = c cA cA n L N M M M M O Q P P P P 1 is a triangular matrix with diagonal elements equal to unity; |V| = (– 1) n for all = i ’s. This proves the result. 12.25 U = [b Ab ... A n–1 b] is a triangular matrix with diagonal elements equal to unity; |U| = (– 1) n for all = i ’s. This proves the result. 12.26 (i) U = [b Ab] = 1 2 0 1 L N M O Q P ; H(U) = 2 Completely controllable V = c cA L N M O Q P = 1 1 3 3 L N M O Q P ; H(V) = 1 Not completely observable. (ii) The system is in Jordan canonical form; it is controllable but not observable. (iii) The system is in controllable companion form. The given system is therefore controllable. We have to test for observability property only. H(V) = H c cA cA 2 L N M M M O Q P P P = 3 The system is completely observable. http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ 1 2 8 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (iv) Given system is in observable companion form. We have to test for controllability property only. H(U) = H[b Ab A 2 b] = 3 The system is completely controllable. 12.27 (i) Observable but not controllable (ii) Controllable but not observable (iii) Neither controllable nor observable. Controllable and observable realization: A = – 1 ; b = 1 ; c = 1 12.28 (i) G(s)= c(sI – A) –1 b = 1 2 s Given state model is in observable companion form. Since there is a pole-zero cancellation, the state model is uncontrollable. (ii) G(s) = s s s 4 2 3 ( )( ) The given state model is in controllable companion form. Since there is a pole-zero cancellation, the model is unobservable. 12.29 (a) |I – A| = ( – 1) ( + 2) ( + 1) The system is unstable. (b) G(s) = c(sI – A) –1 b = 1 1 2 ( )( ) s s The G(s) is stable (c) The unstable mode e t of the free response is hidden from the transfer function representation 12.30 (a) G(s) = 10 2 s s = > = = 2 2 1 2 s s A = 0 1 0 1 L N M O Q P ; b = 0 1 L N M O Q P ; c = [10 0] (b) Let us obtain controllable companion form realization of G(s)= 10 2 1 2 ( ) ( )( ) s s s s = 10 20 3 2 3 2 s s s s = > > = = = 2 3 3 1 2 2 3 s s s s http://ebook29.blogspot.com/ http://ebook29.blogspot.com/ SOLUTION MANUAL 1 2 9 A = 0 1 0 0 0 1 0 2 3 L N M M M O Q P P P ; b = 0 0 1 L N M M M O Q P P P ; c = 20 10 0 (c) Let us obtain observable companion form realization of G(s) = 10 2 1 2 ( ) ( )( ) s s s s A = 0 0 0 1 0 2 0 1 3 L N M M M O Q P P P ; b = 20 10 0 L N M M M O Q P P P ; c = 0 0 1 http://ebook29.blogspot.com/ http://ebook29.blogspot.com/