Solution Manual - Control Systems by Gopal

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(1)http://ebook29.blogspot.com/. CHAPTER 2 2.1. DYNAMIC MODELS AND DYNAMIC RESPONSE. Revisiting Examples 2.4 and 2.5 will be helpful. (a). (b). R(s) = A;. Y ( s) =. y(t) =. KA - t / t e ; t. R(s) =. A ; s. KA t s +1 yss = 0. Y(s) =. KA s (τs + 1). y(t) = KA (1 − e − t / τ ) ; (c). R(s) =. yss = KA. A KA ; Y(s) = 2 s2 s (τ s + 1). y(t) = KA (t − τ + τe − t / τ ) ; (d). KAω (τs + 1)( s 2 + ω 2 ). R(s) =. Aw ; s2 + w 2. y(t) =. KAωτ − t / τ KA sin(ω t + θ ) e + 2 2 2 τ ω +1 τ ω2 +1. θ y( t ) 2.2. yss = KA (t - t ). Y(s) =. = tan −1 ( −ωτ ) t®¥. =. KA. τ ω2 +1 2. sin(ω t + θ ). (a) The following result follows from Eqns (2.57 - 2.58) y(t) = 1 −. e −ζω n t 1−ζ2. F GH. sin ω d t + 2 tan −1. 1−ζ2 ζ. I JK. ωd = ωn 1−ζ2 The final value theorem is applicable: the function sY(s) does not have poles on the jw-axis and right half of s-plane. yss = lim sY ( s ) = 1 s® 0. (b). The following result follows from Review Examples 2.2.. http://ebook29.blogspot.com/.

(2) http://ebook29.blogspot.com/. SOLUTION MANUAL. y(t) = t −. yss = t −. F GH. 2ζ e −ζω n t 1−ζ2 sin ω d t + 2 tan −1 + ωn ωd ζ. 3. I JK. 2ζ ωn. The final value theorem is not applicable; the function sY(s) has a pole on the jw -axis. 2.3. Revisiting Review Example 2.1 will be helpful. E0 (s ) R / 10 4 10 = ; Ei ( s ) = 2 Ei ( s) RCs + 1 s. e0 (t ) = 10 ´. R - t /t ( ); t = RC 4 t -t +te 10. For the output to track the input with a steady-state delay 100 × 10–6 sec, it is necessary that. R = 1, and t = RC = 100 ´ 10-6 10 4 This gives. R = 10 k W ;. C = 0.1 mF. Steady-state error = 10 t - (10 t – 10t ) = 0.001 2.4. Revisiting Review Example 2.2 will be helpful. w n = 100; z = 3; 2z / w n = 6 / 100 sec = 60 msec. FG H. Steady-state error = 25t - 25t - 25 ´. 2z wn. IJ K. = 1.5 2.5. My&&( t ) + By& ( t ) + Ky( t ) = F(t) = 1000m (t) Y(s) =. ωn. 1 s( s + 10 s + 100 ) 2. = 10, ζ = 0.5, wd = 5 3. Using Eqns (2.57)-(2.58), we obtain. http://ebook29.blogspot.com/.

(3) http://ebook29.blogspot.com/. 4. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. LM N. 2 -5 t e sin(5 3 t + tan -1 3 3. y(t) = 0.01 1 -. 2.6. Y ( s) 6 6 = G( s ) = 2 = R( s) s + 7s + 6 ( s + 1)(s + 6) 3. | G( jω )|ω = 2 =. 5 2 3. yss =. 5 2. ; ∠ G(j2) = – 81.87º sin( 2t − 81.87º ). 3 21 sin 2t – cos 2t 50 50. =. 2.7. OP Q. Y (s) s+3 s+3 = G(s) = 2 = R( s ) ( s + 2 )( s + 5) s + 7s + 10 (a). R(s) =. 1 s +1. y(t) =. FH 1 e 2. −t. IK. 1 1 − e −2 t − e −5 t µ ( t ) 3 6. (s2 + 7s + 10) Y(s) – sy(0) – y& (0) – 7y(0). (b). = (s + 3) R(s) – sr(0) Initial conditions before application of the input are y(0–) = 1, y& (0–) =. 2.8. 1 , r(0–) = 0 2. Y(s) =. s + 15 / 2 s+3 + ( s + 2 )( s + 5) ( s + 2 )( s + 5)( s + 1). y(t) =. 1 - t 3 -2t e + e - e -5t 2 2. X = X + x; I = I + i. M. d 2 (X + x) = F ( X + x , I + i ) − Mg dt 2. http://ebook29.blogspot.com/.

(4) http://ebook29.blogspot.com/. SOLUTION MANUAL. FG H. ∂F Mx&& = F ( X + I ) + ∂X. X ,I. 5. IJ x + FG ∂F IJ i − Mg K H ∂I K X ,I. −3 F ( X + I ) = Mg = 8.4 × 10 × 9.8 Newtons. For this value of force, we get from Fig. 2.8b,. X = 0.27cm ; I = 0.6 amps Again from Fig. P 2.8b,. K1 =. ¶F ¶X. X ,I. K2 =. ¶F ¶I. X ,I. x&& =. 2.9. = 0.14 Newtons/cm. = 0.4 Newtons/amp. K1 K X (s ) 47.6 x + 2 i; = 2 M M I (s ) s - 16.67. Mx&& + Bx& + Kx = K1 ( y - x ). Gravitational effect has been eliminated by appropriate choice of the zero position.. X (s) 0.1667 = G( s) = Y (s ) ( 0.0033 s + 1)( 0.0217 s + 1) w = 2p v / l = 11.63 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) 1 F( s) = Ms 2 + Bs + K. Force transmitted to the ground = K X(s) + Bs X(s). http://ebook29.blogspot.com/.

(5) http://ebook29.blogspot.com/. 6. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. =. FH B s + 1IK F(s) K M 2 B s + s +1 K K A sin wt. F(t) =. Peak amplitude of the force transmitted to the ground at steady-state. FH B ω IK K FG1 − Mω IJ + F Bω I H K K HKK 2. A 1+. =. 2. 2. 2. (b) Mx&& + B( x& - y& ) + K ( x - y) = 0. X (s) Bs + K = 2 Y (s ) Ms + Bs + K Peak amplitude of machine vibration. FH B vIK K FG1 - Mv IJ + F Bv I H K K H `K K 2. A 1+. =. 2. 2. 2. 2.11 M1 && y1 + K1 ( y1 − y0 ) + B1 ( y&1 − y&0 ) + K 2 ( y1 − y2 ) + B2 ( y&1 − y&2 ) = 0 M2 && y2 + K 2 ( y2 − y1 ) + B2 ( y&2 − y&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). E0 (s ) 1 + 2 RC2 s + R2 C1C2 s 2 = Ei ( s) 1 + R(C1 + 2C2 )s + R2C1C2 s2. (b). E0 (s ) 1 + 2 R1Cs + R1 R2C 2 s 2 = Ei ( s) 1 + (2 R1 + R2 )Cs + R1 R2 C 2 s 2. 2.13 Refer Example 11.6 2.14 Refer Example 11.7. http://ebook29.blogspot.com/.

(6) http://ebook29.blogspot.com/. SOLUTION MANUAL. 2.15 The following result follows from Example 12.9.. x1 = f, x 2 = f& , x 3 = z, x 4 = z&, r = F(t ) x& = Ax + br. A =. LM 0 MM4.4537 0 MN0.5809. OP LM 0 OP 0 P1P ; b = MM−0.3947 P 0 P MN 0.9211 PQ 0 PQ. 1 0 0 0 0 0 0 0 0. 2.16 Mixing valve obeys the following equations: (Qi + qi ) r cq H + [Q - (Qi + qi )] r cq C = Qr c(Qi + q i ). q i = kv x The perturbation equation is K v (q H - q C ) x(t) = Q q i (t ). x(t) = Kq i (t ); K = Q / [ K v (q H - q C )]. or,. The tank obeys the equations Vr c. dq = Qr c(q id - q ); q id (t ) = q i ( t - t D ) dt. This gives q (s ) e -1. 5s = q i ( s) s +1. 2.17 C1 C2. q1 - q 2 1 dθ 1 = qm (t )l - R ; R1 = UA 1 dt. dq 2 dt. =. q1 - q 2 q i - q 2 + ; R1 R2. C2 = Vrc; R2 =. 1 Qrc. From these equations, we get. θ& 1 = −1.92 θ 1 + 1.92 θ 2 + 4.46 q m θ& 2 = 0.078 q 1 - 0.2 q 2 + 0.125 q i. http://ebook29.blogspot.com/. 7.

(7) http://ebook29.blogspot.com/. 8. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. 2.18 At steady-state 0 =. q1 - q 2 q i - q 2 ; this gives θ 1 = 120 º C + R1 R2. 0 = Qm λ −. θ1 − θ 2 ; this gives Qm = 17.26 kg / min R1. 2.19 Energy balance on process fluid:. V2 ρ 2 c2. d (θ 2 + θ 2 ) = Q2 ρ 2 c2 [θ i 2 − (θ 2 + θ 2 )] − UA[θ 2 + θ 2 − θ 1 − θ 1 ] dt. At steady-state 0 = Q2 r 2 c2 (q i 2 - q 2 ) - UA(q 2 - q 1 ); this gives q1 = 40 º C The perturbation equation is V2r 2c2. dq 2 - Q2 r 2c2q 2 - UA(q 2 - q 1 ) dt. 544.5 +. dq 2 + q 2 = 0.432 q1 dt. This gives. Energy balance on the cooling water: V1r1c1. d q1 + q 1 = (Q1 + q1 )r 1c1 [q i1 - (q 1 + q 1 )] dt. d. i. + UA[q 2 + q 2 - q1 - q 1 ] At steady-state 0 = Q1r1c1[q i1 - q1] + UA[q 2 - q1 ]; this gives –3 3 Q1 = 5.28 ×10 m /sec. The perturbation equation is V1r1c1. dq 1 = (θ i1 − θ 1 ) ρ1c1q1 − Q1 ρ1c1θ 1 + UA (θ 2 − θ 1 ) dt. http://ebook29.blogspot.com/.

(8) http://ebook29.blogspot.com/. SOLUTION MANUAL. This gives 184.55. dq1 + q1 = 0.465 q 1 – 1318.2 q1 dt. Manipulation of the perturbation equation gives q 2 (s ) 557.6 =– 3 2 q 1 ( s) 100.5 × 10 s + 729s + 0.8. 2.20 Tank 1: C1. dp1 = q1 – q10 – q11 dt. q10 = flow through R0; q11 = flow through R1 A1 dh rgh1 rg(h1 - h2 ) ´ rg 1 = q1 R0 R1 dt rg. F GH. I JK. 1 1 rg 1 rg dh1 + h1 + h2 + q1 =A1 R0 R1 A1 R1 A1 dt. or. = – 3h1 + 2h2 + r1 Tank 2: C2. dp2 = q2 + q11 – q20 dt. q20 = flow through R2 A2 dh rg(h1 - h2 ) rgh2 ´ rg 2 = + q2 dt rg R1 R2. FG H. IJ K. 1 1 rg rg 1 dh2 + h1 h2 + q2 = A R A R R A dt 2 1 2 1 2 2. or. = 4h1 – 5h2 + r2 2.21 A. d ( H + h) = Q1 + q1 + Q2 + q2 - Q - q dt. At steady-state 0 = Q1 + Q2 - Q ; this gives Q = 30 litres/sec The perturbation equation is. http://ebook29.blogspot.com/. 9.

(9) http://ebook29.blogspot.com/. 10. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. A. dh = q1 + q 2 – q dt. 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 ) +. = Q+. FG ¶f ( H) H ¶H. K. H= H. IJ (H - H ) K. h(t ). 2 H. Therefore q(t) =. K 2 H. h(t ) =. K H Q h (t ) h(t ) = 2H 2H. 1 1 1 dh(t ) = - q(t ) + q1(t ) + q2 (t ) dt A A A. = – 0.01 h(t) + 0.133 q1(t) + 0.133 q2(t) Mass balance on salt in the tank:. A. d [( H + h( t ) ( C + c ( t ))] = C1[Q1 + q1(t )] + C2 [Q2 + q2 (t )] dt − [ C + c ( t )][ Q + q ( t )]. At steady state, 0 = C1Q1 + C2Q2 - CQ ; this gives C = 15 The perturbation equation is AH. dc( t ) dh( t ) = C1q1(t ) + C2q2 (t ) - Cq(t ) - Q c(t ) + AC dt dt. This gives dc(t ) = – 0.02 c(t) – 0.004 q1(t) + 0.002 q2(t) dt. http://ebook29.blogspot.com/.

(10) http://ebook29.blogspot.com/. SOLUTION MANUAL. 2.22 e − τ D s = 1 - t D s +. t 2D s 2 t 3D s 3 t 4D s 4 t 5D s 5 + + ... 2! 3! 4! 5!. It is easy to calculate with long division that. 1- t Ds / 2 t 2 s2 t 3 s3 = 1 - t D s + D - D + ... 2 4 1+ tDs / 2 1 - t D s / 2 + t D2 s 2 / 12 t D2 s 2 t D3 s 3 = 1 t + s D 2 6 1 + t D s / 2 + t D2 s 2 / 12. +. t D4 s 4 t D5 s 5 + - ... 12 144. http://ebook29.blogspot.com/. 11.

(11) http://ebook29.blogspot.com/. CHAPTER 3. MODELS OF INDUSTRIAL CONTROL DEVICES AND SYSTEMS. 3.1. R +. + –. –. Y. G2 G3 +G4. +. G1. – H2 H1G2 G2G3 + G4. Y (s ) G1 (G2 G3 + G4 ) = R( s ) 1 + (G2 G3 + G4 )(G1 + H2 ) + G1 H1G2. 3.2 H2 – R. G1. +. +. G2. G3. +. – (1–G1)H1 G4. G1G2G3 Y ( s) G4 + = R( s ) 1 + G2 G3 H2 + G2 H1 (1 - G1 ). 3.3. +. Y ( s) ( P D + P D + P3 D 3 + P4 D 4 ) / D R( s ) = 1 1 2 2. P1 = G1G2; P2 = G1G3; P3 = G4H2G1G2; P4 = G4H2G1G3 ∆ = 1 – (– G1G3H1H2 – G1G2H1H2) ∆ 1 = ∆2 = ∆3 = ∆4 = 1. http://ebook29.blogspot.com/. Y.

(12) http://ebook29.blogspot.com/. SOLUTION MANUAL. 13. Y ( s) (1 + G4 H2 )(G2 + G3 )G1 = 1 + G H H (G + G ) R( s ) 1 1 2 2 3. G2 R. G3. G1. 1. Y. 1. H1. –H2 – G4. 3.4 H3 /G 3 R. +. G1. –. –. +. G3. G2. Y. 1+G3 H 2. H1. Y (s) R(s ) W =0. = M(s) =. G1G2 G3 1 + G2 H3 + G3 H2 + G1G2 G3 H1 H3 G 2. – (G 1 H1 G 2+H 2). Y (s ) W ( s). +. = MW (s) = R=0. –. +. W +. G3 (1 + H 3G2 ) 1 + H3G2 + G3 ( G1 H1G2 + H 2 ). http://ebook29.blogspot.com/. G3. Y.

(13) http://ebook29.blogspot.com/. 14. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. 3.5 –1 R1. G1. 1. G2. 1. 1. Y1. G6. 1. Y2. 1 Js. w. H1. H2 R2. G3. G4. 1. G5. –1. Y1 R1. R2 = 0. =. P1D1 D. G1G2 G3 (1 + G4 ). = 1 - (-G G - G + G H G G H ) + G G G 1 2 4 1 2 4 5 1 1 2 4. Y1 R2 Y2 R1. =. G4G5 H1G1G2G3 D. =. G1G4G5G6 H2 Y2 ; R 2 D. R1 = 0. R2 = 0. R1 = 0. =. G4G5G6 (1 + G1G2 ) D. 3.6 er. +. KA. if. +. Kg. –. – e t. 1 Ra. ia. Kb Kt. w (s ) 50 = Er ( s) s + 10.375. For Er(s) = 100/s,. http://ebook29.blogspot.com/. KT.

(14) http://ebook29.blogspot.com/. SOLUTION MANUAL. 100 ´ 50 w (s) = s( s + 10.375) –10.375 t ) w(t ) = 481.8 (1 – e. 3.7 qR + –. KP. 1 sLf + Rf. KA. KT. Js2. 1 + Bs. qM. 1 qL 50. q L (s) 1 q R (s ) = s( 0.1s + 1)(0.2 s + 1) + 1. 3.8. TM = K1 ec + K 2θ& M. =. J&eq q&& M + Beq q& M. K2 = slope of the characteristic lines shown in Fig. P3.8b =. K1 =. 3- 0 = – 0.01 0 - 300. ∆TM ∆ec. J eq = JM +. Beq. θ& M =0. FG N N IJ HN N K. 2. 1 3. 2. FNN I =G H N N JK. J L = 0.0032. 4. 2. 1 3. 2. 3- 2 = 0.1 30 - 20. =. BL = 0.00001. 4. q M (s) 10 1 q L (s ) = ; = s(0.32 s + 1) s(0.32 s + 1) Ec (s ) Es c. F q& I +J G H q& JK F q& I +B G& J Hq K. 2. 3.9 (i). J eq = JM. L. = 0.6. L. M. Beq = BM. L. L. 2. = 0.015. M. KT = Kb. http://ebook29.blogspot.com/. 15.

(15) http://ebook29.blogspot.com/. 16. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. KT q M (s) = s[ J eq s + Beq )( sLa + Ra ) + KT K b ] Ea ( s ) 16.7. = s(s 2 + 100 s + 19.2) (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 = JM = n2 JL = 1.5 × 10 –5; n = 1 B = BM + n2BL = 1 × 10–5. θ M (s) 13.33 = 2 θ R (s) s + 3.5s + 13.33 qR +. Ks. –. +. KA. –. ec. +. K1. TM. 1 Js +B. –. qM. 1 s. K2 Kt. 3.11 qR +. Ks. –. KA. Kg + –. 1 sLf +Rf. qM 1 Js + B. KT Ra. 1 s. qM. Kb. θ M (s) θ (s) = L θ R (s) θ R (s ) 3.12 (a). ω (s) = Er ( s ). =. 15 × 10 3 s + 54 s + 200 s + 15 × 10 3 3. 2. K ; K = 17.2, z = 1.066, w n = 61.2 1 2 2ζ + + s s 1 ωn ω 2n. (b) Modification in the drive system may be made on the lines of Fig. 3.64.. http://ebook29.blogspot.com/. qM.

(16) http://ebook29.blogspot.com/. SOLUTION MANUAL. er +. KA. Kr. –. + –. w. KT Js + B. 1 sLa + Ra. Kb. Kt. 3.13 TW. N(s) qR +. KT. KA. KP. TM. G(s). +. –. TM = JM q&&M + B(q& M - q& L ) + K (q M - q L ). B(q& M - q& L ) + K (q M - q L ) = JL q&&L + TW (JMs2 + Bs + K) θM(s) – (Bs + K) θL(s) = TM(s) (Bs + K) θM (s) – (JLs2 + Bs + K) θL(s) = TW(s) qL ( s) = G(s) TM(s) – N(s) TW(s). K P K A K T G( s ) qL ( s ) = 1 + K P K A K T G( s ) qR ( s) N ( s) qL ( s ) = 1 + K K K G( s ) P A T - TW ( s) G(s) =. Bs + K s [ JM JL s + ( JM + JL ) Bs + ( JM + JL ) K ] 2. 2. http://ebook29.blogspot.com/. –. 17. qL.

(17) http://ebook29.blogspot.com/. 18. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. N(s) =. JM s 2 + Bs + K s 2 [ JM JL s 2 + ( JM + JL ) Bs + ( JM + JL ) K ]. Revisiting Example 2.7 will be helpful. 3.14 The required block diagram easily follows from Fig. 3.49. K1 x - Ay& = K 2 Dp. ( K 2 D p). A && K 2 = My. (r – y) Kp KA K = x. Y ( s) 2 = 2 R( s) s + 0.02 s + 2 K A K v K1 AK r / K2 Y ( s) = 2 Yr ( s) Ms + ( B + A2 / K2 )s + K A K v K1 K s A / K 2. 3.15. Fw. er yr. Kr. ea. +. KA. –. x Kv. K1. + –. A K2. –. +. y 1 Ms + B. 1 s. y. e0 A. Ks. 3.16 Revisiting Example 3.7 will be helpful. From geometry of the linkage, we get b a X ( s) Y (s ) a+b a+b bK r AK1 / ( a + b) K2 Y ( s) = 2 Ms + ( B + A 2 / K 2 )s + K + aAK1 / ( a + b) K2 Yr ( s). E(s) =. http://ebook29.blogspot.com/.

(18) http://ebook29.blogspot.com/. SOLUTION MANUAL. yr. x. Kr. b a+b. e. + –. K1. +. 1 Ms2 + Bs + K. A K2. –. a a+b. 19. y. sA. 3.17 Let z(t) = displacement of the power piston d Y(s) = e - st D Z(s) ; t D = sec v Y ( s) G( s ) = Er ( s ) 1 + G( s ) H ( s ) ( K A K v K1 A / K 2 )e - t D ; H(s) = Ks s( Ms + B + A 2 / K 2 ) s. G(s) =. i er +. KA. –. Fw. x Kv. K1. + –. A K2. +. –. z 1 Ms + B. e0 A. Ks. 3.18. A1. rgh1 dh1 = q1 - qw - R dt 1. A2. rgh1 rgh2 dh2 = R - R dt 1 2. http://ebook29.blogspot.com/. z 1 s. e–stD. y.

(19) http://ebook29.blogspot.com/. 20. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. Qw(s) Q1(s) +. – R2/rg (t 1s + 1) (t 2s + 1). H2 (s). t 1 = A1R1 ; t 2 = A2R2 rg rg hr. er +. Ks. Ka. –. Ke. Kv. q1. h2. h. q. e0 Ks. 3.19 Ct. dq dt. = h-. q -qa Rt. qa. 1/Rt h. +. +. Rt ts+ 1. q. t = Rt Ct er. + –. Rf R. –1. –. eA. R2 R1. Kh. Kt. 3.20 Heater:. C1. q1 - q dq 1 = hR1 dt. http://ebook29.blogspot.com/.

(20) http://ebook29.blogspot.com/. SOLUTION MANUAL. Air:. Vrc. q1 - q dq - Qrc(q - q i ) = R1 dt. These equations give us. 1 t = R1C1, Kg = Qrc , w n = 2zw n =. Q R1C1V. R1C1Qrc + Vrc + C1 VR1C1 rc. qi ts + 1 Kg +. h. Kg. + 1 wn2. qr. Ks. er. +. KA. –. eA. s2. 2z s+1 wn. q. Kh. h. q. e0 Ks. 3.21 Process: V2 r 2 c2. d (q 2 + q 2 ) = Q2 r 2 c2 (q i 2 + q i 2 ) - UA(q 2 + q 2 - q 1 - q 1 ) dt. – Q2 r 2 c2 (q 2 + q 2 ) Linearized equation is. V2 r 2 c2. dq 2 = Q2 r 2 c2q i 2 - UA(q 2 - q 1 ) - Q2 r 2 c2q 2 dt. http://ebook29.blogspot.com/. 21.

(21) http://ebook29.blogspot.com/. 22. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. Rearranging this equation and taking Laplace transform yields (t 2 s + 1)q 2 ( s) = K1′θ 1 ( s ) + K 2′ θ i 2 ( s ) t 2 = 544.5; K1′ = 0.423; K 2′ = 0 .577. Cooling water: V1 r1c1. d (q 1 + q 1 ) = (Q + q ) r c (q i + q i ) + UA(q + q 1 1 1 1 1 1 2 2 dt - q 1 - q 1 ) - (Q1 + q1 )r 1c1 (q 1 + q 1 ). Linearized equation is. V1 r1c1. dq 1 = Q1r 1c1q i1 + q i1 r1c1q1 + UA(q 2 - q 1 ) dt − Q1 ρ 1c1θ 1 − θ 1ρ 1c1q1. From this equation, we obtain (t 1s + 1)q 1 ( s) = − K3′ Q1 ( s ) + K 4′ θ 2 ( s ) + K 5′θ i1 ( s ). 1.82 ´ 1000 ´ 4184 t 1 = 3550 ´ 5.4 + Q ´ 1000 ´ 4184 1 K3′ =. ( -27 + q 1 ) ´ 1000 ´ 4184 3550 ´ 5.4 + Q1 ´ 1000 ´ 4184. 3550 ´ 5.4 K 4′ = 3550 ´ 5.4 + Q ´ 1000 ´ 4184 1 K 5′ =. Q1 ´ 1000 ´ 4184 3550 ´ 5.4 + Q1 ´ 1000 ´ 4184. q 1 and Q1 are obtained from the following energy-balance equations at steady state:. Q1r 1c1 (q i1 - q 1 ) + UA(q 2 - q 1 ) = 0 Q2 r 2 c2 (q i 2 - q 2 ) - UA(q 2 - q 1 ) = 0 These equations yield –3 3 q 1 = 40ºC; Q1 = 5.28 × 10 m /sec. http://ebook29.blogspot.com/.

(22) http://ebook29.blogspot.com/. SOLUTION MANUAL. qi1. qi2. K¢5. K¢2. + q1. –. K¢3. +. q1 1 t1 s + 1. +. +. K¢1. 23. q2 1 t2 s + 1. +. K¢4 qr. +. K1. e. K2. –. K3. K4. q2. q1. K1. With these parameters, t 1 = 184.55 sec, K3¢ = 1318.2, K 4¢ = 0.465, K 5¢ = 0.535 q 2 ( s) q r ( s). =. (5.55 ´ 10 -3 ) K1 K 2 K 3 K 4 s 2 + 7.26 ´ 10 -3 s + 119 . ´ 10 -5. G( s) ; G(s) = 1 + G( s ). Refer Fig. 3.57 for suitable hardware required to implement the proposed control scheme. 3.22 q2. q1 qr. + –. Controller I. + –. Controller II. Process II. Process I. Sensor II. Set-point adjustment Sensor I. Refer Fig. 3.57 for suitable hardware required to implement the proposed control scheme.. http://ebook29.blogspot.com/.

(23) http://ebook29.blogspot.com/. CHAPTER 4. BASIC PRINCIPLES OF FEEDBACK CONTROL. 4.1. SGM =. 1 s( s + 1) = 1 + G( s) H (s) s( s + 1) + 50. | SGM ( jw )|w =1 = 0.0289 SHM =. -50 - G( s ) H ( s ) = 1 + G( s) H ( s) s( s + 1) + 50. | S HM ( jw )|w =1 = 1.02. Y ( s) P1D 1 = M(s) = R( s) D. 4.2. P1 =. FG H. M(s) =. SKM1 = | SKM1 ( jw )|w = 0 = 4.3. IJ K. K 3 K1 KK K K 5 ; D = 1- - 1 2 - 2 3 1 ; D1 = 1 2 s 2 ( s + 1) s ( s + 1) s + 1 s ( s + 1). 5K1 s ( s + 1 + 5K1 ) + 5 K1 + 5 2. s 2 ( s + 1 + 5 K1 ) + 5 - 5 K1s 2 ¶M K1 = 2 ´ s ( s + 1 + 5 K1 ) + 5 K1 + 5 ¶K1 M. 5 5K1 + 5. = 0.5. For G(s) = 20/(s + 1), and R(s) = 1/s,. y ( t )|open-loop = 20 (1 – e–t) y ( t )|closed-loop =. 20 (1 - e -21t ) 21. 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 (1 - e -20.4 t ) 20.4. The transient response of the closed-loop system is less sensitive to variations in plant parameters. http://ebook29.blogspot.com/.

(24) http://ebook29.blogspot.com/. SOLUTION MANUAL. 4.4 For. G(s) = 10/ (t s + 1), and R(s). 25. = 1/s,. ess |open-loop = 0 1 ess |closed-loop = 1 + 10 K p = For. 0.0099. G'(s) = 11/ (t s + 1) , and R(s) = 1/s,. ess |open-loop = – 0.1 ess |closed-loop = 0.009 The steady-state response of the closed-loop system is less sensitive to variations in plant parameters. 4.5. L 2 = = 0.04 sec R 50. t f |open-loop =. For the feedback system,. K A (e f - Ki f ) = L. di f dt. + ( R + Rs )i f. This gives I f ( s). KA = E f (s ) sL + R + Rs + K A K t f |closed-loop =. L 2 = = 0.004 R + Rs + K A K 51 + 90 K. This gives K = 4.99 4.6. The given block diagram is equivalent to a single-loop block diagram given below, W (s). 0.4 s+1. + –. 1. Y(s). 10(K + s) s+1. Y ( s) 0.4 = W ( s) 11s + 10 K + 1. http://ebook29.blogspot.com/.

(25) http://ebook29.blogspot.com/. 26. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. For. W(s) = 1/s, yss = lim sY ( s ) = s →0. 0.4 = 0.01 10 K + 1. This gives K = 3.9 4.7. Case I:. Y ( s) s(t s + 1) = W ( s) s(t s + 1) + K for W(s) = 1/s, yss = 0 Case II:. Y ( s) K = W ( s) s(t s + 1) + K For W(s) = 1/s, yss = 1 The control scheme shown in the following figure will eliminate the error in Case II. W (s) Kc 1 + 1 TIs. –. 4.8. +. + Plant. Y(s). A unity-feedback configuration of the given system is shown below. qr. + –. e. D(s). G (s) =. 200 ¥ 0.02 (s + 1) (s + 2). q. 1 1 E(s ) = ;ess = lim sE ( s ) ; qr ( s) = s s →0 qr ( s ) 1 + D( s )G( s) (i) ess = 1/3 (ii) ess = 0 (iii) ess = 1/3. 4.9. The integral term improves the steady-state performance and the derivative term has no effect on steady-state error. (i) s2 + 1 = 0; oscillatory (ii) s2 + 2s + 1 = 0; stable (iii) s3 + s + 2 = 0; unstable. http://ebook29.blogspot.com/.

(26) http://ebook29.blogspot.com/. SOLUTION MANUAL. 27. The derivative term improves the relative stability; and the integral term has the opposite effect. er = 50 × 0.03 = 1.5 volts. 4.10 (a). V (s ) Er ( s ). (b). = M(s) =. SKMe = (c). V(s) =. 0.65K e ( s + 1)(5s + 1) + 0.0195 K e. ¶M Ke ( s + 1)(5s + 1) ´ = ¶K e M ( s + 1)(5s + 1) + 9.75. 325 19.5( s + 1) Er ( s ) W ( s) ( s + 1)(5s + 1) + 9.75 ( s + 1)(5s + 1) + 9.75. Suppose that the engine stalls when constant percent gradient is x%. Then for Er(s) = 1.5/s and W(s) = x/s,. sV (s) = vss = lim s ®0. 325 ´ 1.5 19.5 ´ x =0 10.75 10.75. This gives x = 25 4.11 (a). The block diagram is shown in the figure below. IL. Ra er +. KA. –. 1 sLf + Rf. Kg. KA. (b). E0 ( s ) Er ( s ). IL =0. =. 10 2 s + 1 + 10 K. For Er(s) = 50/s,. sE0 ( s) = eoss = lim s ®0. 500 = 250 1 + 10 K. This gives K = 0.1 With. K = 0.1, we have. http://ebook29.blogspot.com/. +. –. eo.

(27) http://ebook29.blogspot.com/. 28. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. Eo(s) = When. 10 2s + 1 Er ( s ) I L (s ) 2s + 2 2s + 2. Er(s) = 50/s and IL(s) = 30/s, we get eoss =. 10 ´ 50 30 = 235 V 2 2. Let x V be the reference voltage required to restore the terminal voltage of 250 V. 10 ´ x 30 = 250; this gives x = 53V 2 2. (c). Eo(s) =. 10 Er ( s ) - I L ( s ) 2s + 1. er = 25 V gives eoss = 250 V when IL = 0 er = 25 V gives eoss = 220 V when IL = 30 amps (d) The effect of load disturbance on terminal voltage has been reduced due to feedback action. 4.12 wr. Kt. +. KA. –. Kg. + –. 1 Ra. Kb Kt. K A Kg (a) w o ( s) = M(s) = 2( s + 5) + K A K g w r ( s) For wr(s) = 10/s,. wo(t). =. 125 (1 - e -130t ) ; woss = 9.61 rad/sec 13. (b) When the feedback loop is opened,. w o ( s) 50 K A = w r ( s) 2( s + 5). http://ebook29.blogspot.com/. KT. 1 Js. w0.

(28) http://ebook29.blogspot.com/. SOLUTION MANUAL. KA = 0.192 gives woss = 9.61 for wr = 10. w 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) SKMA =. 2 s + 10 ¶M K A = ´ 2 s + 10 + K A K g ¶K A M Kg = Kωg, where K is a constant.. SwMg =. Kg ¶K g wg ¶M 2 s + 10 ¶M ´ ´ = ´ = M M 2 s + 10 + K A K g ¶K g ¶w g ¶K g. The open-loop transfer function of the system is. G (s) =. K A Kg 2 ( s + 5). Therefore. SKGA = 1 = SwGg S KMA < S KGA ; SwMg < SwGg. 4.13 Tw qR. KP. +– –. KA. 1 sLf + Rf. Kg. + –. 1 Ra. KT. +. –. qM = qC 1 Jeqs2. sKb KP. (b). 5Kg q L ( s) = M (s) = q R ( s) s ( s + 1) ( s + 4 ) + 10 K g SKMg =. Kg s ( s + 1) ( s + 4 ) ¶M ´ = ¶K g M s ( s + 1) ( s + 4 ) + 10 ´ 100. http://ebook29.blogspot.com/. 1 2. qL. 29.

(29) http://ebook29.blogspot.com/. 30. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. SKMg ( jw ) (c). w = 0 .1. ~ 4 ´ 10. -4. q L (s) (s + 4) / 2 = s ( s ) ( s + 4) + 10 K g + 1 - Tw ( s ). Wind gust torque on the load shaft = 100 N-m. Therefore, on the motor shaft Tw (s) = n × 100/s = 50/s θLss = lim sq L ( s ) = 0.1 rad s®0. (d). FG H. Replace KA by D(s) = K c 1 +. 1 TI s. IJ to reduce the steady-state error K. to zero value. 4.14 (a). q (s) q r (s) q (s) q r (s). = G(s) =. K P K A Ke K ( 25s + 1) s( 0.25s 2 + 0.02s + 1). = M(s) =. G( s ) 1 + G( s ). open - loop. closed - loop. s( 0.25s 2 + 0.02 s + 1) s( 0.25s + 0.02 s + 1) + K P K A K e K (25s + 1). M SKG = 1; SK =. 2. SKM < SKG (b) A PD control scheme with proportional controller in the forward path and derivative action realized by feeding back q& , will increase aircraft damping. 4.15. (a). q (s) Tw ( s ). =. 1/ B sK b K T K K J s s +1 + + T A B BRa BRa. FH. IK. For Tw(s) = Kw/s,. θss = lim sθ (s) = s®0. K w Ra KT K A. http://ebook29.blogspot.com/.

(30) http://ebook29.blogspot.com/. SOLUTION MANUAL. (b). q (s) = M (s) = q r (s). SJM. 31. K T K A / BRa sK K K K J s s +1 + b T + T A B BRa BRa. F H. I K. J - s2 ¶M J B ´ = = sK b K T K K J M ¶J s + T A s +1 + B BRa BRa. FH. IK. (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 KA. 4.16 (a). (b). (c). dV = qi − q; V = C h(t), q = h(t)/R dt hr + –. qi. K. R RCs + 1. h. 1 R. q. H (s) KR = M (s) = RCs + 1 + KR Hr ( s ) S RM =. SRM. s=0. =. R 1 ¶M ´ = ¶R M RCs + 1 + KR 1 1 + KR. (d) For Hr(s) = 1/s, hss =. FG H. KR 1 ; ess = 1 + KR 1 + KR. IJ K. 1 reduces the steady-state error to zero; and the Tl s system becomes insensitive to changes in R under steady dc conditions. Replacing K by K c 1 +. 4.17 V. d (C + c ) = (Qi + qi ) ( Ci + ci ) − Q ( C + c ) dt. The perturbation dynamics is given by t. C Q dc ( t ) V + c(t) = i qi ( t ) + i ci (t );t = dt Q Q Q. Block diagram of the closed-loop system is shown below.. http://ebook29.blogspot.com/.

(31) http://ebook29.blogspot.com/. 32. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. ci. Qi /Q. qi. K. –. +. Ci /Q. +. 1. c. ts + 1. Qi / Q C( s) = M(s) = Ci ( s ) τs + 1 + KCi / Q For Ci(s) = A/s, css =. AQi Q + KCi. With integral control, css becomes zero. 4.18 (b). Open-loop case:. KR1 q( s ) ; SKG = 1 = G(s) = R1Cs + 1 Er ( s ). (i). qi. er +. 1/R1. er K K. –. +. +. R1 R1C s + 1. et Kt. http://ebook29.blogspot.com/. q.

(32) http://ebook29.blogspot.com/. SOLUTION MANUAL. (ii). 1 q (s) . For q i ( s) = 1 / s, q ss = 1 = Ri Cs q i (s). (iii) τ = R1C Closed-loop case: (i). KR1 R1Cs + 1 q( s ) = M(s) = ; SKM = R1Cs + 1 + KKt R1 R1Cs + 1 + KKt R1 Er ( s ). (ii). q (s) 1 1 = . For θ i(s) = 1/s, θss = R1Cs + 1 + KKt R1 1 + KK t R1 q i (s). (iii) τ =. (c). R1C 1 + KK t R1. K(er − et ) = C. 1 R1. +. er K K. +. qi. er + –. q - qi q - qa RRC 1 dq ; R2 = ,t = 1 2 + + dt R1 R2 UA R1 + R2 +. +. 1 R2. qa. R1R2 (R1 + R2 ) (ts + 1). q. Kt. q (s) q a (s). = open - loop. R1 ; q ss ( R1 + R2 ) (t s + 1). open - loop =. R1 (R1 + R2). R1 θ (s) = ( R1 + R2 ) (t s + 1) + KK t R1R2 θ a ( s ) closed − loop. θ ss. 4.19 (a). closed − loop. =. R1 R1 + R2 + KK t R1 R2. 0.1 ( s + 1) Y (s) = . For W(s) = 1/s, yss = 0.089 W (s) 2 s 2 + 3s + 1125 .. http://ebook29.blogspot.com/. 33.

(33) http://ebook29.blogspot.com/. 34. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. (b). (c). FG H. IJ K 0.1 ( s + 1) Y (s) LM1 - K K OP = W ( s ) 2 s + 3s + 1125 . N s +1 Q. Replace Kc by K c 1 +. 1 Tl s. f. c. 2. For W(s) = 1/s, yss = 0.089 (1 − Kf Kc) yss = 0 when Kf = 1/Kc = 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 openloop technique; it contains no self-correcting action. If the value of Kc is not accurately known, the gain Kf 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 r. Kt. +. ef +. –. –. KA. 1 sLf + Rf + 1. Kt. Time-constant (without current feedback) = Lf /(Rf + 1). I f (s ) E f (s). =. KA sL f + R f + K A + 1. Time-constant (with current feedback) =. Lf Rf + KA + 1. Position control system is shown below.. http://ebook29.blogspot.com/. if. w KT Js + B.

(34) http://ebook29.blogspot.com/. SOLUTION MANUAL. r. KP. +. +. +. –. –. –. if. KA sLf + Rf + 1. Kt. w. Js + B. 1 s. Kt KP. s2 + Kc TDs + Kc = (s + 1 + j1.732) (s + 1 − j 1.732). 4.21 (a). This gives Kc = 4, TD = 0.5 (b) qr. Alternative control scheme is shown below. +. K1. –. +. 1 s. –. q. 1 s. K2. K1 q (s) = 2 ; K1 = 4, K 2 = 2 q r ( s ) s + K 2 s + K1 4.22 Fig. P4.22a:. q (s ) 25 K = M (s ) = 2 q r (s) s + 5s + 25 K SKM =. s( s + 5) K ¶M ´ = 2 M ¶K s + 5s + 25. S KM ( jw ) w = 5 = 1.41. Fig. P4.22b: KK 2 25 K q (s) = M (s) = 2 = 2 q r (s) s + 5s + 25 K s + (1 + KK1 ) s + KK 2. This gives K2 = 25 and K1 = 4. http://ebook29.blogspot.com/. q. 35. w.

(35) http://ebook29.blogspot.com/. 36. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. M (s) =. SKM =. 25 K s + (1 + 4 K ) s + 25 K 2. s( s + 1) s + 5s + 25 2. | SKM ( jw )|w =5 ≈ 1 This shows the superiority of the two-loop system over a single-loop system. —-. http://ebook29.blogspot.com/.

(36) http://ebook29.blogspot.com/. CHAPTER 5. 5.1. CONCEPTS OF STABILITY AND THE ROUTH STABILITY CRITERION. (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 lefthalf plane. (c) s3-row of the Routh array has zero pivot element, but the entire row is not all zeros. We replace the pivot element by e and then proceed with the construction of the Routh array. As e ® 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) s1-row of the Routh array is an all-zero row. Auxiliary polynomial formed using the elements of s2-row is given by A(s) = s2 + 1 We replace the elements of s1-row with the coefficients of. dA( s ) = 2s + 0 ds 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 s0-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) s3-row of the Routh array is an all-zero row. The auxillary polynomial is A(s) = 9s4 + 0s2 + 36 We replace elements of s -row with the coefficients of 3. dA( s ) = 36s3 + 0s ds The resulting Routh array has s2-row with zero pivot element but the entire row is not all zeros. We replace the pivot element with e and proceed with the construction of the Routh array. Two sign changes are found in the first column; the characteristic polynomial D(s) has. http://ebook29.blogspot.com/.

(37) http://ebook29.blogspot.com/. 38. 5.2. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. two roots in the right-half plane. Three possibilities exist: (i) the 4thorder auxiliary polynomial has all the four purely imaginary roots and the two right-half plane roots are contributed by the factor D(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 D(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 D(s)/A(s) has one root in the right-half plane. Examining A(s) we find that auxiliarypolynomial 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. (a) s1-row of the Routh array is an all-zero row. The auxiliary polynomial is A(s) = s2 + 9 By long division D(s)/A(s) = (s4 + 2s3 + 11s2 + 18s + 18)/A(s) = s2 + 2s + 2 Therefore D(s) = (s2 + 9) (s2 + 2s + 2) = (s + j3) (s – j3) (s + 1 + j1) (s + 1 – j1) (b) s3-row of the Routh array is an all-zero row. The auxiliary polynomial is A(s) = s4 + 24s2 – 25 D(s)/A(s) = (s5 + 2s4 + 24s3 + 48s2 – 25s – 50)/A(s) =s+2 D(s) = (s + 2) (s4 + 24s2 – 25) = (s + 2) (s2 – 1) (s2 + 25) = (s + 2) (s + 1) (s – 1) (s + j5) (s – j5) (c) s3-row of the Routh array is an all-zero row. The auxiliary polynomial is A(s) = s4 + 3s2 + 2 D(s)/A(s) = (s6 + 3s5 + 5s4 + 9s3 + 8s2 + 6s + 4)/A(s) = s2 + 3s + 2 D(s) = (s2 + 3s + 2) (s4 + 3s2 + 2) = (s2 + 3s + 2) (s2 + 1) (s2 + 2) = (s + 1) (s + 2) (s + j1) (s - j1) (s + j 2 ) (s – j 2 ). http://ebook29.blogspot.com/.

(38) http://ebook29.blogspot.com/. SOLUTION MANUAL. 5.3. 39. With s = s$ – 1, the characteristic equation becomes 3 2 s$ + s$ + s$ + 1 = 0 1 s$ -row in the Routh array is an all-zero row. The auxiliary polynomial is. A( s$ ) = s$ 2 + 1 =. ( s$ + j1) ( s$ – j1). 2. We replace the elements of s$ -row with the coefficients of dA( s$ ) ds$. 5.4. 5.5. = 2 s$ + 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. (a) Unstable for all values of K (b) Unstable for all values of K (c) K < 14/9 (d) K > 0.528 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. LM N. G(s) =. 20 / ( s + 1) 4K 2 s + 1 1 + 20 ´ 0 .2 / ( s + 1). H(s) =. 0.05 4s + 1. OP Q. 1 + G(s) H(s) = 0 gives 8s3 + 46s2 + 31s + 5 + 4K = 0. 5.7. 5.8. From the Routh array, we find that the closed-loop system is stable for K < 43.3. (i) s3 + 5s2 + 9s + K = 0; K < 45 (ii) s3 + 5s2 + (9 – K) s + K = 0; K < 7.5 (iii) s4 + 7s3 + 19s2 + (18 – K) s + 2K = 0; K < 10.1 (a) D(s) = s3 + 10s2 + (21 + K)s + 13K = 0 (i) K < 70 (ii) K = 70 (iii) for K = 70, A(s) = s2 + 91 = 0 D(s) = (s2 + 91) (s + 10) = (s + j9.54) (s – j9.54) (s + 10) (b) D(s) = s3 + 5s2 + (K – 6)s + K = 0. http://ebook29.blogspot.com/.

(39) http://ebook29.blogspot.com/. 40. 5.9. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. (i) K > 7.5 (ii) K = 7.5 (iii) For K = 7.5, A(s) = s2 + 1.5 = 0 D(s) = (s2 + 1.5) (s + 5) = (s + j1.223) (s – j1.223) (s + 5) (c) D(s) = s4 + 7s3 + 15s2 + (25 + K)s + 2K = 0 (i) K < 28.1 (ii) K = 28.1 (iii) For K = 28.1, A(s) = s2 + 7.58 = 0 D(s) = (s2 + 7.58) (s2 + 7s + 7.414) = (s + j2.75) (s – j2.75) (s + 5.7) (s + 1.3) (a) D(s) = s4 + 12s3 + 69s2 + 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) = 52.5s2 + 866.25 = 0 This gives s = ± j 4.06 Frequency of oscillations is 4.06 rad/sec when K = 666.25. (b) D(s) = s4 + 3s3 + 12s2 + (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) = 9.57s2 + 23.3 = 0; s = ± j1.56 For K = 35.7, A(s) = 5.43s2 + 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 D(s) = s3 + as2 + (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 =2 a. These equations give a = 0.75, K = 2 5.11 D(s) = 0.02s3 + 0.3s2 + s + K = 0 (a) K < 15 (b) For K = 15, the auxiliary equation is A(s) = s2 + 50 = 0. The oscillation frequency is 7.07 rad/sec at K = 15. (c) With K = 7.5 and s = s$ – 1, D( 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 Kc (b) For stability, TD > t. 5.13 (a) D(s) = s3 + 15s2 + 50s + 25K = 0; K < 30. http://ebook29.blogspot.com/.

(40) http://ebook29.blogspot.com/. SOLUTION MANUAL. 41. (b) D(s) = t s3 + (1 + 5t)s2 + 5s + 50 = 0; t < 0.2 (c) D(s) = t s3 + (1 + 5t)s2 + 5s + 2.5K = 0; K < 5.14 D(s) = s4 + 5s3 + 6s2 + Kcs + Kca = 0; a =. FH 2 + 10IK t. 1 TI. From the Routh array, we find that K < (30 – 25a) results in system stability. Larger the a (smaller the TI), lower is the limit on gain for stability. 5.15 D(s) = s3 + 15s2 + (50 + 100Kt)s + 100K = 0 From the Routh array, we find that for stability K < (7.5 + 15Kt). Limit on K for stability increases with increasing value of Kt.. http://ebook29.blogspot.com/.

(41) http://ebook29.blogspot.com/. CHAPTER 6 6.1. THE PERFORMANCE OF FEEDBACK SYSTEMS. (a) Refer Section 6.3 for the derivation (b) The characteristic equation is s2 + 10s + 100 = s2 + 2zwn s + w 2n = 0 (i) wn = 10 rad/sec; z = 0.5; wd = w n 1 - z 2 = 8.66 rad/sec (ii) tr =. p p - cos -1 z = 0.242 sec; tp = = 0.363 sec wd wd. Mp = e - pz /. 1-z 2. = 16.32%; ts =. 4 = 0.8 sec zw n. (iii) Kp = lim G(s) = ¥ ; Kv = lim sG(s) = 10; s®0. s®0. Ka = lim s2G(s) = 0 s®0. (iv) ess = 6.2. 1 1 1 = 0; ess = = 0.1; ess = =¥ Ka Kv 1+ Kp. Characteristic equation is s2 + 10s + 10KA = s2 + 2zwn s + w 2n = 0 (a) KA = 2.5 gives z = 1 which meets the response requirements. (b) Kp = ¥ , Kv = 2.5 ess =. 5 1 + = 0.4 1 + K p Kv. 6.3 (a) Using Routh criterion, it can be checked that the close-loop system is stable. (i) ess =. 10 10 = =0; ¥ Kv. (ii) ess =. 10 0.2 10 0.2 + + = =2 ¥ 0.1 Kv K a. (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 ( 0.1s + 1)( s + 1)( 0.2 s + 1). http://ebook29.blogspot.com/.

(42) http://ebook29.blogspot.com/. SOLUTION MANUAL. ess|r = 10 =. Y (s) W (s). =. 43. 10 10 = 1+ 5 1+ Kp 0.1s + 1 ( s + 1)( 0.2 s + 1)( 0.1s + 1) + 5. For W(s) = 4/s, yss = 4/6 ess|total = 7/3 6.5. (a) The characteristic equation of the system is 900s3 + 420s2 + 43s + 1 + 0.8 KA = 0 Using Routh criterion, we find that the system is stable for KA < 23.84. (b) Rearrangement of the block diagram of Fig. P6.5 is shown below. qr. +. KA. –. 0.016 3s + 1. 50 30s + 1. q. 1 10s + 1. G(s) =. 0.016 ´ 50 K A ( 3s + 1)(30 s + 1). H(s) =. 1 ; the dc gain of H(s) is unity. 10 s + 1. Kp = lim G(s) = 0.8 KA ; ess = s®0. 10 =1 1+ Kp. This gives KA = 11.25 6.6. (a) The system is stable for Kc < 9.. 1 1 + Kc ess= 0.1 (10%) is the minimum possible value for steady-state error. Therefore ess less than 2% is not possible with proportional compensator.. Kp = lim D(s) G(s) = Kc; ess = s®0. http://ebook29.blogspot.com/.

(43) http://ebook29.blogspot.com/. 44. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. KI . The closed-loop system is stable for s 0 < K1 < 3. Any value in this range satisfies the static accuracy requirements. 1000 Kc = 1000 (a) Kv = 10. (b) Replace Kc by D(s) = 3 +. 6.7. z =. 10 + 1000 KD = 0.5 2 ´ 100. These equations give Kc = 10 and KD = 0.09. (b) The closed-loop poles have real part = – zwn = – 50. The zero is present at s = – Kc/KD = – 111.11. The zero will result in pronounced early peak. z is not an accurate estimate of Mp. 6.8. (a) Kv = KI = 10 (b) With KI = 10, the characteristic equation becomes D(s) = s3 + 10s2 + 100(1 + Kc)s + 1000 = 0 With s = s$ – 1, D( s$ ) = s$ 3 + 7 s$ 2 + [100(1 + Kc) – 17] s$ + 1009 – 100 (1 + Kc) = 0 In Routh array formation, Kc = 0.41 results in auxiliary equation with purely imaginary roots. Therefore Kc = 0.41 results in dominant s-plane poles with real part = – 1. (c) Routh array formation for D( 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. wd = wn 1 - z 2 = 11.14; zwn = 1 These equations give z = 0.09. (d) 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 = – KI/Kc = – 24.39. The zero will not give pronounced early peak. Therefore z approximately represents Mp.. 6.9. (a) With TI = ¥ , D(s)G(s) =. 80 (1 + TD s ) s 2 + 8 s + 80. http://ebook29.blogspot.com/.

(44) http://ebook29.blogspot.com/. SOLUTION MANUAL. 45. The characteristic polynomial becomes D(s) = s2 + (8 + 80TD)s + 160 TD = 0.216 gives z = 1 (b) The zero is at s = – 1/0.216 = – 4.63. Real part of the closed-loop pole = – zwn = – 12.65. Therefore, small overshoot will be observed. (c) The characteristic equation becomes s3 + 25.28s2 + 160s +. 80 =0 TI. TI > 0.0198 for stability. 6.10 (a) G(s) =. K ; Kv = lim sG(s) = K s®0 s( s + 1). K = 10 will give steady-state unit-ramp following error of 0.1. D(s) = s2 + s + K = s2 + 2zwns + w 2n K = 10 gives z = 0.158. (b) G(s) =. 10 s( s + 1 + 10 Kt ). D(s) = s2 + (1 + 10Kt)s + 10 = s2 + 2zwns + w 2n Kt = 0.216 gives z = 0.5. Kv = lim sG(s) = s®0. 10 10 = 3.16 1 + 10 K t. ess = 0.316 (c) G(s) =. Kv =. 10 K A s( s + 1 + 10 Kt ) 10 K A 1 + 10 Kt. D(s) = s2 + (1 + 10Kt)s + 10 KA = s2 + 2zwns + w 2n From these equations, we find that KA = 10 and Kt = 0.9 give ess = 0.1 and z = 0.5.. http://ebook29.blogspot.com/.

(45) http://ebook29.blogspot.com/. 46. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. 6.11 (a) z =. ( In M p ) 2. 2. ( In M p ) 2 + π 2 M p = 0.1 gives z = 0.59 ts =. G(s) =. 4 = 0.05. Therefore, wn = 135.59 zw n. 5 K1 s( 0.2 s + 1 + 100 K2 ). D(s) = 0.2s2 + s(1 + 100K2) + 5K1 = 0 Therefore s2 + 5(1 + 100K2)s + 25K1 = s2 + 2zwns + w 2n This equation gives K1 = 735.39 and K2 = 0.31 5 K1 = 114.9, Ka = 0 1 + 100 K 2 q (s) 4 = 6.12 s( s + 1) + 10 K + 10 Kt s Tw ( s ) 4 For Tw(s) = 1/s, qss = = 0.05 10K. (b) Kp = ¥ , Kv =. This gives K = 8 D(s) = s2 + (1 + 10Kt)s + 10K = s2 + 2zwns + w 2n Kt = 0.79 gives z = 0.5 6.13. Y (s) 1 9 = . For F(s) = 9/s, yss = = 0.03 K Ms 2 + Bs + K F( s ) This gives K = 300 Newtons/m e - pz /. 1-z 2. =. 0.003 = 0.1 0.03. This gives z = 0.59 tp =. p wn 1-z. 2. =. p =2 0.81w n. This gives wn = 1.94 rad/sec From the equation. http://ebook29.blogspot.com/.

(46) http://ebook29.blogspot.com/. SOLUTION MANUAL. s2 +. 47. K B s+ = s2 + 2zwns + w 2n M M. we obtain M = 79.71 kg, B = 182.47 Newtons per m/sec 6.14. KT / J w 2n q (s) = = 2 K B q R (s) s + 2zw n s + w 2n s2 + s + T J J Tw. qR +. KT. –. z2 =. +. – Js2. 1 + Bs. q. ( ln M p ) 2 ( ln M p ) 2 + π 2. For Mp = 0.25, z = 0.4 Steady-state error to unit-ramp input =. 2z = 0.04 wn. This gives wn = 20. q( s ) 1 = 2 - Tw ( s ) Js + Bs + K T qss =. 10 = 0.01 KT. From these equations, we obtain the following values of system parameters. KT = 1000 N-m/rad B = 40 N-m/(rad/sec) J = 2.5 kg-m2. http://ebook29.blogspot.com/.

(47) http://ebook29.blogspot.com/. 48. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. 6.15 J w& + Bw = KT(wr – w). w (s) 45 = 100 s + 50 w r (s ) wr = 50 ×. 2p rad/sec = step input 60. w(t) = 1.5p (1 – e–0.5t) wss = 1.5p rad/sec = 45 rpm; ess = 5 rpm A control scheme employing gain adjustment with integral error compensation will remove the offset. 6.16 Jq&& + Bq& = KT(qr – q). q ( s) 2400 = 150 s 2 + 600 s + 2400 qr ( s ) qr(s) =. q(t) =. =. LM N. p 16 p ; q(s) = 2 3s 3 s ( s + 4 s + 16 ). LM MN p L 1 - 1.1547e 3 MN. OP Q. OP PQ p sin FH 3.46 t + IK O 3 PQ. e -zw nt p 1sin (w d t + cos -1z ) 2 3 1-z. M p = e - pz /. 1-z 2. -2 t. = 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 qR + –. Ks. e. KA. ec. K1. + –. Js2. qM 1 + Bs. sK2. http://ebook29.blogspot.com/. 1 50. qL.

(48) http://ebook29.blogspot.com/. SOLUTION MANUAL. (b) G(s) = Kv. 49. 4KA s( s + 100.5). = 4KA/100.5. For a ramp input qR(t) = p t, the steady-state error = The specification is 5 deg or. p Kv. 5p rad. 180. Therefore. 100.5p p = ; this gives KA = 904.5 180 4KA The characteristic equation is s2 + 100.5s + 4 × 904.5 = 0 wn = 60.15; z = 0.835; Mp = 0.85%; ts = 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.. FH. IK. 1 s ; K = 904.5 G(s) = A s( s + 100.5) 4KA 1 +. The characteristic equation is s3 + 100.5s2 + 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) (s2 + 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 s2 + 99.4508s + 3514.6332 = 0 wn = 59.28; z = 0.84; Mp = 0.772%; ts = 0.08 sec. http://ebook29.blogspot.com/.

(49) http://ebook29.blogspot.com/. 50. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. 6.18. Ket. et. Ia. (const). Rf Lf. KA. e. qR. JL. qL. qR. KP. +. e+. –. –. KA. KT. sKKt KP. (c) The characteristic equation of the system is s2 + 100 KKAs + 6KA = s2 + 2zwn s + w 2n = 0 This gives KA = 2.67; K = 0.024. http://ebook29.blogspot.com/. 1 Jeqs2. qM. 1 qL 50.

(50) http://ebook29.blogspot.com/. SOLUTION MANUAL. 51. 6.19 qR + –. KA. KP. + –. 1. KT Ra. Js2. + Bs. 1 10. qL. sKb. (b) Open-loop transfer function G(s) =. Kv =. 20 K A s( 4 s + 202 ). 20 K A 1 = ; this gives KA = 1010. 0.01 202. The characteristic equation becomes s2 + 50.5s + 5050 = 0 = s2 + 2zwn s + w 2n This gives z = 0.355; Mp = 30.33% (c) G(s) =. 20( K A + sK D ) s( 4 s + 202) Kv =. 20 K A = 100; no effect on ess. 202. Characteristic equation of the system is s2 + (50.5 + 5KD)s + 5050 = 0 = s2 + 2zwns + w 2n This gives KD = 6.95. For z = 0.6, Mp = 9.5% (d) System zero: s =. - KA = – 145.32 KD. Real part of closed-loop poles = – zwn = – 42.636. The zero will affect the overshoot. The peak overshoot will be slightly more than 9.5%. 6.20 (a) G(s) =. 3KA s( 0.3s + 1)(1 + 2 s ). http://ebook29.blogspot.com/.

(51) http://ebook29.blogspot.com/. 52. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. Kv = 3KA; ess =. 0.03 0.01 = 3K A KA. Y (s) 0.3(1 + 2 s ) = Fw ( s ) s( 0.3s + 1)(1 + 2 s ) + 3 K A. (b). yss =. 0.1 KA. (c) Characteristic equation is 0.6s3 + 2.3s2 + s + 3KA = 0 For stability, KA < 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. –. u. 1. i. 47.6 16.67. s 2–. Power amp. e. 100. (a) Characteristic equation of the system is s2 – 16.67s + 4760Kc = 0 Unstable or oscillatory for all Kc > 0. (b) U(s) = – Kc(1 + sTD) E(s) Characteristic equation becomes s2 + 4760 KcTDs + 4760Kc – 16.67 = 0 For stability, Kc > 3.5 × 10–3 and TD > 0. (c) z2 =. [ ln ( 0.2)]2 ; z = 0.456 [ ln ( 0.2)]2 + π 2 ts =. 4 = 0.4; wn = 21.93 zw n. http://ebook29.blogspot.com/. x.

(52) http://ebook29.blogspot.com/. SOLUTION MANUAL. s2 + 4760 KcTDs + 4760 Kc – 16.67 = s2 + 2zwn s + w 2n This gives Kc = 0.1045, TD = 0.04. 6.22 (a) System-type number = 2 (b). H (s) 0.5s( 0.1s + 1) = 2 - Qw ( s ) s ( 0.1s + 1) + 0.5( Kc + sK D ) hss = -. 0.5 ; Kc > 10 0.5Kc. The characteristic equation is 0.1s3 + s2 + 0.5KDs + 0.5Kc = 0 For stability, KD > 0.1Kc 6.23 qr + –. K1 +. K2 s. 1 s+1. Characteristic equation is s2 + (1 + K1)s + K2 = 0 = s2 + 2zwn s + w 2n Mp = 20% ® z = 0.456; ts = 2 ® wn = 4.386 K2 = 19.237; K1 = 3 ——. http://ebook29.blogspot.com/. q. 53.

(53) http://ebook29.blogspot.com/. CHAPTER 7. 7.1. COMPENSATOR DESIGN USING ROOT LOCUS PLOTS. (a) – sA = – 2 ; fA = 60º, 180º, 300º ; intersection with jw-axis at ± j5.042; angle of departure fp from – 1 + j4 = 40º. (b) –sA = – 1.25 ; fA = 45º, 135º, 225º, 315º ; intersection with jw-axis at ± j1. 1; angle of departure fp from – 1 + j1 = – 71.6º; multiple roots at s = – 2.3. (c) Angle of arrival fz at – 3 + j4 = 77.5º ; multiple roots at s = – 0.45. (d) – sA = –1.33 ; fA = 60º; 180º, 300º; intersection with jw-axis at ± j 5 ; angle of departure fp from – 2 + j1 = – 63.43º; multiple roots at s = – 1, – 1.67. (e) – sA = – 1.5 ; fA = 90º, 270º (f) – sA = – 5.5 ; fA = 90º, 270º; multiple roots at s = – 2.31, – 5.18. (g) – sA = 0.5 ; fA = 90º, 270º; intersection with jw-axis at ± j2.92.. 7.2. (i) – sA = – 2; fA = 60º, 180º, 300º; intersection with jw-axis at ± j 10 ; angle of departure fp 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. – 3 + j1. s. –2. (i). http://ebook29.blogspot.com/.

(54) http://ebook29.blogspot.com/. SOLUTION MANUAL. 55. (ii) Intersection with jw-axis at ± j2 3 ; angle of departure fp from – 3 + j 3 = – 60º. Three root loci approach the point s = – 2, and then break away in directions 120º apart. jw. jw – 3 + j ÷3. j2÷3. 60° s –2. (ii). –9. s. –4. (iii). (iii) – sA = – 4 ; fA = 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) – sA = – 1 ; fA = 45º, 135º, – 45º, – 135º; intersection with the jw axis at ± j 1.58; angle of departure fp 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/.

(55) http://ebook29.blogspot.com/. 56. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. – 1 + j2. jw. – 1 + j1. jw. 45° s. –2. (iv). 7.3. –2. 45°. s. (v). – sA = – 1 ; fA = 60º, 180º, 300º; intersection with the jw-axis at ± j 2 ; multiple roots at s = – 0.42. The z = 0.5 loci passes through the origin and makes an angle of q = cos–1 z = 60º with the negative real axis. The point sd = – 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 ( s + 0.33 + j 0.58)( s + 0.33 - j 0.58)( s + 2.33). 7.4. – sA = – 3 ; fA = 45º, 135º, 225º, 315º; intersection with the jw-axis at ± j3.25; departure angle fp from – 4 + j4 = 225º; multiple roots at s = – 1.5. At the intersection of the z = 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 jw-axis is approximately four times as large as that of the pair near jw-axis. Therefore, the transient response term due to the pair away from the jw-axis will decay much more rapidly than the transient response term due to the pair near jw-axis.. 7.5. – sA = – 2.67 ; fA = 60º, 180º, 300º ; intersection with jw-axis at ± j 32 ; departure angle fp from – 4 + j4 = – 45º. The point sd = – 2 + j3.4 on the z = 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. – sA = – 2; fA = 60º, 180º, 300º; intersection with the jw-axis at ± j 5 ; multiple roots at s = – 0.473.. http://ebook29.blogspot.com/.

(56) http://ebook29.blogspot.com/. SOLUTION MANUAL. 57. 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 jw-axis. From the Routh array formulation it is found that the greatest value of K is 30 and 7.7 7.8. 7.9. at this gain continuous oscillations of frequency 5 rad/sec occur. The proof given in Example 7.1. 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 z = 0.82 and the value of K at the point of tangency is 2.1. The proof given in Example 7.2. 7.10 s = s + jw tan–1. w +1 w -1 + tan -1 s s. = tan -1. w w + tan -1 ± 180 º (2 q + 1) s s +2. Taking tangents on both sides of this equation, and noting that. LM N. tan tan −1. w ω , ± 180 º = s +2 σ +2. OP Q. we obtain. w +1 w - 1 + s s w +1 w -1 1s s. FH. IK FH. IK. w w + s s +2 = w w 1s s +2. FH IK FH. IK. Manipulation of this equation gives. LMF s - 1 I NH 2 K. 2. +w2 -. OP Q. 5 =0 4. There exists a circular root locus with centre at s = ½, w = 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 z = 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 ts; therefore we choose K = 5(sd = – 3 + j3).. http://ebook29.blogspot.com/.

(57) http://ebook29.blogspot.com/. 58. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. For z = 0.707, Mp = e - pz / ts =. 1-z 2. = 4.3%. 4 4 = sec 3 zw n. G(s) =. K (s + 4) ( s + 2)( s - 1). E(s) =. ( s + 2 )( s + 1) 1 R( s ) R(s) = ( s + 2 )( s - 1) + K ( s + 4 ) 1 + G( s). For R(s) = 1/s (assuming selection of K that results in stable closed-loop poles) ess = lim sE ( s ) = s ®0. -2 -2 + 4K. For K = 5, ess = – 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 z = 1. For minimum steadystate 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+. or. 1+. 0.8 K =0 (10 s + 1)(30 s + 1)( 3s + 1) K¢. FH s + 1 IK FH s + 1 IK FH s + 1 IK 10 30 3. = 0 ; K' = 0.000888 K. – sA = – 0.155 ; fA = 60º, 180º, 300º; intersection with the jw-axis at ± j 0.22; multiple roots at s = – 0.063 The point on the z = 0.707 line that satisfies the angle criterion is sd = – 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/.

(58) http://ebook29.blogspot.com/. SOLUTION MANUAL. 59. 7.14 (a) Ideal feedback sensor: 1+. K 10 K =1+ =0 s( 0.1s + 1) s ( s + 10 ). K = 10 gives z = 0.5. (b) Sensor with appreciable time-constant:. or. 1+. K =0 s( 0.1s + 1)( 0 .02 s + 1). 1+. K¢ = 0 ; K ¢ = 500 K s( s + 10 )( s + 20). 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 z = 0.4. 7.15 Characteristic equation is given by D(s) = s3 + 6s2 + 8s + 0.1K(s + 10) = 0 which can be arranged as 1+. K ¢( s + 10) = 0 ; K ¢ = 0.1K s( s + 2 )( s + 4). – sA = 2 ; fA = 90º, – 90º; intersection with the jw-axis at ± j 20 ; multiple roots at s = – 0.8951. The point on z = 0.5 line that satisfies the angle criterion is sd = – 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 s3 + s + 10Kts + 10 = 0 which can be rearranged as 1+. Ks = 0 ; K = 10Kt s 2 + s + 10. 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 z = 0.7 line that satisfies the angle criterion is sd = – 2.214 + j2.258.. http://ebook29.blogspot.com/.

(59) http://ebook29.blogspot.com/. 60. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. The gain K corresponding to this point is 3.427. Hence the desired value of velocity feedback gain Kt is 0.3427. 7.17 Characteristic equation of the system is D(s) = s(s + 1) (s + 4) + 20Kts + 20 = 0 which may be rearranged as 1+. Ks = 0 ; K = 20Kt ( s + j 2)( s - j 2)(s + 5). – sA = – 2.5 ; fA = 90º, – 90º; departure angle fp from s = j2 is 158.2º. The following two points on z = 0.4 line satisfy the angle criterion: s1 = – 1.05 + j2.41, s2 = – 2.16 + j4.97 The value of K at s1 is 0.449, and at s2 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) 20 = s( s + 1)( s + 4 ) + 20(1 + Kt s ) R( s ) 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 = 20Kt appears as a multiplying factor. Kt = 0.449 satisfies the dominance condition to a reasonable extent. For Kt = 1.413, complex-conjugate poles are not dominant. 7.18 The characteristic equation of the system is s(s + 1) (s + 3) + 2s + 2a = 0 which may be rearranged as 1+. K = 0 ; K = 2a s ( s + 4 s + 5) 2. – sA = – 1.3333 ; fA = 60º, – 60º, 180º ; intersection with the jw-axis at ± j 5 ; departure angle fp from complex pole in the upper half of s-plane = – 63.43º; multiple roots at s = – 1, – 1.666. The point on z = 0.5 line that satisfies the angle criterion is sd = – 0.63 + j1.09. The value of K at this point, obtained by magnitude criterion, is 4.32. Therefore a = 2.16. The third pole for K = 4.32 is at s = – 2.75.. http://ebook29.blogspot.com/.

(60) http://ebook29.blogspot.com/. SOLUTION MANUAL. 61. 7.19 The characteristic equation of the system is s3 + 9s2 + 18s + 10a = 0 which may be rearranged as 1+. K = 0 ; K = 10a s( s + 3)( s + 6). – sA = – 3 ; fA = 60º, – 60º, 180º; intersection with the jw-axis at ± j3 2 ; multiple roots at s = – 1.268. The point on z = 0.5 line that satisfies the angle criterion is sd = – 1 + j1.732. The value of K at this point is 28. Therefore a = 2.8. The third pole for K = 28 is at s = – 7. 7.20 (a) 1 +. K =0 s( s - 2 ). 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 + 1) =0 s( s − 2 )( s + 8). From the Routh array, we find that the loci cross the jw-axis when K = 19.2. For K > 19.2, the closed-loop poles are always in the lefthalf plane and the system is stable.. E( s ) s ( s - 2 )( s + 8) 1 = = s( s - 2)( s + 8) + Ks + K R( s ) 1 + G( s) For R(s) = 1/s and K > 19.2 (required for stability), ess = lim sE(s) = 0 s®0. 2. For R(s) = 1/s and K > 19.2, ess = lim sE(s) = s®0. -16 K. 7.21 – sA = – 2 ; fA = 60º, 180º, 300º; intersection with the jw-axis at ± j 5 ; multiple roots at s = – 0.473. The value of K at the point of intersection of the root locus and z = 0.3 line is 7.0.. http://ebook29.blogspot.com/.

(61) http://ebook29.blogspot.com/. 62. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. Kv = lim sG(s) = K/5 = 1.4 ; ts = 4/zwn = 11.6 sec. s®0. With the compensator, the characteristic equation becomes 1+. K (10 s + 1) =0 s(100 s + 1)( s + 1)( s + 5). 1+. K ¢( s + 0.1) = 0 ; K¢ = 0.1K s( s + 0.01)(s + 1)( s + 5). or. The value of K¢ at the point of intersection of z = 0.3 line and the root locus is 6.0. Therefore K = 60. Kv = lim sD(s) G(s) = K/5 = 12, ; ts = 4/zwn = 12.7 sec. s®0. 7.22 1 +. K ( s + 2.5) =0 s( s + 1)( s + a ). Desired closed-loop pole; sd = – 1.6 + j4. Angle criterion at sd is satisfied when a = 5.3. The value of K at sd is 23.2. The third pole is at s = – 3.1. 7.23 (a) 1 +. K =0 s( s + 2 ). z = 0.707 ; ts =. 4 = 2. Therefore, zwn = 2 zw n. Proportional control does not meet this requirement. (b) 1 +. K c + sK D K ( s + Kc / K D ) =1+ D s(s + 2) s( s + 2). At the points of intersection of z = 0.707 line and zwn = 2 line, the angle criterion is satisfied when Kc/KD = 4. The root locus plot of 1+. K D (s + 4) =0 s(s + 2). has circular complex-root branches (refer Problem 7.7). At the point of intersection of z = 0.707 line and zwn = 2 line, we find by magnitude criterion that KD = 2. Therefore Kc = 8. 7.24 G(s) =. K ( s + 0.1) ; K = 4000 s( s 2 + 0.8s + 4). http://ebook29.blogspot.com/.

(62) http://ebook29.blogspot.com/. SOLUTION MANUAL. D(s) =. 63. s 2 + 0.8s + 4 ( s + 0.384)( s + 10.42). D(s) improves transient response considerably, and there is no effect on steady-state performance. 7.25 z = 0.45 (specified). Let us select the real part of the desired roots as zwn = 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( s + 4 ) s( s + 2 )( s + 10.6 ). Kv = lim sD(s) G(s) = 18.2. s®0. The Kv 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 sd = – 1.588 + j3.152. s +1 , s + 10 angle criterion is met. Magnitude criterion gives K = 147. The compensated system has Kv = 2.94. It meets the requirements on Mp and settling time. With D(s) =. 7.27 z = 0.707; 4/zwn = 1.4 ® zwn = 2.85 wn = 4 ; w n 1 - z 2 = 2.85 sd = – 2.85 + j2.85. ( s + 1.5) 2 , angle criterion is satisfied at sd. The value of K, ( s + 3.5) 2 found by magnitude criterion, is 37.. With D(s) =. 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 z is 0.5 and that of wn is 0.66. The velocity error constant is 0.53. The desired dominant closed-loop poles are – 0.33 ± j0.58 with a Kv of 5. The lag compensator. http://ebook29.blogspot.com/.

(63) http://ebook29.blogspot.com/. 64. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. s + 0.1 s + 0.001. D¢(s) =. gives Kv 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 z = 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. 0.925( s + 0.1) s + 0.01. D(s) =. 7.29 The dominant closed-loop poles of uncompensated system are located at s = –3.6 ± j4.8 with z = 0.6. The value of K is found as 820. Therefore Kv = lim sG(s) = 820/200 = 4.1. s®0. s + 0.25 gives a Kv boost of the factor of 10. s + 0.025 The angle contribution of this compensator at – 3.6 + j4.8 is – 1.8º, which is acceptable in the present problem.. Lag compensator D(s) =. 7.30 (a) 1 +. K K =1+ =0 ( 2 s + 1)( 0.5s + 1) ( s + 0.5)( s + 2 ). From the root locus we find that no value of K will yield zwn = 0.75.. FG H. (b) D(s) = K c 1 +. D(s)G(s) =. 1 TI s. IJ = K + K /s = K F s + K / K I H s K K I. c. I. c. c. Kc (s + K I / Kc ) s( s + 0.5)( s + 2). At the point corresponding to z = 0.6 and zwn = 0.75, the angle criterion is satisfied when KI /Kc = 0.75. The gain at the desired point is Kc = 2.06. This yields KI = 1.545. 7.31 Kv = 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 Kv requirement is satisfied are a long way from satisfying the z requirement. It will be difficult to bring the dominant. http://ebook29.blogspot.com/.

(64) http://ebook29.blogspot.com/. SOLUTION MANUAL. 65. roots from the jw-axis to the z = 0.707 line by using a lead compensator. Therefore, we will attempt to satisfy the Kv and z requirements by a lag compensator. From the uncompensated root locus we find that corresponding to z = 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| 2000 = = 8.5 236 | p| We will choose z = 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 + 1 / t 1 )(s + 1/t 2 ) s( s + 1)(s + 5)(s + 1/ at 1 )(s + a /t 2 ). 1 1 1 a = 0.05 ; =1; = 0.005 ; = 10 t2 t1 t2 at 1 D(s)G(s) =. K ( s + 0.05) s( s + 0.005)( s + 5)( s + 10). For z = 0.45, we obtain sd = – 1.571 + j3.119, K = 146.3, Kv = 29.3, wn = 3.49. FG H. 7.33 D(s) = Kc + KDs = Kc 1 +. FG H. KD s Kc s2 + 1. Kc 1 +. (a) D(s)G(s) =. KD s Kc. IJ K. IJ K. s2 + 1 + Kc + KDs = (s + 1 + j 3 ) (s + 1 – j 3 ) This equation gives Kc = 3 and KD = 2 Kp = lim D(s)G(s) = 3. s®0. (b) With PID controller D(s) =. K ( s + 0.2)( s + 1.3332 ) s. http://ebook29.blogspot.com/.

(65) http://ebook29.blogspot.com/. 66. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. the angle criterion at sd = – 1 + j 3 is satisfied. The gain K is 2.143. 7.34 (a) Uncompensated system: Mp = 20% ® z = 0.45 At the point of intersection of z = 0.45 line and the root locus, zwn = 1.1 and K = 56 (b) ts = 4/zwn = 3.6 sec Kv = lim sG(s) = 2.07 s®0. (c) Compensated system: M p = 15% ® z = 0.55 ts = wn =. 3.6 = 1.45 ; zwn = 4/1.45 = 2.75 2.5 2.75 = 5 ; sd = – 2.75 + j4.18 0.55. ³ 20. Kv. Lead-lag compensated system: D1(s) D2(s) G(s) = 7.35 G(s) =. K ( s + 3)( s + 0.9) ; K = 506 s( s + 0.15)(s + 3)( s + 9)(s + 13.6 ). K s( s + 2 + KKt ) E( s ) s( s + 2 + KK t ) 1 = = R( s ) s( s + 2 + KK t ) + K 1 + G( s). For R(s) = 1/s2, ess =. 2 + KKt £ 0.35 K. z ³ 0.707 ; ts £ 3 sec. The characteristic equation of the system is s2 + 2s + KKts + K = 0 which may be rearranged as. http://ebook29.blogspot.com/.

(66) http://ebook29.blogspot.com/. SOLUTION MANUAL. 1+. 67. KK t s =0 s + 2s + K 2. The locus of the roots as K varies (set KKt = 0) is determined from the following equation: 1+ For. K =0 s( s + 2 ). K = 20, the roots are – 1 ± j4.36.. Then the effect of varying KKt is determined from the locus equation 1+. KKt s =0 ( s + 1 + j 4.36)( s + 1 - j 4.36). Note that complex root branches follow a circular path. At the intersection of the root locus and z = 0.707 line, we obtain KKt = 4.3. The real part of the point of intersection is s = 3.15, and therefore the settling time is 1.27 sec. 7.36 The characteristic equation of the system is 1+. KK t s =0 s + 2s + K. 1+. KK t s =0 ( s + 1 + j ( K - 1 )( s + 1 - j ( K - 1). 2. or. The root contour plotted for various values of K with K¢t = = KKt varying from 0 to ¥ are shown in the figure below.. http://ebook29.blogspot.com/.

(67) http://ebook29.blogspot.com/. 68. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. jw K¢t = 0. K=5 K=2. K¢t = 0. j2. j1. • ¨ K¢t – ÷5. – ÷2. K¢t = •. –1. K¢t = 0. K¢t = 0. 7.37 G(s) =. s. – j1. – j2. AK s( s + 1)( s + 5) + KK t s. The characteristic equation is s(s + 1) (s + 5) + KKt (s + A/Kt) = 0 which may be rearranged as 1+. KKt ( s + A / K t ) =0 s( s + 1)( s + 5). The angle criterion at s = – 1 + j2 is satisfied when A/Kt = 2.5. Let us take A = 2 and Kt = 0.8. Kv = lim sG(s) = s®0. With. AK 5 + KK t. K = 10, Kv = 1.54. The closed-loop transfer function of the system is. Y (s) 20 = ( s + 1 + j 2 )( s + 1 - j 2 )( s + 4) R( s ) 7.38 The characteristic equation of the system is s(s + 1) (s + 5) + KKt s + AK = 0. http://ebook29.blogspot.com/.

(68) http://ebook29.blogspot.com/. SOLUTION MANUAL. 69. or s(s + 1) (s + 5) + bs + a = 0 ; b = KKt ; a = AK The root locus equation as a function of a with b = 0 is 1+. a =0 s ( s + 1)( s + 5 ). Sketch a root locus plot for a varying from 0 to ¥ . The points on these loci become the open-loop poles for the root locus equation 1+. bs =0 s( s + 1)( s + 5) + a. For some selected values of a, the loci for b are sketched (refer Fig. 7.44). e–s = -. 7.39. 1–. (s - 2 ) s+2. K (s - 2) = 0 = 1 – F(s) ( s + 1)( s + 2 ) F(s) =. =. K (s + jw - 2 ) (s + jw + 1)(s + jw + 2 ) K (s - 2 + jw ) (s + 1)(s + 2 ) - w 2 + jw ( 2s + 3). Ð F(s) = tan–1. = tan–1. w (2s + 3) w - tan -1 s -2 (s + 1)(s + 2 ) - w 2. R| w - w (2s + 3) S| s -w2 L(s + 1w)((s2s+ +2)3)- w |T 1 + s - 2 MN (s + 1)(s + 2) - w. Ð F(s) = 0º if. w ( 2s + 3) w =0 s - 2 (s + 1)(s + 2 ) - w 2 Manipulation of this equation gives (s – 2)2 + w2 = 12 Centre = (2, 0) and radius = 12 = 3.464. http://ebook29.blogspot.com/. 2. 2. U| OP V| Q |W.

(69) http://ebook29.blogspot.com/. 70. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. It can easily be verified that at every point on this circle, Ð F(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 £ 0. Centre at zero at s = 0 Radius = a 2 + b 2 = ( 0.5) 2 + ( 0.5) 2 = 0.707 The root locus plot is shown in the figure below.. j0.5. K=0. KÆ –ñ •. KÆ –ñ •. – 0.5. K=0. 7.41 G(s) =. 20.7( s + 3) s( s + 2 )( s + 8). (i) a = a0 + Da = 8 + Da The characteristic equation is s(s + 2) (s + 8) + Da s(s + 2) + 20.7 (s + 3) = 0 When Da = 0, the roots are (may be determined by root-locus method or the Newton-Raphson method) l1,2 = – 2.36 ± j2.48, l3 = – 5.27 The root locus for Da is determined using the root locus equation 1+. Da s( s + 2 ) =0 ( s - l 1 )( s - l 2 )( s - l 3 ). http://ebook29.blogspot.com/.

(70) http://ebook29.blogspot.com/. SOLUTION MANUAL. 71. The angle of departure at s = l1, is – 80º. Near s = l1, the locus may be approximated by a line drawn from – 2.36 + j2.48 at an angle of – 80º. For a change of Dl1 = 0.2 Ð – 80º along the departure line, Da is determined by magnitude criterion: Da = 0.48 Therefore the sensitivity at l1 is Sal +1 =. Dl 1 0.2 Ð - 80 º = = 3.34 Ð – 80º Da / a 0 0. 48 / 8. (ii) Sensitivity of the root s = l1 to a change in zero at s = – 3 is determined as follows. b = b0 + Db = 3 + Db The characteristic equation is s(s + 2) (s + 8) + 20.7 (s + 3 + Db) = 0 or 1+. 20.7 Db =0 ( s - l 1 )( s - l 2 )( s - l 3 ). The angle of departure of l1 is 50º. For a change of Dl1 = 0.2 Ð + 50º, we obtain Db = 0.21. Therefore. Sbl1 =. Dl 1 0.2 Ð + 50 º = = 2.84 Ð + 50º Db / b 0 0.21 / 3. 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 = l1. 7.42 (a) The characteristic equation is 1+. K ( s + 5) =0 s( s + 2 )( s + 3). It can easily be verified from the root locus plot that for K = 8 the closed-loop poles are l1,2 = – 0.5 ± j 3.12, l3 = – 4 (b) s(s + 2 + d) (s + 3) + 8(s + 5) = 0. http://ebook29.blogspot.com/.

(71) http://ebook29.blogspot.com/. 72. CONTROL SYSTEMS: PRINCIPLES AND DESIGN. or 1+. or. 1+. d s( s + 3) =0 s( s + 2)( s + 3) + 8(s + 5). d s( s + 3) =0 s( s - l 1 )(s - l 2 )(s - l 3 ). Case I: d > 0 From the root locus plot we find that the direction of departure from the pole at s = l1, is away from the jw-axis. Case II: d < 0 From the root locus plot we find that the direction of departure from the pole at s = l1 is towards the jw-axis. Therefore the variation d < 0 is dangerous.. http://ebook29.blogspot.com/.

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