Solution Manual for Modern Control Systems, 13th Edition

C H A P T E R 2 Mathematical Models of Systems Exercises E2.1 We have for the open-loop T a his th nd wo o eir is rk w r sa co pro is u v p y ill de le o rse ide rot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) y = r2 and for the closed-loop e = r − y and y = e2 . So, e = r − e2 and e2 + e − r = 0 . 40 35 30 25 20 15 10 open−loop 5 closed−loop 0 0 1 2 3 r 4 5 6 FIGURE E2.1 Plot of open-loop versus closed-loop. For example, if r = 1, then e2 + e − 1 = 0 implies that e = 0.618. Thus, y = 0.382. A plot y versus r is shown in Figure E2.1. 22 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 23 Exercises E2.2 Define f (T ) = R = R0 e−0.1T and ∆R = f (T ) − f (T0 ) , ∆T = T − T0 . Then, ∆R = f (T ) − f (T0 ) = ∂f ∆T + · · · ∂T T =T0 =20◦ where T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ∂f = −0.1R0 e−0.1T0 = −135, ∂T T =T0 =20◦ when R0 = 10, 000Ω. Thus, the linear approximation is computed by considering only the first-order terms in the Taylor series expansion, and is given by ∆R = −135∆T . E2.3 The spring constant for the equilibrium point is found graphically by estimating the slope of a line tangent to the force versus displacement curve at the point y = 0.5cm, see Figure E2.3. The slope of the line is K ≈ 1. 2 1.5 Spring breaks 1 0.5 Force (n) 0 -0.5 -1 -1.5 -2 -2.5 -3 -2 Spring compresses -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 y=Displacement (cm) FIGURE E2.3 Spring force as a function of displacement. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 24 CHAPTER 2 E2.4 Mathematical Models of Systems Since R(s) = 1 s we have Y (s) = 6(s + 50) . s(s + 30)(s + 10) The partial fraction expansion of Y (s) is given by Y (s) = A1 A2 A3 + + s s + 30 s + 10 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) where A1 = 1 , A2 = 0.2 and A3 = −1.2 . Using the Laplace transform table, we find that y(t) = 1 + 0.2e−30t − 1.2e−10t . The final value is computed using the final value theorem: 6(s + 50) lim y(t) = lim s =1. 2 t→∞ s→0 s(s + 40s + 300) E2.5 The circuit diagram is shown in Figure E2.5. R2 v+ A + vin – + v0 – R1 FIGURE E2.5 Noninverting op-amp circuit. With an ideal op-amp, we have vo = A(vin − v − ), © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 25 Exercises where A is very large. We have the relationship v− = R1 vo . R1 + R2 Therefore, vo = A(vin − R1 vo ), R1 + R2 and solving for vo yields vo = A 1 1 + RAR 1 +R2 vin . T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) 1 1 Since A ≫ 1, it follows that 1 + RAR ≈ RAR . Then the expression for 1 +R2 1 +R2 vo simplifies to vo = E2.6 R1 + R2 vin . R1 Given y = f (x) = ex and the operating point xo = 1, we have the linear approximation y = f (x) = f (xo ) + ∂f (x − xo ) + · · · ∂x x=xo where df = e, dx x=xo =1 f (xo ) = e, and x − xo = x − 1. Therefore, we obtain the linear approximation y = ex. E2.7 The block diagram is shown in Figure E2.7. R(s) Ea(s) + G1(s) G2(s) I(s) – H(s) FIGURE E2.7 Block diagram model. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 26 CHAPTER 2 Mathematical Models of Systems Starting at the output we obtain I(s) = G1 (s)G2 (s)E(s). But E(s) = R(s) − H(s)I(s), so I(s) = G1 (s)G2 (s) [R(s) − H(s)I(s)] . Solving for I(s) yields the closed-loop transfer function G1 (s)G2 (s) I(s) = . R(s) 1 + G1 (s)G2 (s)H(s) E2.8 The block diagram is shown in Figure E2.8. T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) H2(s) – R(s) K – E(s) – G1(s) W(s) – A(s) G2(s) Z(s) 1 s Y(s) H3(s) H1(s) FIGURE E2.8 Block diagram model. Starting at the output we obtain Y (s) = 1 1 Z(s) = G2 (s)A(s). s s But A(s) = G1 (s) [−H2 (s)Z(s) − H3 (s)A(s) + W (s)] and Z(s) = sY (s), so 1 Y (s) = −G1 (s)G2 (s)H2 (s)Y (s) − G1 (s)H3 (s)Y (s) + G1 (s)G2 (s)W (s). s Substituting W (s) = KE(s) − H1 (s)Z(s) into the above equation yields Y (s) = −G1 (s)G2 (s)H2 (s)Y (s) − G1 (s)H3 (s)Y (s) 1 + G1 (s)G2 (s) [KE(s) − H1 (s)Z(s)] s © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 27 Exercises and with E(s) = R(s) − Y (s) and Z(s) = sY (s) this reduces to Y (s) = [−G1 (s)G2 (s) (H2 (s) + H1 (s)) − G1 (s)H3 (s) 1 1 − G1 (s)G2 (s)K]Y (s) + G1 (s)G2 (s)KR(s). s s Solving for Y (s) yields the transfer function Y (s) = T (s)R(s), where T (s) = E2.9 KG1 (s)G2 (s)/s . 1 + G1 (s)G2 (s) [(H2 (s) + H1 (s)] + G1 (s)H3 (s) + KG1 (s)G2 (s)/s From Figure E2.9, we observe that and T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Ff (s) = G2 (s)U (s) FR (s) = G3 (s)U (s) . Then, solving for U (s) yields U (s) = 1 Ff (s) G2 (s) FR (s) = G3 (s) U (s) . G2 (s) and it follows that Again, considering the block diagram in Figure E2.9 we determine Ff (s) = G1 (s)G2 (s)[R(s) − H2 (s)Ff (s) − H2 (s)FR (s)] . But, from the previous result, we substitute for FR (s) resulting in Ff (s) = G1 (s)G2 (s)R(s)−G1 (s)G2 (s)H2 (s)Ff (s)−G1 (s)H2 (s)G3 (s)Ff (s) . Solving for Ff (s) yields G1 (s)G2 (s) Ff (s) = R(s) . 1 + G1 (s)G2 (s)H2 (s) + G1 (s)G3 (s)H2 (s) © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 28 CHAPTER 2 Mathematical Models of Systems H2(s) + – R(s) U(s) G2(s) Ff (s) U(s) G3(s) FR(s) G1(s) – H2(s) FIGURE E2.9 Block diagram model. The shock absorber block diagram is shown in Figure E2.10. The closedloop transfer function model is T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) E2.10 T (s) = Gc (s)Gp (s)G(s) . 1 + H(s)Gc (s)Gp (s)G(s) Controller Gear Motor Plunger and Piston System Gc(s) Gp(s) G(s) + R(s) Desired piston travel – Y(s) Piston travel Sensor H(s) Piston travel measurement FIGURE E2.10 Shock absorber block diagram. E2.11 Let f denote the spring force (n) and x denote the deflection (m). Then K= ∆f . ∆x Computing the slope from the graph yields: (a) xo = −0.14m → K = ∆f /∆x = 10 n / 0.04 m = 250 n/m (b) xo = 0m → K = ∆f /∆x = 10 n / 0.05 m = 200 n/m (c) xo = 0.35m → K = ∆f /∆x = 3n / 0.05 m = 60 n/m © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 29 Exercises E2.12 The signal flow graph is shown in Fig. E2.12. Find Y (s) when R(s) = 0. -K Td(s) 1 1 K2 G(s) Y (s) -1 FIGURE E2.12 Signal flow graph. T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) The transfer function from Td (s) to Y (s) is Y (s) = G(s)(1 − K1 K2 )Td (s) G(s)Td (s) − K1 K2 G(s)Td (s) = . 1 − (−K2 G(s)) 1 + K2 G(s) If we set K1 K2 = 1 , then Y (s) = 0 for any Td (s). E2.13 The transfer function from R(s), Td (s), and N (s) to Y (s) is 1 K K R(s)+ 2 Td (s)− 2 N (s) Y (s) = 2 s + 10s + K s + 10s + K s + 10s + K Therefore, we find that Y (s)/Td (s) = E2.14 1 s2 + 10s + K and K Y (s)/N (s) = − 2 s + 10s + K Since we want to compute the transfer function from R2 (s) to Y1 (s), we can assume that R1 = 0 (application of the principle of superposition). Then, starting at the output Y1 (s) we obtain Y1 (s) = G3 (s) [−H1 (s)Y1 (s) + G2 (s)G8 (s)W (s) + G9 (s)W (s)] , or [1 + G3 (s)H1 (s)] Y1 (s) = [G3 (s)G2 (s)G8 (s)W (s) + G3 (s)G9 (s)] W (s). Considering the signal W (s) (see Figure E2.14), we determine that W (s) = G5 (s) [G4 (s)R2 (s) − H2 (s)W (s)] , © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 30 CHAPTER 2 Mathematical Models of Systems H1(s) + G1(s) R1(s) + G7(s) R2(s) G4(s) – + G2(s) G3(s) + Y1(s) G9(s) G8(s) + + G6(s) G5(s) Y2(s) W(s) – FIGURE E2.14 Block diagram model. or T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) H2(s) [1 + G5 (s)H2 (s)] W (s) = G5 (s)G4 (s)R2 (s). Substituting the expression for W (s) into the above equation for Y1 (s) yields Y1 (s) G2 (s)G3 (s)G4 (s)G5 (s)G8 (s) + G3 (s)G4 (s)G5 (s)G9 (s) = . R2 (s) 1 + G3 (s)H1 (s) + G5 (s)H2 (s) + G3 (s)G5 (s)H1 (s)H2 (s) E2.15 For loop 1, we have di1 1 R1 i1 + L1 + dt C1 Z (i1 − i2 )dt + R2 (i1 − i2 ) = v(t) . And for loop 2, we have 1 C2 E2.16 Z di2 1 i2 dt + L2 + R2 (i2 − i1 ) + dt C1 Z (i2 − i1 )dt = 0 . The transfer function from R(s) to P (s) is P (s) 4.2 = 3 . 2 R(s) s + 2s + 4s + 4.2 The block diagram is shown in Figure E2.16a. The corresponding signal flow graph is shown in Figure E2.16b for P (s)/R(s) = 4.2 s3 + 2s2 + 4s + 4.2 . © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 31 Exercises v1(s) R(s) v2(s) 7 – q(s) 0.6 s 1 s2+2s+4 P(s) (a) R(s ) 1 V1 V2 7 1 s2 + 2 s + 4 0.6 s P (s) T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) -1 (b) FIGURE E2.16 (a) Block diagram, (b) Signal flow graph. E2.17 A linear approximation for f is given by ∆f = ∂f ∆x = 2kxo ∆x = k∆x ∂x x=xo where xo = 1/2, ∆f = f (x) − f (xo ), and ∆x = x − xo . E2.18 The linear approximation is given by ∆y = m∆x where m= ∂y . ∂x x=xo (a) When xo = 1, we find that yo = 2.4, and yo = 13.2 when xo = 2. (b) The slope m is computed as follows: m= ∂y = 1 + 4.2x2o . ∂x x=xo Therefore, m = 5.2 at xo = 1, and m = 18.8 at xo = 2. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 32 CHAPTER 2 E2.19 Mathematical Models of Systems The output (with a step input) is Y (s) = 28(s + 1) . s(s + 7)(s + 2) The partial fraction expansion is Y (s) = 2 4.8 2.8 − + . s s+7 s+2 Taking the inverse Laplace transform yields y(t) = 2 − 4.8e−7t + 2.8e−2t . E2.20 The input-output relationship is where T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Vo A(K − 1) = V 1 + AK K= Z1 . Z1 + Z2 Assume A ≫ 1. Then, Vo K−1 Z2 = =− V K Z1 where Z1 = R1 R1 C 1 s + 1 and Z2 = R2 . R2 C 2 s + 1 Therefore, Vo (s) R2 (R1 C1 s + 1) 2(s + 1) =− =− . V (s) R1 (R2 C2 s + 1) s+2 E2.21 The equation of motion of the mass mc is mc ẍp + (bd + bs )ẋp + kd xp = bd ẋin + kd xin . Taking the Laplace transform with zero initial conditions yields [mc s2 + (bd + bs )s + kd ]Xp (s) = [bd s + kd ]Xin (s) . So, the transfer function is bd s + kd 0.65s + 1.8 Xp (s) = = 2 . Xin (s) mc s2 + (bd + bs )s + kd s + 1.55s + 1.8 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 33 Exercises E2.22 The rotational velocity is ω(s) = 2(s + 4) 1 . 2 (s + 5)(s + 1) s Expanding in a partial fraction expansion yields ω(s) = 81 1 1 3 13 1 1 + − − . 2 5 s 40 s + 5 2 (s + 1) 8 s+1 Taking the inverse Laplace transform yields ω(t) = E2.23 8 1 3 13 + e−5t − te−t − e−t . 5 40 2 8 The closed-loop transfer function is E2.24 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Y (s) K1 K2 = T (s) = 2 . R(s) s + (K1 + K2 K3 + K1 K2 )s + K1 K2 K3 Let x = 0.6 and y = 0.8. Then, with y = ax3 , we have 0.8 = a(0.6)3 . Solving for a yields a = 3.704. A linear approximation is y − yo = 3ax2o (x − xo ) or y = 4x − 1.6, where yo = 0.8 and xo = 0.6. E2.25 The closed-loop transfer function is Y (s) 10 = T (s) = 2 . R(s) s + 21s + 10 E2.26 The equations of motion are m1 ẍ1 + k(x1 − x2 ) = F m2 ẍ2 + k(x2 − x1 ) = 0 . Taking the Laplace transform (with zero initial conditions) and solving for X2 (s) yields X2 (s) = k (m2 s2 + k)(m1 s2 + k) − k 2 F (s) . Then, with m1 = m2 = k = 1, we have X2 (s)/F (s) = 1 . s2 (s2 + 2) © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 34 CHAPTER 2 E2.27 Mathematical Models of Systems The transfer function from Td (s) to Y (s) is Y (s)/Td (s) = E2.28 G2 (s) . 1 + G1 G2 H(s) The transfer function is Vo (s) R2 R4 R2 R4 C s+ = 46.08s + 344.91 . = V (s) R3 R1 R3 E2.29 (a) If G(s) = 1 s2 + 15s + 50 and H(s) = 2s + 15 , T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) then the closed-loop transfer function of Figure E2.28(a) and (b) (in Dorf & Bishop) are equivalent. (b) The closed-loop transfer function is T (s) = E2.30 1 s2 + 17s + 65 . (a) The closed-loop transfer function is T (s) = 15 G(s) 1 = 1 + G(s) s s(s2 + 5s + 30) where G(s) = 15 . s2 + 5s + 15 0.7 0.6 Amplitude 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 Time (seconds) 2 2.5 FIGURE E2.30 Step response. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 35 Exercises (b) The output Y (s) (when R(s) = 1/s) is Y (s) = or 0.5 −0.25 + 0.1282j −0.25 − 0.1282j + + s s + 2.5 − 4.8734 s + 2.5 + 4.8734j 1 Y (s) = 2 1 s+5 − 2 s s + 5s + 30 (c) The plot of y(t) is shown in Figure E2.30. The output is given by y(t) = 0.5(1 − 1.1239e−2.5t sin(4.8734t + 1.0968)); E2.31 The partial fraction expansion is a b + s + p1 s + p2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) V (s) = where p1 = 4 − 22j and p2 = 4 + 22j. Then, the residues are a = −11.37j b = 11.37j . The inverse Laplace transform is v(t) = −11.37je(−4+22j)t + 11.37je(−4−22j)t = 22.73e−4t sin 22t . © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 36 CHAPTER 2 Mathematical Models of Systems Problems P2.1 The integrodifferential equations, obtained by Kirchoff’s voltage law to each loop, are as follows: R2 i1 + 1 C1 Z i1 dt + L1 d(i1 − i2 ) + R1 (i1 − i2 ) = v(t) dt R3 i2 + 1 C2 Z i2 dt + R1 (i2 − i1 ) + L1 (loop 1) and P2.2 d(i2 − i1 ) =0 dt (loop 2) . The differential equations describing the system can be obtained by using a free-body diagram analysis of each mass. For mass 1 and 2 we have T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) M1 ÿ1 + k12 (y1 − y2 ) + bẏ1 + k1 y1 = F (t) M2 ÿ2 + k12 (y2 − y1 ) = 0 . Using a force-current analogy, the analagous electric circuit is shown in Figure P2.2, where Ci → Mi , L1 → 1/k1 , L12 → 1/k12 , and R → 1/b . FIGURE P2.2 Analagous electric circuit. P2.3 The differential equations describing the system can be obtained by using a free-body diagram analysis of each mass. For mass 1 and 2 we have M ẍ1 + kx1 + k(x1 − x2 ) = F (t) M ẍ2 + k(x2 − x1 ) + bẋ2 = 0 . Using a force-current analogy, the analagous electric circuit is shown in Figure P2.3, where C→M L → 1/k R → 1/b . © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 37 Problems FIGURE P2.3 Analagous electric circuit. P2.4 (a) The linear approximation around vin = 0 is vo = 0vin , see Figure P2.4(a). T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n vo . e W eb ) (b) The linear approximation around vin = 1 is vo = 2vin − 1, see Figure P2.4(b). (a) 0.4 (b) 4 3.5 0.3 3 0.2 2.5 vo 0.1 2 0 1.5 linear approximation 1 -0.1 0.5 -0.2 0 -0.3 -0.4 -1 linear approximation -0.5 -0.5 0 vin 0.5 1 -1 -1 0 1 2 vin FIGURE P2.4 Nonlinear functions and approximations. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 38 CHAPTER 2 P2.5 Mathematical Models of Systems Given Q = K(P1 − P2 )1/2 . Let δP = P1 − P2 and δPo = operating point. Using a Taylor series expansion of Q, we have Q = Qo + ∂Q (δP − δPo ) + · · · ∂δP δP =δPo where Qo = KδPo1/2 K ∂Q = δPo−1/2 . ∂δP δP =δPo 2 and T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Define ∆Q = Q − Qo and ∆P = δP − δPo . Then, dropping higher-order terms in the Taylor series expansion yields ∆Q = m∆P where m= P2.6 K 1/2 2δPo . From P2.1 we have and R2 i1 + 1 C1 Z i1 dt + L1 R3 i2 + 1 C2 Z i2 dt + R1 (i2 − i1 ) + L1 d(i1 − i2 ) + R1 (i1 − i2 ) = v(t) dt d(i2 − i1 ) =0. dt Taking the Laplace transform and using the fact that the initial voltage across C2 is 10v yields [R2 + 1 + L1 s + R1 ]I1 (s) + [−R1 − L1 s]I2 (s) = 0 C1 s and [−R1 − L1 s]I1 (s) + [L1 s + R3 + 1 10 + R1 ]I2 (s) = − . C2 s s Rewriting in matrix form we have   R2 + C11 s + L1 s + R1 −R1 − L1 s −R1 − L1 s L1 s + R3 + C12 s + R1   I1 (s) I2 (s)   = 0 −10/s   . © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 39 Problems Solving for I2 yields      1 0 R1 + L 1 s 1  L1 s + R3 + C2 s + R1  .  =  1 ∆ −10/s I2 (s) R1 + L 1 s R2 + C1 s + L1 s + R1 I1 (s) or I2 (s) = −10(R2 + 1/C1 s + L1 s + R1 ) s∆ where P2.7 1 1 + L1 s + R1 )(L1 s + R3 + + R1 ) − (R1 + L1 s)2 . C1 s C2 s T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ∆ = (R2 + Consider the differentiating op-amp circuit in Figure P2.7. For an ideal op-amp, the voltage gain (as a function of frequency) is V2 (s) = − Z2 (s) V1 (s), Z1 (s) where Z1 = R1 1 + R1 Cs and Z2 = R2 are the respective circuit impedances. Therefore, we obtain V2 (s) = − R2 (1 + R1 Cs) V1 (s). R1 Z 1 Z C R2 + R1 2 + + V1(s) V2(s) – – FIGURE P2.7 Differentiating op-amp circuit. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 40 CHAPTER 2 P2.8 Mathematical Models of Systems Let ∆= G1 + Cs −Cs −G1 −Cs G2 + 2Cs −Cs −G1 −Cs Cs + G1 . Then, Vj = ∆ij I1 ∆ V3 ∆13 I1 /∆ = . V1 ∆11 I1 /∆ or Therefore, the transfer function is T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) −Cs 2Cs + G2 T (s) = ∆13 V3 = = V1 ∆11 −G1 −Cs 2Cs + G2 −Cs −Cs Cs + G1 Pole-zero map (x:poles and o:zeros) 3 2 o Imag Axis 1 0 x x -1 -2 -3 -8 o -7 -6 -5 -4 -3 -2 -1 0 Real Axis FIGURE P2.8 Pole-zero map. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 41 Problems = C 2 R1 R2 s2 + 2CR2 s + 1 . C 2 R1 R2 s2 + (2R2 + R1 )Cs + 1 Using R1 = 1.0, R2 = 0.5, and C = 0.5, we have T (s) = (s + 2 + 2j)(s + 2 − 2j) s2 + 4s + 8 √ √ . = 2 s + 8s + 8 (s + 4 + 8)(s + 4 − 8) The pole-zero map is shown in Figure P2.8. P2.9 From P2.3 we have T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) M ẍ1 + kx1 + k(x1 − x2 ) = F (t) M ẍ2 + k(x2 − x1 ) + bẋ2 = 0 . Taking the Laplace transform of both equations and writing the result in matrix form, it follows that  M s2 + 2k  −k M s2 + bs + k −k   X1 (s) X2 (s)   = F (s) 0   , Pole zero map 0.4 0.3 0.2 Imag Axis 0.1 0 – 0.1 -0.2 -0.3 -0.4 -0.03 -0.025 -0.02 -0.015 Real Axis -0.01 -0.005 0 FIGURE P2.9 Pole-zero map. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 42 CHAPTER 2 Mathematical Models of Systems or      k F (s) 1  M s2 + bs + k  =   2 ∆ k X2 (s) M s + 2k 0 X1 (s) where ∆ = (M s2 + bs + k)(M s2 + 2k) − k 2 . So, G(s) = M s2 + bs + k X1 (s) = . F (s) ∆ When b/k = 1, M = 1 , b2 /M k = 0.04, we have G(s) = s2 + 0.04s + 0.04 . s4 + 0.04s3 + 0.12s2 + 0.0032s + 0.0016 P2.10 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) The pole-zero map is shown in Figure P2.9. From P2.2 we have M1 ÿ1 + k12 (y1 − y2 ) + bẏ1 + k1 y1 = F (t) M2 ÿ2 + k12 (y2 − y1 ) = 0 . Taking the Laplace transform of both equations and writing the result in matrix form, it follows that  or   M1 s2 + bs + k1 + k12 M2 s2 + k12 −k12  −k12   Y1 (s) Y2 (s)   =  F (s)  0    k12 F (s) 1  M2 s2 + k12  =   ∆ Y2 (s) k12 M1 s2 + bs + k1 + k12 0 Y1 (s) where 2 ∆ = (M2 s2 + k12 )(M1 s2 + bs + k1 + k12 ) − k12 . So, when f (t) = a sin ωo t, we have that Y1 (s) is given by Y1 (s) = aM2 ωo (s2 + k12 /M2 ) . (s2 + ωo2 )∆(s) For motionless response (in the steady-state), set the zero of the transfer function so that (s2 + k12 ) = s2 + ωo2 M2 or ωo2 = k12 . M2 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 43 Problems P2.11 The transfer functions from Vc (s) to Vd (s) and from Vd (s) to θ(s) are: K1 K2 , and (Lq s + Rq )(Lc s + Rc ) Km . θ(s)/Vd (s) = 2 (Js + f s)((Ld + La )s + Rd + Ra ) + K3 Km s Vd (s)/Vc (s) = The block diagram for θ(s)/Vc (s) is shown in Figure P2.11, where θ(s)/Vc (s) = K1 K2 Km θ(s) Vd (s) = , Vd (s) Vc (s) ∆(s) where Vc 1 L cs+R c Ic K1 Vq T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ∆(s) = s(Lc s + Rc )(Lq s + Rq )((Js + b)((Ld + La )s + Rd + Ra ) + Km K3 ) . 1 L qs+R q Iq K2 Vd + 1 (L d+L a)s+R d+R a Id Tm Km – 1 Js+f w 1 s q Vb FIGURE P2.11 Block diagram. P2.12 K3 The open-loop transfer function is Y (s) K = . R(s) s + 50 With R(s) = 1/s, we have Y (s) = K . s(s + 50) The partial fraction expansion is K Y (s) = 50 1 1 − , s s + 50 and the inverse Laplace transform is y(t) = K 1 − e−50t , 50 As t → ∞, it follows that y(t) → K/50. So we choose K = 50 so that y(t) © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 44 CHAPTER 2 Mathematical Models of Systems approaches 1. Alternatively we can use the final value theorem to obtain y(t)t→∞ = lim sY (s) = s→0 K . 50 It follows that choosing K = 50 leads to y(t) → 1 as t → ∞. P2.13 The motor torque is given by Tm (s) = (Jm s2 + bm s)θm (s) + (JL s2 + bL s)nθL (s) = n((Jm s2 + bm s)/n2 + JL s2 + bL s)θL (s) where But T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) n = θL (s)/θm (s) = gear ratio . Tm (s) = Km Ig (s) and Ig (s) = and 1 Vg (s) , (Lg + Lf )s + Rg + Rf Vg (s) = Kg If (s) = Kg Vf (s) . Rf + L f s Combining the above expressions yields θL (s) Kg Km = . Vf (s) n∆1 (s)∆2 (s) where ∆1 (s) = JL s2 + bL s + Jm s2 + bm s n2 and ∆2 (s) = (Lg s + Lf s + Rg + Rf )(Rf + Lf s) . P2.14 For a field-controlled dc electric motor we have ω(s)/Vf (s) = Km /Rf . Js + b © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 45 Problems With a step input of Vf (s) = 80/s, the final value of ω(t) is ω(t)t→∞ = lim sω(s) = s→0 80Km = 2.4 Rf b or Km = 0.03 . Rf b Solving for ω(t) yields ω(t) = 80Km −1 1 L Rf J s(s + b/J) = 80Km (1−e−(b/J)t ) = 2.4(1−e−(b/J)t ) . Rf b At t = 1/2, ω(t) = 1, so ω(1/2) = 2.4(1 − e−(b/J)t ) = 1 implies b/J = 1.08 sec . Therefore, P2.15 0.0324 . s + 1.08 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ω(s)/Vf (s) = Summing the forces in the vertical direction and using Newton’s Second Law we obtain ẍ + k x=0. m The system has no damping and no external inputs. Taking the Laplace transform yields X(s) = x0 s , 2 s + k/m where we used the fact that x(0) = x0 and ẋ(0) = 0. Then taking the inverse Laplace transform yields x(t) = x0 cos P2.16 s k t. m (a) For mass 1 and 2, we have M1 ẍ1 + K1 (x1 − x2 ) + b1 (ẋ3 − ẋ1 ) = 0 M2 ẍ2 + K2 (x2 − x3 ) + b2 (ẋ3 − ẋ2 ) + K1 (x2 − x1 ) = 0 . (b) Taking the Laplace transform yields (M1 s2 + b1 s + K1 )X1 (s) − K1 X2 (s) = b1 sX3 (s) −K1 X1 (s) + (M2 s2 + b2 s + K1 + K2 )X2 (s) = (b2 s + K2 )X3 (s) . (c) Let G1 (s) = K2 + b2 s © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 46 CHAPTER 2 Mathematical Models of Systems G2 (s) = 1/p(s) G3 (s) = 1/q(s) G4 (s) = sb1 , where p(s) = s2 M2 + sf2 + K1 + K2 and q(s) = s2 M1 + sf1 + K1 . The signal flow graph is shown in Figure P2.16. X3 G1 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) G4 G2 G3 X1 K1 K1 FIGURE P2.16 Signal flow graph. (d) The transfer function from X3 (s) to X1 (s) is X1 (s) K1 G1 (s)G2 (s)G3 (s) + G4 (s)G3 (s) = . X3 (s) 1 − K12 G2 (s)G3 (s) P2.17 Using Cramer’s rule, we have  1 1.5 x1   or  2 4   x1 x2   =  6 11     1  4 −1.5   6   = ∆ −2 x2 1 11 where ∆ = 4(1) − 2(1.5) = 1 . Therefore, x1 = 4(6) − 1.5(11) = 7.5 1 and x2 = −2(6) + 1(11) = −1 . 1 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 47 Problems The signal flow graph is shown in Figure P2.17. 11 1/4 1 6 -1/2 X2 X1 -1.5 FIGURE P2.17 Signal flow graph. So, P2.18 6(1) − 1.5( 11 4 ) = 7.5 3 1− 4 Z2 Va Y3 Ia Z4 V2 Y1 -Y 3 -Z 2 -Y 1 FIGURE P2.18 Signal flow graph. 11( 14 ) + −1 2 (6) = −1 . 3 1− 4 The signal flow graph is shown in Figure P2.18. I1 V1 and x2 = T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) x1 = The transfer function is V2 (s) Y 1 Z2 Y 3 Z4 = . V1 (s) 1 + Y 1 Z2 + Y 3 Z2 + Y 3 Z4 + Y 1 Z2 Z4 Y 3 P2.19 (a) Assume Rg ≫ Rs and Rs ≫ R1 . Then Rs = R1 + R2 ≈ R2 , and vgs = vin − vo , where we neglect iin , since Rg ≫ Rs . At node S, we have vo = gm vgs = gm (vin − vo ) or Rs vo gm Rs = . vin 1 + gm Rs (b) With gm Rs = 20, we have vo 20 = = 0.95 . vin 21 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 48 CHAPTER 2 Mathematical Models of Systems (c) The block diagram is shown in Figure P2.19. vin(s) vo(s) gmRs – FIGURE P2.19 Block diagram model. P2.20 From the geometry we find that l2 l1 − l2 (x − y) − y . l1 l1 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ∆z = k The flow rate balance yields A dy = p∆z dt which implies Y (s) = p∆Z(s) . As By combining the above results it follows that Y (s) = p l1 − l2 l2 k (X(s) − Y (s)) − Y (s) . As l1 l1 Therefore, the signal flow graph is shown in Figure P2.20. Using Mason’s -1 (l 1 – l 2)/l 1 k X DZ p/As Y 1 -l 2 / l 1 FIGURE P2.20 Signal flow graph. gain formula we find that the transfer function is given by k(l1 −l2 )p Y (s) K1 l1 As = = , k(l −l )p l p X(s) s + K2 + K1 1+ 2 + 1 2 l1 As l1 As where K1 = k(l1 − l2 )p p l1 A and K2 = l2 p . l1 A © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 49 Problems P2.21 (a) The equations of motion for the two masses are 2 L L (θ1 − θ2 ) = f (t) 2 2 2 L M L2 θ¨2 + M gLθ2 + k (θ2 − θ1 ) = 0 . 2 M L2 θ¨1 + M gLθ1 + k a – F (t) w1 1/s 1/s T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) 1/2ML (a) q1 b w2 1/s 1/s q2 a Imag(s) + j (b) g k L + 4M + j g k L + 2M + j g L X O X Re(s) FIGURE P2.21 (a) Block diagram. (b) Pole-zero map. With θ˙1 = ω1 and θ˙2 = ω2 , we have g k ω˙1 = − + L 4M θ1 + k f (t) θ2 + 4M 2M L © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 50 CHAPTER 2 Mathematical Models of Systems ω˙2 = k θ1 − 4M k g + L 4M θ2 . (b) Define a = g/L + k/4M and b = k/4M . Then θ1 (s) 1 s2 + a . = F (s) 2M L (s2 + a)2 − b2 (c) The block diagram and pole-zero map are shown in Figure P2.21. For a noninverting op-amp circuit, depicted in Figure P2.22a, the voltage gain (as a function of frequency) is P2.22 Z1 (s) + Z2 (s) Vin (s), Z1 (s) T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Vo (s) = where Z1 (s) and Z2 (s) are the impedances of the respective circuits. In Z2 Z1 + vin (a) v0 + vin v0 (b) FIGURE P2.22 (a) Noninverting op-amp circuit. (b) Voltage follower circuit. the case of the voltage follower circuit, shown in Figure P2.22b, we have Z1 = ∞ (open circuit) and Z2 = 0. Therefore, the transfer function is Vo (s) Z1 = = 1. Vin (s) Z1 P2.23 The input-output ratio, Vce /Vin , is found to be β(R − 1) + hie Rf Vce = . Vin −βhre + hie (−hoe + Rf ) P2.24 (a) The voltage gain is given by vo RL β1 β2 (R1 + R2 ) = . vin (R1 + R2 )(Rg + hie1 ) + R1 (R1 + R2 )(1 + β1 ) + R1 RL β1 β2 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 51 Problems (b) The current gain is found to be ic2 = β1 β2 . ib1 (c) The input impedance is vin (R1 + R2 )(Rg + hie1 ) + R1 (R1 + R2 )(1 + β1 ) + R1 RL β1 β2 = , ib1 R1 + R2 and when β1 β2 is very large, we have the approximation vin RL R1 β1 β2 ≈ . ib1 R1 + R2 The transfer function from R(s) and Td (s) to Y (s) is given by 1 (G(s)R(s) + Td (s)) + Td (s) + G(s)R(s) G(s) T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) P2.25 Y (s) = G(s) R(s) − = G(s)R(s) . Thus, Y (s)/R(s) = G(s) . Also, we have that Y (s) = 0 . when R(s) = 0. Therefore, the effect of the disturbance, Td (s), is eliminated. P2.26 The equations of motion for the two mass model of the robot are M ẍ + b(ẋ − ẏ) + k(x − y) = F (t) mÿ + b(ẏ − ẋ) + k(y − x) = 0 . Taking the Laplace transform and writing the result in matrix form yields   M s2 + bs + k −(bs + k) −(bs + k) ms2 + bs + k   X(s) Y (s)   = F (s) 0   . Solving for Y (s) we find that 1 Y (s) mM (bs +k) . = b F (s) s2 [s2 + 1 + m s+ k ] M m m © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 52 CHAPTER 2 P2.27 Mathematical Models of Systems The describing equation of motion is i2 mz̈ = mg − k 2 . z Defining f (z, i) = g − ki2 mz 2 leads to z̈ = f (z, i) . The equilibrium condition for io and zo , found by solving the equation of motion when is T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ż = z̈ = 0 , ki2o = zo2 . mg We linearize the equation of motion using a Taylor series approximation. With the definitions ∆z = z − zo and ∆i = i − io , ˙ = ż and ∆z ¨ = z̈. Therefore, we have ∆z ¨ = f (z, i) = f (zo , io ) + ∂f z=z ∆z + ∂f z=z ∆i + · · · ∆z ∂z i=ioo ∂i i=ioo But f (zo , io ) = 0, and neglecting higher-order terms in the expansion yields 2 ¨ = 2kio ∆z − 2kio ∆i . ∆z mzo3 mzo2 Using the equilibrium condition which relates zo to io , we determine that ¨ = 2g ∆z − g ∆i . ∆z zo io Taking the Laplace transform yields the transfer function (valid around the equilibrium point) ∆Z(s) −g/io = 2 . ∆I(s) s − 2g/zo © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 53 Problems P2.28 The signal flow graph is shown in Figure P2.28. -d G B +c +b P D +a -m +e -k M +g +f +h S C T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) FIGURE P2.28 Signal flow graph. (a) The PGBDP loop gain is equal to -abcd. This is a negative transmission since the population produces garbage which increases bacteria and leads to diseases, thus reducing the population. (b) The PMCP loop gain is equal to +efg. This is a positive transmission since the population leads to modernization which encourages immigration, thus increasing the population. (c) The PMSDP loop gain is equal to +ehkd. This is a positive transmission since the population leads to modernization and an increase in sanitation facilities which reduces diseases, thus reducing the rate of decreasing population. (d) The PMSBDP loop gain is equal to +ehmcd. This is a positive transmission by similar argument as in (3). P2.29 Assume the motor torque is proportional to the input current Tm = ki . Then, the equation of motion of the beam is J φ̈ = ki , where J is the moment of inertia of the beam and shaft (neglecting the inertia of the ball). We assume that forces acting on the ball are due to gravity and friction. Hence, the motion of the ball is described by mẍ = mgφ − bẋ © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 54 CHAPTER 2 Mathematical Models of Systems where m is the mass of the ball, b is the coefficient of friction, and we have assumed small angles, so that sin φ ≈ φ. Taking the Laplace transfor of both equations of motion and solving for X(s) yields X(s)/I(s) = P2.30 gk/J s2 (s2 + b/m) . Given H(s) = k τs + 1 where τ = 5µs = 4 × 10−6 seconds and 0.999 ≤ k < 1.001. The step response is k 1 k k · = − . τs + 1 s s s + 1/τ T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Y (s) = Taking the inverse Laplace transform yields y(t) = k − ke−t/τ = k(1 − e−t/τ ) . The final value is k. The time it takes to reach 98% of the final value is t = 19.57µs independent of k. P2.31 From the block diagram we have Y1 (s) = G2 (s)[G1 (s)E1 (s) + G3 (s)E2 (s)] = G2 (s)G1 (s)[R1 (s) − H1 (s)Y1 (s)] + G2 (s)G3 (s)E2 (s) . Therefore, Y1 (s) = G1 (s)G2 (s) G2 (s)G3 (s) R1 (s) + E2 (s) . 1 + G1 (s)G2 (s)H1 (s) 1 + G1 (s)G2 (s)H1 (s) And, computing E2 (s) (with R2 (s) = 0) we find G4 (s) E2 (s) = H2 (s)Y2 (s) = H2 (s)G6 (s) Y1 (s) + G5 (s)E2 (s) G2 (s) or E2 (s) = G4 (s)G6 (s)H2 (s) Y1 (s) . G2 (s)(1 − G5 (s)G6 (s)H2 (s)) Substituting E2 (s) into equation for Y1 (s) yields Y1 (s) = G1 (s)G2 (s) R1 (s) 1 + G1 (s)G2 (s)H1 (s) © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 55 Problems + G3 (s)G4 (s)G6 (s)H2 (s) Y1 (s) . (1 + G1 (s)G2 (s)H1 (s))(1 − G5 (s)G6 (s)H2 (s)) Finally, solving for Y1 (s) yields Y1 (s) = T1 (s)R1 (s) where T1 (s) = G1 (s)G2 (s)(1 − G5 (s)G6 (s)H2 (s)) (1 + G1 (s)G2 (s)H1 (s))(1 − G5 (s)G6 (s)H2 (s)) − G3 (s)G4 (s)G6 (s)H2 (s) . . Similarly, for Y2 (s) we obtain T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Y2 (s) = T2 (s)R1 (s) . where T2 (s) = P2.32 G1 (s)G4 (s)G6 (s) (1 + G1 (s)G2 (s)H1 (s))(1 − G5 (s)G6 (s)H2 (s)) − G3 (s)G4 (s)G6 (s)H2 (s) The signal flow graph shows three loops: L1 = −G1 G3 G4 H2 L2 = −G2 G5 G6 H1 L3 = −H1 G8 G6 G2 G7 G4 H2 G1 . The transfer function Y2 /R1 is found to be Y2 (s) G1 G8 G6 ∆1 − G2 G5 G6 ∆2 = , R1 (s) 1 − (L1 + L2 + L3 ) + (L1 L2 ) where for path 1 ∆1 = 1 and for path 2 ∆ 2 = 1 − L1 . Since we want Y2 to be independent of R1 , we need Y2 /R1 = 0. Therefore, we require G1 G8 G6 − G2 G5 G6 (1 + G1 G3 G4 H2 ) = 0 . © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 56 CHAPTER 2 P2.33 Mathematical Models of Systems The closed-loop transfer function is G3 (s)G1 (s)(G2 (s) + K5 K6 ) Y (s) = . R(s) 1 − G3 (s)(H1 (s) + K6 ) + G3 (s)G1 (s)(G2 (s) + K5 K6 )(H2 (s) + K4 ) P2.34 The equations of motion are m1 ÿ1 + b(ẏ1 − ẏ2 ) + k1 (y1 − y2 ) = 0 m2 ÿ2 + b(ẏ2 − ẏ1 ) + k1 (y2 − y1 ) + k2 y2 = k2 x Taking the Laplace transform yields (m1 s2 + bs + k1 )Y1 (s) − (bs + k1 )Y2 (s) = 0 (m2 s2 + bs + k1 + k2 )Y2 (s) − (bs + k1 )Y1 (s) = k2 X(s) T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Therefore, after solving for Y1 (s)/X(s), we have Y2 (s) k2 (bs + k1 ) = . 2 X(s) (m1 s + bs + k1 )(m2 s2 + bs + k1 + k2 ) − (bs + k1 )2 P2.35 (a) We can redraw the block diagram as shown in Figure P2.35. Then, T (s) = K1 /s(s + 1) K1 = 2 . 1 + K1 (1 + K2 s)/s(s + 1) s + (1 + K2 K1 )s + K2 (b) The signal flow graph reveals two loops (both touching): L1 = −K1 s(s + 1) and L2 = −K1 K2 . s+1 Therefore, T (s) = K1 /s(s + 1) K1 = 2 . 1 + K1 /s(s + 1) + K1 K2 /(s + 1) s + (1 + K2 K1 )s + K1 (c) We want to choose K1 and K2 such that s2 + (1 + K2 K1 )s + K1 = s2 + 20s + 100 = (s + 10)2 . Therefore, K1 = 100 and 1 + K2 K1 = 20 or K2 = 0.19. (d) The step response is shown in Figure P2.35. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 57 Problems R(s ) K1 s (s+1) + Y (s) 1 +K 2s 1 0.9 0.8 0.7 <—- time to 90% = 0.39 sec y(t) 0.6 0.5 0.4 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 time(sec) FIGURE P2.35 The equivalent block diagram and the system step response. P2.36 (a) Given R(s) = 1/s2 , the partial fraction expansion is Y (s) = 30 s2 (s + 5)(s2 + 4s + 6) = 0.1091 0.7576(s + 3.4) 1 0.8667 − 2 + 2− . s+5 s + 4s + 6 s s Therefore, using the Laplace transform table, we determine that the ramp response for t ≥ 0 is ! √ √ √ 6 −5t 25 −2t 7 2 13 y(t) = e + e cos 2t + sin 2t + t − . 55 33 10 15 (b) For the ramp input, y(t) ≈ 0.25 at t = 1 second (see Figure P2.36a). (c) Given R(s) = 1, the partial fraction expansion is Y (s) = 30 30 1 30 s−1 = − . 2 2 (s + 5)(s + 4s + 6) 11 s + 5 11 s + 4s + 6 Therefore, using the Laplace transform table, we determine that the © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 58 CHAPTER 2 Mathematical Models of Systems impulse response for t ≥ 0 is ! √ √ √ 30 −5t 30 −2t 3 2 e − e cos 2t − sin 2t) . y(t) = 11 11 2 (d) For the impulse input, y(t) ≈ 0.73 at t = 1 seconds (see Figure P2.36b). (a) Ramp input (b) Impulse input 1.2 3 1 2.5 0.8 2 0.6 1.5 0.4 1 0.2 0.5 0 y(t) T an his th d wo or eir is p rk w sa co ro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n y(t) . e W eb ) 3.5 0 0 1 2 Time (s) 3 −0.2 4 0 1 2 Time (s) 3 4 FIGURE P2.36 (a) Ramp input response. (b) Impulse input response. P2.37 The equations of motion are m1 d2 x = −(k1 + k2 )x + k2 y dt2 and m2 d2 y = k2 (x − y) + u . dt2 When m1 = m2 = 1 and k1 = k2 = 1, we have d2 x = −2x + y dt2 and d2 y =x−y+u . dt2 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 59 Problems P2.38 The equation of motion for the system is J dθ d2 θ + b + kθ = 0 , 2 dt dt where k is the rotational spring constant and b is the viscous friction coefficient. The initial conditions are θ(0) = θo and θ̇(0) = 0. Taking the Laplace transform yields J(s2 θ(s) − sθo ) + b(sθ(s) − θo ) + kθ(s) = 0 . Therefore, θ(s) = (s + Jb θo ) (s2 + Jb s + K J) = (s + 2ζωn )θo . s2 + 2ζωn s + ωn2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Neglecting the mass of the rod, the moment of inertia is detemined to be J = 2M r 2 = 0.5 kg · m2 . Also, s ωn = k = 0.02 rad/s J and ζ = b = 0.01 . 2Jωn Solving for θ(t), we find that q θo −ζωn t θ(t) = p e sin(ω 1 − ζ 2 t + φ) , n 1 − ζ2 where tan φ = p 1 − ζ 2 /ζ). Therefore, the envelope decay is θo θe = p e−ζωn t . 2 1−ζ So, with ζωn = 2 × 10−4 , θo = 4000o and θf = 10o , the elapsed time is computed as t= P2.39 1 θo ln p = 8.32 hours . ζωn 1 − ζ 2 θf When t < 0, we have the steady-state conditions 6 i1 (0) = A , 7 va (0) = 12 V 7 and vc (0) = 36 V , 7 where vc (0) is associated with the 0.75F capacitor. After t ≥ 0, we have 1.5 di1 + 2i1 + 5(i1 − i2 ) = 10e−2t dt © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 60 CHAPTER 2 Mathematical Models of Systems and 0.75 Z i2 dt + 10i2 + 5(i2 − i1 ) − i1 = 0 . Taking the Laplace transform (using the initial conditions) yields 1.5(sI1 (s) − i1 (0)) + 2I1 (s) + 5I1 (s) − 5I2 (s) = 10 s+2 or s+ 10 18s + 176 14 I1 (s) − I2 (s) = 3 3 21(s + 2) and or 1 I2 (s) − vc (0) + 10I2 (s) + 5(I2 (s) − I1 (s)) = I1 (s) s T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) 3 4 −24sI1 (s) + (60s + 3)I2 (s) = 108 s. 7 Solving for I2 (s) yields I2 (s) = 4s(27s2 + 216s + 604) . 7(s + 2)(60s2 + 203s + 14) Then, Vo (s) = 10I2 (s). P2.40 The equations of motion are J1 θ̈1 = K(θ2 − θ1 ) − b(θ̇1 − θ̇2 ) + T and J2 θ̈2 = b(θ̇1 − θ̇2 ) . Taking the Laplace transform yields (J1 s2 + bs + K)θ1 (s) − bsθ2 (s) = Kθ2 (s) + T (s) and (J2 s2 + bs)θ2 (s) − bsθ1 (s) = 0. Solving for θ1 (s) and θ2 (s), we find that θ1 (s) = (Kθ2 (s) + T (s))(J2 s + b) ∆(s) and θ2 (s) = b(Kθ2 (s) + T (s)) , ∆(s) where ∆(s) = J1 J2 s3 + b(J1 + J2 )s2 + J2 Ks + bK. P2.41 Assume that the only external torques acting on the rocket are control torques, Tc and disturbance torques, Td , and assume small angles, θ(t). Using the small angle approximation, we have ḣ = V θ © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 61 Problems J θ̈ = Tc + Td , where J is the moment of inertia of the rocket and V is the rocket velocity (assumed constant). Now, suppose that the control torque is proportional to the lateral displacement, as Tc (s) = −KH(s) , where the negative sign denotes a negative feedback system. The corresponding block diagram is shown in Figure P2.41. Td K + Tc + + – 1 Js 2 V s H( s) T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) H desired=0 FIGURE P2.41 Block diagram. P2.42 (a) The equation of motion of the motor is J dω = Tm − bω , dt where J = 0.1, b = 0.06, and Tm is the motor input torque. (b) Given Tm (s) = 1/s, and ω(0) = 0.7, we take the Laplace transform of the equation of motion yielding sω(s) − ω(0) + 0.6ω(s) = 10Tm or ω(s) = 0.7s + 10 . s(s + 0.6) Then, computing the partial fraction expansion, we find that ω(s) = A B 16.67 15.97 + = − . s s + 0.6 s s + 0.6 The step response, determined by taking the inverse Laplace transform, is ω(t) = 16.67 − 15.97e−0.6t , t≥0. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 62 CHAPTER 2 P2.43 Mathematical Models of Systems The work done by each gear is equal to that of the other, therefore Tm θm = TL θL . Also, the travel distance is the same for each gear, so r1 θ m = r2 θ L . The number of teeth on each gear is proportional to the radius, or r1 N 2 = r2 N 1 . So, and T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) θm r2 N2 = = , θL r1 N1 N1 θ m = N2 θ L N1 θL = θm = nθm , N2 where n = N1 /N2 . Finally, Tm θL N1 = = =n. TL θm N2 P2.44 The inertia of the load is JL = πρLr 4 . 2 Also, from the dynamics we have T2 = JL ω̇2 + bL ω2 and T1 = nT2 = n(JL ω̇2 + bL ω2 ) . So, T1 = n2 (JL ω̇1 + bL ω1 ) , © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 63 Problems since ω2 = nω1 . Therefore, the torque at the motor shaft is T = T1 + Tm = n2 (JL ω̇1 + bL ω1 ) + Jm ω̇1 + bm ω1 . P2.45 Let U (s) denote the human input and F (s) the load input. The transfer function is P (s) = G(s) + KG1 (s) Gc (s) + KG1 (s) U (s) + F (s) , ∆(s) ∆(s) where P2.46 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ∆ = 1 + GH(s) + G1 KBH(s) + Gc E(s) + G1 KE(s) . Consider the application of Newton’s law ( mv we obtain P F = mẍ). From the mass mv ẍ1 = F − k1 (x1 − x2 ) − b1 (ẋ1 − ẋ2 ). Taking the Laplace transform, and solving for X1 (s) yields X1 (s) = b1 s + k1 1 F (s) + X2 (s), ∆1 (s) ∆1 (s) where ∆1 := mv s2 + b1 s + k1 . From the mass mt we obtain mt ẍ2 = −k2 x2 − b2 ẋ2 + k1 (x1 − x2 ) + b1 (ẋ1 − ẋ2 ). Taking the Laplace transform, and solving for X2 (s) yields X2 (s) = b1 s + k1 X1 (s), ∆2 (s) where ∆2 := mt s2 + (b1 + b2 )s + k1 + k2 . Substituting X2 (s) above into the relationship fpr X1 (s) yields the transfer function ∆2 (s) X1 (s) = . F (s) ∆1 (s)∆2 (s) − (b1 s + k1 )2 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 64 CHAPTER 2 P2.47 Mathematical Models of Systems Using the following relationships h(t) = Z (1.6θ(t) − h(t))dt ω(t) = θ̇(t) J ω̇(t) = Km ia (t) va (t) = 50vi (t) = 10ia (t) + vb (t) θ̇ = Kvb we find the differential equation is Km d3 h + 1+ 3 dt 10JK d2 h Km dh 8Km + = vi . 2 dt 10JK dt J (a) The transfer function is T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) P2.48 V2 (s) (1 + sR1 C1 )(1 + sR2 C2 ) = . V1 (s) R1 C 2 s (b) When R1 = 250 kΩ, R2 = 200 kΩ, C1 = 2 µF and C2 = 0.1 µF , we have V2 (s) 0.4s2 + 20.8s + 40 = . V1 (s) s (c) The partial fraction expansion is 40 V2 (s) = 20.8 + + 0.4s . V1 (s) s P2.49 (a) The closed-loop transfer function is T (s) = G(s) 5000 = 3 . 2 1 + G(s) s + 20s + 1000s + 5000 (b) The poles of T (s) are s1 = −5.43 and s2,3 = −7.28 ± j29.46. (c) The partial fraction expansion (with a step input) is Y (s) = 1 1.06 0.0285 + 0.0904j 0.0285 − 0.0904j − + + , s s + 5.43 s + 7.28 − j29.46 s + 7.28 + j29.46 and y(t) = 1 − 1.06e−5.43t + 0.06e−7.28t (cos 29.46t − 3.17 sin 29.46t) ; (d) The step response is shown in Figure P2.49. The real and complex roots are close together and by looking at the poles in the s-plane we © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 65 Problems 1 0.9 0.8 Amplitude 0.7 0.6 0.5 0.4 0.3 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 1.2 Time (seconds) 1.4 1.6 1.8 2 FIGURE P2.49 Step response. have difficulty deciding which is dominant. However, the residue at the real pole is much larger and thus dominates the response. P2.50 (a) The closed-loop transfer function is T (s) = 14000 s3 + 45s2 + 3100s + 14500 . (b) The poles of T (s) are s1 = −5 and s2,3 = −20 ± j50. (c) The partial fraction expansion (with a step input) is Y (s) = 0.9655 1.0275 0.0310 − 0.0390j 0.0310 + 0.0390j − + + . s s+5 s + 20 + j50 s + 20 − j50 (d) The step response is shown in Figure P2.50. The real root dominates the response. (e) The final value of y(t) is yss = lim sY (s) = 0.9655 . s→0 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 66 CHAPTER 2 Mathematical Models of Systems 1 0.9 0.8 Amplitude 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) 0.2 Time (secs) FIGURE P2.50 Step response. P2.51 Consider the free body diagram in Figure P2.51. Using Newton’s Law and summing the forces on the two masses yields M1 ẍ(t) + b1 ẋ(t) + k1 x(t) = b1 ẏ(t) M2 ÿ(t) + b1 ẏ(t) + k2 y(t) = b1 ẋ(t) + u(t) k1x M1 k1 x . . b1(x – y) k2 M1 x . . b1(y – x) k2 y b1 M2 M2 y y u(t) u(t) FIGURE P2.51 Free body diagram. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 67 Advanced Problems Advanced Problems AP2.1 The transfer function from V (s) to ω(s) has the form ω(s) Km = . V (s) τm s + 1 In the steady-state, ωss = lim s s→0 Km 5 = 5Km . τm s + 1 s So, Km = 70/5 = 14 . T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Also, ω(t) = Vm Km (1 − e−t/τm ) where V (s) = Vm /s. Solving for τm yields τm = −t . ln(1 − ω(t)/ωss ) When t = 2, we have τm = −2 = 3.57 . ln(1 − 30/70) Therefore, the transfer function is ω(s) 14 = . V (s) 3.57s + 1 AP2.2 The closed-loop transfer function form R1 (s) to Y2 (s) is Y2 (s) G1 G4 G5 (s) + G1 G2 G3 G4 G6 (s) = R1 (s) ∆ where ∆ = [1 + G3 G4 H2 (s)][1 + G1 G2 H3 (s)] . If we select G5 (s) = −G2 G3 G6 (s) then the numerator is zero, and Y2 (s)/R1 (s) = 0. The system is now decoupled. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 68 CHAPTER 2 AP2.3 Mathematical Models of Systems (a) Computing the closed-loop transfer function: G(s)Gc (s) R(s) . Y (s) = 1 + Gc (s)G(s)H(s) Then, with E(s) = R(s) − Y (s) we obtain E(s) = 1 + Gc (s)G(s)(H(s) − 1) R(s) . 1 + Gc (s)G(s)H(s) If we require that E(s) ≡ 0 for any input, we need 1 + Gc (s)G(s)(H(s) − 1) = 0 or H(s) = Gc (s)G(s) − 1 n(s) = . Gc (s)G(s) d(s) T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Since we require H(s) to be a causal system, the order of the numerator polynomial, n(s), must be less than or equal to the order of the denominator polynomial, d(s). This will be true, in general, only if both Gc (s) and G(s) are proper rational functions (that is, the numerator and denominator polynomials have the same order). Therefore, making E ≡ 0 for any input R(s) is possible only in certain circumstances. (b) The transfer function from Td (s) to Y (s) is Gd (s)G(s) Y (s) = Td (s) . 1 + Gc (s)G(s)H(s) With H(s) as in part (a) we have Gd (s) Y (s) = Td (s) . Gc (s) (c) No. Since Y (s) = Gd (s)G(s) Td (s) = T (s)Td (s) , 1 + Gc (s)G(s)H(s) the only way to have Y (s) ≡ 0 for any Td (s) is for the transfer function T (s) ≡ 0 which is not possible in general (since G(s) 6= 0). AP2.4 (a) With q(s) = 1/s we obtain τ (s) = 1/Ct s + QS+1/R Ct · 1 . s Define α := QS + 1/R Ct and β := 1/Ct . © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 69 Advanced Problems Then, it follows that τ (s) = β 1 −β/α β/α · = + . s+α s s+α s Taking the inverse Laplace transform yields τ (t) = β −β −αt β e + = [1 − e−αt ] . α α α 1 (b) As t → ∞, τ (t) → αβ = Qs+1/R . (c) To increase the speed of response, you want to choose Ct , Q, S and R such that α := Qs + 1/R Ct AP2.5 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) is ”large.” Considering the motion of each mass, we have M3 ẍ3 + b3 ẋ3 + k3 x3 = u3 + b3 ẋ2 + k3 x2 M2 ẍ2 + (b2 + b3 )ẋ2 + (k2 + k3 )x2 = u2 + b3 ẋ3 + k3 x3 + b2 ẋ1 + k2 x1 M1 ẍ1 + (b1 + b2 )ẋ1 + (k1 + k2 )x1 = u1 + b2 ẋ2 + k2 x2 In matrix form the three equations can be written as   M1   0   0 0 M2 0    M3  ẍ3   0 −b3 b3  k1 + k2 0  −k2 −k2 −k3 k3  +   AP2.6   0   ẍ1   b1 + b2 −b2 0   ẋ1            0  b2 + b3 −b3    ẍ2  +  −b2   ẋ2  0  ẋ3         x1      u1    x3 u3 Considering the cart mass and using Newton’s Law we obtain M ẍ = u − bẋ − F sin ϕ where F is the reaction force between the cart and the pendulum. Considering the pendulum we obtain m      k2 + k3 −k3    x 2  =  u2  . d2 (x + L sin ϕ) = F sin ϕ dt2 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.  70 CHAPTER 2 Mathematical Models of Systems m d2 (L cos ϕ) = F cos ϕ + mg dt2 Eliminating the reaction force F yields the two equations (m + M )ẍ + bẋ + mLϕ̈ cos ϕ − mLϕ̇2 sin ϕ = u mL2 ϕ̈ + mgL sin ϕ + mLẍ cos ϕ = 0 If we assume that the angle ϕ ≈ 0, then we have the linear model (m + M )ẍ + bẋ + mLϕ̈ = u mL2 ϕ̈ + mgLϕ = −mLẍ The transfer function from the disturbance input to the output is T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) AP2.7 Y (s) = 1 Td (s) . s + 40 + K When Td (s) = 1, we obtain y(t) = e−(40+K)t . Solving for t when y(t) 2.3 . 40 + K When t = 0.05 and y(0.05) = 0.1, we find K = 6.05. AP2.8 The closed-loop transfer function is T (s) = 200K(0.25s + 1) (0.25s + 1)(s + 1)(s + 8) + 200K The final value due to a step input of R(s) = A/s is v(t) → A 200K . 200K + 8 We need to select K so that v(t) → 50. However, to keep the percent overshoot to less than 10%, we need to limit the magnitude of K. Figure AP2.8a shows the percent overshoot as a function of K. Let K = 0.06 and select the magnitude of the input to be A = 83.3. The inverse Laplace transform of the closed-loop response with R(s) = 83.3/s is v(t) = 50 + 9.85e−9.15t − e−1.93t (59.85 cos(2.24t) + 11.27 sin(2.24t)) © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 71 Advanced Problems The result is P.O. = 9.74% and the steady-state value of the output is approximately 50 m/s, as shown in Figure AP2.8b. 25 15 10 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr Amplitude de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) Percent Overshoot (%) 20 5 0 0 0.01 0.02 0.03 0.04 0.05 K 0.06 0.07 0.08 0.09 0.1 Step Response 60 System: untitled1 Peak amplitude: 54.9 Overshoot (%): 9.74 At time (sec): 1.15 50 40 30 20 10 0 0 0.5 1 1.5 2 2.5 Time (sec) FIGURE AP2.8 (a) Percent overshoot versus the gain K. (b) Step response. AP2.9 The transfer function is Vo (s) Z2 (s) =− , Vi (s) Z1 (s) © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 72 CHAPTER 2 Mathematical Models of Systems where Z1 (s) = R1 R1 C 1 s + 1 and Z2 (s) = R2 C 2 s + 1 . C2 s Then we can write KI Vo (s) = Kp + + KD s Vi (s) s where R1 C 1 +1 , R2 C 2 KI = − 1 , R1 C 2 KD = −R2 C1 . T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) KP = − © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 73 Design Problems Design Problems CDP2.1 The model of the traction drive, capstan roller, and linear slide follows closely the armature-controlled dc motor model depicted in Figure 2.18 in Dorf and Bishop. The transfer function is T (s) = rKm , s [(Lm s + Rm )(JT s + bm ) + Kb Km ] where T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) JT = Jm + r 2 (Ms + Mb ) . Va(s) – 1 JTs+bm Km Lms+Rm 1 s q r X(s) Kb Back EMF DP2.1 w The closed-loop transfer function is Y (s) G1 (s)G2 (s) = . R(s) 1 + G1 (s)H1 (s) − G2 (s)H2 (s) When G1 H1 = G2 H2 and G1 G2 = 1, then Y (s)/R(s) = 1. Therefore, select G1 (s) = DP2.2 1 G2 (s) and H1 (s) = G2 (s)H2 (s) = G22 (s)H2 (s) . G1 (s) At the lower node we have 1 1 v + + G + 2i2 − 20 = 0 . 4 3 Also, we have v = 24 and i2 = Gv . So 1 1 v + + G + 2Gv − 20 = 0 4 3 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 74 CHAPTER 2 Mathematical Models of Systems and G= DP2.3 20 − v 1 1 4 + 3 3v = 1 S. 12 Taking the Laplace transform of 3 1 1 y(t) = e−t − e−2t − + t 4 4 2 yields Y (s) = 1 1 3 1 − − + 2 . s + 1 4(s + 2) 4s 2s Similarly, taking the Laplace transform of the ramp input yields 1 . s2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) R(s) = Therefore G(s) = DP2.4 Y (s) 1 = . R(s) (s + 1)(s + 2) For an ideal op-amp, at node a we have vin − va vo − va + =0, R1 R1 and at node b vin − vb = C v̇b , R2 from it follows that 1 1 + Cs Vb = Vin . R2 R2 Also, for an ideal op-amp, Vb − Va = 0. Then solving for Vb in the above equation and substituting the result into the node a equation for Va yields 1 + Cs Vo 2 1 = 1 − R2 Vin R2 2 R2 + Cs ” # or Vo (s) R2 Cs − 1 =− . Vin (s) R2 Cs + 1 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 75 Design Problems For vin (t) = At, we have Vin (s) = A/s2 , therefore 2 2 vo (t) = A e−βt + t − β β where β = 1/R2 C. DP2.5 The equation of motion describing the motion of the inverted pendulum (assuming small angles) is ϕ̈ + g ϕ=0. L Assuming a solution of the form ϕ = k cos ϕ, taking the appropriate derivatives and substituting the result into the equation of motion yields the relationship g . L T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ϕ̇ = r If the period is T = 2 seconds, we compute ϕ̇ = 2π/T . Then solving for L yields L = 0.99 meters when g = 9.81 m/s2 . So, to fit the pendulum into the grandfather clock, the dimensions are generally about 1.5 meters or more. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 76 CHAPTER 2 Mathematical Models of Systems Computer Problems CP2.1 The m-file script is shown in Figure CP2.1. pq = 1 10 p=[1 8 12]; q=[1 2]; % Part (a) pq=conv(p,q) % Part (b) P=roots(p), Z=roots(q) % Part (c) value=polyval(p,-1) 28 24 P= -6 -2 Z= -2 value = 5 CP2.2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) FIGURE CP2.1 Script for various polynomial evaluations. The m-file script and step response is shown in Figure CP2.2. numc = [1]; denc = [1 1]; sysc = tf(numc,denc) numg = [1 2]; deng = [1 3]; sysg = tf(numg,deng) % part (a) sys_s = series(sysc,sysg); sys_cl = feedback(sys_s,[1]) % part (b) step(sys_cl); grid on Transfer function: s+2 ————s^2 + 5 s + 5 Step Response From: U(1) 0.4 0.35 0.3 To: Y(1) Amplitude 0.25 0.2 0.15 0.1 0.05 0 0 0.5 1 1.5 2 2.5 3 3.5 4 Time (sec.) FIGURE CP2.2 Step response. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 77 Computer Problems CP2.3 Given ÿ + 6ẏ + 5y = u with y(0) = ẏ = 0 and U (s) = 1/s, we obtain (via Laplace transform) Y (s) = 1 s(s2 + 6s + 5) = 1 . s(s + 5)(s + 1) Expanding in a partial fraction expansion yields Y (s) = 1 1 1 − − . 5s 20(s + 5) 4(s + 1) Taking the inverse Laplace transform we obtain the solution T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) y(t) = 0.2 + 0.05e−5t − 0.25e−t . The m-file script and step response is shown in Figure CP2.3. Step Response 0.2 0.18 0.16 0.14 n=[1]; d=[1 6 5]; sys = tf(n,d); t=[0:0.1:5]; y = step(sys,t); ya=0.2+0.05*exp(-5*t)-0.25*exp(-t); plot(t,y,t,ya); grid; title(‘Step Response’); xlabel(‘Time (s)’); ylabel(‘Amplitude’); Amplitude 0.12 0.1 0.08 0.06 0.04 0.02 0 0 0.5 1 1.5 2 2.5 Time (s) 3 3.5 4 4.5 5 FIGURE CP2.3 Step response. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 78 CHAPTER 2 CP2.4 Mathematical Models of Systems The mass-spring-damper system is represented by mẍ + bẋ + kx = f . Taking the Laplace transform (with zero initial conditions) yields the transfer function X(s)/F (s) = 1/m s2 + bs/m + k/m . The m-file script and step response is shown in Figure CP2.4. T a his th nd To:wY(1) o o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) m=10; k=1; b=0.5; num=[1/m]; den=[1 b/m k/m]; sys = tf(num,den); t=[0:0.1:150]; step(sys,t) Step Response From: U(1) 1.8 1.6 1.4 Amplitude 1.2 1 0.8 0.6 0.4 0.2 0 0 50 100 150 Time (sec.) FIGURE CP2.4 Step response. CP2.5 The spacecraft simulations are shown in Figure CP2.5. We see that as J is decreased, the time to settle down decreases. Also, the overhoot from 10o decreases as J decreases. Thus, the performance seems to get better (in some sense) as J decreases. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 79 Computer Problems Nominal (solid); Off-nominal 80% (dashed); Off-nominal 50% (dotted) 18 16 Spacecraft attitude (deg) 14 12 10 8 6 4 2 0 0 10 20 30 40 50 60 70 80 90 100 Time (sec) Th a is th nd wo or eir is p rk w sa co ro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) %Part (a) a=1; b=8; k=10.8e+08; J=10.8e+08; num=k*[1 a]; den=J*[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[1]); % % Part (b) and (c) t=[0:0.1:100]; % % Nominal case f=10*pi/180; sysf=sys_cl*f ; y=step(sysf,t); % % Off-nominal case 80% J=10.8e+08*0.8; den=J*[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[1]); sysf=sys_cl*f ; y1=step(sysf,t); % % Off-nominal case 50% J=10.8e+08*0.5; den=J*[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[1]); sysf=sys_cl*f ; y2=step(sysf,t); % plot(t,y*180/pi,t,y1*180/pi,’–‘,t,y2*180/pi,’:’),grid xlabel(‘Time (sec)’) ylabel(‘Spacecraft attitude (deg)’) title(‘Nominal (solid); Off-nominal 80% (dashed); Off-nominal 50% (dotted)’) FIGURE CP2.5 Step responses for the nominal and off-nominal spacecraft parameters. © 2017 Pearson Education, Inc., Hoboken, NJ. 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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 80 CHAPTER 2 CP2.6 Mathematical Models of Systems The closed-loop transfer function is T (s) = 4s6 + 8s5 + 4s4 + 56s3 + 112s2 + 56s , ∆(s) p= 7.0709 -7.0713 1.2051 + 2.0863i 1.2051 – 2.0863i 0.1219 + 1.8374i 0.1219 – 1.8374i -2.3933 -2.3333 -0.4635 + 0.1997i -0.4635 – 0.1997i T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) num1=[4]; den1=[1]; sys1 = tf(num1,den1); num2=[1]; den2=[1 1]; sys2 = tf(num2,den2); num3=[1 0]; den3=[1 0 2]; sys3 = tf(num3,den3); num4=[1]; den4=[1 0 0]; sys4 = tf(num4,den4); num5=[4 2]; den5=[1 2 1]; sys5 = tf(num5,den5); num6=[50]; den6=[1]; sys6 = tf(num6,den6); num7=[1 0 2]; den7=[1 0 0 14]; sys7 = tf(num7,den7); sysa = feedback(sys4,sys6,+1); sysb = series(sys2,sys3); sysc = feedback(sysb,sys5); sysd = series(sysc,sysa); syse = feedback(sysd,sys7); sys = series(sys1,syse) poles % pzmap(sys) % p=pole(sys) z=zero(sys) z= 0 1.2051 + 2.0872i 1.2051 – 2.0872i -2.4101 -1.0000 + 0.0000i -1.0000 – 0.0000i Polezero map 2.5 2 1.5 1 Imag Axis 0.5 0 -0.5 -1 -1.5 -2 -2.5 -8 -6 -4 -2 0 2 4 6 8 Real Axis FIGURE CP2.6 Pole-zero map. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 81 Computer Problems where ∆(s) = s10 + 3s9 − 45s8 − 125s7 − 200s6 − 1177s5 − 2344s4 − 3485s3 − 7668s2 − 5598s − 1400 . CP2.7 The m-file script and plot of the pendulum angle is shown in Figure CP2.7. With the initial conditions, the Laplace transform of the linear system is θ(s) = θ0 s . 2 s + g/L To use the step function with the m-file, we can multiply the transfer function as follows: s2 θ0 , 2 s + g/L s T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) θ(s) = which is equivalent to the original transfer function except that we can use the step function input with magnitude θ0 . The nonlinear response is shown as the solid line and the linear response is shown as the dashed line. The difference between the two responses is not great since the initial condition of θ0 = 30◦ is not that large. 30 L=0.5; m=1; g=9.8; theta0=30; % Linear simulation sys=tf([1 0 0],[1 0 g/L]); [y,t]=step(theta0*sys,[0:0.01:10]); % Nonlinear simulation [t,ynl]=ode45(@pend,t,[theta0*pi/180 0]); plot(t,ynl(:,1)*180/pi,t,y,’–‘); xlabel(‘Time (s)’) ylabel(‘theta (deg)’) 20 θ (deg) 10 0 -10 function [yd]=pend(t,y) L=0.5; g=9.8; yd(1)=y(2); yd(2)=-(g/L)*sin(y(1)); yd=yd’; -20 -30 0 2 4 6 8 10 Time (s) FIGURE CP2.7 Plot of θ versus xt when θ0 = 30◦ . © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 82 CHAPTER 2 The system step responses for z = 5, 10, and 15 are shown in Figure CP2.8. T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) CP2.8 Mathematical Models of Systems FIGURE CP2.8 The system response. CP2.9 (a,b) Computing the closed-loop transfer function yields T (s) = G(s) s2 + 2s + 1 = 2 . 1 + G(s)H(s) s + 4s + 3 The poles are s = −3, −1 and the zeros are s = −1, −1. (c) Yes, there is one pole-zero cancellation. The transfer function (after pole-zero cancellation) is T (s) = s+1 . s+3 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 83 Computer Problems Pole?Zero Map 1 0.8 0.6 0.4 Imaginary Axi s 0.2 0 ?-0.2 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) ?-0.4 ?-0.6 ?-0.8 ?-1 ?-3 ?-2.5 ?-2 ?-1.5 ?-1 ?-0.5 0 Real Axi s ng=[1 1]; dg=[1 2]; sysg = tf(ng,dg); nh=[1]; dh=[1 1]; sysh = tf(nh,dh); sys=feedback(sysg,sysh) % pzmap(sys) % pole(sys) zero(sys) poles >> Transfer function: s^2 + 2 s + 1 ————s^2 + 4 s + 3 p= -3 -1 zeros z= -1 -1 FIGURE CP2.9 Pole-zero map. CP2.10 Figure CP2.10 shows the steady-state response to a unit step input and a unit step disturbance. We see that K = 1 leads to the same steady-state response. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 84 CHAPTER 2 Mathematical Models of Systems 0.35 0.3 0.25 Steady−state response K=[0.1:0.1:10]; sysg=tf([1],[1 20 20]); for i=1:length(K) nc=K(i); dc=[1];sysc=tf(nc,dc); syscl=feedback(sysc*sysg,1); systd=feedback(sysg,sysc); y1=step(syscl); Tf1(i)=y1(end); y2=step(systd); Tf2(i)=y2(end); end plot(K,Tf1,K,Tf2,’–‘) xlabel(‘K’) ylabel(‘Steady-state response’) Input Response Steady-State 0.2 0.15 0.1 Disturbance Response Steady-State 0.05 K=1 0 1 2 3 T a his th nd wo o eir is rk w r sa co pro is ill le u vi pr de o rse de ot st f a s d s ec ro n an o te y y p d le d th a a ly by e rt ss fo U in o e r te f t ss th nite gr hi in e ity s w g us d S of or stu e o tat th k ( de f i es e in nt ns co w cl le tr p or ud a uc y r k an ing rnin tors igh d on g. in t la is w D no the iss tea s t p W em ch er or in ing m ld a itt W tio ed id n . e W eb ) 0 4 5 K 6 7 8 9 10 FIGURE CP2.10 Gain K versus steady-state value. © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.