Solution Manual for Feedback Control of Dynamic Systems, 8th Edition

2000 Solutions Manual: Chapter 2 8th Edition Feedback Control of Dynamic Systems . . Gene F. Franklin . J. David Powell . Abbas Emami-Naeini . . . . Assisted by: H. K. Aghajan H. Al-Rahmani P. Coulot P. Dankoski S. Everett R. Fuller T. Iwata V. Jones F. Safai L. Kobayashi H-T. Lee E. Thuriyasena M. Matsuoka Copyright (c) 2019 Pearson Education All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of the publisher. Chapter 2 Dynamic Models Problems and Solutions for Section 2.1 1. Write the di¤erential equations for the mechanical systems shown in Fig. 2.43. For (a) and (b), state whether you think the system will eventually decay so that it has no motion at all, given that there are non-zero initial conditions for both masses, and give a reason for your answer. Also, for part (c), answer the question for F=0. 2001 2002 CHAPTER 2. DYNAMIC MODELS Fig. 2.43 Mechanical systems Solution: The key is to draw the Free Body Diagram (FBD) in order to keep the signs right. For (a), to identify the direction of the spring forces on the object, let x2 = 0 and …xed and increase x1 from 0. Then the k1 spring will be stretched producing its spring force to the left and the k2 spring will be compressed producing its spring force to the left also. You can use the same technique on the damper forces and the other mass. x2 x1 . b1x1 m1 k1x1 k2(x1 – x2) k2(x1 – x2) m2 Free body diagram for Problem 2.1(a) (a) m1 x •1 m2 x •2 = = k1 x1 b1 x_ 1 k2 (x2 x1 ) k2 (x1 k3 (x2 x2 ) y) There is friction a¤ecting the motion of mass 1 which will continue to take energy out of the system as long as there is any movement of x1 :Mass 2 is undamped; therefore it will tend to continue oscillating. However, its motion will drive mass 1 through the spring; therefore, the entire system will continue to lose energy and will eventually decay to zero motion for both masses. k3(x2 – y) 2003 x2 x1 m1 k1x1 k2(x1 – x2) m2 k2(x1 – x2) k3x2 . b1x2 Free body diagram for Problem 2.1(b) m1 x •1 m2 x •2 = = k1 x1 k2 (x1 x2 ) k2 (x2 x1 ) b1 x_ 2 k3 x2 Again, there is friction on mass 2 so there will continue to be a loss of energy as long as there is any motion; hence the motion of both masses will eventually decay to zero. x2 x1 m1 k1x1 . . b1(x1 – x2) . . b1(x1 – x2) k2(x1 – x2) k2(x1 – x2) m2 Free body diagram for Problem 2.1 (c) m1 x •1 m2 x •2 = = k1 x1 k2 (x1 x2 ) b1 (x_ 1 x_ 2 ) F k2 (x2 x1 ) b1 (x_ 2 x_ 1 ) The situation here is similar to part (a). It is clear that the relative motion between mass 1 and 2 would decay eventually, but as long as mass 1 is oscillating, it will drive some relative motion of the two masses and that will cause energy loss in the damper. So the entire system will eventually decay to zero. 2. Write the di¤erential equations for the mechanical systems shown in Fig. 2.44. State whether you think the system will eventually decay so that it has no motion at all, given that there are non-zero initial conditions for both masses, and give a reason for your answer. F 2004 CHAPTER 2. DYNAMIC MODELS Fig. 2.44 Mechanical system for Problem 2.2 Solution: The key is to draw the Free Body Diagram (FBD) in order to keep the signs right. To identify the direction of the spring forces on the left side object, let x2 = 0 and increase x1 from 0. Then the k1 spring on the left will be stretched producing its spring force to the left and the k2 spring will be compressed producing its spring force to the left also. You can use the same technique on the damper forces and the other mass. x1 . . b2(x1 – x2) x2 . . b2(x1 – x2) m2 m1 k1x1 k2(x1 – x2) Free body diagram for Probelm 2.2 Then the forces are summed on each mass, resulting in m1 x •1 m2 x •2 = = k1 x1 k2 (x1 x2 ) b1 (x_ 1 x_ 2 ) k2 (x1 x2 ) b1 (x_ 1 x_ 2 ) k1 x2 The relative motion between x1 and x2 will decay to zero due to the damper. However, the two masses will continue oscillating together without decay since there is no friction opposing that motion and ‡exure of the end springs is all that is required to maintain the oscillation of the two masses. However, note that the two end springs have the same spring constant and the two masses are equal If this had not been true, the two masses would oscillate with di¤erent frequencies and the damper would be excited thus taking energy out of the system. 3. Write the equations of motion for the double-pendulum system shown in Fig. 2.45. Assume the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal. The pendulum k1 x2 2005 rods are taken to be massless, of length l, and the springs are attached 3/4 of the way down. Fig. 2.45 Double pendulum Solution: θ1 3 l 4 θ2 k m m 3 l sin θ1 4 3 l sin θ 2 4 De…ne coordinates If we write the moment equilibrium about the pivot point of the left pendulem from the free body diagram, M= mgl sin 1 3 k l (sin 1 4 3 sin 2 ) cos 1 l = ml2 •1 4 2006 CHAPTER 2. DYNAMIC MODELS ml2 •1 + mgl sin 1 + 9 2 kl cos 1 (sin 1 16 sin 2 ) = 0 Similary we can write the equation of motion for the right pendulem 3 mgl sin 2 + k l (sin 1 4 3 sin 2 ) cos 2 l = ml2 •2 4 As we assumed the angles are small, we can approximate using sin 1 1, and cos 2 1. Finally the linearized equations 1 ; sin 2 2 , cos 1 of motion becomes, 9 kl ( 1 16 9 ml•2 + mg 2 + kl ( 2 16 ml•1 + mg 1 + 2) = 0 1) = 0 Or •1 + g 1 + 9 k ( 1 l 16 m g •2 + 2 + 9 k ( 2 l 16 m 2) = 0 1) = 0 4. Write the equations of motion of a pendulum consisting of a thin, 2-kg stick of length l suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec? (The inertia I of a thin stick about an endpoint is 13 ml2 . Assume is small enough that sin = .) Solution: Let’s use Eq. (2.14) M =I ; 2007 l 2 O θ mg De…ne coordinates and forces Moment about point O. MO = mg = 1 2• ml 3 l sin = IO • 2 • + 3g sin = 0 2l As we assumed is small, • + 3g = 0 2l The frequency only depends on the length of the rod !2 = 3g 2l s T = 2 =2 ! 2l =2 3g l = 3g = 0:3725 m 8 2 2008 CHAPTER 2. DYNAMIC MODELS Grandfather clocks have a period of 2 sec, i.e., 1 sec for a swing from one side to the other. This pendulum is shorter because the period is faster. But if the period had been 2 sec, the pendulum length would have been 1.5 meters, and the clock itself would have been about 2 meters to house the pendulum and the clock face. q 2l (a) Compare the formula for the period, T = 2 3g with the well known formula for the period of a point mass hanging with a string with q l length l. T = 2 g. (b) Important! In general, Eq. (2.14) is valid only when the reference point for the moment and the moment of inertia is the mass center of the body. However, we also can use the formular with a reference point other than mass center when the point of reference is …xed or not accelerating, as was the case here for point O. 5. For the car suspension discussed in Example 2.2, plot the position of the car and the wheel after the car hits a “unit bump” (i.e., r is a unit step) using Matlab. Assume that m1 = 10 kg, m2 = 350 kg, kw = 500; 000 N=m, ks = 10; 000 N=m. Find the value of b that you would prefer if you were a passenger in the car. Solution: The transfer function of the suspension was given in the example in Eq. (2.12) to be: (a) kw b ks Y (s) m1 m2 (s + b ) : = 4 ks ks kw kw ks R(s) s + ( mb1 + mb2 )s3 + ( m +m +m )s2 + ( mk1wmb 2 )s + m 1 2 1 1 m2 This transfer function can be put directly into Matlab along with the numerical values as shown below. Note that b is not the damping ratio, but damping. We need to …nd the proper order of magnitude for b, which can be done by trial and error. What passengers feel is the position of the car. Some general requirements for the smooth ride will be, slow response with small overshoot and oscillation. While the smallest overshoot is with b=5000, the jump in car position happens the fastest with this damping value. 2009 From the …gures, b 3000 appears to be the best compromise. There is too much overshoot for lower values, and the system gets too fast (and harsh) for larger values. 2010 CHAPTER 2. DYNAMIC MODELS % Problem 2.5 Using state space methods % Can also be done using the transfer function above clear all, close all m1 = 10; m2 = 350; kw = 500000; ks = 10000; Bd = [ 1000 3000 4000 5000]; t = 0:0.01:2; for i = 1:4 b = Bd(i); A=[0 1 0 0;-( ks/m1 + kw/m1 ) -b/m1 ks/m1 b/m1; 0 0 0 1; ks/m2 b/m2 -ks/m2 -b/m2 ]; B=[0; kw/m1; 0; 0 ]; C=[ 1 0 0 0; 0 0 1 0 ]; D=0; y=step(A,B,C,D,1,t); subplot(2,2,i); plot( t, y(:,1), ’:’, t, y(:,2), ’-’); legend(’Wheel’,’Car’); ttl = sprintf(’Response with b = %4.1f ’,b ); title(ttl); end 6. For the quadcopter shown in Figs. 2.13 and 2.14: (a) Determine the appropriate commands to rotor #s 1, 2, 3, & 4 so a pure vertical force will be applied to the quadcopter, that is, a force that will have no e¤ect on pitch, roll, or yaw. (b) Determine the transfer function between Fh , and altitude, h. That is, …nd h(s)=Fh (s). Solution: (a) To increase the lifting force, we want to increase the speed of all four rotors. Since two rotors are rotating CW and two are rotating CCW, there will be no net torque about the yaw axis. Since rotor #s 1 and 3 are rotating CW, we want to apply a positive torque to those two rotors, i.e., F T1 = T3 = +KFh ; where Fh is the command for an increased lift for motion in the upward vertical direction and K is the scale factor between torque and lift. Likewise, since rotor #s 2 and 4 are rotating in the CCW direction, we want tp apply a negative torque to those two rotors, i.e., 2011 T2 = T4 = KFh ; (b) Assuming the mass of the quadrotor is m, the transfer function would be h(s) 1 = Fh (s) ms2 7. Automobile manufacturers are contemplating building active suspension systems. The simplest change is to make shock absorbers with a changeable damping, b(u1 ): It is also possible to make a device to be placed in parallel with the springs that has the ability to supply an equal force, u2; in opposite directions on the wheel axle and the car body. (a) Modify the equations of motion in Example 2.2 to include such control inputs. (b) Is the resulting system linear? (c) Is it possible to use the forcer, u2; to completely replace the springs and shock absorber? Is this a good idea? Solution: (a) The FBD shows the addition of the variable force, u2 ; and shows b as in the FBD of Fig. 2.5, however, here b is a function of the control variable, u1 : The forces below are drawn in the direction that would result from a positive displacement of x. u2 ks(x-y) . . b(x-y) x y m2 m1 kw(x-r) ks(x-y) u2 . . b(x-y) Free body diagram m1 x • = m2 y• = b (u1 ) (y_ x) _ + ks (y x) kw (x ks (y x) b (u1 ) (y_ x) _ + u2 r) u2 2012 CHAPTER 2. DYNAMIC MODELS (b) The system is linear with respect to u2 because it is additive. But b is not constant so the system is non-linear with respect to u1 because the control essentially multiplies a state element. So if we add controllable damping, the system becomes non-linear. (c) It is technically possible. However, it would take very high forces and thus a lot of power and is therefore not done. It is a much better solution to modulate the damping coe¢ cient by changing ori…ce sizes in the shock absorber and/or by changing the spring forces by increasing or decreasing the pressure in air springs. These features are now available on some cars… where the driver chooses between a soft or sti¤ ride. 8. In many mechanical positioning systems there is ‡exibility between one part of the system and another. An example is shown in Figure 2.7 where there is ‡exibility of the solar panels. Figure 2.46 depicts such a situation, where a force u is applied to the mass M and another mass m is connected to it. The coupling between the objects is often modeled by a spring constant k with a damping coe¢ cientb, although the actual situation is usually much more complicated than this. (a) Write the equations of motion governing this system. (b) Find the transfer function between the control input, u; and the output, y: Fig. 2.46 Schematic of a system with ‡exibility Solution: (a) The FBD for the system is 2013 y x k(x – y) k(x – y) m M . . b(x – y) u . . b(x – y) Free body diagrams which results in the equations m• x = M y• = k (x y) b (x_ y) _ u + k (x y) + b (x_ y) _ or k b k b x + x_ y y_ m m m m b k b k x x_ + y• + y+ y_ M M M M x •+ = 0 = 1 u M (b) If we make Laplace Transform of the equations of motion s2 X + k X M k b k b X + sX Y sY m m m m b k b sX + s2 Y + Y + sY M M M = 0 = 1 U M = 0 U In matrix form, ms2 + bs + k (bs + k) (bs + k) M s2 + bs + k X Y From Cramer’s Rule, ms2 + bs + k 0 (bs + k) U 2 ms + bs + k (bs + k) (bs + k) M s2 + bs + k det Y = det = Finally, ms2 + bs + k (ms2 + bs + k) (M s2 + bs + k) 2U (bs + k) 2014 CHAPTER 2. DYNAMIC MODELS Y U = = ms2 + bs + k 2 (ms2 + bs + k) (M s2 + bs + k) (bs + k) ms2 + bs + k 4 mM s + (m + M )bs3 + (M + m)ks2 9. Modify the equation of motion for the cruise control in Example 2.1, Eq(2.4), so that it has a control law; that is, let u = K(vr v); where vr K = = reference speed constant: This is a ‘proportional’control law where the di¤erence between vr and the actual speed is used as a signal to speed the engine up or slow it down. Put the equations in the standard state-variable form with vr as the input and v as the state. Assume that m = 1500 kg and b = 70 N s= m; and …nd the response for a unit step in vr using Matlab. Using trial and error, …nd a value of K that you think would result in a control system in which the actual speed converges as quickly as possible to the reference speed with no objectional behavior. Solution: v_ + substitute in u = K (vr b 1 v= u m m v) v_ + b 1 K v= u= (vr m m m v) Rearranging, yields the closed-loop system equations, v_ + K K b v + v = vr m m m A block diagram of the scheme is shown below where the car dynamics are depicted by its transfer function from Eq. 2.7. 2015 Block diagram The transfer function of the closed-loop system is, K V (s) m = b Vr (s) s+ m +K m so that the inputs for Matlab are num = K m den = [1 For K = 100; 500; 1000; 5000 We have, b K + ] m m 2016 CHAPTER 2. DYNAMIC MODELS We can see that the larger the K is, the better the performance, with no objectionable behaviour for any of the cases. The fact that increasing K also results in the need for higher acceleration is less obvious from the plot but it will limit how fast K can be in the real situation because the engine has only so much poop. Note also that the error with this scheme gets quite large with the lower values of K. You will …nd out how to eliminate this error in chapter 4 using integral control, which is contained in all cruise control systems in use today. For this problem, a reasonable compromise between speed of response and steady state errors would be K = 1000; where it responds in 5 seconds and the steady state error is 5%. 2017 % Problem 2.9 clear all, close all % data m = 1500; b = 70; k = [ 100 500 1000 5000 ]; % Overlay the step response hold on t=0:0.2:50; for i=1:length(k) K=k(i); num =K/m; den = [1 b/m+K/m]; sys=tf(num,den); y = step(sys,t); plot(t,y) end hold o¤ 2018 CHAPTER 2. DYNAMIC MODELS 10. Determine the dynamic equations for lateral motion of the robot in Fig. 2.47. Assume it has 3 wheels with the front a single, steerable wheel where you have direct control of the rate of change of the steering angle, Usteer , with geometry as shown in Fig. 2.48. Assume the robot is going in approximately a straight line and its angular deviation from that straight line is very small Also assume that the robot is traveling at a constant speed, Vo . The dynamic equations relating the lateral velocity of the center of the robot as a result of commands in Usteer is desired. Fig. 2.47 Robot for delivery of hospital supplies Source: AP Images . Fig. 2.48 Model for robot motion 2019 Solution: This is primarily a problem in kinematics. First, we know that the control input, Usteer ; is the time rate of change of the steering wheel angle, so _ s = Usteer When s is nonzero, the cart will be turning, so that its orientation wrt the x axis will change at the rate _ = Vo s : L as shown by the diagram below. Diagram showing turning rate due to s T The actual change in the carts lateral position will then be proportional to according to y_ = Vo as shown below. 2020 CHAPTER 2. DYNAMIC MODELS Lateral motion as a function of These linear equations will hold providing and s stay small enough that sin ‘ ; and sin s ‘ s :Combining them all, we obtain, … Vo2 y = Usteer L Note that no dynamics come into play here. It was assumed that the velocity is constant and the front wheel angle time rate of change is directly commanded. Therefore, there was no need to invoke Eqs (2.1) or (2.10). As you will see in future chapters, feedback control of such a system with a triple integration is tricky and needs signi…cant damping in the feedback path to achieve stability. 11. Determine the pitch, yaw, and roll control equations for the hexacopter shown in Fig. 2.49 that are similar to those for the quadcopter given in Eqs. (2.18) to (2.20). Fig. 2.49 Hexacopter 2021 Assume that rotor #1 is in the direction of ‡ight and the remaining rotors are numbered CW from that rotor. In other words, rotors #1 and #4 will determine the pitch motion. Rotor #s 2, 3, 5, & 6 will determine roll motion. Pitch, roll and yaw motions are de…ned by the coordinate system shown in Fig. 2.14 in Example 2.5. In addition to developing the equations for the 3 degrees of freedom in terms of how the six rotor motors should be commanded (similar to thoseforthe quadrotorinEqs. (2.18)– (2.20)), it will also be necessary to decide which rotors are turning CW and which ones are turning CCW. The direction of rotation for the rotors needs to be selected so there is no net torque about the vertical axis; that is, the hexicopter will have no tendancy for yaw rotation in steadystate. Furthermore, a control action to a¤ect pitch should have no e¤ect on yaw or roll. Likewise, a control action for roll should have no e¤ect on pitch or yaw, and a control action for yaw should have no e¤ect on pitch or roll. In other words, the control actions should produce no cross-coupling between pitch, roll, and yaw just as was the case for the quadcopter in Example 2.5. Solution: For starters, the instructions above give us the following definitions Hexacopter coordinate system de…nition To obtain some symmetry in the rotations, let’s assign rotor #s 1, 3, & 5 to rotate CW, while rotor #s 2, 4, & 6 rotate CCW. Now let’s start out by examining the yaw torque e¤ects. If we assume the same control action as we found for the quadrotor in Eq. (2.20), we would apply a negative torque on all six rotors. This will clearly produce a positive yawing torque on the hexacopter with no net change in vertical 2022 CHAPTER 2. DYNAMIC MODELS lift for the same reasons given on pages 38 and 39. However, it not necessarily produce zero torque about the pitch and roll axes. So let’s check that and adjust things if required. Unlike the quadrotor where the two rotors on the x-axis were both turning CW, rotors1 and 4 are turning in the opposite direction. So in this case, giving those two rotors an equal negative torque will speed one up rotor 4 and slow down,rotor 1, thus producing a negative torque about the y-axis. However, note that rotors 2 and 6 will speed up, while rotors 3 and 5 will slow down, thus producing a positive torque.about the y-axis. These two torques cancel each other because the angle between rotors 2, 3, 5 & 6 and the y-axis are all 30o between the arms and the y-axis. Since the sine of 30o = 1/2, that means that the torque about the y-axis from rotors 2, 3, 5 & 6 will o¤set the torque from rotors 1 & 4. For no roll moment, we require a net zero torque about the x-axis. Rotors 1 & 4 are on the x-axis, so no torque from them. Rotors 2 & 6 are both CCW, so they produce no net torque about the x-axis, and the same applies for rotors 3 & 5; hence when we apply an equal torque on all six rotors in the same direction along the z-axis, we obtain a yaw torque with no e¤ect on pitch or roll or a vertical force. So, the control using: Yaw control: T1 = T2 = T3 = T4 = T5 = T6 = will produce motion in yaw, with no e¤ect on pitch, roll, motion. T ; or vertical For pitch control, we want an increase in the speed and lift of the CW rotor 1 (ie a positive T1 ) and a decrease in the speed of the CCW rotor 4, which requires a positive T4 ; so T1 = T4 = +T . But note that this action will produce a yawing torque, which.can be canceled if we apply an equal and opposite yawing torque from rotors 2, 3, 5, 6. This accomplished by T2 = T3 = T5 = T6 = T =2: In fact, this additional control will also add to the torque about the pitch (y-axis) by 50% while having no e¤ect on roll. Thus the control using: Pitch control: T1 = 2 T2 = 2 T3 = T4 = 2 T5 = 2 T6 = 2 T ; 3 will produce motion in pitch, with no e¤ect on yaw, roll, or vertical motion. For roll control, by increasing the speed of rotors 5 & 6, and decreasing the speed of rotors 2 & 3, we will obtain a positive roll torque about the x-axis. No change is needed for rotors 1 & 4. To accomplish these speed changes, we need to apply a positive torque to the CW rotor 5 and a negative torque to the CCW rotor 6. Likewise, we need a positive torque to the CCW rotor 2 and a negative torque to the CW rotor 3. 2023 Since the torques applied are half positive and half negative, there will be no yaw torque. Furthermore, there will be no pitch torque of change in overall lift with these torque applications. However, the moment arm is not 1 arm length; rather it is cos(30o ), so to keep the scale factor for the two arms consistent with the pitch axis, we need to divide by 2 cos(30o ): Therefore, the control using: Roll control: T2 = T3 = T5 = T6 = T =(2 cos(30o )); will produce motion in roll, with no e¤ect on pitch, yaw, or vertical motion. 12. In most cases, quadcopters have a camera mounted that does not swivel in the x y plane and its direction of view is oriented at 45o to the arms supporting the rotors. Therefore, these drones typically ‡y in a direction that is aligned with the camera rather than along an axis containing two of the rotors. To simplify the ‡ight dynamics, the x-direction of the coordinate system is aligned with the camera direction. Based on the coordinate de…nitions for the axes in Fig. 2.14, assume the x-axis lies half way between rotors # 1 and 2 and determine the rotor commands for the four rotors that would accomplish independent motion for pitch, roll, and yaw. Solution: The orientation of the coordinate system and arrangement of the rotors is 2024 CHAPTER 2. DYNAMIC MODELS Coorinate system and rotor arrangement It should be clear from the discussion on pages 38 and 39 of the book (plus the discussion here for Problem 2.11) that the following commands to the rotors produces the desired independent motion: Roll control: Pitch control: Yaw control: T1 = T2 = T3 = T4 = T ; T1 = T2 = T3 = T4 = T ; T1 = T2 = T3 = T4 = T : Problems and Solutions for Section 2.2 13. A …rst step toward a realistic model of an op amp is given by the equations below and shown in Fig. 2.50. Vout = i+ = 107 [V+ s+1 i =0 V ] 2025 Fig. 2.50 Circuit for Problem 2.13 Find the transfer function of the simple ampli…cation circuit shown using this model. Solution: As i = 0, (a) Vin V Rin V = Vout = = = = V Vout Rf Rf Rin Vin + Vout Rin + Rf Rin + Rf 107 [V+ V ] s+1 107 Rf V+ Vin s+1 Rin + Rf 107 s+1 Rin Vout Rin + Rf Rf Rin Vin + Vout Rin + Rf Rin + Rf R f 107 Rin +R Vout f = in Vin s + 1 + 107 R R+R in f 14. Show that the op amp connection shown in Fig. 2.51 results in Vo = Vin if the op amp is ideal. Give the transfer function if the op amp has the non-ideal transfer function of Problem 2.13. 2026 CHAPTER 2. DYNAMIC MODELS Fig. 2.51 Circuit for Problem 2.14 Solution: Ideal case: Vin V+ V = = = V+ V Vout Non-ideal case: Vin = V+ ; V = Vout but, V+ 6= V instead, Vout = = 107 [V+ s+1 107 [Vin s+1 V ] Vout ] so, 107 Vout 107 107 = s+1107 = = 7 Vin s + 1 + 10 s + 107 1 + s+1 2027 15. A common connection for a motor power ampli…er is shown in Fig. 2.52. The idea is to have the motor current follow the input voltage and the connection is called a current ampli…er. Assume that the sense resistor, Rs is very small compared with the feedback resistor, R and …nd the transfer function from Vin to Ia : Also show the transfer function when Rf = 1: Node A Node B Fig. 2.52 Op Amp circuit for Problem 2.15 with nodes marked. Solution: At node A, Vin 0 Vout 0 VB 0 =0 + + Rin Rf R At node B, with Rs Ia + (201) R 0 VB 0 VB + R Rs = VB = VB 0 (202) RRs Ia R + Rs R s Ia The dynamics of the motor is modeled with negligible inductance as Jm •m + b _ m Jm s + b = = Kt I a Kt I a (203) 2028 CHAPTER 2. DYNAMIC MODELS At the output, from Eq. (202). Eq. (203) and the motor equation Va = I a R a + Ke s Vo = I a R s + Va = I a R s + I a R a + Ke Kt I a Jm s + b Substituting this into Eq.(201) Vin 1 Kt I a Ia R s + I a R s + I a R a + Ke + =0 Rin Rf Jm s + b R This expression shows that, in the steady state when s ! 0; the current is proportional to the input voltage. If fact, the current ampli…er normally has no feedback from the output voltage, in which case Rf ! 1 and we have simply Ia = Vin R Rin Rs 16. An op amp connection with feedback to both the negative and the positive terminals is shown in Fig 2.53. If the op amp has the non-ideal transfer function given in Problem 13, give the maximum value possible for the r positive feedback ratio, P = in terms of the negative feedback r+R Rin ratio,N = for the circuit to remain stable. Rin + Rf Fig. 2.53 Op Amp circuit for Problem 2.16 2029 Solution: Vin V Vout V + Rin Rf Vout V+ 0 V+ + R r V 0 = 0 Rf Rin Vin + Vout Rin + Rf Rin + Rf = (1 N ) Vin + N Vout r Vout = P Vout = r+R = V+ Vout = = = 107 [V+ V ] s+1 107 [P Vout (1 s+1 Vout Vin = = = 0 P < T 1 > T 2 > T 3 and, with a same cold air ‡ow into every room, the ones with some sun load will be hotest. Let’s rede…nce the resistances Ro = resistance to heat ow through one unit of outer wall Ri = resistance to heat ow through one unit of inner wall Room type 1: 2050 CHAPTER 2. DYNAMIC MODELS T_1 = = qout = qin = 2 (T1 Ri 2 (To Ro 1 (qin qout ) C 1 2 (To T1 ) C Ro T2 ) + q T1 ) 2 (T1 Ri T2 ) 2 (T1 Ri T2 ) q Room type 2: 1 T_2 = C qin = qout = 1 (To Ro 1 (To Ro 1 (T2 Ri T2 ) + T2 ) + T3 ) + q 2 (T1 Ri T2 ) 1 (T2 Ri T3 ) q Room type 3: qin = qout = 1 T_3 = C 4 (T2 Ri q 4 (T2 Ri T3 ) T3 ) q 27. For the two-tank ‡uid-‡ow system shown in Fig. 2.59, …nd the di¤erential equations relating the ‡ow into the …rst tank to the ‡ow out of the second tank. 2051 Fig. 2.60 Two-tank ‡uid-‡ow system for Problem 27 Solution: This is a variation on the problem solved in Example 2.21 and the de…nitions of terms is taken from that. From the relation between the height of the water and mass ‡ow rate, the continuity equations are m _1 m _2 = = A1 h_ 1 = win w A2 h_ 2 = w wout Also from the relation between the pressure and outgoing mass ‡ow rate, w = wout = 1 1 ( gh1 ) 2 R1 1 1 ( gh2 ) 2 R2 Finally, h_ 1 = h_ 2 = 1 1 1 ( gh1 ) 2 + win A1 R 1 A1 1 1 1 1 ( gh1 ) 2 ( gh2 ) 2 : A2 R 1 A2 R2 28. A laboratory experiment in the ‡ow of water through two tanks is sketched in Fig. 2.61. Assume that Eq. (2.96) describes ‡ow through the equal-sized holes at points A, B, or C. (a) With holes at B and C but none at A, write the equations of motion for this system in terms of h1 and h2 . Assume that when h2 = 15 cm, the out‡ow is 200 g/min. (b) At h1 = 30 cm and h2 = 10 cm, compute a linearized model and the transfer function from pump ‡ow (in cubic centimeters per minute) to h2 . 2052 CHAPTER 2. DYNAMIC MODELS (c) Repeat parts (a) and (b) assuming hole B is closed and hole A is open. Assume that h3 = 20 cm, h1 > 20 cm, and h2 < 20:cm. Fig. 2.61 Two-tank ‡uid-‡ow system for Problem 28 Solution: (a) Following the solution of Example 2.21, and assuming the area of both tanks is A; the values given for the heights ensure that the water will ‡ow according to WB Win WB = WC = WC WB = = 1 [ g (h1 R 1 1 [ gh2 ] 2 R Ah_ 2 Ah_ 1 1 h2 )] 2 From the out‡ow information given, we can compute the ori…ce resistance, R; noting that for water, = 1 gram/cc and g = 981 cm/sec2 ' 1000 cm/sec2 : WC = R = = 1p 1p gh2 = g 15 cm 200 g= mn = R p R p 1 g= cm3 1000 cm= s2 g 10 cm = 200 g= mn 200 g=60 s s 1 1 122:5 g cm2 s2 60 = 36:7 g 2 cm 2 200 cm3 s2 g2 15 cm 2053 (b) The nonlinear equations from above are h_ 1 = h_ 2 = 1 1 p g (h1 h2 ) + Win AR A 1 p 1 p g (h1 h2 ) gh2 AR AR The square root functions need to be linearized about the nominal heights. In general the square root function can be linearized as below p x0 + x = s = p x0 1 + x x0 x0 1 + 1 x 2 x0 So let’s assume that h1 = h10 + h1 and h2 = h20 + h2 where h10 = 30 cm and h20 = 10 cm. And for round numbers, let’s assume the area of each tank A = 100 cm2 : The equations above then reduce to h_ 1 = h_ 2 = p 1 1 (1)(1000) (30 + h1 10 h2 ) + Win (1)(100)(36:7) (1)(100) p p 1 1 (1)(1000) (30 + h1 10 h2 ) (1)(1000)(10 + h2 ) (1)(100)(36:7) (1)(100)(36:7) which, with the square root approximations, is equivalent to, p 2 1 1 1 h_ 1 = (1 + h1 h2 ) + Win 36:7 40 40 100 p 2 1 1 1 1 (1 + h1 h2 ) (1 + h2 ) h_ 2 = 36:7 40 40 36:7 20 p 10 The nominal in‡ow Wnom = 3:67 2cc/sec is required in order for the system to be in equilibrium, as can be seen from the …rst equation. So we will de…ne the total in‡ow to be Win = Wnom + W: Including the nominal in‡ow, the equations become p 2 1 _h1 = ( h1 h2 ) + W 1468 100 p p p 1 2 2 2 1 h1 + ( ) h2 + h_ 2 = 1468 1468 734 36:7 However, holding the nominal ‡ow rate maintains h1 at equilibrium, but h2 will not stay at equilibrium. Instead, there will be a constant term increasing h2 : Thus the standard transfer function will not result. 2054 CHAPTER 2. DYNAMIC MODELS (c) With hole B closed and hole A open, the relevant relations are Win WA h_ 1 = h_ 2 = WA = WA = WC = WC = Ah_ 1 1p g(h1 R Ah_ 2 1p gh2 R h3 ) 1 p 1 g(h1 h3 ) + Win AR A 1 p 1 p g(h1 h3 ) gh2 AR AR For the value of R, we will use the same calculation from part (a), since that value has not changed with di¤erent hole openings, so we still have 122:5 R= 60 200 s g cm2 s2 = 36:7 g cm3 s2 g2 1 2 cm 1 2 . With the same de…nitions for the perturbed quantities as for part (b), plus h3 = 20 cm,.we obtain h_ 1 = h_ 2 = p 1 1 Win (1)(1000)(30 + h1 20) + (1)(100)(36:7) (1)(100) p 1 (1)(1000)(30 + h1 20) (1)(100)(36:7) p 1 (1)(1000)(10 + h2 ) (1)(100)(36:7) which, with the linearization carried out, reduces to h_ 1 = h_ 2 = 1 1 1 1 (1 + h1 ) + Wnom + W (36:7) 20 (100) (100) 1 1 1 1 (1 + h1 ) (1 + h2 ) (36:7) 20 (36:7) 20 10 and with the nominal ‡ow rate of Win = 3:67 removed h_ 1 = h_ 2 = 1 1 h1 + W 734 100 1 1 h1 h2 734 734 2055 Taking the Laplace transform of these two equations, and solving for the desired transfer function (in cc/sec) yields H2 (s) 1 0:01 : = W (s) 734 (s + 1=734)2 which becomes, with the in‡ow in grams/min, H2 (s) 0:000817 1 (0:01)(60) =: = 2 W (s) 734 (s + 1=734) (s + 1=734)2 29. The equations for heating a house are given by Eqs. (2.81) and (2.82) and, in a particular case can be written with time in hours as C dTh = Ku dt Th To R where (a) C is the Thermal capacity of the house, BT U=o F (b) Th is the temperature in the house, o F (c) To is the temperature outside the house, o F (d) K is the heat rating of the furnace, = 90; 000 BT U=hour (e) R is the thermal resistance, o F per BT U=hour (f) u is the furnace switch, =1 if the furnace is on and =0 if the furnace is o¤. It is measured that, with the outside temperature at 32 o F and the house at 60 o F , the furnace raises the temperature 2 o F in 6 minutes (0.1 hour). With the furnace o¤, the house temperature falls 2 o F in 40 minutes. What are the values of C and R for the house? Solution: For the …rst case, the furnace is on which means u = 1. C dTh dt = K T_h = K C 1 (Th To ) R 1 (Th To ) RC and with the furnace o¤, T_h = 1 (Th RC To ) In both cases, it is a …rst order system and thus the solutions involve exponentials in time. The approximate answer can be obtained by simply 2056 CHAPTER 2. DYNAMIC MODELS looking at the slope of the exponential at the outset. This will be fairly accurate because the temperature is only changing by 2 degrees and this represents a small fraction of the 30 degree temperature di¤erence. Let’s solve the equation for the furnace o¤ …rst Th = t 1 (Th RC To ) plugging in the numbers available, the temperature falls 2 degrees in 2/3 hr, we have 1 2 = (60 32) 2=3 RC which means that RC = 28=3 For the second case, the furnace is turned on which means K Th = t C 1 (Th RC To ) and plugging in the numbers yields 2 90; 000 = 0:1 C 1 (60 28=3 32) and we have C = R = 90; 000 = 3913 23 RC 28=3 = = 0:00239 C 3913