(Solution Manual) Feedback Control of Dynamic Systems

Franklin, 4

Engineering

Solution Manual

4th Edition

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Solution Manual for Feedback Control of Dynamic Systems, The downloadable file consists of 672 pages, and the total size of the file is 10.23 MB. Purchase now only for highly discounted price.

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Chapter 2 Dynamic Models Problems and Solutions for Section 2.1 1. Write the di fferential equations for the mechanical systems shown in Fig. 2.38. 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 x 2=0a n d Þxed and increase x1from 0. Then the k1spring will be stretched producing its spring force to the left and the k2spring 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. (a) m1¨x1=−k1x1−b1úx1−k2(x1−x2) m2¨x2=−k2(x2−x1)−k3(x2−y)−b2úx2 1112 CHAPTER 2. DYNAMIC MODELS Figure 2.38: Mechanical systems 13 m1¨x1=−k1x1−k2(x1−x2)−b1úx1 m2¨x2=−k2(x2−x1)−k3x2 m1¨x1=−k1x1−k2(x1−x2)−b1(úx1−úx2) m2¨x2=F−k2(x2−x1)−b1(úx2−úx1) 2. Write the equations of motion of a pendulum consisting of a thin, 2-kg stick of length lsuspended from a pivot. How long should the rod be in order for the period to be exactly 2 secs? (The inertia Iof a thin stick about an endpoint is1 3ml2.A s s u m e θis small enough that sin θ∼=θ.) Solution: Let’s use Eq. (2.14) M=Iα,14 CHAPTER 2. DYNAMIC MODELS /G20 mg/G712lO Moment about point O. MO=−mg×l 2sinθ=IO¨θ =1 3ml2¨θ ¨θ+3g 2lsinθ=0 As we assumed θis small, ¨θ+3g 2lθ=0 T h ef r e q u e n c yo n l yd e p e n d so nt h el e n g t ho ft h er o d ω2=3g 2l T=2π ω=2πs 2l 3g=2 l=3g 2π2=1.49m <Notes >15 Figure 2.39: Double pendulum (a) Compare the formula for the period, T=2πq 2l 3gwith the well known formula for the period of a point mass hanging with a string with length l.T=2πq l 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 thebody. However, we also can use the formular with a reference point other than mass center when the point of reference is Þxed or not accelerating, as was the case here for point O. 3. Write the equations of motion for the double-pendulum system shown in Fig. 2.39. Assume the displacement angles of the pendulums are smallenough to ensure that the spring is always horizontal. The pendulum rods are taken to be massless, of length l, and the springs are attached 3/4 of the way down. Solution:16 CHAPTER 2. DYNAMIC MODELS 1/G712/G71 k 1 sin43/G71l2 sin43/G71lm ml43 If we write the moment equilibrium about the pivot point of the left pen- dulem from the free body diagram, M=−mglsinθ1−k3 4l(sinθ1−sinθ2)c o s θ13 4l=ml2¨θ1 ml2¨θ1+mglsinθ1+9 16kl2cosθ1(sinθ1−sinθ2)=0 Similary we can write the equation of motion for the right pendulem −mglsinθ2+k3 4l(sinθ1−sinθ2)c o s θ23 4l=ml2¨θ2 As we assumed the angles are small, we can approximate using sin θ1≈ θ1,sinθ2≈θ2,c o s θ1≈1, and cos θ2≈1. Finally the linearized equations of motion becomes, ml¨θ1+mgθ1+9 16kl(θ1−θ2)=0 ml¨θ2+mgθ2+9 16kl(θ2−θ1)=0 Or17 ¨θ1+g lθ1+9 16k m(θ1−θ2)=0 ¨θ2+g lθ2+9 16k m(θ2−θ1)=0 4. Write the equations of motion for a body of mass Msuspended from a Þxed point by a spring with a constant k. Carefully de Þne where the body’s displacement is zero. Solution:Some care needs to be taken when the spring is suspended vertically in t h ep r e s e n c eo ft h eg r a v i t y . W ed e Þnex=0t ob ew h e nt h es p r i n gi s unstretched with no mass attached as in (a). The static situation in (b) results from a balance between the gravity force and the spring. From the free body diagram in (b), the dynamic equation results m¨x=−kx−mg. We can manipulate the equation m¨x=−k³ x+m kg´ , so if we replace xusing y=x+m kg, m¨y=−ky m¨y+ky =018 CHAPTER 2. DYNAMIC MODELS The equilibrium value of xincluding the e ffect of gravity is at x=−m kg andyrepresents the motion of the mass about that equilibrium point. An alternate solution method, which is applicable for any problem involving vertical spring motion, is to de Þne the motion to be with respect to the static equilibrium point of the springs including the e ffect of gravity, and then to proceed as if no gravity was present. In this problem, we would de Þneyto be the motion with respect to the equilibrium point, then the FBD in (c) would result directly in m¨y=−ky. 5. For the car suspension discussed in Example 2.2, (a) write the equations of motion (Eqs. (2.10) and (2.11)) in state-variable form. Use the state vector x=[xúxy úy]T. (b) Plot the position of the car and the wheel after the car hits a “unit bump” (i.e., ris a unit step) using MATLAB. Assume that m1= 10 kg, m2= 350 kg, kw= 500 ,000 N /m,ks=1 0,000 N /m. Find the value of bthat you would prefer if you were a passenger in the car. Solution: (a) We can arrange the equations of motion to be used in the state- variable form ¨x=−ks m1x−b m1úx+ks m1y+b m1úy−kw m1x+kw m1r =−µks m1+kw m1¶ x−b m1úx+ks m1y+b m1úy+kw m1r ¨y=ks m2x+b m2úx−ks m2y−b m2úy So, for the given sate vector of x=[xúxy úy]T, the state-space form will be,  úx ¨x úy ¨y = 01 0 0 −³ ks m1+kw m1´ −b m1ks m1b m1 00 0 1 ks m2b m2−ks m2−b m2  x úx y úy + 0 kw m1 0 0 r (b) Note that bis not the damping ratio, but damping. We need to Þnd the proper order of magnitude for b, which can be done by trial and19 error. What passengers feel is the position of the car. Some general requirements for the smooth ride will be, slow response with smallovershoot and oscillation. 0 0.5 1 1.5 200.511.5Response with b = 1000.0 Wheel Car 0 0.5 1 1.5 200.511.5Response with b = 2000.0 WheelCar 0 0.5 1 1.5 200.511.5Response with b = 3000.0 Wheel Car 0 0.5 1 1.5 200.511.5Response with b = 4000.0 WheelCar From the Þgures, b≈3000 would be acceptable. There is too much overshoot for lower values, and the system gets too fast (and harsh) for larger values.20 CHAPTER 2. DYNAMIC MODELS %P r o b l e m2 . 5b clear all, close all m1 = 10; m2 = 350;kw = 500000; ks = 10000; B = [ 1000 2000 3000 4000 ];t = 0:0.01:2; for i = 1:4 b=B ( i ) ;F = [ 0 1 0 0; -( ks/m1 + kw/m1 ) -b/m1 ks/m1 b/m1; 0001 ;k s / m 2b / m 2- k s / m 2- b / m 2] ; G=[0 ;k w / m 1 ;0 ;0] ; H=[1000 ;0010] ;J=0 ; y=s t e p (F ,G ,H ,J ,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. Automobile manufacturers are contemplating building active suspension systems. The simplest change is to make shock absorbers with a change-able damping, b(u 1).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 con- trol 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 bis a function of the control variable, u1.The forces below are drawn in the direction that would result from a positive displacement of x.21 m1¨x=b(u1)(úy−úx)+ks(y−x)−kw(x−r)−u2 m2¨y=−ks(y−x)−b(u1)(úy−úx)+u2 (b) The system is linear with respect to u2because it is additive. But bis not constant so the system is non-linear with respect to u1be- cause the control essentially multiplies a state element. So if we addcontrollable 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 bet- ter solution to modulate the damping coe fficient by changing ori Þce sizes in the shock absorber and/or by changing the spring forces byincreasing or decreasing the pressure in air springs. These featuresare now available on some cars... where the driver chooses between as o f to rs t i ffride. 7. 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(v r−v) (124) where vr= reference speed (125) K= constant . (126) This is a ‘proportional’ control law where the di fference between vrand 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 vras the input andvas the state. Assume that m= 1000 kg and b=5 0 N·s/m,and Þnd the response for a unit step in vrusing MATLAB. Using trial and error, Þnd 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:22 CHAPTER 2. DYNAMIC MODELS úv+b mv=1 mu substitute in u=K(vr−v) úv+b mv=1 mu=K m(vr−v) A block diagram of the scheme is shown below where the car dynamics are depicted by its transfer function from Eq. 2.7. mbsm /G2b1 K /G53uvrv /G2d/G2b T h es t a t e - v a r i a b l ef o r mo ft h ee q u a t i o n si s , úv=−µb m+K m¶ v+K mvr y=v so that the matrices for Matlab are F=−µb m+K m¶ G=K m H=1 J=0 ForK= 100 ,500,1000,5000 We have,23 Time (sec.)Step Response 0 5 10 15 20 25 30 35 4000.10.20.30.40.50.60.70.80.91From: U(1)K=100 K=500 K=1000 K=5000 We can see that the larger the Kis, 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 theplot but it will limit how fast Kc a nb ei nt h er e a ls i t u a t i o nb e c a u s et h e 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 Þnd out how to eliminate this error in chapter 4 using integral control, which is containedin 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 is 5 seconds and the steady state error is 5%.24 CHAPTER 2. DYNAMIC MODELS %P r o b l e m2 . 7 clear all, close all %d a t a m = 1000;b = 50; k = [ 100 500 1000 5000 ]; % Overlay the step response hold on for i=1:length(k) K=k(i);F=- ( b + K ) / m ;G=K / m ; H=1 ; J=0 ; step( F,G,H,J); end Problems and Solutions for Section 2.2 8. In many mechanical positioning systems there is ßexibility between one part of the system and another. An example is shown in Figure 2.6 where there is ßexibility of the solar panels. Figure 2.40 depicts such a situation, where a force uis applied to the mass Mand another mass mis connected to it. The coupling between the objects is often modeled by a spring constant kwith a damping coe fficient b, although the actual situation is usually much more complicated than this. (a) Write the equations of motion governing this system, identify appro- priate state variables, and express these equations in state-variable form. (b) Find the transfer function between the control input, u,and the output, y. Solution: (a) The FBD for the system is25 Figure 2.40: Schematic of a system with ßexibility which results in the equations m¨x=−k(x−y)−b(úx−úy) M¨y=u+k(x−y)+b(úx−úy) Let the state-space vector x=[xúxy úy]T ¨x=−k mx−b múx+k my+b múy ¨y=k Mx+b Múx−k My−b Múy+1 Mu So,  úx ¨x úy ¨y = 0100 −k m−b mk mb m 0001 k Mb M−k M−b M  x úx y úy + 0 00 1 M u (b) We have the complete state-variable form from part a. We will learn the systematic way to convert from the state-variable form to transfer function later in chapter 7. If we make Laplace Transform of the equations of motion s2X+k mX+b msX−k mY−b msY =0 −k MX−b MsX+s2Y+k MY+b MsY =1 MU In matrix form, ·ms2+bs+k−(bs+k) −(bs+k)Ms2+bs+k¸·X Y¸ =·0 U¸26 CHAPTER 2. DYNAMIC MODELS From Cramer’s Rule, Y=det·ms2+bs+k0 −(bs+k)U¸ det·ms2+bs+k−(bs+k) −(bs+k)Ms2+bs+k¸ =ms2+bs+k (ms2+bs+k)(Ms2+bs+k)−(bs+k)2U Finally, Y U=ms2+bs+k (ms2+bs+k)(Ms2+bs+k)−(bs+k)2 =ms2+bs+k mMs4+(m+M)bs3+(M+m)ks2 9. For the inverted pendulum, Eqs. (2.34), (a) Try to put the equations of motion into state-variable form using the state vector x=[θúθxúx]T. Why is it not possible? (b) Write the equations in the “descriptor” form E˙x=F0x+G0u, and de Þne values for E,F0,a n d G0(note that Eis a 4 ×4m a t r i x ) . Then show how you would compute FandGfor the standard state- variable description of the equations of motion.] Solution: (a) It is impossible because the acceleration terms are coupled. (b)  1000 0I+mpl20−mpl 0010 0−mpl0mt+mp  úθ ¨θ úx ¨x = 01 0 0 mpgl00 0 00 0 1 00 0 −b  θ úθ x úx + 0 0 0 1 u E˙x=F0x+G0u ˙x=E−1F0x+E−1G0u F=E−1F0,G=E−1G027 10. The longitudinal linearized equations of motion of a Boeing 747 are given in Eq. (9.28). Using MATLAB or other computer aid: (a) Determine the response of the altitude hfor a 2-sec pulse of the ele- vator with a magnitude of 2◦.Note that, since Eq. (9.28) represents a set of linearized equations, the state variables actually representthe deviation of the state from the nominal operating point. For example, hrepresents the amount the altitude of the aircraft di ffers from 20,000 ft. (b) Consider using the feedback law δ e=Khh+δe,ext (127) where the elevator input angle is the sum of a term proportional to the error in altitude hplus an external input (a disturbance or command input). Note from part (a) that a positive change inelevator causes a negative change in altitude, so that the proposed proportional feedback law has the logical sign to anticipate a stable system provided K h>0.By trial and error, try to Þnd a value for the feedback gain Khsuch that a 2◦pulse of 2 sec on δe,ext yields a more stable altitude response. (c) If you have trouble Þnding a value of Khthat produces a stable response, try modifying the feedback law to include information onpitch rate q: δ e=Khh+Kqq+δe,ext (128) Use trial and error to pick appropriate values for both KhandKq. A s s u m et h es a m et y p eo fp u l s ei n p u tf o r δe,ext as in part (b). (d) Show that the further introduction of pitch-angle feedback, θ,such that δe=Khh+Kqq+Kθθ+δe,ext (129) allows you to decrease the time it takes for the altitude to settle back to its nominal value, as well as to decrease the value of Khrequired for a stable response. Note that, although Kh= 0 produces stable altitude behavior, we require Kh>0 in order to guarantee that h→0 (so there will be no steady-state error). Solution:28 CHAPTER 2. DYNAMIC MODELS 0 1 2 3 4 5 6 7 8 9 10-350-300-250-200-150-100-50050 2 deg. pulse applied between t = 1 and t =3 sec. (a) G /G53exte,/G64 /G2b/G2b De/G64hqwu ,,,, /G71 (b) Requires 0 <K h<∼10−6for stability, but yields very large oscilla- tory altitude variations. (c) Numbers like Kh<0.001 and Kq>300 yields somewhat reasonable altitude behavior ( altitude variations are stable and on the order of a few hundred feet), but settling time is on the order of tens of minutes. (d)Kh=0.005,Kθ= 25, and Kq= 10 (for example) yield maximum altitude deviation of less than 50ft, plus a settling time of about 3029 second. clear all, close all F = [ -0.00643 0.0263 0 -32.2 0; -0.0941 -0.624 820 0 0;-0.000222 -0.00153 -0.668 0 0;00100 ; 0- 108 3 00] ; G = [ 0; -32.7; -2.08; 0; 0 ];J=0 ;x0 = [ 0; 0; 0; 0; 0 ]; %p a r ta . H h=[00001] ;[n u m ,d e n]=s s 2 t f (F ,G ,H h ,J) ; sysa = tf( num, den ); T = 0:0.01:10;u = (2/180*pi)*rectpuls( (T-2)/2 );ya = lsim( sysa, u, T, x0 ); Þgure(1) plot(T,ya)title(’ 2 deg. pulse applied between t = 1 and t =3 sec. ’); Problems and Solutions for Section 2.3 11. A Þrst step toward a realistic model of an op amp is given by the equations below and shown in Fig. 2.41. Vout =107 s+1[V+−V−] i+=i−=0 (a) Find the transfer function of the simple ampli Þcation circuit shown using this model. Solution: (a) As i−=0 , Vin−V− Rin=V−−Vout Rf V−=Rf Rin+RfVin+Rin Rin+RfVout30 CHAPTER 2. DYNAMIC MODELS Figure 2.41: Circuit for Problem 11. Figure 2.42: Circuit for Problem 12. Vout =107 s+1[V+−V−] =107 s+1µ V+−Rf Rin+RfVin−Rin Rin+RfVout¶ =−107 s+1µRf Rin+RfVin+Rin Rin+RfVout¶ Vout Vin=−107Rf Rin+Rf s+1+1 07Rin Rin+Rf 12. Show that the op amp connection shown in Fig. 2.42 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 2.11. Solution:31 Figure 2.43: Circuit for Problem 13. Ideal case: Vin=V+ V+=V− V−=Vout Non-ideal case: Vin=V+,V−=Vout but, V+6=V− instead, Vout =107 s+1[V+−V−] =107 s+1[Vin−Vout] so, Vout Vin=107 s+1 1+107 s+1=107 s+1+1 07∼=107 s+1 07 13. Show that, with the non-ideal transfer function of Problem 2.11, the op amp connection shown in Fig. 2.43 is unstabl e.32 CHAPTER 2. DYNAMIC MODELS Figure 2.44: Op Amp circuit for Problem 14. Solution: Vin=V−,V+=Vout Vout =107 s+1[V+−V−] =107 s+1[Vout−Vin] Vout Vin=107 s+1 107 s+1−1=107 −s−1+1 07∼=−107 s−107 The transfer function has a denominator with s−107,and the minus sign means the exponential time function is increasing, which means that it has an unstable root. 14. A common connection for a motor power ampli Þer is shown in Fig. 2.44. The idea is to have the motor current follow the input voltage and theconnection is called a current ampli Þer. Assume that the sense resistor, R sis very small compared with the feedback resistor, Rand Þnd the transfer function from VintoIa. Solution:33 [Note to Instructors: You might want to assign this problem with Rf=∞,meaning no feedback from the output voltage.] At node A, Vin−0 Rin+Vout−0 Rf+VB−0 R= 0 (130) At node B, with Rs¿R Ia+0−VB R+0−VB Rs= 0 (131) VB=RRs R+RsIa VB≈RsIa The dynamics of the motor is modeled with negligible inductance as Jm¨θm+búθm=KtIa (132) JmsΩ+bΩ=KtIa At the output, from Eq. 131. Eq. 132 and the motor equation Va= IaRa+KesΩ Vo=IaRs+Va =IaRs+IaRa+KeKtIa Jms+b Substituting this into Eq.130 Vin Rin+1 Rf· IaRs+IaRa+KeKtIa Jms+b¸ +IaRs R=0 This expression shows that, in the steady state when s→0,the current is proportional to the input voltage. If fact, the current ampli Þer normally has no feedback form the output voltage, in which case Rf→∞ and we have simply Ia Vin=−R RinRs 15. An op amp connection with feedback to both the negative and the positive terminals is shown in Fig 2.45. If the op amp has the non-ideal transfer function given in Problem 11, give the maximum value possible for the positive feedback ratio, P=r r+Rin terms of the negative feedback ratio, N=Rin Rin+Rffor the circuit to remain stable.34 CHAPTER 2. DYNAMIC MODELS Figure 2.45: Op Amp circuit for Problem 15. Solution: Vin−V− Rin+Vout−V− Rf=0 Vout−V+ R+0−V+ r=0 V−=Rf Rin+RfVin+Rin Rin+RfVout =( 1−N)Vin+NVout V+=r r+RVout=PVout Vout =107 s+1[V+−V−] =107 s+1[PVout−(1−N)Vin−NVout]35 Vout Vin=107 s+1(1−N) 107 s+1P−107 s+1N−1 =107(1−N) 107P−107N−(s+1 ) =−107(1−N) s+1−107P+1 07N 0<1−107P+1 07N P<N +1 0−7 16. Write the dynamic equations and Þnd the transfer functions for the circuits s h o w ni nF i g .2 . 4 6 . (a) lead circuit (b) lag circuit. (c) notch circuit Solution: (a) lead circuit inVoutV CV 1R2RfR Vin−V R2+0−V R1+Cd dt(0−V) = 0 (133) Vin−V R2=0−Vout Rf(134) We need to eliminate V. From Eq. 134, V=Vin+R2 RfVout