1 x(k +1)=(Φ LH) x(k) = T 1 x 2 (k) x1 (0) 1 T x 2(0) T x 1 (0) x 2 (0) x(1) = x(2) = x(3) =


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1 567 This is often referred to as Þnite settling time or deadbeat design because the dynamics will settle in a Þnite number of sample periods. This estimator always drives the error to zero in time 2T or less as veriþed below. The estimatorerror equation : x(k +)=(Φ LH) x(k) = T x (k) T x 2 (k) For k =,, 2, 3,... x() = x(2) = x(3) = x () T x 2() T x () x 2 (). (c) Yes, since one state element is measured directly, one can estimate the other element with a Þrstorder estimator following the development for reduced order estimators for contintuous systems in Section From the discrete statespace representation, x 2 (k +)=Tx (k)+x 2 (k)+ T 2 2 u(k) = x (k) = T x 2 (k +) x 2 (k) T 2 2 u(k) {z } known measurement and input For K =, x () = x 2 () x 2 () T 2 T 2 u() = y() y() T 2 T 2 u() x 2 () = x 2 () = y() 4. Singleaxis Satellite Attitude Control: Satellites often require attitude control for proper orientation of antennas and sensors with respect to Earth.
2 568 CHAPTER 8. DIGITAL CONTROL Figure 8.23: Satellite control schematic for Problem 4 Figure 2.6 shows a communication satellite with a threeaxis attitudecontrol system. To gain insight into the threeaxis problem we often consider one axis at a time. Figure 8.23 depicts this case where motion is only allowed about an axis perpendicular to the page. The equations of motion of the system are given by where I θ = M C + M D, I = moment of inertia of the satellite about its mass center, M C =control torque applied by the thrusters, M D =disturbance torques, θ =angle of the satellite axis with respect to an inertial reference with no angular acceleration. We normalize the equations of motion by deþning and obtain u = M C I, w d = M D I, θ = u + w d. Taking the Laplace transform yields which with no disturbance becomes θ(s) = s 2 [u(s)+w d(s)], θ(s) u(s) = s 2 = G (s).
3 569 Inthediscretecasewhereuis applied through a ZOH, we can use the methods described in this chapter to obtain the discrete transfer function G (z) = θ(z) u(z) = T 2 z + 2 (z ) 2. (a) Sketch the root locus of this system by hand assuming proportional control. (b) Draw the root locus using MATLAB to verify the hand sketch. (c) Add a lead network to your controller so that the dominant poles correspond to ζ =.5 andω n =3π/(T ). (d) What is the feedback gain if T = sec? If T =2sec. (e) Plot the closedloop step response and the associated control time history for T =sec. Solution (a) The hand sketch will show that the loci branches depart vertically from z = ; therefore, the system is marginally stable or unstable foranyvalueofgain. (b) The Matlab version below conþrms the situation. G (z) = T 2 (z +) 2 (z ) 2 = K (z +) (z ) 2 (c) To obtain the desired damping and frequency, Fig. 8.4 shows that a root locus branch should go through the desired poles at
4 57 CHAPTER 8. DIGITAL CONTROL z =.44 ±.44j. After some trial and error, you can Þnd that this can be accomplished with the lead compensation : (z.63) D(z) =K (z +.44) The speciþc value of K that yields the closedloop poles are at : z =.44 ±.44j,.3 is K =.692 K. The secondorder pair give : ω n =.97 T ζ =.59 rad/sec which is close enough to the desired speciþcations. locus for the compensated design is: The root (d) K =.692 T 2 2 ½ ¾.383 for T =sec =.3458 for T =2sec
5 u theta 57 (e) Closedloop step response : ClosedLoop Step Response (T=s) Time (sec).5 Time History of Control Effort (T=s) Time (sec) 5. In this problem you will show how to compute Φ by changing states so
6 a b (c) Let T = c d, TA = FT = 2a = c, b = d 573 = Choosing c = d =,thena =.5, b= c = d =.5 = T = This T satisþes TAT = F. Note : Let T = v v 2, TFT = A = v v 2 F v v 2 = 2 ½ ¾ Fv =( )v = Fv 2 =( 2)v 2 (d) Thus, the columns v and v 2 are the eigenvectors of F. e FT = Te AT T.5 = e T 2 2 e 2T 2 e = T +2e 2T e T e 2T 2e T +2e 2T 2e T e 2T 6. It is possible to suspend a mass of magnetic material by means of an electromagnet whose current is controlled by the position of the mass (Woodson and Melcher, 968). The schematic of a possible setup is shown in Fig. 8.24, and a photo of a working system at Stanford University is shown in Fig The equations of motion are mẍ = mg + f(x, I), where the force on the ball due to the electromagnet is given by f(x, I). At equilibrium the magnet force balances the gravity force. Suppose we let I represent the current at equilibrium. If we write I = I + i, expand f about x =andi = I, and neglect higherorder terms, we obtain the linearized equation mẍ = k x + k 2 i. () Reasonable values for the constants in Eq. () are m =.2 kg, k =2 N/m, and k 2 =.4 N/A. (a) Compute the transfer function from I tox, and draw the (continuous) root locus for the simple feedback i = Kx.
7 574 CHAPTER 8. DIGITAL CONTROL Figure 8.24: Schematic of magnetic levitation device for Problems6 and 7 (b) Assume the input is passed through a ZOH, and let the sampling period be.2 sec. Compute the transfer function of the equivalent discretetime plant. (c) Design a digital control for the magnetic levitation device so that the closedloop system meets the following speciþcations: t r.sec, t s.4 sec, and overshoot 2%. (d) Plot a root locus with respect to k foryourdesign,anddiscussthe possibility of using your closedloop system to balance balls of various masses. (e) Plot the step response of your design to an initial disturbance displacement on the ball, and show both x and the control current i. If the sensor can measure x only over a range of ±/4cm and the ampliþer can only provide a current of A, what is the maximum displacement possible for control, neglecting the nonlinear terms in f(x, I)? Solution: (a) G(s) = X(s) I(s) = k 2/m s 2 k /m 2 = s 2 (2)
8 575 (b) T =.2 sec, G(z) = ½ ¾ G(s) ( z )Z s = z (z.533)(z.8822) (c) The speciþcations imply that : t r. sec= ω n.8 = 8 rad/sec. t s.4 sec= σ 4.6 =.5 rad/sec.4 = r = z e.5.2 =.7945 ( z = e st ) M p 2% = ζ.48 Thus, the closedloop poles must be pulled into the unit circle near r =.8 andζ =.5. Using the template of Fig. 8.4, we experiment with lead compensation and select, D(z) = 6 z.533 z.93 The closedloop poles are : z =.75 ±.39j,.53 Performance : t r =.72 t s =.397 M p = 8.3% which meet all the speciþcations. The root locus is below.
9 x [m] 576 CHAPTER 8. DIGITAL CONTROL The step response shows M p 8 ClosedLoop Step Response Time (sec)
10 i [A] Time History of Control Effort Time (sec) (d) As can be seen from Eq. (2), the loop gain and the open loop pole locations depend on the mass of the ball. Changing the mass will therefore affect the dynamic characteristics of the system and may render it unstable. A root locus of the closedloop poles versus k shows how the locus changes as a function of the mass: The closedloop system becomes unstable for k 24. Since a small increase in k makes the system unstable and a decrease
11 i [A] x [m] 578 CHAPTER 8. DIGITAL CONTROL in m has the same effect on the system, it is difficult to balance balls of smaller masses. (e) The response to an initial x displacement is shown : ClosedLoop Initial Response Time (sec) Time History of Control Effort Time (sec)
12 579 The assumption here is that an allowable transient must stay in the range of the sensor and not require more than the limit of the current. From I(z) = D(z)( X(z)), we have a difference equation : i(k).93i(k ) = 6{x(k).533x(k )} For k =, i() = 6x(). We see that if x() = then i() = 6. Note that i() = D( )x(). Thus, if i is to be kept below A then x()mustbekeptbelow /6 =.862 m =.862 cm =.339 inch, which is greater than the sensor range. The current control can handle any displacement in the range of ±.25 inch. 7. In Problem 6 we described an experiment in magnetic levitation described by Eq. () which reduces to Let the sampling time be. sec. ẍ =x +2i. (a) Use pole placement to design a controller for the magnetic levitator so that the closedloop system meets the following speciþcations: settling time, t s.25 sec, and overshoot to an initial offset in x that is less than 2%. (b) Plot the step response of x, x, andi to an initial displacement in x. (c) Plot the root locus for changes in the plant gain, and mark the pole locations of your design. (d) Introduce a command reference input r (as discussed in Section 7.8) that does not excite the estimate of x. Measure or compute the frequency response from r to the system error r x and give the highest frequency for which the error amplitude is less than 2% of the command amplitude. Solution (Note: part (d) of this problem is a stretch for the students. The text doesn t cover the relevant discrete development and the student would be helped by referring to Digital Control of Dynamic Systems, by Franklin, Powell, and Workman.) (a) Statespace representation of the plant : x x x = =,y= x, u = i úx d x dt x 2 x 2 = = Fx + Gu x x u
13 y y 66 CHAPTER 8. DIGITAL CONTROL.2 Step Response (T=.).8 continuous tustin and MPZ time (sec).5 Step Response (T=.).5 Continuous Tustin and MPZ time (sec) 22. Repeat Problem 5.29 by constructing discrete root loci and performing the designs directly in the zplane. Assume that the output y is sampled,
14 67 the input u is passed through a ZOH as it enters the plant, and the sample rate is 5 Hz. Solution (a) The most effective discrete design method is to start with some idea what the continuous design looks like, then adjust that as necessary with the discrete model of the plant and compensation. We refer to the solution for Probelm 5.29 for the starting point. It shows that the specs can be met with a lead compensation, and a lag compensation, (s +) D (s) =K (s + 6) D 2 (s) = (s +.4) (s +.32). Although it is stated in the solution to Problem 5.29 that a gain, K = 24 will satisfy the constraints, in fact, a gain of about K = 27 is actually required to meet the rise time constraint of t r.4 sec. So we will assume here that our reference continuous design is (s +) (s +.4) D (s) = 27 (s + 6) (s +.32) It yields a rise time, t r =.38,Mp = 5%, and Kv = (27)(.4 6 )(.32 )= 56. So all specs are met. For interest, use of the damp function shows that ω n =6.4 rad/sec for the dominant roots, and those roots have a damping ratio, ζ =.7. For the discrete case with T =5Hz, we should expect some degradation in performance, especially the damping, because the sample rate is approximately 5 ω n. The discrete transfer function for the plant described by G(s) and preceeded by a ZOH is: ½ ¾ G(s) G(z) = ( z )Z s = z Z z This is most easily determined via Matlab, sysc = tf([],[ ]); T=/5; sysd = c2d(sysc,t, zoh ); which produces: ½ s(s +)(s + ) ¾
15 68 CHAPTER 8. DIGITAL CONTROL (z +3.36)(z +.22) G(z) =.4424 (z )(z.9355)(z.534 The essential elements of the compensation are that the lead provides velocity feedback with a T D = and the lag provides some high frequency gain. The discrete equivalent of the proportional plus lead would be (see Eq. 8.42): D (z) =K( + T D T ( z )) = K ( + T D/T )z T D /T z which for T =/5 =.667 and T D = reduces to 6z 5z D (z) = 27 z = 432 z z The lag equivalent is best introduced by use of one of the approximation techniques, such as the matched polezero: D 2 (z) = z e.4t z.9737 = z e.32t z.9979 as its whole function is to raise the gain at very low frequencies for error reduction. Examining the resulting discrete root locus and picking roots with rlocfind to yield the required damping shows that the gain, K = 6. While the use of damp indicates that the frequency and damping are acceptable, the time response shows an overshoot of about 2% and the rise time is slightly below spec. We therefore need to increase the lead (move the lead zero closer to z =+)to decrease the overshoot and increase gain to speed up the rise time. Several iterations on these two quantities indicates that moving the lead zero from z =.9375 to z =.96 and increasing the gain from K = 6toK = 65 meets both specs. The velocity coefficient is found from and is also satisþed. K v = lim (z )D(z)G(z) z Tz ThetimeresponseoftheÞnal design below shows that all specs are met.
16 y 69.2 Step Response time (sec) 23. Design a digital controller for the antenna servo system shown in Figs and3.64anddescribedinproblem3.3. Thedesignshouldprovideastep response with an overshoot of less than % and a rise time of less than 8 sec. (a) What should the sample rate be? (b) Use emulation design with the matched polezero method. (c) Use discrete design and the zplane root locus. Solution (a) The equation of motion is : J θ + Bθ = T c where J = 6 kg.m 2,B= 2 N.m.sec If we deþne : a = B J = 3,U= T c B after Laplace transform, we obtain : G(s) = θ(s) u(s) = s(3s +)
17 y 69.2 Step Response time (sec) 23. Design a digital controller for the antenna servo system shown in Figs and3.64anddescribedinproblem3.3. Thedesignshouldprovideastep response with an overshoot of less than % and a rise time of less than 8 sec. (a) What should the sample rate be? (b) Use emulation design with the matched polezero method. (c) Use discrete design and the zplane root locus. Solution (a) The equation of motion is : J θ + Bθ = T c where J = 6 kg.m 2,B= 2 N.m.sec If we deþne : a = B J = 3,U= T c B after Laplace transform, we obtain : G(s) = θ(s) u(s) = s(3s +)
18 6 CHAPTER 8. DIGITAL CONTROL From the speciþcations, µ M p < % = M P = ζ = ζ >.54.6 t r < 8 sec = t r.8 = < 8 = ω n = ωbw >.225 ω n Note that ω pn = /3 < ωn. If designing by emulation, a sample rate of 2 times the bandwidth is recommended. If using discrete design, the sample rate can be lowered somewhat to perhaps as slow as times the bandwidth. However, to reject random disturbances, best results are obtained by sampling at 2 times the closedloop bandwidth or faster. Thus, for both design methods, we choose : T = sec ω s =.628 rad/sec, which is > 2ω n =.45 rad/sec (b) Continuous design : Use a proportional controller : u(s) =D(s)(θ r (s) θ(s)) = K(θ r (s) θ(s)) Root locus : Choose K =.2. The closedloop pole location in splane : s =.67 ±.25j The corresponding natural frequency and damping : ω n =.265, ζ =.6299
19 6 Digitized the continuous controller with matched polezero method : Performance : D(z) =.2 T c (z) = Bu(z) = 42(θ r (z) θ(z)) M p =.9 t r = 67.3 sec (c) With u(k) applied through a ZOH, the transfer function for an equivalent discretetime system is : where z + b G(z) =K (z )(z e at ) K = at +e at,b= e at at e at a at +e at = z G(z) =.4959 (z )(z.765) Use a proportional control of the form : Root locus : D(z) =K
20 62 CHAPTER 8. DIGITAL CONTROL The speciþcation can be achieved with the proportional control. However, we try to achieve the same closedloop poles as the emulation design (part (b)) for comparison. These closedloop pole locations are denoted by + in the root locus. UseaPDcontroloftheform: D(z) =K z α z Root locus : Choose K =.294, α =.3. The resulting zplane roots : z =.828 ±.725j,.65 This corresponds to the splane roots : s =.67±.25j (the design point of emulation design),.44 which satisfy the speciþcation : The control law : ω n =.265,.44 ζ =.632,. D(z) =.294 z.3 z T c (z) = Bu(z) = 588 z.3 z
21 u (Tc/B) x (m) 63 Performance : M p =.79 t r = 7.3 sec Step response :.2 Step Response emulation design discrete design time (sec).4 Time History of Control Effort.3 discrete design.2 emulation design time (sec) 24. The system G(s) = (s +.)(s +3)
22 64 CHAPTER 8. DIGITAL CONTROL is to be controlled with a digital controller having a sampling period of T =. sec. Using a zplane root locus, design compensation that will respond to a step with a rise time t r secandanovershootm p 5%. What can be done to reduce the steadystate error? Solution (a) Continuous plant : G(s) =, Type system (s +.)(s +3) Discrete model of G(s) preceeded by a ZOH (T =. sec) : z +.99 G(z) =.45 (z.748)(z.99) SpeciÞcations : t r sec ω n.8 rad/sec M p 5% ζ.7 Discrete design : A simple proportional feedback, D(z) = K = 4., will bring the closedloop poles to : which are inside the specs region. Root locus : z =.8564 ±.278j ω n =2.7 rad/sec, ζ =.7
23 u y 65 Step response : ClosedLoop Step Response Time (sec) Time History of Control Effort Thestepresponseshowsthat: Time (sec) t r =.2 sec M p = 4.7% However, since the system is type, steadystate error exists and is 7% in this case. An integral control of the form, D(z) = K T I Tz z
24 66 CHAPTER 8. DIGITAL CONTROL can be added to the proportional control to reduce the steadystate error, but this typically occurs at the cost of reduced stability. 25. The transfer function for pure derivative control is z D(z) =KT D Tz, where the pole at z = adds some destabilizing phase lag. Can this phase lag be removed by using derivative control of the form (z ) D(z) =KT D? T Support your answer with the difference equation that would be required, and discuss the requirements to implement it. Solution: (825) (a) No, we cannot use derivative control of the form : D(z) =KT D z T to remove the phase lag. The difference equation corresponding to D(z) =K p T D z T = U(z) E(z) is e(k +) e(k) u(k) =K p T D T This is not a causal system since it needs the future error signal to compute the current control. In real time applications, it is not possible to implement a noncausal system.
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