Lecture Notes on Systems and Control for ASEN 2003 Dynamics and Systems (Spring 2002)

Transcription

1 STUDY MATERIALS FOR THE NEXT TWO WEEKS: You have one or both of the following two options in preparing for the next two weeks in control. There will be a reading quiz next Tuesday, 26 February 22. Option : Read the following in Modeling and Simulation of Dynamics Systems that is reserved in Engineering Library.. Read pages 23-2 of Chapter Read pages of Chapter Read pages of Appendix E. 4. Read pages of Appendix E. Option 2: Read your instructor s lecture note that begins in the next page. Welcome! Now you have made a courageous choice to read the note in lieu of the materials in the reserved book. Here re steps you may follow to ease your pain in reading through this note..first reading: Skip the mathematical details and concentrate on the narratives and figures. 2.Second reading: Concentrate on Sections, 2, and 3; continue to read the rest of the sections while paying attention only to what each section is devoted to. 3.Third reading: Devote yourself to the Laplace transform section while going over equations (9), (), (4)-(2). Then, continue to read section 5. Finish sections 6 and 7 but you don t need to grasp everything contained in those sections. 4.Fourth reading: Comb through section -6 and you are now ready to converse with your instructor on the subject of PPD control system! 5.MATLAB Practice: Save each MATLAB code and run each at a time to see what you will be able to do with those codes. Now you are beginning to practice what you have read in the note. Option 3: Read both the recommended sections in the reserved book and this lecture note.

2 LECTURE NOTES ON SYSTEMS AND CONTROL. Introduction A system in this class denotes either a vehicle or a device. Rarely a typical modern device or vehicle is operated without controls. Control in mechanical systems is invoked to achieve, to name a few, one or more of the following objectives: to move a mechanical arm from one position to the final position; to attenuate vibrations in the structure; to regulate the fluid flow in time; and, to position a satellite in space in a specific attitude and orientation. Human heart valves are an example of remarkable evolutionary control design. There has been an intense effort by scientists, medical professionals and engineers over the years to imitate the human heart valves but none has been as good and reliable as the ones in our hearts. Traditionally, the main part of a mechanical system was designed with minimal control. Control devices were then added or integrated into the system. The steam engine of James Watts is perhaps the first example to which a controller named governor was introduced to regulate the steam pressure, which in turn regulated the speed of a turbine engine. For this historical reasons and for the simplicity of mathematical modeling, the concept of systems has been used to model a complex mechanical design with controls. It goes like this: Identify the building blocks of a system under consideration, characterize each building block by a mathematical model, arrange the building blocks in such a way to reflect the functionality or information flow or input/output relations of the system. Once a total mathematical system model is constructed using the building block models, then desirable control element(s) are added into the system model ( imagine when the steering wheel (a controller) was finally assembled into Ford s 98 Model T). 2. Model Equation for Control So far we have studied a damped spring-mass model equation described as m ẍ(t) + c ẋ(t) + kx(t) f (t) () where m is the mass, c is the damping coefficient; k is the linear spring constant; x(t) is the displacement of the mass point measured from its static equilibrium position; ẋ(t) and ẍ(t) are the velocity and acceleration, respectively; and, f (t) is the applied force. Note that the applied force is a function of time only, for which we have considered two categories as given by Sinusoidal forcing function: f (t) P m sin (ω f t) Step input forcing function: f (t) P m u(t), u(t) {, t <, t > (2) Suppose now the model equation is undamped (i.e., c ), and a step input forcing function is applied. This occurs when a robotic arm is to be moved from its original position, say, θ o toa 2

3 Robotic Arm Slewing Problem Final desired position Κ θ /2 J θ Κ θ /2 Initial position Fig. A robotic arm slewing problem desired position, θ D π in an assembly plant as shown in Fig.. The governing equation for this can be expressed from () and (2) as Uncontrolled model equation J θ(t) + K θ θ(t) M m u(t) (N.m) of a robot arm: M m K θ θ D (N.m) (3) where J θ is the rotational moment of inertia (Kgm 2 ), K θ is the torsional spring constant (Nm/rad), and M m is the the applied moment chosen to be M m K θ θ D (Nm) such that the final desired rotation will be θ D. 7 6 Response of undamped system under unit step input ω n 3.46, ζ. 5 Rotation angle (radian) Desired final angular position Time (sec) Fig. 2 Response of a robotic arm with no system damping 3

4 Dividing the above equation by J θ and substituting M m by K θ θ D, we arrive at the following equation: θ(t) + ω 2 n θ(t) ω2 n θ D u(t) (rad/sec2 ), ω n K θ /J θ (rad/sec) (4) Figure 2 illustrates the response of the arm angle governed by the model equation(4) with the following choice of parameters: ω n π(rad/s), θ D π (5) Observe that the arm oscillates around the desired angle π. Hence, the robotic arm cannot reach the desired position π without system damping. The reality is that the robotic arm is designed with a fine quality joint which is almost devoid of damping. 3. The Derivative Control In order to have the arm to reach the desired position and settle at that position without prolonged oscillations, it is necessary to inject damping into the system. This active damping is called in the control community as derivative control as the control torque, f d, must be proportional to the angular velocity θ or the velocity ẋ. Mathematically, a derivative control is expressed as Derivative control torque : f d ( θ) K d θ 2ζ d J θ ω n θ, K d 2ζ d J θ ω n (Nm) (6) where K d is parameterized in terms of the derivative control damping ratio ζ d. When this derivative controller is applied to the undamped system(3), the resulting model equation becomes J θ(t) + K θ θ(t) K θ θ D u(t) + f d ( θ) (N.m) with f d ( θ) K d θ (7) J θ(t) + K θ θ(t) K θ θ D u(t) K d θ (rad) Finally, moving the term K d θ in the above equation to the left-hand side and dividing by J θ,we obtain: Derivative control system: θ(t) + 2ζ d ω n θ(t) + ωn 2 θ(t) ω2 n θ D u(t) (rad) (8) where K d 2ζ d J θ ω n from (6) is used. It should be recalled that θ D in the above equation is the desired final angular position of the arm. 4

5 5 Response of damped system under unit step input Rotationa angle (rad) ω 3.46,.25 n ζ d Desired final angle Time (sec) Fig. 3 Response of a robotic arm with actively controlled damping 8 Active damping force for position slewing problem under unit step input Active damping force /(2 ζ J θ ω ) n ω 3.46,.25 n ζ d Time (sec) Fig. 4 Active damping torque vs. time Figure 3 shows the response of the actively damped model equation(8) with the control-damping ratio ζ d.25. Note that the arm settles to the desired position (θ D π) after about 6 seconds. 5

6 To gain insight into the role of the active damping torque f d 2ζ d J θ ω n history of f d vs time as shown in Figure 4. θ, let us plot the time It is interesting to realize that the active damping torque f d is proportional to the angular velocity θ, which is generated in practice by a variable torque motor. However, without measuring the angular velocity at every instant during the slewing maneuver, the active damping torque motor cannot generate the desired control torque as shown in Fig. 4. In other words, the angular velocity must be measured and fed back into the active torque generating motor at every instant during the arm maneuvering. For this reason, the mathematical model equation (7)or (8) is called a feedback system model equation. In order for us to take advantage of feedback system theory, its design experience, and data processing, and real-time control which we will carry out during the upcoming control experiments, it is necessary for us to be acquainted with the mathematical tools well suited for system synthesis and design. We will devote the next two sections for a tutorial of basic mathematical tools widely used in the control community. 4. The Laplace Tranform So far in the previous section, the three figures that we have generated are characterized as functions of time, viz., a time-domain approach. Suppose now you want to see the effects of changes in system model parameters on the system performance, such as J θ, K θ, θ D, and the active control (feedback) torque damping ratio ζ d. The system performance may include how fast your customer like to maneuver the arm and how rapidly the initial oscillations should be dampened out. If you were to employ the time domain approach, you may need to repeat the corresponding time-domain simulations for every parameter change as well as maneuvering strategy. Fortunately, there is an alternative approach that we can utilize with clarity in carrying out the feedback system design synthesis without having to resort to time-domain approach: the Laplace transform. As this is a mathematical tool, we summarize its essential features that are pertinent to control system analysis and synthesis that we will study in this course. The Laplace transform of a time-domain function x(t) is defined by L [x(t)] e st x(t) dt X (s) (9) where the symbol L is called the Laplace transform operator. Using this definition, we transform ẋ(t) as L [ẋ(t)] e st ẋ(t) dt () To carry out the necessary integration of the above transform, we utilize the following integrationby-part formulas for two functions (g, p): 6

7 d(gp) dg dt dt g dp d(gp) dt dt g dp dt dt g dp dt p + g dp dt dg dt p d(gp) dt dt dt (gp) t t dg dt Now, let us apply the last expression of the above equation with dg dt pdt pdt () {g e st, dg dt se st } and {p x(t), dp dt ẋ(t)} (2) and carry out the indicated integration of (): L [ẋ(t)] e st ẋ(t) dt, [ Us the formula: L [ẋ(t)] (e st x(t)) t t ( s e st )x(t) dt, L [ẋ(t)] {e s x( ) (e x()) }+s L [ẋ(t)] x() + s [ e st x(t) dt g dp dt dt ] [Formula used: (gp) t t e st x(t) dt], as e s and e dg dt pdt] (3) Note that the term in the bracket([])in the aboveequation is, by definition of the Laplace transform (9), the very Laplace transform of x(t), viz., L [x(t)] X (s). Hence, we have L [ẋ(t)] e st ẋ(t) dt x() + s L [ẋ(t)] sx(s) x() (4) For the Laplace transform of the acceleration ẍ(t), applying the above process twice we obtain L [ẍ(t)] e st ẍ(t) dt s 2 X (s) sx() ẋ() (5) 7

8 We now summarize the three Laplace transform formulas: L [x(t)] X (s) L [ẋ(t)] sx(s) x() L [ẍ(t)] s 2 X (s) sx() ẋ() (6) Let us now apply the above three formulas to Laplace-transform the damped spring-mass model equation (). To this end, we apply the Laplace transform to both sides of () to read: L [mẍ(t) + cẋ(t) + kx(t)] L [ f (t)] (7) First, we decompose the Laplace transform operator L to each of the three terms in the left-hand side of the above equation: m L [ẍ(t)] + c L [ẋ(t)] + k L [x(t)] L [ f (t)] (8) Applying the three formulas (6), we have m {s 2 X (s) sx() ẋ()}+c {s X(s) x()}+kx(s) F(s) (s 2 m + sc + k) X (s) F(s) + sx() +ẋ() (9) where we have used the Laplace transform definition given by () to obtain L [ f (t)] F(s) (2) For most applications and control systems experiments we will be carrying out in this course, we may assume that the initial displacement (angle) and velocity (angular velocity) are zero, i.e., x() ẋ(). For this special case equation (9) may be expressed as X (s) F(s) G(s), G(s) (s 2 m + sc + k) (2) where G(s) is called the transfer function (TF). Finally, if desired, one may obtain the time-domain solution of the Laplace transform equation by the following inverse Laplace transform definition: x(t) L [X (s)] e st X (s) ds (22) As an example, the time-domain solution of X (s) given in (2) can be obtained, first, by expressing X (s) as X (s) G(s) F(s) (23) 8

9 TABLE : LAPLACE TRANSFORM PAIRS x(t) X (s) δ(t) (Dirac delta function) u(t) (unit step function) s t n n! s (n+) e ωt s+ω sin ωt ω s 2 +ω 2 cos ωt s s 2 +ω 2 ω d e ζωt sin ω d t, ω d ( ζ 2 )ω s 2 +2ζωs+ω 2 e ζωt [ cos ω d t + ζω ω d sin ω d t ], ω d ( ζ 2 )ω s+2ζωs s 2 +2ζωs+ω 2 and substituting the above definition into (22) to obtain x(t) L [X (s)] e st {G(s) F(s)} ds (24) Several useful formulas for the Laplace transform and the inverse Lapalce transform pairs are listed in Table. The Laplace transform pairs listed in Table can be used as follows. Take x(t) sin ωt. From 9

10 Table, we have (look for the corresponding second column entry of sin ωt) L [x(t)] X (s) L [sin ωt] ω s 2 + ω 2 (25) The inverse Laplace transform of X (s) ω column entry of )by s 2 +ω 2 ω s 2 +ω 2 is likewise given (pick the corresponding first L [X (s)] x(t) L ω [ ] sin ωt (26) s 2 + ω2 The Laplace transform and inverse Laplace transform of other functions can be similarly obtained using Table. 5. Application of the Laplace transform: Frequency Response Function The time-domain responses shown in Figs. 2 and 3 are the solutions of the undamped model equation (3) and the derivative control model equation (8), respectively. Note that the forcing function for the two problems is characterized as a step input, f (t) K θ θ D. Hence, if different forcing functions were applied to the two model equations, one needs to carry out new time-domain response analyses. For the control system designers as well as mechanical system designers, it is important to grasp the intrinsic characteristics of the system that is independent of different forcing functions. The Laplace transform given by (2) provides the intrinsic characterics of the system without having to solve for different forcing functions. That equation is recalled below: X (s) F(s) G(s), G(s) (s 2 m + sc + k) (2) where G(s) is called the system transfer function. Observe that the transfer function G(s) is independent of the forcing function F(s) as well as the initial conditions, (x(), ẋ()). The transfer functions provide a convenient means for the control systems designers to analyze, synthesize and evaluate the overall system performance. For example, by examining the roots of G(s) one can determine whether the system is stable, how fast the system responds, and the overshoot the system will exhibits, etc. In this course, we will restrict ourselves to one usage: the overshoot of the system which indicates how much the arm will swing beyond the desired final position. To do this, we first substitute the Laplace variable s by s jω f (27) where j is the imaginary number j 2 and ω f is the frequency of the forcing function f (t). Substituting this into the transfer function G(s) in (2) we obtain H(ω f ) G( jω f ) ( ω 2 f m + jω f c + k) (28)

11 which can be expressed, with ω 2 n k/m and c 2ζω nm,as H(ω f ) /k ( ω 2 f m/k + jω f c/k + ) [divide by k both nominator and denominator] /k ( ω 2 f /ω2 n + jω f 2ζω n /(k/m) + ) [substitute c 2ζ mω n and factor k m ] /k ( ω 2 f /ω2 n + jω f 2ζω n /ωn 2 + ) [substitute k/m by ω 2 n ] H(ω f ) /k ( ω 2 f /ω2 n + j 2ζω f /ω n + ) [final desired form!] (29) 3 Frequency response function of a damped system 2.5 Magnification factor 2.5 ζ. ζ.2 ζ.4.5 ζ.6 ζ Frequency ratio ( ω f / ω n ) Fig. 5 Frequency response function of a damped system An important application of the frequency response function H(ω f ) is to predict the maximum amplitude when the system is subjected to a persistent frequency-dependent forcing function such as due to the engine excitations in vehicles and airplane. For this case, instead of plotting H(ω f ) given by (29), we normalize it by the static amplitude. Note that H(ω f ) becomes for ω f : H() /k ( /ω 2 n + j 2ζ /ω n + ) /k (3)

12 Physically, the value of H() corresponds to the static deflection subjected to a unit applied force. In other words, when the forcing function is independent of time, the applied force is a static force! The frequency response function normalized by its static magnitude is called the amplification factor A(ω f ), which is written as A(ω f ) H(ω f ) H() ( ω 2 f /ω2 n + j 2ζ(ω f /ω n ) + ) (3) Figure 5 shows the amplification factor vs. the frequency ratio (ω f /ω n ). Observe that A(ω f /ω n ) indicates the static equilibrium line. Hence, the amount of overshoot over the static equilibrium deflection is easily determinedd without having to simulate the time-domain responses for different damping ratios. It is this information that the frequency response functions can provide. A MATLAB code that has generated the curves is listed below. % FRF.m % % computes FRFs using the transfer function generated by % the laplace transform method % % programmed by kc park, 4 feb 22 % clear all; % input data % omega_n : natural frequency omega_n pi; % Define some useful numbers: Hz2rps2*pi; rps2hz/2/pi; % Conversions to/from Hz and rad/s omega_hertz omega_n* rps2hz; % loop over different damping ratio, zeta for zeta:.2:.8, % construct the laplace domain transfer function % tf generates the transfer function sys tf (, [/omega_n^2 2*zeta/omega_n ]) % pick a range of frequencies to look at from. Hz to Hz % with discrete point data wf logspace(-2,, 5)*Hz2rps; % Find the frequency response Hfreqresp(sys,wf); % This ends up being a 3 dimensional array! Hreshape(H,size(wf)); % this makes it the same size as wf % Plot the amplitude as a function of frequency in Hz 2

13 figure() % semilogx(wf*rps2hz/omega_hertz,abs(h)), plot(wf*rps2hz/omega_hertz,abs(h)), grid on; xlabel( Frequency ratio ( omega_f / omega_n ) (Hz) ) ylabel( Magnification factor ) title( Frequency response function of a damped system ); legend([ omega_n, num2str(omega_n),, zeta, num2str(zeta)]); hold on; end; % adjust the curve amgnitude axis([. 3. 3]); % end of the program 6. Application of the Inverse Laplace transform One problem of interest for this course is the time-domain solution of the robot arm maneuvering subjected to a step input as governed by equation (8). When that equation is divided by ω 2 n, the following normalized equation results: Derivative control system (normalized) : ω 2 n θ(t) + 2ζ d ω n θ + θ(t) θ D u(t) (rad) (32) The Laplace transform of the above equation with the initial conditions (θ() θ() ) can be obtained by using (6) and Table as Derivative control system: X (s) G(s) F(s) G(s) ω 2 n s 2 + 2ζ d ω n s + (33) F(s) θ D s In order to obtain the time-domain response from the Laplace-transformed equation such as equation (33), there are two approaches available to us: a MATLAB-based solution and analytical solution as described below. 6. Solution by MATLAB Solution via MATLAB uses two functions, tf (num.den) and step (sys, time) whose usage can be found by typing help tf and help step on MATLAB command window. The MATLAB function tf generates the Transfer Function (TF). For example, Figure 3 have been generated using the following MATLAB code: % response_step.m % 3

14 % obtain the response histories for damped system under step input % using the transfer function generated by the Laplace transform method % % programmed by kc park, 4 feb 22 % clear all; % input data % omega_n : natural frequency % zeta_d : damping ratio % T_final : response time desired % theta_d : final position omega_n pi; zeta_d.25; T_final (2*pi/omega_n)*5 theta_d pi; % five periods % construct the laplace domain transfer function % tf generates the transfer function % define a system in transfer function form sys tf (, [/omega_n^2 2*zeta_d/omega_n ]) % generate unit step response [y, time] step(sys, T_final); % multiply the output y by the final position angle, theta_d ytheta_d*y; figure() plot(time, y); grid on; xlabel( Time (sec) ) ylabel( Displacement ) title( Response of damped system under unit step input ); legend([ omega_n, num2str(omega_n),, zeta, num2str(zeta_d)]); % end of the program You are strongly encouraged to experiment with the above code or your own code that produces the time-domain solution of a system with a derivative control torque. 6.2 Analytical Solution via the Inverse Laplace transform You may skip this section in your initial reading! In orde rto obtain the analytical time-domain solution of θ(t) from the Laplace-transformed equation (33), we express X (s) as 4

15 X (s) G(s) F(s) s 2 + 2ζ d ω n ω 2 n ω 2 n s + θd s s 2 + 2ζ d ω n s + ω 2 n θd s (34) s 2 +2ζ d ω n s+ω 2 n By consulting with Table, we find that we have formula for and in the third and s seventh row, respectively. This suggests that we need to expand (34) into the following form: ω 2 n X (s) θd s 2 + 2ζ d ω n s + ωn 2 s as + b + c s 2 + 2ζ d ω n s + ωn 2 s (as + b)s + c(s2 + 2ζ d ω n s + ωn 2) (s 2 + 2ζ d ω n s + ωn 2) (s) (a + c)s2 + (b + 2ζ d ω n c) s + ωn 2c (s 2 + 2ζ d ω n s + ωn 2) (s) where (a, b, c) are constants to be determined while satisfying the partial fraction expression identity. Comparing the first and the third expressions, we find that (35) (36) (a + c), (b + 2ζ d ω n c), ωn 2 θ D ωn 2 c a θ D, b 2ζ d ω n θ D, c θ D Substituting these coefficients into the second expression of (35), we find: X (s) (s + 2ζ dω n )θ D s 2 + 2ζ d ω n s + ωn 2 + θ D s (37) X (s) θ D s Note that from the third row of Table that (s + 2ζ dω n )θ D s 2 + 2ζ d ω n s + ω 2 n L [ θ D s ] θ D u(t) (38) 5

16 and from the last row of Table we find L (s + 2ζ d ω n )θ D [ s 2 + 2ζ d ω n s + ωn 2 ] e ζωt ( cos ω d t + ζ dω n sin ω d t )θ D, ω d ( ζd 2 ω )ω n (39) d Hence, the inverse Laplace transform of X (s) of (37) is obtained by combining (38) and (39) as x(t) [u(t) e ζ dω n t ( cos ω d t + ζ dω ω d sin ω d t ) ] θ D (4) A MATLAB code has been written to compute x(t) vs. time using the analytical solution given by (4) below. % response_analytical.m % step input response of a damped spring-mass system % by the analytical solution % %input data % zeta_d : damping coefficient c 2 * zeta * m (or J) * omega_n % omega_n : natural frequency omega_n % theta_d : finbal position desired omega_n pi; zeta_d.25; theta_d pi; % final position to reach % time to be simulated T_total (2*pi/omega_n) * 6; % ten times the undamped period increment T_total/5; % time increment response zeros(,); % storage for response timezeros(,); % time axis to be stored for t :increment: T_total, omega_d sqrt(-zeta_d^2) * omega_n; omegad_t omega_d*t; factor zeta_d*omega_n/omega_d; x - exp(-zeta_d*omega_n*t) * (cos(omegad_t) + factor*sin(omegad_t)); response[response x]; time[time t]; end; % multiply theta_d to response to get the total position vs time response theta_d * response; figure() plot(time, response); xlabel( Time ); 6

17 ylabel( Response ); title( Response of damped spring-mass system under step input ); legend([ natural frequency, num2str(omega_n),, dampin ratio,... num2str(zeta_d) ]); grid on; % end of the code Naturally, the figure generated by the analytical solution (4) is identical with the MATLAB-based solution. 7. Proportional Plus Derivative (PPD) Control The derivative control introduced in (8) achieves an important objective: it brings the arm to the desired position as it effectively modifies the undamped model equation (3) to the damped model equation (8). A closer comparison of Figures 2 and 3 reveals that by introducing damping the response period is increased! This can be seen by comparing the periods of undamped vs. damped cases: τ damped τ undamped where we have used ω d 2π ω d 2π ω n ω n ω d For example, if ζ d / 2,wehave ω n ζ 2 d ω n ζ 2 d ω n from Table or (39). τ damped τ undamped ζd 2 >, ζ d > (4) (42) That is, the period of the derivative control system is increased by over 4 percent! Other cases are illustrated in Figure 6. As the damping is further increased, the longer it will take for the arm to reach the desired position. In practice, regardless of the natural period of the arm it is highly desirable within physically feasible means to have faster maneuvering speed. This objective is accomplished by requiring that the control motor not only produces the output torque proportional to the angular velocity but also proportional to the angular position itself. Mathematically, this is accomplished by introducing the active control torque as f p K p (θ D θ(t))u(t), K p > (43) Note that a major difference between the derivative control f d given in (6) and the proportional control f p given in the above equation is that the torque constant K p is proportional to the error or the difference betwen the target position θ D and the present position θ(t). When this controller f p (θ) is applied to the derivative control system as given by (6), the resulting model equation reads: 7

18 2.5 Period elongation of damped system compared to undamped system 2 Period ratio Damping ratio Fig. 6 Period elongation of damped system compared to undamped system J θ(t) + K θ θ(t) K θ θ D u(t) + f d ( θ)+ f p (θ), f d K d θ(t) (Nm) (44) where f d has the same form as given in (6), but not necessarily the same magnitude as denoted by K d Substituting f p from (43), we obtain the so-called proportional plus derivative (PPD) control system, similar to equation(6), as J θ(t) + K d θ(t) + (K θ + K P )θ(t) (K θ + K p )θ D u(t) (Nm) (45) Dividing both sides of the above equation by (K θ + K p ), we obtain PPD control θ(t) + 2ζ ppd θ(t) + θ(t) θ ω n 2 ω D u(t) (rad) n (46) system: ω n 2 (K θ + K p )/J θ, K d 2 ζ ppd J θ ω n where ω n is the natural frequency of the PPD system, and ζ ppd is the damping ratio of the PPD system. Let us compare the PPD control system (46) with the derivative control system given by 8

19 Derivative control system: ω 2 n θ(t) + 2ζ d θ(t) + θ(t) θ ω D u(t) (rad) n ζ d K d /(2J θ ω n ) (8) Suppose we would like to have both systems to have the same amplification factor A(ω f /ω n ) (3). It can be shown that the maximum amplification factors for two systems are given, respectively, by Derivative system: max( A d ), ζ d < / 2 2ζ d ( ζd 2) K θ ω n, ζ d K d /(2J θ ω n ) J θ PPD system: max( A ppd ), ζ ppd < / 2 2ζ ppd ( ζppd 2 ) K θ + K p ω n, ζ ppd K d /(2J θ ω n ) J θ (47) The two systems will have the same amplification factor if ζ d ζ ppd K d 2J θ ω n K d 2J θ ω n K d K θ + K p K θ K d (48) which means that the control torque f d for the PPD case must be larger than required for the derivative control case. However, the maneuvering time is faster with the PPD system than the derivative control system. This is because we have with ζ d ζ ppd τ ppd τ d ω d ω ppd ( ζ 2 d )ω n ( ζ 2 ppd ) ω n ω n ω n + K p /K θ < (49) Hence, by judiciously selecting the proportional gain K p and the derivative control parameter K d ( ζ ppd ), the PPD system may achieve both faster response than is possible by the derivative control system, while achieving the same overshoot. This is illustrated in Figure 7. 9

20 5 4.5 Comparison of derivative and PPD control under unit step input w 3.46,.25, n z d Rotationa angle (rad) PPD conrol Desired final angle Derivative control z ppd.25, Kp / K q Time (sec) Fig. 7 Response comparison of Derivative vs. PPD Control Systems Note that the PPD system experiences the same overshoot yet settles to the desired position faster. The price that has been paid for the faster response is the higher damping torque f d and the addition of the proportional control f p. This aspect will be evaluated in the experiment. Finally, a MATLAB code that has generated the PPD part in Fig. 7 is listed below. % response_ppd_step.m % % obtain the response histories for damped system under step input % using the transfer function generated by the laplace transform method % % programmed by kc park, 4 feb 22 % clear all; % input data % omega_n : natural frequency of the derivative control system % zeta : damping ratio % T_final : response time desired % theta_d : final position % kappa : ratio of the system spring constant K_theta to the PPD K_p omega_n pi; kappa.; omega_nbar sqrt(+kappa)*omega_n; 2

21 zeta_ppd.25; % zeta ; T_final (2*pi/omega_n)*5 % five periods theta_d pi; % construct the laplace domain transfer function % tf generates the transfer function % define a system in transfer function form sys tf (, [/omega_nbar^2 2*zeta_ppd/omega_nbar ]) % generate unit step response [y, time] step(sys, T_final); % multiply the output y by the final position angle, theta_d ytheta_d*y; %figure(2) plot(time, y); grid on; xlabel( Time (sec) ) ylabel( Displacement ) title( Response of PPD system under unit step input ); legend([ omega_nbar, num2str(omega_nbar),..., zeta, num2str(zeta_ppd)]); % end of the program 2