Positioning Servo Design Example


 Magdalene Willis
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1 Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pickandplace robot to move the link of a robot between two positions. Usually one would be given some design specifications, then select control hardware to satisfy those specifications, but I am going to do the reverse; I ll select a DC servomotor and a servo amplifier, and a length of steel square tubing (to model the robot link), and design a control system to move the link as fast as possible through a resttorest maneuver of 9. 2 Hardware. 2.1 Actuator. Pittman 7214 series DC motor, winding number # Amplifier. Inland model EM1825A Low Cost Wide Band Linear D.C. Servo Amplifier Motor & Amplifier Characteristics. The motor and amplifier are a good match: Motor peak voltage 19.1V Amplifier output voltage ±2V Motor peak (stall) current 4.92A Amplifier output current ±5A continuous, ±1A peak 2.3 Load: The robot link will be modeled by a 12inch length of steel square tubing with.49inch wall thickness. The vendor states the weight of this square tubing is.634 lb/ft. This link will be driven in the horizontal plane by a shaft attached to its end, thus we ll need the mass moment inertia about the end of the link. The inertia properties are: The bearings which support the load pivot have viscous friction coefficient of L = 12 in =.348 meters (1) W =.634 lb = m =.288 kg (2) J L = 1 3 ml2 = kgm 2 (3) b L = Nms (4) Thus the quotient J L /b L = 1 sec this is the time constant of the link and bearing in free rotation. Thus if give an initial velocity to the link, in 1 sec it will have slowed down 63%. 1
2 2.3.1 Load Position Measurement. Since the servo amplifier requires a supply voltage of ±24V  ±32V, let s assume we have a power supply of ±25V available. We ll install a 1turn potentiometer on the load shaft, with a voltage of ±25V across it, and the wiper of the potentiometer is the voltage indicating load position. Thus with the load position θ and the potentiometer output v f, the feedback transfer function is H(s) = K f = V f (s) θ(s) = 5 2π = V rad (5) 3 Trajectory. As I said during the course, nobody uses step inputs in positioning control systems like this. We will specify a cubic polynomial trajectory for the desired angle θ r (t). This trajectory is of the form θ r (t) = a o + a 1 t + a 2 t 2 + a 3 t 3 (6) A desirable feature of a cubic trajectory for a resttorest motion is that it yields zero velocity at the end points. If the initial value is θ o =, and the final value is θ f = π, and the trajectory has duration t f, then it can be shown that a o = (7) a 1 = (8) a 2 = 3π 2t 2 f (9) a 3 = π t 3 f (1) 3.1 Gear Ratio Selection. Small DC motors like we are using are invariably used along with a gear train. Selection of the gear ration n is very important. If the gear ratio is too low (n is large), the velocity of the link will not be very high. If the gear ratio is too high (n is small), the torque required will be too high, and the motor can t generate it. So we want a compromise gear ratio Maximum Velocity and Acceleration. One way to look at gear ratio selection is to find the maximum velocity and maximum acceleration required for a given motion duration t f. The acceleration can be converted into a torque using Newton s 2 nd law. We can then adjust motion duration t f and gear ratio n to find acceptable values. The reference position, velocity, and acceleration expressions are θ(t) = 3π 2t 2 t 2 π f t 3 t 3 rad (11) f θ(t) = 3π t 2 f θ(t) = 3π t 2 f t 3π t 3 t 2 rad/s (12) f 6π t 3 t rad/s 2 (13) f The maximum velocity occurs when the acceleration is zero, 3π t 2 f = 6π t 3 t = t = t f f 2 (14) The peak velocity occurs at the midpoint of the motion, and is ω max = θ max = θ(t f /2) = 3π 4t f rad/s (15) 2
3 The peak acceleration occurs at the beginning (and the end) of the motion, and is θ max = θ() = 3π t 2 f Finally, the maximum required torque can be found by using the moment of inertia J: rad/s 2 (16) T max = 3πJ L t 2 f Nm (17) Effect of Gear Train. Next we add the gear train, which is both a speed reducer and a torque multiplier. With a gear ratio of n (where n > 1) we have the maximum required motor torque and motor speed as T max = 3πJ L nt 2 f (18) ω max = 3πn 4t f (19) Motor Specifications. From the motor data sheet, the maximum motor torque (at stall, so a good match for our trajectory, which requires max torque at zero speed) and maximum motor speed (noload speed) are given as T max = 17 mnm =.17 Nm (2) ω max = 56 rpm = 53 rad/s (21) Optimal Trajectory. We want to move as fast as possible, but not exceed the limits of the motor. The two unknowns are trajectory duration t f and gear ratio n. We can take the motor data from (18) and (19) and substitute into the two equations (16) and (17) and solve for both n and t f. From the speed equation (17) we have Substituting this result for n into torque equation (16) yields 3πJ L 225t 3 f ω max = 3πn 4t f = 53 = n 225t f (22) =.17 = t 3 f = 3πJ L 38 Let s be a little conservative, and relax the duration a little, so we ll select =.22 = t f =.13 sec (23) t f =.2 sec (24) n = 225t f = = n = 3 (25) Cubic Trajectory. Now we have expressions for the position, velocity, and acceleration of the trajectory. From equations (5) (8) the coefficients of the cubic will be a o = (26) a 1 = (27) a 2 = 3π 2t 2 f = (28) a 3 = π t 3 f = (29) 3
4 Plots of the position, velocity, and acceleration are shown in Figure 1a, 1b, and 1c Position (RAD) Velocity (RAD/SEC) Time (sec) Time (sec) (a) Position of link. (b) Velocity of link Acceleration (RAD/SEC 2 ) Time (sec) (c) Acceleration of link. Figure 1: Desired trajectory of the robot link. 4 Controller Design. The first question to consider is What ω n and ζ should we try to achieve? We probably want to pick a reasonable damping ratio ζ, but it s not clear how the smooth cubic polynomial trajectory relates to natural frequency ω n. We can use the Fourier Transform to find the frequency spectrum of the cubic trajectory, and this may give us some insight. 4.1 Frequency Spectrum of Trajectory. Figure 2 shows the frequency spectrum of the cubic trajectory. This was obtained with the MATLAB fft() function. 4
5 Magnitude (RAD) Frequency (Hz) Figure 2: Frequency Spectrum of Cubic Trajectory Examination of the spectrum in Figure 2 shows that most of the content is below 5 Hz, so let s initially attempt to obtain a system natural frequency of 5 Hz. Also, let s specify a damping ratio of ζ =.77. Thus the design specifications are ω n = 5(2π) = 1π = 314 rad/sec (3) ζ =.77 (31) 4.2 Amplifier Mode. The Inland Model #EM182 can be used either as a voltage amplifier (voltage input and voltage output) or as a current amplifier (voltage input and current output). Since the motor dynamics are simpler when we use the amplifier as a current amplifier, we ll use that mode. The way a current amplifier works is by using a current sensing resistor to measure the amplifier output current. There is actually a closedloop feedback system within the amplifier to maintain output current at the desired value. The nominal gain of the current amplifier is Although the gain is adjustable, we ll leave it at the nominal setting. K a =.5 A V. (32) 4.3 Plant Model. The plant consists of the amplifier, motor, gear train, and load inertia. We need to include the viscous friction of both the motor bearings and the load bearings Total inertia and viscous friction. As we have learned, the effect of a gear train is to reduce the load inertia and viscous friction by a factor of n 2. The motor armature inertia and viscous friction coefficient listed in the spec sheet as damping (infinite source impedance) are J m = kgm 2 (33) b m = mnms = Nms, (34) 5
6 and the total inertia and viscous friction are J t = J m + J L n 2 = = kgm 2 (35) b t = J m + b L n 2 = = Nms (36) Thus the total inertia is about ten times the motor s own inertia, and the total viscous friction is not much larger than the motor s own viscous friction Plant block diagram. The motor torque constant (WDG #3) K t is given by K t = 35.7 mnm/a =.357 Nm/A (37) A block diagram of the plant with input of voltage v a and output θ L is shown in Figure 3. torque total inertia gear train amplifier constant v a i a T m θ m θ m K a K t +  J t s s n θ L b t viscous friction Figure 3: Block diagram of plant. Reducing the block diagram of Figure 3, and substituting numerical values (I used MATLAB ) one obtains G(s) = θ L(s) V a (s) = K ak t /J t n s (s + b/j t ) = s(s +.322) rad V (38) If we multiply the transfer function of (38) by s to convert from θ L to ω L, we obtain the DC gain as [ ] ωl 161 rad/s rpm 17 V V V a ss This DC gain looks reasonable this DC gain was done as a reality check and I think we pass. (39) 4.4 System Block Diagram. Figure 4 shows a block diagram of the overall system. Note the input block K f so the input and output have the same units. θ r V r V e V a θ L K f D(s) G(s) + V f K f Figure 4: Block diagram of system. The block D(s) in Figure 4 will contain the controller, which will be initially P, then PD, finally PID. 6
7 4.5 Proportional Control. As usual, proportional control is the first thing to try. Letting D(s) = K p, the closedloop transfer function (don t forget the input block K f ) will be T (s) = θ L(s) θ r (s) = 413K p s(s +.322) + 413K p = We can pick off the natural frequency and damping ratio as Since we want to have ω n = 314 rad/s, then we find that 413K p s s + 413K p (4) ω n = 413K p (41) ζ = K p (42) K p = (43) ζ =.51 (44) This is essentially an undamped system, and there is really no point in trying it on the cubic polynomial trajectory. We know that we need more damping. For this we ll add derivative action, resulting in a PD controller. 4.6 PD Control. PD control will add damping, which we badly need. It will also tend to speed up the response, because of the zero it will add in the forward path. To follow a rapidly changing input, we will need that fast response. A PD controller is given by ( D(s) = K P + K D s = K D s + K ) P (45) K D For our desired characteristics of ω n = 314, and ζ =.77, the desired closedloop pole locations are The rootlocus form of the characteristic equation in (4) is and the design rootlocus diagram is shown in Figure 5. s = 222 ± j222 (46) K p 1 s(s +.322) = (47) Vectors from the two poles (one at the origin, the other very near) to the desired pole location are drawn. The angle condition for this system is approximately θ = ±18 = θ = 9 (48) So the PD zero should be right in line with the desired pole locations, at s = 222. Then, the PD controller is D(s) = K D (s + 222) (49) A rootlocus of the system with PD controller installed will allow the determination of gain K D. transfer function for this system is L(s) = D(s)G(s)H(s) = 413K D s s(s +.322) The rootlocus (5) The MATLAB rootlocus diagram of the PD system is shown in Figure 6. The two branches leave the poles at (well, one at and one near) the origin and bend to the left as the zero attracts them. They do indeed go through the desired point, and the derivative gain that places them there is K D = (51) 7
8 jω root locus branches going straight up and down vectors from both poles to the desired pole location j222 desired pole location PD zero location 222 vector from PD zero to desired pole location σ Figure 5: PD design using root locus. The proportional gain from before still applies, so the two gains for the PD controller are K P = 8.72 (52) K D = (53) and the controller transfer function D(s) is D(s) = 1.742(s + 222) (54) 4.7 PID Control. Sorry, but my physical condition has prevented me from doing a PID controller sorry! 5 System Performance. Having finished the PD controller design, the obvious question is how well does it work? The closedloop transfer function (including the input transfer function) is: This closedloop system has: θ L (s) θ r (s) which is identical to our desired characteristics. 444(s + 222) = s 2 rad/rad (55) + 444s ζ =.78 (56) ω n = 314 rad/sec (57) 8
9 Root Locus Poles when Kd = Imaginary Axis Real Axis Figure 6: Root locus of PDcontrolled system. 5.1 Step Response. Let s first consider the unit step response, which is shown in Figure 7. This step response looks pretty good, although note the relatively large overshoot this is because of the zero in the PD controller. This zero makes the response faster and gives it more overshoot. The overshoot isn t much of a concern because our input is that nice gentle cubic polynomial trajectory Step response characteristics. From the sytem ζ and ω n in (55)(56), the step response characteristics using the design relations (inside back cover of text) are: t r 1.8 =.57 sec actual system is faster because of zero ω n (58) M p 5% actual system has more overshoot because of zero (59) 2%t s 3.9 =.176 sec actual system is pretty close! σ (6) 5.2 Tracking the Cubic Polynomial Input The real task for this controller is to follow the cubic polynomial input. Using the system transfer function of (54), and finding the response with the MATLAB lsim function, the response of the system is shown in Figure 8a. Note the added.5 seconds for a settling region. 9
10 Step Response Overshoot (%): 2.7 At time (sec): Amplitude.8.6 Rise Time (sec):.27 Settling Time (sec): Time (sec) Figure 7: Unit step response of PDcontrolled system. The tracking performance looks excellent! There are actually two plots in Figure 8a, the reference input (dashed line) and the response (solid line) but they are virtual overlays. To better show the tracking performance, a plot of the error is shown in Figure 8b. The peak error is about.13, which is.23 radian, which is equal to.273 inches at the tip of the link, or less then 1/32 inch! Note that the error doesn t settle to exactly zero until about t =.21 seconds, or.1 seconds after the reference trajectory is over. This isn t long, however Motor current during trajectory. It s important to check the motor current during this motion to check for motor saturation the current cannot exceed 4.92 A (see motor specs). Figure 9 shows a plot of motor current vs time during the cubic trajectory. The motor current was obtained by redrawing the block diagram to expose motor current as the output. The maximum current is which is about half of the maximum. So motor saturation is not a problem. (i a ) max = 2.24 A (61) 6 FrequencyDomain Analysis. 6.1 OpenLoop Bode Plot. The openloop transfer for this system is K D D(s)G(s)H(s). (62) 1
11 A Bode plot of the openloop transfer function with the numerical values from the PD Design is shown in Figure 1. The phase margin is which is very close to our design specification. P M = 65 = ζ.65 (63) 6.2 ClosedLoop Bode Plot. A Bode plot of the closedloop transfer function (see equation (55)) is shown in Figure 11. A plot of the closedloop transfer function illustrates the system bandwidth. It looks like the BANDWIDTH of the system is around 7 rad/sec (11 Hz). This is somewhat higher than expected...maybe I made a mistake! 7 Conclusion. Hopefully this design example gives you some idea of a somewhat realistic control design problem. 11
12 Angle (degrees) Cubic Polynomial Input Actual Response Time (sec) (a) Trajectory tracking Error Angle (degrees) Time (sec) (b) Error while tracking cubic. Figure 8: System response to cubic polynomial trajectory. 12
13 Current (A) Time (sec) Figure 9: Motor current during trajectory 15 Bode Diagram 1 Magnitude (db) Phase Margin (deg): 65.6 At frequency (rad/sec): 488 Phase (deg) Frequency (rad/sec) Figure 1: Openloop Bode plot. 13
14 1 Bode Diagram Magnitude (db) Frequency where magnitude is 3dB from the LF magnitude Phase (deg) Frequency (rad/sec) Figure 11: Closedloop Bode plot. 14
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