Experiment # 7 Servo Motor Position Control 1. Objective: ο· To become familiar with the fundamentals of servo technology. ο· To become familiar with position control. ο· Investigate the effect ofPID-Controller on the performance of Servo Motor system. 2. Theory and Background: Servo system is a system to control mechanical instruments in compliance with variation of position or speed target value. The word “servo” derives from Greek “servant”. The system is called “servo system” as it responds faithfully to command. The major difference of servo motor compared with general use motors is that it has a detector to detect rotation speed and position. DC servo motors have beenwidely used as an actuator for motion control anddirect-drive applications. Examples are as roboticand actuator for automation process, mechanicalmotion and others. The DC servo motors have beenextensively applying in many servome chanisms. Therefore, it is very important to study about theposition control of the DC servo motors.Consider the servo system shown in Figure 1. The objectiveof this system is to control the position of the mechanical load in accordance withthe reference position. Figure 1: Schematic diagram of servo system. The operation of this system is as follows: A pair of potentiometersacts as an errormeasuring device. They convert the input and output positions into proportional electric signals. The command input signal determines the angular positionr of the wiper arm of the input potentiometer.The angular position r is the reference inputto the system, and the electric 1 potential of the arm is proportional to the angularposition of the arm. The output shaft position determines the angular position c of thewiper arm of the output potentiometer. The difference between the input angular positionr and the output angular position c is the error signal e, or π =π−π The potential difference ππ − ππ = ππ£ is the error voltage, where e, is proportional to rand ππ is proportional to c; that is, ππ = πΎ0 π and ππ = πΎ0 π, where πΎ0 is a proportionalityconstant. The error voltage that appears at the potentiometer terminals is amplified bythe amplifier whose gain constant is πΎ1 .The output voltage of this amplifier is appliedto the armature circuit of the dc motor.A fixed voltage is applied to thefield winding. If an error exists, the motor develops a torque to rotate the output loadin such a way as to reduce the error to zero. For constant field current, the torque developedby the motor is π = πΎ2 ππ whereπΎ2 is the motor torque constant and ππ is the armature current.When the armature is rotating, a voltage proportional to the product of the flux andangular velocity is induced in the armature. For a constant flux, the induced voltage ππ is directly proportional to the angular velocity ππ/ππ‘, or ππ = πΎ3 ππ ππ‘ whereππ is the back emf, πΎ3 is the back emf constant of the motor, and π is the angulardisplacement of the motor shaft. The transfer function between themotor shaft displacement and the error voltage can be simplified and get the following: πΊ(π ) = πΎ π½π 2 + π΅π Where π½is the moment of inertia referred to the output shaft.π΅ = [π0 + (πΎ2 πΎ3 /π π )]/π2is the viscous-friction coefficient referred to the output shaft. πΎ = πΎ0 πΎ1 πΎ2 /ππ π . Figure 2: block diagram for the servo system; (c) simplified block diagram. 2 3. Components Required: 1 Power Supply +/- 15 V 72686 1 PID Controller 734064 1 Power Amplifier 73413 1 Servo Set point potentiometer 73410 1 DC-Servo 73414 1 CASSY– interface with Computer 4. Procedures: a) Characterization of the Set point potentiometer ο§ Connect the Set point potentiometer to the DC power supply 72686. ο§ Adjust the set point integrator to ∞. ο§ Adjust the setpoint value to 0° (360° ), and the output voltage to 0.0 v using the “zero” adjustor. ο§ Adjust the setpoint value to 90° (360° ), and the output voltage to 5.0 v using the “scale” adjustor. ο§ Start changing the set value clock-wise and counter clock-wise and record the output voltages as in table 1. Table 1: Angular displacement π ° of the set point potentiometer vs. the output voltage π1. ° π½ ππ ο§ Counter Clock-wise Clock-wise 0 10 20 30 60 90 120 150 0 10 20 30 60 90 120 150 0 -5 0 5 Draw the input output characteristic curve. Is it linear? Comment on your results. b) Step response of servo system: ο§ Set the experiment according to the connection diagram shown in Figure 3. ο§ With the Switch S1 kept off, repeat the 2nd, 3rd and 4th steps in part a. ο§ Adjust πΎπ = 1 and switch off Ki and Kd. ο§ Set the angular displacement of the set point potentiometer π ° = 90° . ο§ Set Channel input B in CASSY to the output of system (i.e. o/p of DC servo block). ο§ At the same time, pushthe switch S1 (ON) and start the CASSY, F9 key. 3 c) P- Control: ο§ [switch off Ki and Kd] For πΎπ = 1, πΎπ = 5, πΎπ = 10, change the set value of the potentiometer (i.e. system input) as indicated in table 2 and record the output angle of the DC servo motor. Table 2: Input and output values of servo system in different cases for P-controller. π²π· = π ° πππ ° πππ’π‘ Counter Clock-wise 0 50 90 0 ° πππ ° πππ’π‘ Counter Clock-wise 0 50 90 0 150 Clock-wise 0 50 0 90 150 Clock-wise 0 50 0 90 150 Clock-wise 0 50 0 90 150 π²π· = π 150 π²π· = ππ ° πππ ° πππ’π‘ ο§ Counter Clock-wise 0 50 90 0 150 Plot the step response, fix the input angle to π ° = 90°, and change πΎπ = 1, πΎπ = 5, πΎπ = 10. ο§ Comment, analyze and discuss the effects of each response and choose the best one. ο§ Evaluate the performance in terms of speed, steady state error, overshoot and oscillation. d) PI- Control:[Fix π²π· = ππ and switch off Kd] ο§ Repeat the first step in part (b) when π²π = π πππ π²π = π and complete table π 3.Note:π²π = π» π ο§ Plot the above two responses and put them on the same graph. Table 3: Input and output values of Servo system in different cases for PI-controller. ° πππ ° πππ’π‘ 0 0 π²π· = ππ and π²π = π Counter Clock-wise 50 90 150 0 0 4 Clock-wise 50 90 150 ° πππ ° πππ’π‘ 0 0 π²π· = ππ and π²π = π Counter Clock-wise 50 90 150 0 0 Clock-wise 50 90 150 e) PID Control design: ο§ Repeat the first step in part (b) when π²π· = ππ, π²π = π πππ π²π« = π. π and complete table 4. Table 4: Input and output values of Servo system for PID-controller cases. ° πππ ° πππ’π‘ ο§ 0 0 π²π· = ππ, π²π = π πππ π²π« = π. π Counter Clock-wise Clock-wise 50 90 150 0 50 90 0 Plot the above response and put them on the same graph. Switch S1 Figure 3: Experiment connection diagram. 5 150