Dynamic systems-analysis-4
- 1. Dynamic System Analysis Lecture “4” Dr. Sameh Farid Saad
- 2. Agenda 1. Mathematical Modeling of Pneumatic systems in state space. 2. Mathematical Modeling of Hydraulic servo systems in state space. 3. Modeling of Mixed system.
- 3. Pneumatic Systems • The working medium in a pneumatic device is a compressible fluid, most commonly air. • The availability of air is an advantage for pneumatic devices, because it can be exhausted to the atmosphere at the end of the device’s work cycle, thus eliminating the need for return lines. • On the other hand, because of the compressibility of the working fluid, the response can be slower and more oscillatory than that of hydraulic systems.
- 4. Resistance and Capacitance of Pressure Systems
- 5. Modeling of Pressure Systems
- 6. Pneumatic Nozzle–Flapper Amplifiers • Converts displacement into a pressure signal. • A large power output can be controlled by the very little power that is needed to position the flapper.
- 7. Pneumatic Relays • In practice, in a pneumatic controller, a nozzle–flapper amplifier acts as the first-stage amplifier and a pneumatic relay as the second stage amplifier. • The pneumatic relay is capable of handling a large quantity of airflow.
- 8. Flapper as a Lever There are two small movements (𝒆 and 𝒚) in opposite directions, Consider such movements separately and add up the results of two movements into one displacement 𝒙.
- 9. Bellows Acts Like a Spring 𝐹 = 𝑃𝑐 𝐴 𝐹 = 𝑘𝑦 𝑃𝑐 𝐴 = 𝑘𝑦 𝒀 𝑷 𝒄 = 𝑨 𝒌
- 10. Example (1)
- 11. Example (2)
- 12. Hydraulic Systems Hydraulic Servo System
- 13. Hydraulic Servo System Neglecting Time constant (T)
- 15. Dashpots The force acting on the piston must balance the spring force.
- 16. Hydraulic Proportional-Plus-Integral Control Action
- 17. Example (1) For the aircraft elevator control system, find the transfer function 𝜙 𝑠 𝜃 𝑠
- 18. Process Control System Input device Reference input Input potentiometer output potentiometer Feedback signal Error measuring device Amplifier Motor Gear train Load
- 19. Error Measuring Device 𝒆 = 𝒓 − 𝒄 ⟹ 𝑬 𝒔 = 𝑹 𝒔 − 𝑪(𝒔) • The angular position 𝒓 is the reference input to the system, • The electric potential of the arm 𝑒 𝑟 is proportional to the angular position of the arm 𝒓. 𝑒 𝑟 = 𝐾0 𝑟 • The output shaft position determines the angular position 𝒄 of the wiper arm of the output potentiometer. 𝑒 𝑐 = 𝐾0 𝑐 • The potential difference is the error voltage, 𝑒 𝑣 𝑒 𝑣 = 𝑒 𝑟 − 𝑒 𝑐 = 𝐾0 𝑟 − 𝐾0 𝑐 = 𝐾0(𝑟 − 𝑐) 𝒆 𝒗 = 𝑲 𝟎 𝒆 ⟹ 𝑬 𝒗 𝒔 = 𝑲 𝟎 𝑬(𝒔)
- 20. Amplifier • The error voltage that appears at the potentiometer terminals is amplified by the amplifier whose gain constant is 𝐾1. 𝑒 𝑎 = 𝐾1 𝑒 𝑣 ⟹ 𝑬 𝒂 𝒔 = 𝑲 𝟏 𝑬 𝒗(𝒔) • The output voltage of this amplifier is applied to the armature circuit of the dc motor.
- 21. Motor • For constant field current, the torque developed by the motor is 𝑇 = 𝐾2 𝑖 𝑎 ⟹ 𝑻 𝒔 = 𝑲 𝟐 𝑰 𝒂(𝒔) • The induced voltage 𝒆 𝒃 is directly proportional to the angular velocity 𝒅𝜽 𝒅𝒕 𝑒 𝑏 = 𝐾3 𝑑𝜃 𝑑𝑡 ⟹ 𝑬 𝒃 𝒔 = 𝑲 𝟑 𝒔 𝜣(𝒔) • The differential equation for the armature circuit is 𝐿 𝑎 𝑑𝑖 𝑎 𝑑𝑡 + 𝑅 𝑎 𝑖 𝑎 + 𝑒 𝑏 = 𝑒 𝑎 • The equation for torque equilibrium is
- 22. Gear Train • We assume that the gear ratio of the gear train is such that the output shaft rotates n times for each revolution of the motor shaft. Thus, 𝑪(𝑺) = 𝒏𝜣(𝒔)
- 23. Block Diagram
- 24. Report Draw the block diagram, and compute the transfer function 𝐺(𝑠) = 𝑋(𝑠)/𝐸1(𝑠) for the position control system
- 25. Thanks for your Attention