Meeting w6 chapter 2 part 3

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Meeting w6 chapter 2 part 3

  1. 1. Chapter 2 – Analog Control System (cont.) <ul><li>Reduction of Multiple Subsystem </li></ul><ul><li>System Response – Poles/ Zeros, Second Order System, Steady State Error, Stability Analysis </li></ul>
  2. 2. 7. Reduction of Multiple Subsystem <ul><li>Subsystem is represented as a block with an input, output and transfer function. </li></ul><ul><li>Many systems are composed of multiple subsystems. </li></ul><ul><li>When multiple subsystem are interconnected, new element must be added like summing junction and pickoff points. </li></ul><ul><li>The main purpose of reduction multiple subsystem is to reduce more complicated systems to a single block. </li></ul>
  3. 3. Component of a block diagram for a linear, time-invariant system
  4. 5. Example 1 <ul><li>Cascade form </li></ul>
  5. 6. Example 2 <ul><li>Parallel form </li></ul>
  6. 7. Example 3
  7. 8. Example 4
  8. 9. 8. System response <ul><li>Time is an independent variable mostly used in a control system evaluation, also referred as time response. </li></ul><ul><li>In order to find a system response, 2 steps are commonly used: </li></ul><ul><li>1. differential equation </li></ul><ul><li>2. inverse Laplace transformation </li></ul><ul><li>In general, system response contains 2 parts: </li></ul><ul><li>1. Transient response (natural response) </li></ul><ul><ul><ul><li>Part of the time response that goes to zero as time becomes very large. </li></ul></ul></ul><ul><ul><li>2. Steady state response (forced response) </li></ul></ul><ul><ul><ul><li>Part of the time response that remains after the transient has died out </li></ul></ul></ul>
  9. 10. Example 1 <ul><li>Find the response of the system for a step input . </li></ul><ul><li>Solution : </li></ul>
  10. 11. Poles, Zeros and System Response <ul><li>Poles is the values of the Laplace transform variable( s ) that cause the transfer function to become infinite or any roots of the denominator of the transfer function . </li></ul><ul><li>Zeros is the values of the Laplace transform variable( s ) that cause the transfer function to become zero , or any roots of the numerator of the transfer function . </li></ul>
  11. 12. Example 1 <ul><li>Given the transfer function G ( s ) in Figure below, a pole exists at s = -5, and a zero exists at -2. These values are plotted on the complex s-plane, using x for the pole and О for the zero. </li></ul>
  12. 13. cont. <ul><li>Solution: </li></ul><ul><li>To show the properties of the poles and zeros, lets applied unit step response of the system. Multiplying the transfer function by a step function yields : </li></ul><ul><li>Thus : </li></ul>
  13. 14. cont. <ul><li>From the development summarize, the following conclusions can be drawn: </li></ul>
  14. 15. Second Order System <ul><li>A system where the closed loop transfer function possesses two poles is called a second-order system </li></ul>
  15. 16. cont.
  16. 17. cont. <ul><li>Step input </li></ul>
  17. 18. cont.
  18. 19. cont.
  19. 20. cont.
  20. 21. Example 1 <ul><li>The transfer function of a position control system is given by: </li></ul><ul><li>Determine: </li></ul><ul><ul><li>Rise time, t r </li></ul></ul><ul><ul><li>Peak time, t p </li></ul></ul><ul><ul><li>Percent overshoot, %M p </li></ul></ul><ul><ul><li>Settling time t s for 2% criterion </li></ul></ul>
  21. 22. cont. <ul><li>Solution: </li></ul>27.39 rad/s 2 x 27.39
  22. 23. cont.
  23. 24. Steady State Error <ul><li>Steady-state error is the difference between the input and the output for prescribed test input as t   . </li></ul><ul><li>Test inputs used for steady-state error analysis are: </li></ul><ul><ul><li>Step Input </li></ul></ul><ul><ul><li>Ramp Input </li></ul></ul><ul><ul><li>Parabolic Input </li></ul></ul>
  24. 25. cont. <ul><li>Figure has a step input and two possible output. Output 1 has zero steady-state error, and Output 2 has a finite steady-state error e2(  ) . </li></ul>
  25. 26. cont. <ul><li>The formula to obtain steady-state error is as follow: </li></ul>
  26. 27. Step input
  27. 28. Ramp Input
  28. 29. Parabolic Input
  29. 30. Stability Analysis <ul><li>A system is stable if every bounded input yields a bounded output . </li></ul><ul><li>A system is unstable if any bounded input yields an unbounded output. </li></ul><ul><li>A system is marginally stable/neutral if the system is stable for some bounded input and unstable for the others ( as undamped ) but remains constant or oscillates. </li></ul>
  30. 31. Stability Analysis in the Complex s-Plane <ul><li>There have 3 condition of poles location that indicates transient response, which is: </li></ul><ul><ul><ul><li>Stable systems have closed-loop transfer function with poles only in the left half-plane (LHP). </li></ul></ul></ul><ul><ul><ul><li>Unstable systems have closed loop transfer function with at least one pole in the right half-plane (RHP). </li></ul></ul></ul><ul><ul><ul><li>Marginally stable systems have closed loop transfer function with only imaginary axis poles . </li></ul></ul></ul>
  31. 32. cont.
  32. 34. Example 1 <ul><li>Determine whether the unity feedback system below is stable or unstable. </li></ul><ul><li>Solution: </li></ul><ul><li>Let equation = 0 to find the poles, </li></ul>
  33. 35. cont. <ul><li>Thus, the roots of characteristic equation are </li></ul><ul><li>-2.672 </li></ul><ul><li>-0.164 + j1.047 </li></ul><ul><li>-0.164 - j1.047 </li></ul><ul><li>Plot pole-zero on the s-plane and sketch the response of the system > all of the roots are locate at left half-plane, therefore the system is stable. </li></ul>
  34. 36. Example 2 <ul><li>Determine the stability of the system shown below. </li></ul><ul><li>Solution: </li></ul>
  35. 37. cont. <ul><li>Let equation = 0 to find the poles, </li></ul><ul><li>Thus, the roots of characteristic equation are, </li></ul><ul><li>-3.087 </li></ul><ul><li>0.0434 + j1.505 </li></ul><ul><li>0.164 - j1.505 </li></ul><ul><li>Plot pole-zero on the s-plane and sketch the response of the system > only one of the roots are locate at left half-plane and the others 2 roots locate at right half-plane, therefore the system is unstable </li></ul>

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