Meeting w6   chapter 2 part 3
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    Meeting w6   chapter 2 part 3 Meeting w6 chapter 2 part 3 Presentation Transcript

    • Chapter 2 – Analog Control System (cont.)
      • Reduction of Multiple Subsystem
      • System Response – Poles/ Zeros, Second Order System, Steady State Error, Stability Analysis
    • 7. Reduction of Multiple Subsystem
      • Subsystem is represented as a block with an input, output and transfer function.
      • Many systems are composed of multiple subsystems.
      • When multiple subsystem are interconnected, new element must be added like summing junction and pickoff points.
      • The main purpose of reduction multiple subsystem is to reduce more complicated systems to a single block.
    • Component of a block diagram for a linear, time-invariant system
    •  
    • Example 1
      • Cascade form
    • Example 2
      • Parallel form
    • Example 3
    • Example 4
    • 8. System response
      • Time is an independent variable mostly used in a control system evaluation, also referred as time response.
      • In order to find a system response, 2 steps are commonly used:
      • 1. differential equation
      • 2. inverse Laplace transformation
      • In general, system response contains 2 parts:
      • 1. Transient response (natural response)
          • Part of the time response that goes to zero as time becomes very large.
        • 2. Steady state response (forced response)
          • Part of the time response that remains after the transient has died out
    • Example 1
      • Find the response of the system for a step input .
      • Solution :
    • Poles, Zeros and System Response
      • 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 .
      • 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 .
    • Example 1
      • 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.
    • cont.
      • Solution:
      • 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 :
      • Thus :
    • cont.
      • From the development summarize, the following conclusions can be drawn:
    • Second Order System
      • A system where the closed loop transfer function possesses two poles is called a second-order system
    • cont.
    • cont.
      • Step input
    • cont.
    • cont.
    • cont.
    • Example 1
      • The transfer function of a position control system is given by:
      • Determine:
        • Rise time, t r
        • Peak time, t p
        • Percent overshoot, %M p
        • Settling time t s for 2% criterion
    • cont.
      • Solution:
      27.39 rad/s 2 x 27.39
    • cont.
    • Steady State Error
      • Steady-state error is the difference between the input and the output for prescribed test input as t   .
      • Test inputs used for steady-state error analysis are:
        • Step Input
        • Ramp Input
        • Parabolic Input
    • cont.
      • 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(  ) .
    • cont.
      • The formula to obtain steady-state error is as follow:
    • Step input
    • Ramp Input
    • Parabolic Input
    • Stability Analysis
      • A system is stable if every bounded input yields a bounded output .
      • A system is unstable if any bounded input yields an unbounded output.
      • 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.
    • Stability Analysis in the Complex s-Plane
      • There have 3 condition of poles location that indicates transient response, which is:
          • Stable systems have closed-loop transfer function with poles only in the left half-plane (LHP).
          • Unstable systems have closed loop transfer function with at least one pole in the right half-plane (RHP).
          • Marginally stable systems have closed loop transfer function with only imaginary axis poles .
    • cont.
    •  
    • Example 1
      • Determine whether the unity feedback system below is stable or unstable.
      • Solution:
      • Let equation = 0 to find the poles,
    • cont.
      • Thus, the roots of characteristic equation are
      • -2.672
      • -0.164 + j1.047
      • -0.164 - j1.047
      • 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.
    • Example 2
      • Determine the stability of the system shown below.
      • Solution:
    • cont.
      • Let equation = 0 to find the poles,
      • Thus, the roots of characteristic equation are,
      • -3.087
      • 0.0434 + j1.505
      • 0.164 - j1.505
      • 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