PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Chapter 10 – The Design of Feedback Control SystemsPID Compensation NetworksDifferent Types of Feedback ControlOn-Off ControlThis is the simplest form of control.Proportional ControlA proportional controller attempts to perform better than the On-off type by applying power in proportion to the difference in temperature between the measured and the set-point. As the gain is increased the system responds faster to changes in set-point but becomes progressively underdamped and eventually unstable. The final temperature lies below the set-point for this system because some difference is required to keep the heater supplying power.Proportional, Derivative Control The stability and overshoot problems that arise when a proportional controller is used at high gain can be mitigated by adding a term proportional to the time-derivative of the error signal. The value of the damping can be adjusted to achieve a critically damped response.Proportional+Integral+Derivative ControlAlthough PD control deals neatly with the overshoot and ringing problems associated with proportional control it does not cure the problem with the steady-state error. Fortunately it is possible to eliminate this while using relatively low gain by adding an integral term to the control function which becomesThe Characteristics of P, I, and D controllersA proportional controller (Kp) will have the effect of reducing the rise time and will reduce, but never eliminate, the steady-state error. An integral control (Ki) will have the effect of eliminating the steady-state error, but it may make the transient response worse. A derivative control (Kd) will have the effect of increasing the stability of the system, reducing the overshoot, and improving the transient response.Proportional ControlBy only employing proportional control, a steady state error occurs.Proportional and Integral Control The response becomes more oscillatory and needs longer to settle, the error disappears.Proportional, Integral and Derivative ControlAll design specifications can be reached.CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERRORKp Decrease Increase Small Change DecreaseKi Decrease Increase Increase EliminateKd Small Change Decrease Decrease Small ChangeThe Characteristics of P, I, and D controllersTips for Designing a PID Controller1. Obtain an open-loop response and determine what needs to be improved 2. Add a proportional control to improve the rise time 3. Add a derivative control to improve the overshoot 4. Add an integral control to eliminate the steady-state error 5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. Lastly, please keep in mind that you do not need to implement all three controllers (proportional, derivative, and integral) into a single system, if not necessary. For example, if a PI controller gives a good enough response (like the above example), then you don't need to implement derivative controller to the system. Keep the controller as simple as possible.num=1; den=[1 10 20]; step(num,den) Open-Loop Control - ExampleG s( )1s210s 20Proportional Control - ExampleThe proportional controller (Kp) reduces the rise time, increases the overshoot, and reduces the steady-state error. MATLAB ExampleKp=300;num=[Kp];den=[1 10 20+Kp];t=0:0.01:2;step(num,den,t)Time (sec.)AmplitudeStep Response0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 200.20.40.60.811.21.4From: U(1)To: Y(1)T s( )Kps210 s 20 Kp( )Time (sec.)AmplitudeStep Response0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.70.80.91From: U(1)To: Y(1)K=300K=100Time (sec.)AmplitudeStep Response0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 200.20.40.60.811.21.4From: U(1)To: Y(1)Kp=300;Kd=10;num=[Kd Kp];den=[1 10+Kd 20+Kp];t=0:0.01:2;step(num,den,t)Proportional - Derivative - ExampleThe derivative controller (Kd) reduces both the overshoot and the settling time.MATLAB ExampleT s( )Kd s Kps210 Kd( ) s 20 Kp( )Time (sec.)AmplitudeStep Response0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.70.80.91From: U(1)To: Y(1)Kd=10Kd=20Proportional - Integral - ExampleThe integral controller (Ki) decreases the rise time, increases both the overshoot and the settling time, and eliminates the steady-state errorMATLAB ExampleTime (sec.)AmplitudeStep Response0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 200.20.40.60.811.21.4From: U(1)To: Y(1)Kp=30;Ki=70;num=[Kp Ki];den=[1 10 20+Kp Ki];t=0:0.01:2;step(num,den,t)T s( )Kp s Kis310 s2 20 Kp( ) s KiTime (sec.)A m plitudeStep Response0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 200.20.40.60.811.21.4From: U(1)To: Y(1)Ki=70Ki=100Syntax rltoolrltool(sys)rltool(sys,comp)RLTOOLRLTOOLRLTOOLRLTOOLRLTOOLConsider the following configuration: Example - PracticeThe design a system for the following specifications: · Zero steady state error · Settling time within 5 seconds · Rise time within 2 seconds · Only some overshoot permitted Example -
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