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    Chapter 10 The Design of Feedback Control SystemsPID .ppt

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    Chapter 10 The Design of Feedback Control SystemsPID .ppt

    1、Chapter 10 The Design of Feedback Control SystemsPID Compensation Networks,Different Types of Feedback ControlOn-Off ControlThis is the simplest form of control.,Proportional Control A proportional controller attempts to perform better than the On-off type by applying power in proportion to the diff

    2、erence 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

    3、to keep the heater supplying power.,Proportional, Derivative ControlThe 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

    4、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

    5、using relatively low gain by adding an integral term to the control function which becomes,The 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

    6、) 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 Control By only employing p

    7、roportional 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 Control All design specifications can be reached.,The Characteristics of P, I, and D control

    8、lers,Tips 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 Adjust each of Kp

    9、, 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

    10、example), then you dont 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 - Example,Proportional Control - ExampleThe proportional controller (Kp) reduces the rise time, increases the overshoot, and r

    11、educes the steady-state error. MATLAB Example,Kp=300; num=Kp; den=1 10 20+Kp; t=0:0.01:2; step(num,den,t),K=300,K=100,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

    12、 time.MATLAB Example,Kd=10,Kd=20,Proportional - Integral - ExampleThe integral controller (Ki) decreases the rise time, increases both the overshoot and the settling time, and eliminates the steady-state errorMATLAB Example,Kp=30; Ki=70; num=Kp Ki; den=1 10 20+Kp Ki; t=0:0.01:2; step(num,den,t),Ki=7

    13、0,Ki=100,Syntax rltool rltool(sys) rltool(sys,comp),RLTOOL,RLTOOL,RLTOOL,RLTOOL,RLTOOL,Consider the following configuration:,Example - Practice,The design a system for the following specifications: Zero steady state error Settling time within 5 seconds Rise time within 2 seconds Only some overshoot

    14、permitted,Example - Practice,Lead or Phase-Lead Compensator Using Root LocusA first-order lead compensator can be designed using the root locus. A lead compensator in root locus form is given by where the magnitude of z is less than the magnitude of p. A phase-lead compensator tends to shift the roo

    15、t locus toward the left half plane. This results in an improvement in the systems stability and an increase in the response speed. When a lead compensator is added to a system, the value of this intersection will be a larger negative number than it was before. The net number of zeros and poles will

    16、be the same (one zero and one pole are added), but the added pole is a larger negative number than the added zero. Thus, the result of a lead compensator is that the asymptotes intersection is moved further into the left half plane, and the entire root locus will be shifted to the left. This can inc

    17、rease the region of stability as well as the response speed.,In Matlab a phase lead compensator in root locus form is implemented by using the transfer function in the form numlead=kc*1 z;denlead=1 p;and using the conv() function to implement it with the numerator and denominator of the plant newnum

    18、=conv(num,numlead);newden=conv(den,denlead);,Lead or Phase-Lead Compensator Using Root Locus,Lead or Phase-Lead Compensator Using Frequency ResponseA first-order phase-lead compensator can be designed using the frequency response. A lead compensator in frequency response form is given by In frequenc

    19、y response design, the phase-lead compensator adds positive phase to the system over the frequency range. A bode plot of a phase-lead compensator looks like the following,Lead or Phase-Lead Compensator Using Frequency Response,Additional positive phase increases the phase margin and thus increases t

    20、he stability of the system. This type of compensator is designed by determining alfa from the amount of phase needed to satisfy the phase margin requirements, and determining tal to place the added phase at the new gain-crossover frequency. Another effect of the lead compensator can be seen in the m

    21、agnitude plot. The lead compensator increases the gain of the system at high frequencies (the amount of this gain is equal to alfa. This can increase the crossover frequency, which will help to decrease the rise time and settling time of the system.,In Matlab, a phase lead compensator in frequency r

    22、esponse form is implemented by using the transfer function in the form numlead=aT 1;denlead=T 1; and using the conv() function to multiply it by the numerator and denominator of the plant newnum=conv(num,numlead);newden=conv(den,denlead);,Lead or Phase-Lead Compensator Using Frequency Response,Lag o

    23、r Phase-Lag Compensator Using Root LocusA first-order lag compensator can be designed using the root locus. A lag compensator in root locus form is given by where the magnitude of z is greater than the magnitude of p. A phase-lag compensator tends to shift the root locus to the right, which is undes

    24、irable. For this reason, the pole and zero of a lag compensator must be placed close together (usually near the origin) so they do not appreciably change the transient response or stability characteristics of the system. When a lag compensator is added to a system, the value of this intersection wil

    25、l be a smaller negative number than it was before. The net number of zeros and poles will be the same (one zero and one pole are added), but the added pole is a smaller negative number than the added zero. Thus, the result of a lag compensator is that the asymptotes intersection is moved closer to t

    26、he right half plane, and the entire root locus will be shifted to the right.,It was previously stated that that lag controller should only minimally change the transient response because of its negative effect. If the phase-lag compensator is not supposed to change the transient response noticeably,

    27、 what is it good for? The answer is that a phase-lag compensator can improve the systems steady-state response. It works in the following manner. At high frequencies, the lag controller will have unity gain. At low frequencies, the gain will be z0/p0 which is greater than 1. This factor z/p will mul

    28、tiply the position, velocity, or acceleration constant (Kp, Kv, or Ka), and the steady-state error will thus decrease by the factor z0/p0. In Matlab, a phase lead compensator in root locus form is implemented by using the transfer function in the form numlag=1 z;denlag=1 p; and using the conv() func

    29、tion to implement it with the numerator and denominator of the plant newnum=conv(num,numlag);newden=conv(den,denlag);,Lag or Phase-Lag Compensator Using Root Locus,Lag or Phase-Lag Compensator using Frequency ResponseA first-order phase-lag compensator can be designed using the frequency response. A

    30、 lag compensator in frequency response form is given by The phase-lag compensator looks similar to a phase-lead compensator, except that a is now less than 1. The main difference is that the lag compensator adds negative phase to the system over the specified frequency range, while a lead compensato

    31、r adds positive phase over the specified frequency. A bode plot of a phase-lag compensator looks like the following,In Matlab, a phase-lag compensator in frequency response form is implemented by using the transfer function in the form numlead=a*T 1;denlead=a*T 1; and using the conv() function to im

    32、plement it with the numerator and denominator of the plant newnum=conv(num,numlead);newden=conv(den,denlead);,Lag or Phase-Lag Compensator using Frequency Response,Lead-lag Compensator using either Root Locus or Frequency Response A lead-lag compensator combines the effects of a lead compensator wit

    33、h those of a lag compensator. The result is a system with improved transient response, stability and steady-state error. To implement a lead-lag compensator, first design the lead compensator to achieve the desired transient response and stability, and then add on a lag compensator to improve the steady-state response,Problem 10.36Determine a compensator so that the percent overshoot is less than 20% and Kv (velocity constant) is greater than 8.,


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