计算机控制技术-24-25

1、How to select sampling cycle timenControl performance, anti-disturbance quick responsenActuator characteristics, computing load cost Demands from different viewpoints-faster-slowerTheoretically: Shannons theorem To restore the original continuous signal, the sampling rate should satisfy the eqt.max2

2、sPractically, to select the sampling interval, the following influential factors should be taken into account 1.frequency of main disturbance to the plant.2.dominant time const. of the plant. 3.response speed of actuator.-shorter-shorter(e.g. servo valve 100Hz bandwidth.- Loop time10 ms)4. Required

3、accuracy/response speed (Ts performance )5.Ratio of performance to cost6.Computing load (number of loops Ts )Therefore, determining sampling rate is a trade-off option. Normally, we choose:51101TsTime const. of the plantif no time delay or Should find a way to solve the problem.1) what causes the ri

4、pples? Take the previous design problem as an example: The structure of the system is shown as the following diagram: sesGTsh1)(D(z)Gh(s)Gs(s)r(t)+V(s)Ty(z)-k1k2E(z)r(s)y(t)Where: U(z)T)368. 01)(1 ()718. 01 (68. 3)(1111zzzzzGk)718. 01)(1 ()368. 01)(5 . 01 (543. 0)(1111zzzzzD21)1 ()(zzGE212)(zzzGBAnd

5、 we have already known that there are output ripples.From the system diagram, we have:)()()(zrzGzEE21121)1 ()1 (zTzz11zTz)()()(zEzDzU21321718. 0282. 011 . 04713. 0543. 0zzzzz43213916. 05798. 03175. 0543. 0zzzzComparing the two equations, we realize that !0)(, 0)(, 2kTubutkTekwhenLeading to that.)(co

6、nsttvAs illustrated in the following diagram:diagramBy analysis, we know that: fluctuation of v(t) causes ripples between sampling instants, while the fluctuation of v(t) is the result of that: 20)(kafterkTu)(zGE)(zGB2) Method for avoiding output ripplesWeve already known that in a deadbeat control

7、system with : 11 z21)1 ( z31)1 ( z1z212 zz32133zzz respectively, error e(kT)=0 after a number of intervals. A notable feature of the system is that GB(z) is a finite terms polynomial of z More generally, for a specific input signal, e(kT) could be made to zero after a limited number of intervals as

8、long as we choose GB(z) a polynomial of z with finite terms . From the system diagram, we have:)()()(zrzGzYB)()()(zUzGzYk)()()()(zGzGzrzUkBTherefore:Let:)()()(zMzNzGkkkWhere Nk(z) is zero poly. of Gk(z), Mk(z) is pole poly. of Gk(z)Substitute the Eqt. into)()()(zMzNzGkkk)()()()(zGzGzrzUkBthen:)()()(

9、)()()()(zNzMzGzGzGzrzUkkBkB)()()(zFzNzGkB)(1)()()()()()(zGzGzGzGzGzGzDBkBEkBIf choose GB(z) to make U(z) a finite terms poly. of z, then U(kT) will arrive at its steady state value after a number of intervals.GB(z) retains all zeros of Gk(z)make the system ripple freeConclusion: mathematically choos

10、eWhere: F(z) is a finite terms poly. of z-1 to be decidedAnd the controller is:)()(1)()()()(zFzNzFzNzNzMkkkk)()(1)()(zFzNzFzMkkExample: Given the plant )1(10)(sssGs)368. 01)(1 ()718. 01 (68. 3) 1(101)(1111zzzzssseZzGTsk)()1 ()(11zFzzGEZero order hold is used, sampling interval T=1 secDesign a ripple

11、 free deadbeat controller with respect to unit step inputSolution: There is an unstable pole z=1, and a delay factor z-1 following the design procedure in 3.2.3 1) choose 2) For unit step input, should choose11)(zzGE3) To design a ripple free controller, GB(z) should retain all zeros of Gk(z) )()718

12、. 01 ()(211zFzzzGB- 3.3.1- 3.3.2Notice that Eqt. 3.3.1 is the more generalized expression of Eqt. 3.3.2, therefore,)()1 ()(11zFzzGE)()718. 01 ()(211zFzzzGBFurther, choose F1(z) and F2(z) that conform to the previous principles,111)(azzFbzF)(2Then we get:A=0.418, b=0.582The controller is,11418. 01)36

13、8. 01 (1582. 0)()()()(zzzGzGzGzDEkBTo verify the output of the system ripple free,)()()(zEzDzU)()()(zrzGzDE1111111)418. 01)(1 (418. 01)368. 01 (1582. 0zzzzz1058.01582.0zTherefore, the output sequence0)3()2(,058. 0) 1 (,1582. 0)0(uuuui.e., after two sampling intervals, u(kT) reaches its steady state freerippleconsttv.)(Figuratively,For