模糊PID控制器的設(shè)計與仿真——設(shè)計步驟(修改)

1、模糊PID控制器的設(shè)計與仿真設(shè)計模糊PID控制器時，首先要將精確量轉(zhuǎn)換為模糊量，并且要把轉(zhuǎn)換后的模糊量映射到模糊控制論域當(dāng)中，這個過程就是精確量模糊化的過程。模糊化的主要功能就是將輸入量精確值轉(zhuǎn)換成為一個模糊變量的值，最終形成一個模糊集合。 本次設(shè)計系統(tǒng)的精確量包括以下變量：變化量e ，變化量的變化速率ec 還有參數(shù)整定過程中的輸出量KP，KD，KI，在設(shè)計模糊PID 的過程中，需要將這些精確量轉(zhuǎn)換成為模糊論域上的模糊值。本系統(tǒng)的誤差與誤差變化率的模糊論域與基本論域?yàn)椋篍=-6,-4,-2,0,2,4,6;Ec=-6,-4,-2,0,2,4,6。模糊PID控制器的設(shè)計選用二維模糊控制器。以給定

2、值的偏差e和偏差變化ec為輸入；KP，KD，KI為輸出的自適應(yīng)模糊PID控制器，見圖1。圖1模糊PID控制器（1）模糊變量選取 輸入變量E和EC的模糊化將一定范圍(基本論域)的輸入變量映射到離散區(qū)間(論域)需要先驗(yàn)知識來確定輸入變量的范圍。就本系統(tǒng)而言，設(shè)置語言變量取七個，分別為 NB，NM，NS，ZO，PS，PM，PB。(2) 語言變量及隸屬函數(shù)根據(jù)控制要求，對各個輸入，輸出變量作如下劃定：e，ec論域：-6，-5，-4，-3，-2，-1，0，1，2，3，4，5，6KP，KD，KI論域：-6，-5，-4，-3，-2，-1，0，1，2，3，4，5，6應(yīng)用模糊合成推理PID參數(shù)的整定算法。第k個

3、采樣時間的整定為式中為經(jīng)典PID控制器的初始參數(shù)。設(shè)置輸入變量隸屬度函數(shù)如圖2所示，輸出變量隸屬度函數(shù)如圖3所示。圖2 輸入變量隸屬度函圖3 輸出變量隸屬度函（3） 編輯模糊規(guī)則庫根據(jù)以上各輸出參數(shù)的模糊規(guī)則表，可以歸納出49條控制邏輯規(guī)則，具體的控制規(guī)則如下所示：1. If (e is NB) and (ec is NB) then (kp is NB)(ki is PB)(kd is NS)(1)2. If (e is NB) and (ec is NM) then (kp is NB)(ki is PB)(kd is PS)(1)3. If (e is NB) and (ec is NS

4、) then (kp is NM)(ki is PM)(kd is PB)(1)4. If (e is NB) and (ec is ZO) then (kp is NM)(ki is PM)(kd is PB)(1)5. If (e is NB) and (ec is PS) then (kp is NS)(ki is PS)(kd is PB)(1)6. If (e is NB) and (ec is PM) then (kp is ZO)(ki is ZO)(kd is PM)(1)7. If (e is NB) and (ec is PB) then (kp is ZO)(ki is

5、ZO)(kd is NS)(1)8. If (e is NM) and (ec is NB) then (kp is NB)(ki is PB)(kd is NS)(1)9. If (e is NM) and (ec is NM) then (kp is NB)(ki is PB)(kd is PS)(1)10. If (e is NM) and (ec is NS) then (kp is NM)(ki is PM)(kd is PB)(1)11. If (e is NM) and (ec is ZO) then (kp is NS)(ki is PS)(kd is PM)(1)12. If

6、 (e is NM) and (ec is PS) then (kp is NS)(ki is PS)(kd is PM)(1)13. If (e is NM) and (ec is PM) then (kp is ZO)(ki is ZO)(kd is PS)(1)14. If (e is NM) and (ec is PB) then (kp is PS)(ki is ZO)(kd is ZO)(1)15. If (e is NS) and (ec is NB) then (kp is NM)(ki is PB)(kd is ZO)(1)16. If (e is NS) and (ec i

7、s NM) then (kp is NM)(ki is PM)(kd is PS)(1)17. If (e is NS) and (ec is NS) then (kp is NM)(ki is PS)(kd is PM)(1)18. If (e is NS) and (ec is ZO) then (kp is NS)(ki is PS)(kd is PM)(1)19. If (e is NS) and (ec is PS) then (kp is ZO)(ki is ZO)(kd is PS)(1)20. If (e is NS) and (ec is PM) then (kp is PS

8、)(ki is NS)(kd is PS)(1)21. If (e is NS) and (ec is PB) then (kp is PS)(ki is NS)(kd is ZO)(1)22. If (e is ZO) and (ec is NB) then (kp is NM)(ki is PM)(kd is ZO)(1)23. If (e is ZO) and (ec is NM) then (kp is NM)(ki is PM)(kd is PS)(1)24. If (e is ZO) and (ec is NS) then (kp is NS)(ki is PS)(kd is PS

9、)(1)25. If (e is ZO) and (ec is ZO) then (kp is ZO)(ki is ZO)(kd is PS)(1)26. If (e is ZO) and (ec is PS) then (kp is PS)(ki is NS)(kd is PS)(1)27. If (e is ZO) and (ec is PM) then (kp is PM)(ki is NM)(kd is PS)(1)28. If (e is ZO) and (ec is PB) then (kp is PM)(ki is NM)(kd is ZO)(1)29. If (e is PS)

10、 and (ec is NB) then (kp is NS)(ki is PM)(kd is ZO)(1)30. If (e is PS) and (ec is NM) then (kp is NS)(ki is PS)(kd is ZO)(1)31. If (e is PS) and (ec is NS) then (kp is ZO)(ki is ZO)(kd is ZO)(1)32. If (e is PS) and (ec is ZO) then (kp is PS)(ki is NS)(kd is ZO)(1)33. If (e is PS) and (ec is PS) then

11、 (kp is PS)(ki is NS)(kd is ZO)(1)34. If (e is PS) and (ec is PM) then (kp is PM)(ki is NM)(kd is ZO)(1)35. If (e is PS) and (ec is PB) then (kp is PM)(ki is NB)(kd is ZO)(1)36. If (e is PM) and (ec is NB) then (kp is NS)(ki is ZO)(kd is NB)(1)37. If (e is PM) and (ec is NM) then (kp is ZO)(ki is ZO

12、)(kd is PS)(1)38. If (e is PM) and (ec is NS) then (kp is PS)(ki is NS)(kd is NS)(1)39. If (e is PM) and (ec is ZO) then (kp is PM)(ki is NS)(kd is NS)(1)40. If (e is PM) and (ec is PS) then (kp is PM)(ki is NM)(kd is NS)(1)41. If (e is PM) and (ec is PM) then (kp is PM)(ki is NB)(kd is NS)(1)42. If

13、 (e is PM) and (ec is PB) then (kp is PB)(ki is NB)(kd is NB)(1)43. If (e is PB) and (ec is NB) then (kp is ZO)(ki is ZO)(kd is NB)(1)44. If (e is PB) and (ec is NM) then (kp is ZO)(ki is ZO)(kd is NM)(1)45. If (e is PB) and (ec is NS) then (kp is PM)(ki is NS)(kd is NM)(1)46. If (e is PB) and (ec i

14、s ZO) then (kp is PM)(ki is NM)(kd is NM)(1)47. If (e is PB) and (ec is PS) then (kp is PM)(ki is NM)(kd is NS)(1)48. If (e is PB) and (ec is PM) then (kp is PB)(ki is NB)(kd is NS)(1)49. If (e is PB) and (ec is PB) then (kp is PB)(ki is NB)(kd is NB)(1)把這49條控制邏輯規(guī)則，鍵入到模糊規(guī)則庫中，如圖4。圖4 模糊規(guī)則庫（5）模糊PID控制器仿

15、真利用MATLAB軟件中的Simulink仿真環(huán)境，可以對模糊PID控制器系統(tǒng)進(jìn)行模擬仿真實(shí)驗(yàn)，來檢驗(yàn)設(shè)計是否達(dá)到要求。針對本次控制器設(shè)計，我們設(shè)置被控對象為，根據(jù)被控對象，設(shè)置相應(yīng)的PID參數(shù)為：=6;=3;=2。圖5為控制器系統(tǒng)在Simulink中的仿真模型。為了方便與傳統(tǒng)PID控制器進(jìn)行比較，在Simulink仿真環(huán)境中作出傳統(tǒng)PID控制以便于對模糊PID進(jìn)行比較。在傳統(tǒng)PID控制器中設(shè)置相應(yīng)的PID參數(shù)為：=6;=3;=2。圖6是傳統(tǒng)PID與模糊PID控制器在Simulink中的階躍仿真波形比較。圖5傳統(tǒng)PID與模糊PID控制器在Simulink中的仿真模型圖6傳統(tǒng)PID與模糊PID控制器在Simulink中的階躍仿真波形比較圖6中，黃色線為輸入的階躍信號，紫色為輸出的傳統(tǒng)PID控制信號，青色為輸出的模糊PID控制信號，通過圖1-7中傳統(tǒng)PID控制方式與模糊PID控制控制曲線的對比結(jié)果可以看出，模糊控制的控制性能要明顯好于傳統(tǒng)的PID控制效果。我們把輸入信號