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PID Control Implementation and Tuning Part 9

Tham khảo tài liệu 'pid control implementation and tuning part 9', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Multi-Loop PID Control Design by Data-Driven Loop-Shaping Method 153 The gain plot of s is shown in Fig. 3. Since the peak gain is taken at w Wi 1 a2 0 5 this filter can be used for extracting this frequency component. Let us consider the filtering from the viewpoint of the wavelet transform Addison 2002 . In the last decade wavelet transform has become popular as a time-frequency analysis tool. Wavelet transform is useful to get important information regarding the frequency properties lies locally in the time-domain from the non-stationary signals e y . If we denote the impulse response of Fi s ỹ s Wi as L 1 Fi s Wi Wit then the correspondence a b t ty t t . 48 Wi is satisfied between the filtering yi t Fi s y t 49 Wi Ị Wi t z y z dT. 50 and the integral wavelet transform Wpy b a a 1iTO p T - y T dT. 51 TO a The impulse response t of s with a 0.5 is shown in Fig. 4 and the graph of pđb1ũ t is shown in Fig. 5 for the Daubechies wavelet db10 fdbi0 t . From the uncertainty principle in the wavelet analysis there is a trade-off between the time window and the frequency window. The time-frequency window can be tuned by the parameter a. a 0.5 is the value with which t can be close to pdb10 t . By the way since Fi s has four zeros at s 0 Fi s e t 0 for e t a0 1t 2t2 3t3. Namely the output becomes zero for this class of smooth inputs. For step or ramp inputs their time-derivatives have discontinuity and so we have nonzero outputs. For the response e t y t shown in Fig. 6 the responses filtered by Fi s are shown in Fig. 7. 4.3 Design procedure Step 1 Measure the input output responses ej t yj t t e 0 T j 1 2 . m by exciting the system at the steady state. If the response has bias eliminate it. Step 2 Set Wi i 1 2 . nF as logarithmically equally spaced nF points in the important frequency range for control. Generate the fictitious responses eij t yij t t e 0 T i 1 2 . nF. 52 from ej t yj t t e 0 T j 1 2 . m by 42 and 43 . Set the value of 71. Set the value of 72 if .