Presentation is loading. Please wait.

Presentation is loading. Please wait.

Automatic control systems II

Similar presentations


Presentation on theme: "Automatic control systems II"— Presentation transcript:

1 Automatic control systems II
Automatic control systems II. Basic knowledge of compensation Brief overview

2 Compensation While the variety of compensation schemes is great the classical three-term controller design has been found to be particularly simple and effective. This compensation scheme has historically implemented using analogue mechanical, pneumatically and electrical controllers. Today the compensation is implemented in a software because the controllers are microprocessor based. An extra steps is required to convert the analogue signal to digital information and the digital information to analogue signal. Nowadays in the industrial area the most popular A/D and D/A converter contain 12 bits or 11 bits.

3 The sample time If the sampling time is much more less than the sum of time constants of the process than the theory and design methodology is the same in digital and analogue technologies. If one wants the digital PIDT1 compensator seems to be such as analogue must choose appropriate sample time period. Grey model (Tj are the time constants of the model.) Black model energizing step signal (Ts is the settling time or time constant of the approximate model.) Relationship between Ts and Tj :

4 Structure of classical three-term controller

5 Structure of classical three-term controller in sampled controller

6 Real combination of the three kinds
Proportional P Proportional-Integral PI Proportional-Derivative PD Integral I Proportional-Integral-Derivative PID

7 Tuning controllers Which type of PID is the best solution?
Which parameters are optimal ? The type of PID depend on transfer function of the plant (process model) The optimal parameters depend on which is the required performance specification of the feedback control. (Looking for these parameters in time domain the step response of process field and in frequency domain the Bode or the Nichols diagram are used) If the simple feedback loop with compensation in cascade is unsatisfactory for the required performance specification than must modify the structure of control loop or exchange the classical PID scheme of control algorithm. (The most popular are the cascade and the feedforward control.)

8 Compensation techniques
Depends on the identification of the process field transfer function Step response of the process field: Black-box model and recommendations for the controller type and value of the parameters. Frequency response of the process field: Black-box model, and fitting of P, PI, PDT1, or PIDT1 controller‘s frequency response to the process field transfer function. Pole replacement: Grey/box model. Poles of the process field transfer function is known.

9 Modelled the process from reaction curve by dead-time first order transfer function HPT1
3.3 7.8 55 I PID PI P or PI

10 Suggested parameters for HPT1 minimum settling time and without overshoot system performance with following trajectory of reference signal Recommended by Chien-Hrones-Reswick This parameters based on the absolute integral criterion of performance. There are any other parameter commendation

11 The identified HPT1 model from the measured step response of process field
The files are in the TDcompensationLecture2 library. The model seems to be self-adjusting nature. (Without integral effect) First we define the final value. (The measured value saved as FinalValueStepresponse_SA.fig) Then using denser sampling the time constants can be edit. (stepresponse_SA.fig) Parameters: Kp=0.72, Tg =10.7sec., Tu =0.9sec. The suggested compensation is PI, because the ratio of Tg/Tu is almost 12. (Recommendation of Piwinger) (The result is saved as FeedbackStepResponse_SA.fig)

12 Modelled the process from reaction curve by “n” order transfer function PTn

13 Determination of system parameters

14 Suggested parameters for PTn minimum settling time and without overshoot system performance with following trajectory of reference signal

15 The identified nT1 model from the measured step response of process field
The files are in the TDcompensationLecture2 library. The model seems to be self-adjusting nature. (Without integral effect) First we define the final value. (The measured value saved as FinalValueStepresponse_SA.fig) Then using denser sampling to define the time values which belong to special response values. (nTstepresponse_SA.fig) Parameters: t110%=1.94sec., t230% =4.04sec., t370% =10.1sec. The counted order n =2. dominant time constant T =14.5sec., suggested compensation is PI,. (n less than 3.) (The result is saved as nTfeedbackStepresponse_SA.fig)

16 Modelled the process from reaction curve by dead-time integral first order transfer function HIT1

17 Suggested parameter for IT1 from Friedlich minimum settling time and without overshoot system performance with following trajectory of reference signal If the measured value has got noise, then the suggested compensation is P, other case PDT1. The PIDT1 is only used if it is necessary to use second type feedback control.

18 The identified IT1 model from the measured step response of process field
The files are in the TDcompensationLecture2 library. The model seems to be integrating nature. (With integral effect) First we define the time constants. (stepresponse_I.fig) Parameters: Tg=6.5sec., TI =2.7sec. Assuming that the measured value noise-free, the suggested compensation is PDT1, the counted compensation parameters: KC =0.21, TD =6.5sec., T =0.7sec. (The result is saved as FeedbackStepresponse_I.fig)

19 Analysis of Bode diagram (The files in the ExampleFDcompansatinLecture2 library)
The Bode plot is popular, because many properties of process field are well read. Recorded the measured values or the identified model the following compensation technique are also used. It must be concluded if the process field has or hasn't got integral effect. It possible from the phase plot. It is necessary to determine how many time constants belongs to the process field and how close to each other these time constants. It possible from the Bode plot.

20 Without integral effect
The integral effect is possible to determine from the phase plot of process field transfer function. (without integral effect if sufficiently low frequencies the phase shift is nearly zero.) In this case the most commonly used structure is the PI, if the process field model has got more than three poles relatively close to each other (inside two decade), then offered to use the PIDT structure. The quality properties of the closed loop depends on the phase margin of the G0(s) open loop transfer function. (If the phase margin “pm” higher or equal 90 degree the closed loop system without overshoot, if pm between 70 and 90 degree and there are three or more time constant and none of them are dominant, then the closed loop system can have got a small overshoot, but otherwise not.)

21 Transfer function of PI compensation
European structure There are two variables. First step we chose the KC = 1, and TI = 1 rad/sec. values! In case of PIDT must be: TI > 4TD and TD > 5T conditions.

22 PI compensation Tenfold value of ωI the amplitude gain nearly 0. and the phase shift -5,7°

23 The PI compensation process
Have to plotted the process field Bode plot. On the phase plot must be looking for the future gain-crossover frequency which is corresponding the following phase shift: ps = pm + 5, (The suggested phase margin if the process field order higher than 3 and the break point are relatively close to each other is minimum 70 degree.) A tenth of this frequency is ωI, and TI is the reciprocal of ωI. Plots the Bode diagram of (grey model). On this g0 Bode plot have to be looking for the frequency which is corresponding the chosen phase margin, next have to read the gain at this frequency. The KC controller gain is equal the reciprocal value of the readings gain. (In dB the readings value changes the sign)

24 Bode plot of GPF(jω) It can be seen the rounding is permitted, but must be documented! (Saved as DiagBodePF_SA) 24

25 Black model plot GPF(jω) and GPI(jω)
Based on the above figure 10wI = 0.5 rad/sec., and so wI = rad/sec. Creating the reciprocal value: TI = 20 sec. The g0 is figured with KC = 1: The gain of the controller converted from dB: KC = 5.8 25

26 Step response of the closed-loop
There aren’t overshoot, and steady-state error 26

27 With integral effect PDT1 compensation
European structure Case PDT1 compensation must be TD > 5T. There are three variables. The first step you choose KC = 1, TD = 0.9 rad/sec, and T = 0.1 rad/sec.

28 Bode plot of PDT1 AD = 9, and so φmax = 54.9°.
The φmax phase shift depends on the AD differential gain. Present example AD = 9, and so φmax = 54.9°.

29 The principle of compensation of PDT
First we chose a considers appropriate phase margin! On the phase plot of process field must look for the chosen phase margin and the corresponding frequency will be the future ωC gain-crossover frequency. At the chosen phase margin the phase shift is ps = pm – 54.9° - 180°. (Put this frequency equate with ωmax, which belongs to the maximum positive phase shift of GPDT(jω) and assuming AD =9, then ωD =ωmax/3 ωT =3*ωmax.) The controller gain KC value to be chosen so, that at the future ωC gain-crossover frequency will be unit the K0 loop-gain. (The reciprocal value of the amplitude gain at the future gain-crossover frequency on the amplitude plot of the g0 (the g0 is the G0(jω) open-loop transfer function with KC = 1 value) will be the actual KC.)

30 The GPF(jω) Bode diagram
The and assuming that AD = 9, so ωT= rad/sec and ωD= rad/sec. 30

31 Determination of TD and T with AD═9
Based on above figure wD = rad/sec. and wT = rad/sec. and so TD = 11.9 sec., and T = 1.3 sec. The open-loop transfer function with KC = 1 g0 is : On the Bode plot of g0 should look for the kC gain which corresponding the phase margin = 65°. (Grey model) The next figure (black model we put each other the GC(jω) and GPF(jω) bode plots.) 31

32 Bode plot of GPF(jω) and GPDT(jω)
The gain ═3.95dB, KC = -3.95dB which is converted KC = 0.63 32

33 Step response of feedback system
Possible, but not necessary additional tuning. 33


Download ppt "Automatic control systems II"

Similar presentations


Ads by Google