Tài liệu Mechatronics (Hugh 2001) P2 pdf

page 31 4.4.3 Cascade • By adding extra feedback loops we can add levels of robustness to a control system. • In this case the control loops appear to be nested. • Consider a cascade control system for disturbances, c n r n = + - + - + + d n r n G p r n G c 2 G c 1 page 32 • Now consider a cascade controller for a multistage process, such as a sewage treatment plant with sequential tanks with varying flow rates. Here the disturbance would be the sewage flow- ing into the first tank. the two process blocks are the treatment tanks. The output is water returned to rivers The controllers vary the flow rates by looking at parameters for both tanks. • To do design/analysis for these controllers we do an analysis of the innermost feedback loop. This then becomes a process, and we do design/analysis at the next higher loop. • Select controllers for the system pictured below, + - + - + + r n d n G c 1 G c 2 G p 1 G p 2 r n H f page 33 + - + - + + d n r n 5B 1 B– r n G c 2 G c 1 1. Design for Gc1 as if the disturbance and Gc2 are not present 2. Design for disturbance minimization with Gc2 page 34 4.5 SAMPLE TIME • In deadbeat type controllers we do not typically see mentions of the sample period, but in most practical systems we see a mention of the sample time ‘T’. • We can use the system transfer function to calculate the acceptable range of controller time. In all cases the lower sample period is 0, but for practical reasons we may want to make this larger. • A shorter sample time requires a faster computer, and steals cycle time from other processes. • The general procedure is, 1. Find the characteristic equation of the overall system transfer function. 2. Determine the roots (symbolically) of the equation. 3. Select values of ‘T’ that ensure that all of the roots have a value of 1 or more. 4. Select a sample time that is less than the longest time. • Consider the example below, GB() 4TB 2– B 2 TB 2T 2 –+ = Find the sample time T required for stability, B 2 TB 2T 2 –+0= First, we find the characteristic equation, and solve for the roots, when the roots are B T()– T() 2 41() 2T 2 –()–± 21() T 2 1–18+±() T 2 1–3±()1==== T 1 0.5,= Based on these results we can now determine the stable region. page 35 4.6 SUMMARY • The following is a map of techniques that show typical uses of the mathematical technique cov- ered so far. real system differential equations difference equations estimated values realizability transfer function controller equation block diagram desired behavior design techniques system transfer function pick sample time stability analysis input function output function input functions from data input function equation output value table error functions steady state b-shift table b-shift algebra difference equation B-shift table partial fractions or long division transfer function feedback equation transfer function final value theory characteristic equation transfer function table calculations page 36 4.7 PRACTICE PROBLEMS 1. a) We are developing a freight elevator control system. The first major task is developing a height positioning controller. The elevator is moved to a new floor by issuing a step function for the new floor height. The transfer function below relates the voltage supplied to the eleva- tor motor to the height of the elevator. Develop a controller transfer function that has a first order response, and draw a complete block diagram. b) Develop the discrete equation for the controller if the time constant for the first order response is 5 seconds and the sampling time is 1 second. 2. c) Write a computer program (in the computer language of your choice) that implements the controller in question 1. Assume the input and output values are set using the two functions below. 3. Redraw the following system and add a feedforward controller. Develop the function used for the feed forward controller. 4. The control system below will be used for positioning the height of an elevator. G p B 1+ 1 0.5B– = INPUT() - This function will return the elevator height in feet (floating point) OUTPUT(value) - This function will set the output voltage at ‘value’ + _ 10 B–() B 1 B–() 2 B 5+ B 1– B 2+ page 37 a) Find a controller transfer function that gives a first order response for a time step of T=1 and a time constant of 3. b) Develop the discrete equation and determine if the controller is realizable. c) Develop a transfer function for the final system and determine if the system is stable. 4. We want to design a control system to minimize the effects of disturbance. Given a step input of magnitude 1, we are willing to accept a maximum error of 0.5 for one time step. a) Find a controller transfer function. b) Develop the discrete equation for the system and determine if the controller is realizable? 1 B– B G c r n c n + - ANS. a) G c 0.283B 2 12B– B 2 + = b) realizable c) G 0.283B 2 1 0.717B– = stable G p 10B 1 B– = ANS. a) G c 10 0.5 1 B–() 2 – 5B 1 B–() = b) not realizable m n m n 1– 9.5 5 e n 1+ 1 5 e n 0.5 5 e n 1– –++= page 38 5. DISCRETE SYSTEMS • When dealing with computers we will sample data values from the real world. These sampled values can then be used to estimate how a system will behave. • The term ‘discrete’ refers to the use of single sampled values, instead of a continuous functions. You will see that the difference is between weighted sums and integrals. 5.1 DISCRETE SYSTEM MODELLING WITH EQUATIONS • We can write a differential equation, and then manipulate it to put in terms of time steps of length ‘T’ • In review consider how we can approximate derivatives using two/three points on a line. T - Sample Period sampled values page 39 • Consider the example, T T y i 1– y i 1+ t i y i t i 1– t i 1+ yt() yt i () t i 1– t i ∫ y i y i 1– + 2   t i t i 1– –()≈ T 2 y i y i 1– +()= d dt yt i () y i y i 1– – t i t i 1– –   ≈ y i 1+ y i – t i 1+ t i –   1 T y i y i 1– –() 1 T y i 1+ y i –()== = d dt   2 yt i () 1 T y i 1+ y i –() 1 T y i y i 1– –()– T ≈ y i 1+ y i 1– 2y i –+ T 2 = If we read the flow rate of oil in a pipeline once every hour for three hours. The read- ings in order are 1003, 1007, 1010. What is the integral and first and second deriva- tives for the values? page 40 5.1.1 Getting a Discrete Equation • First, consider the example of a simple differential equation, • We can continue the example and use this equation to simulate the system. Here the ‘x’ values are given, and the first ‘y’ value is assumed to be 0. Assume T=0.5 and K=0.2. d dt   yt() Kx t()= yT() Kx t() td ∞– T ∫ Kx t()td ∞– 0 ∫ Kx t()td 0 T ∫ + y 0 Kx t()td 0 T ∫ +== = We can solve the differential equation at time T, assuming that the slope is constant and known, The equation above is developed for a time T after zero. We can manipulate it so that it will be valid for any point in time, yT() y 0 Kx 0 ()T+= y n ∴ y n 1– Kx n 1– ()T+= y n y n 1– –∴ Kx n 1– ()T= yT() y 0 Kx 0 ()T+= Now we can assume that the velocity ‘x(t)’ is constant, ASIDE: The approximations above can be shown using a slope approxi- mated with two points on a curve. for this particular differential ∆yx 0 t= equation we can assume say d dt AV=