ENGG 1203 Tutorial Systems and Control 12 April Learning Objectives Difference Equation Construction (1) Newton’s law of cooling states that: The change in an object’s temperature from one time step to the next is proportional to the difference (on the earlier step) between the temperature of the object and the temperature of the environment, as well as to the length of the time step. Let o[n] be temperature of object, s[n] be temperature of environment, T be the duration of a time step, K be the constant of proportionality Difference Equations Z-transform Poles Ack.: MIT OCW 6.01, 6.003 1 2 Difference Equation Construction (2) The difference equation for Newton’s law of cooling i) the difference (on the earlier step) between the temperature of the object and the temperature of the environment ii) as well as to the length of the time step Be sure the signs are such that the temperature of the object will eventually equilibrate with that of the environment. Grow, baby, grow (1) The system function corresponding to this equation 3 In each time period, every cell in the bioreactor divides to yield itself and one new daughter cell. However, due to aging, half of the cells die after reproducing. Po : The number of cells at each time step. → Po[n] = 2Po[n−1] − 0.5Po[n−2] Suppose that Po[0]= 10 and Po[n]= 0 if n<0. Po[0] = 10 Po[1] = 2Po[0] − 0.5Po[-1] = 20 + 0 = 20 Po[2] = 2Po[1] − 0.5Po[0] = 40 − 5 = 35 Po[3] = 2Po[2] − 0.5Po[1] = 70 − 10 = 60 4 Grow, baby, grow (2) Your goal is to create a constant population of cells, that is, to keep Po constant at some desired level Pd. You are to design a proportional controller that can add or remove cells as a function of the difference between the actual and desired number of cells. Assume that any additions/deletions at time n are based on the measured number of cells at time n−1. Denote the number of cells added or removed at each step Pinp. The difference equations Po[n] = 2Po[n − 1] − 0.5Po[n − 2] + Pinp[n] Pinp[n] = k(Pd[n] − Po[n − 1]) Grow, baby, grow (3) Draw a block diagram that represents the system Po[n] = 2Po[n − 1] − 0.5Po[n − 2] + Pinp[n] Pinp[n] = k(Pd[n] − Po[n − 1]) Delays + Adders + Gains 5 6 Stepping Up and Down (1) Use a small number of delays, gains, and 2-input adders (and no other types of elements) to implement a system whose response (starting at rest) to a unit-step signal [1,1,1,…] is First find a system whose unit-sample response is the desired sequence. The periodicity of 3 suggests that y[n] depends on y[n − 3]. The resulting difference equation is y[n] = y[n − 3] + w[n] + 2w[n − 1] + 3w[n − 2]. [1,0,0,…] [1,2,3,1,2,3,1,…] Stepping Up and Down (2) [1,2,3,1,2,3,1,…] Draw a block diagram of your system. 7 8 Stepping Up and Down (3) Z Transform (1) Next compute w[n] which is the first difference of x[n]: w[n] = x[n] − x[n − 1]. The result is the cascade of the first difference and the previous result. [1,0,0,…] [1,2,3,1,2,3,1,…] Z transform is discrete-time analog of Laplace transform. Example: Fibonacci system y[n] = x[n] + y[n − 1] + y[n − 2] Difference equation: Y = X + RY + R 2Y Operator expansion: System functional: Y 1 = X 1− R − R2 Unit-sample response: h[n] :1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... [1,1,1,…] What is the relation between system functional and h[n]? 9 10 Z Transform (2) System functional: Unit-sample response: Z Transform (3) Y 1 = X 1− R − R2 h[n] :1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... Example: Fibonacci system Difference equation: Operator expansion: y[n] = x[n] + y[n − 1] + y[n − 2] Y = X + RY + R 2Y System functional: Y 1 = X 1− R − R2 Unit-sample response: h[n] :1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... What is the relation between system functional and h[n]? Y = ∑ h[n]R n X n 11 12 Z Transform (4) Z Transform (5) Example: Fibonacci system Series expansion of system functional: y[n] = x[n] + y[n − 1] + y[n − 2] Difference equation: Y = ∑ h[n]R n X n Y = X + RY + R 2Y Operator expansion: System functional: Y 1 = X 1− R − R2 Unit-sample response: h[n] :1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... Substitute R → Y = ∑ h[n]R n X n 1 z : H ( z ) = ∑ h[n]z − n Å Z transform ! n What is the relation between H(z) and h[n]? 13 14 Z Transform (7) Z Transform (6) Z transform maps a function of discrete time n to a function of z. Multiple representations of discrete-time systems: X ( z ) = ∑ x[n]z − n n 15 16 Z Transform (9) Z Transform (8) 17 Poles (1) 18 Poles (2) 19 20 Z Transform Example (1) Z Transform Example (2) Solve difference equations with Z transforms 21 Z Transform Example (3) 23 22 Z Transform Example (4) 24 Z Transform Example (5) 25 Z Transform Example (7) 27 Z Transform Example (6) 26 Z Transform Example (8) 28 Z Transform Example (9) Z Transform Example (10) 29 30 Z Transform Example (11) Partial Fractions 1 x + 2x − 3 The denominator splits into two distinct linear factors : f ( x) = 2 q ( x) = x 2 + 2 x − 3 = ( x + 3)( x − 1) A B 1 ⇒ f ( x) = 2 = + x + 2x − 3 x + 3 x −1 Multiplying through by x 2 + 2 x − 3, we have the polynomial identity 1 = A( x − 1) + B( x + 3) Substituting x = -3 into this equation gives A = -1/ 4, and substituting x = 1 gives B = 1 / 4, so that f ( x) = 31 1 −1/ 4 1/ 4 + = x + 2x − 3 x + 3 x −1 2 32