Difference Equation Construction (1) ENGG 1203 Tutorial. Difference Equation Construction (2) Grow, baby, grow (1)
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1 ENGG 03 Tutorial Differece Equatio Costructio () Systems ad Cotrol April Learig Objectives Differece Equatios Z-trasform Poles Ack.: MIT OCW 6.0, Newto s law of coolig states that: The chage i a object s temperature from oe time step to the ext is proportioal to the differece (o the earlier step) betwee the temperature of the object ad the temperature of the eviromet, as well as to the legth of the time step. Let o[ be temperature of object, s[ be temperature of eviromet, T be the duratio of a time step, K be the costat of proportioality Differece Equatio Costructio () Grow, baby, grow () The differece equatio for Newto s law of coolig i) the differece (o the earlier step) betwee the temperature of the object ad the temperature of the eviromet ii) as well as to the legth of the time step Be sure the sigs are such that the temperature of the object will evetually equilibrate with that of the eviromet. The system fuctio correspodig to this equatio I each time period, every cell i the bioreactor divides to yield itself ad oe ew daughter cell. However, due to agig, half of the cells die after reproducig. Po : The umber of cells at each time step. P o [ P o [ ] 0.5P o [ ] Suppose that P o [0] 0 ad P o [ 0 if <0. P o [0] 0 P o [] P o [0] 0.5P o [-] P o [] P o [] 0.5P o [0] P o [3] P o [] 0.5P o []
2 Grow, baby, grow () Grow, baby, grow (3) our goal is to create a costat populatio of cells, that is, to keep P o costat at some desired level P d. ou are to desig a proportioal cotroller that ca add or remove cells as a fuctio of the differece betwee the actual ad desired umber of cells. Assume that ay additios/deletios at time are based o the measured umber of cells at time. Deote the umber of cells added or removed at each step P ip. The differece equatios P o [ P o [ ] 0.5P o [ ] + P ip [ P ip [ k(p d [ P o [ ]) Draw a block diagram that represets the system P o [ P o [ ] 0.5P o [ ] + P ip [ P ip [ k(p d [ P o [ ]) Delays + Adders + Gais 5 6 Steppig Up ad Dow () Steppig Up ad Dow () Use a small umber of delays, gais, ad -iput adders (ad o other types of elemets) to implemet a system whose respose (startig at rest) to a uit-step sigal [,,, ] First fid a system whose uit-sample respose is the desired sequece. The periodicity of 3 suggests that y[ depeds o y[ 3]. The resultig differece equatio is y[ y[ 3] + w[ + w[ ] + 3w[ ]. is [,,3,,,3,, ] [,0,0, ] [,,3,,,3,, ] Draw a block diagram of your system. 7 8
3 Steppig Up ad Dow (3) Next compute w[ which is the first differece of x[: w[ x[ x[ ]. The result is the cascade of the first differece ad the previous result. [,0,0, ] [,,3,,,3,, ] [,,, ] Z Trasform () Z trasform is discrete-time aalog of Laplace trasform. Example: Fiboacci system Differece equatio: Operator expasio: System fuctioal: Uit-sample respose: y[ x[ ] ] + R + R R R h[ :,,, 3, 5, 8,3,, 34, 55, 89,... What is the relatio betwee system fuctioal ad h[? 9 0 Z Trasform () System fuctioal: R R Uit-sample respose: h[ :,,, 3, 5, 8,3,, 34, 55, 89,... Z Trasform (3) Example: Fiboacci system Differece equatio: y[ x[ ] ] + R + R Operator expasio: System fuctioal: Uit-sample respose: R R h[ :,,, 3, 5, 8,3,, 34, 55, 89,... What is the relatio betwee system fuctioal ad h[? h[ R
4 Example: Fiboacci system Differece equatio: y[ x[ ] ] Operator expasio: System fuctioal: Uit-sample respose: h[ R Z Trasform (4) + R + R R R h[ :,,, 3, 5, 8,3,, 34, 55, 89,... Series expasio of system fuctioal: h[ R Substitute R : z H Z Trasform (5) ( z) h[ z Z trasform! What is the relatio betwee H(z) ad h[? 3 4 Z Trasform (6) Multiple represetatios of discrete-time systems: Z Trasform (7) Z trasform maps a fuctio of discrete time to a fuctio of z. ( z) x[ z 5 6
5 Z Trasform (8) Z Trasform (9) 7 8 Poles () Poles () 9 0
6 Z Trasform Example () Solve differece equatios with Z trasforms Z Trasform Example () Z Trasform Example (3) Z Trasform Example (4) 3 4
7 Z Trasform Example (5) Z Trasform Example (6) 5 6 Z Trasform Example (7) Z Trasform Example (8) 7 8
8 Z Trasform Example (9) Z Trasform Example (0) 9 30 Z Trasform Example () q( x) x f ( x) x + x 3 Multiplyig through by A( x ) + B Substitutig x ad f ( x) x substitutig Partial Fractios f ( x) x + x 3 The deomiator splits ito two distict liear factors : + x 3 ( x + 3)( x ) + x ( x + 3) -3ito this x gives / 4 3 x + 3 x A B + x + 3 x + x 3, we have the polyomialidetity equatio gives A -/ 4, B / 4, so that + / 4 x 3 3
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