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EE451_Chapter2_Notes_F09_L4

# EE451_Chapter2_Notes_F09_L4 - EE451/551 Digital Control...

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EE451/551: Digital Control Chapter 2: Discrete Time Systems

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Difference Equations d h d Assume y and u are the system output and input, respectively; a general n th order linear time invariant (LTI) difference eqn is given as: (LTI) difference eqn. is given as: 1 1 0 ( ) ( 1) ( 1) ( ) ( ) ( 1) ( 1) ( ) n y k n a y k n a y k a y k b k b k b k b k + + + + + + + + + + + + + + " Wh t d LTI ? Wh i thi i t t? 1 1 0 where , , for all are constant n n i i b u k n u k n b u k b u k a b i = " What does LTI mean? Why is this important? If u(k) is: zero, y(k) is the homogeneous soln. an impulse, y(k) is the impulse response a step, y(k) is the step response
Solving Difference Equations Solve the diff eqn. with zero ICs, i.e., y( 2)=y( 1)=0, for a unit step input, i.e., u(k)=1 for all k>=0: ( 2) 0.8 ( 1) 0.1 ( ) ( ) y k y k y k u k + + + + = Using recursion in the time domain: ( ) 0.8 ( 1) 0.1 ( 2) ( 2) k y k y k y k u k = − + 0 (0) 0.8 ( 1) 0.1 ( 2) ( 2) 0 1 (1) 0.8 (0) 0.1 ( 1) ( 1) 0 y y y u y y y u = − + = = − + = 2 (2) 0.8 (1) 0.1 (0) (0) 1 3 (3) 0.8 (2) 0.1 (1) (1) 0.2 y y y u y y y u = − + = = − + = 4 (4) 0.8 (3) 0.1 (2) (2) 0.8(0.2) 0.1(1) y y y u = − + = − 1 0.74 + = 5 (5) 0.8 (4) 0.1 (3) (3) 0.8(0.74) 0.1(0.2) 1 0.388 y y y u = − + = − + =

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Solving Diff. Eqns. Using Matlab ( 2) 0 8 ( 1) 0 1 ( ) ( ) k k k k + + + + The recur.m function (on class site) can be used to solve 0.8 ( 0.1 ( ) y k y k y k u k = diff. eqns. with non zero initial conditions, as shown: a = [0.8 0.1]; b = [0 0 1]; [ ] [ ] x0 = [0 0]; y0 = [0 0]; k = 0:5; x = ones(1 6); x = ones(1,6); y = recur(a,b,n,x,x0,y0) stem(k,y);xlabel('k');ylabel('y(k)'); stem(k,y);xlabel( k );ylabel( y(k) ); title('Stem plot of system step response‘) Alternately, you can use convolution to solve for the output. See this link for details on both methods: http://users.ece.gatech.edu/~bonnie/book/TUTORIAL/tut_4.html
Solving Diff. Eqns. Using Matlab

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Z Transform (Frequency Domain) h f l f h l f d ff The z transform simplifies the soln. of LTI diff eqns. to algebraic eqns. l ti i ti b lt i f convolution in time becomes mult. in frequency Casual signals encountered in control systems have zero values for negative time and lend themselves to one sided transforms, e.g., for the casual sequence { u 0 , u 1 , … u k , … }, the z transform is defined as }, the z transform is defined as 1 0 1 ( ) k k k k U z u u z u z u z = + + + + = " " 0 - where can be viewed as a delay operator of samples k k z k =
Z Transform (Frequency Domain) The z transform can also be defined in terms of a continuous time signal sampled by an impulse train, defined: e.g., sampling a sinusiod: ( ) in solid grey, ( ) in red u t u t ( ) ( ) ( ) ( ) ( ) u t u t u t T u t kT u t kT δ δ δ δ = + + + + = " " The Laplace transform of the above signal is given by: 0 1 0 k k k = ( ) 0 1 0 ( ) k sT ksT sT k k k U s u u e u e u e = = + + + + = " " It is easy to see that ( ) ( ) , where d i l l d i i sT z e U z U s = = denotes an impluse sampled continuous time system

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EE451_Chapter2_Notes_F09_L4 - EE451/551 Digital Control...

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