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 Assume y and u are the system output and input, respectively; a general n th order linear time invariant TI) difference eqn is given as: (LTI) difference eqn. is given as: 11 0 () ( 1 ) ( 1 ) ( ) ) ( 1 ) ( 1 ) ( ) n yk n a yk n ayk k b k b k bk ++ + + + + + + " 0 where , , for all are constant nn ii b u k n u k n b u kb u k ab 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 . 1 ( ) ( ) yk uk ++ + + = Using recursion in the time domain: ( ) 0.8 ( 1) 0.1 ( 2) ( 2) ky 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 yy y u y u − + − = 2 (2) 0.8 (1) 0.1 (0) (0) 1 3 (3) 0.8 (2) 0.1 (1) 0.2 y u y u + = + = 4 (4) 0.8 (3) 0.1 (2) 0.8(0.2) 0.1(1) y u + 10 . 7 4 += 5 (5) 0.8 (4) 0.1 (3) 0.8(0.74) 0.1(0.2) 1 0.388 y u + + =

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Solving Diff. Eqns. Using Matlab 2 ) 0 8( 1 ) 0 1() () kk k k + + + The recur.m function (on class site) can be used to solve (2 . . 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; =ones(16); x = ones(1,6); y = recur(a,b,n,x,x0,y0) em y abel 'k'); abel '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) The z transform simplifies the soln. of LTI diff eqns. to algebraic eqns. convolution in time becomes mult. in frequency Casual signals encountered in control systems have zero lues for negative time and lend themselves to one values for negative time and lend themselves to one sided transforms, e.g., for the casual sequence u the z ansform is defined as { u 0 , u 1 , … u k , … }, the z transform is defined as 1 01 () kk Uz u u z u z u z −− =+ + + += "" 0 - where can be viewed as a delay operator of samples k k zk =
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 ut u t ) () ( ) ( ) ( ) t u t u tT u tk T u T + + + + = " The Laplace transform of the above signal is given by: 01 0 kk k u δδ δ = =+ "" k sT ksT sT Us u u e u e ue ∗− + + += 0 It is easy to see that ( ) ( ) , where k sT ze Uz U s = = denotes an impluse sampled continuous time system

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## This note was uploaded on 12/20/2009 for the course ECE 451 taught by Professor Staff during the Fall '09 term at Clarkson University .

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EE451_Chapter2_Notes_F09_L4 - EE451/551: Digital Control...

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