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EE3TP4_34_DT_Using_ZT_to_Solve_v3_Lecture 36

# EE3TP4_34_DT_Using_ZT_to_Solve_v3_Lecture 36 - Solving...

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Solving Difference Equations using the Z Transform These overheads were originally developed by Mark Fowler at Binghamton University, State University of New York.

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Given a difference equation that models a D-T system we may want to solve it: -with IC’s -with IC’s of zero Apply ZT to the Difference Equation Use the Transfer Function Approach Note … the ideas here are very much like what is done with the Laplace Transform for CT systems. We’ll consider the ZT/Difference Eq. approach first … Z Transform for Difference Equations
Given: y [ n ] ay [ n 1 ]= bx [ n ] IC = y [− 1 ] x [ n ] for n = 0 , 1 , 2 , ... Solve for: y [ n ] for n = 0, 1, 2, … Solving a First-order Difference Equation using the ZT Take ZT of differential equation: Z { y [ n ] ay [ n 1 ] } = Z { bx [ n ] } Z { y [ n ] } aZ { y [ n 1 ] } = bZ { x [ n ] } Use Linearity of ZT Y ( z ) X ( z ) Need Right-Shift Property… but which one? Because of the non-zero IC we need to use the non -causal form: Z { y [ n 1 ] } = z 1 Y z  y [− 1 ]

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Y z  a [ z 1 Y z  y [− 1 ] ] = bX z Using these results gives: Which is an algebraic equation that can be solved for Y ( z ): Y z = ay [− 1 ] 1 az 1 b 1 az 1 X z Not the best form for doing Inverse ZT… we want things in terms of z not z -1 Y z =− ay [− 1 ] z z a bz z a X z Multiply each term by z / z H z = bz z a On ZT Table y [ n ]=− ay [− 1 ]− a n u [ n ] Z 1 { H z X z } Part due to input signal modified by Transfer Function If | a | < 1 this dies out as n , its an IC-driven transient (ZIR) If the ICs are zero, this is all we have! (ZSR)
Ex.: Solving a Difference Equation using ZT : 1 st -Order System w/ Step Input For x [ n ]= u [ n ] X z = z z 1 Then using our general results we just derived we get: Y z = ay [− 1 ] z z a bz z a  z z 1

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