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Linear Constant-Coefficient Difference Equations

Module by: Richard Baraniuk

Summary: A module concering the concepts of linear constant-coefficient difference equations.

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  • remember linear differential equations? ddtyt−yt=xt t y t y t x t
  • A difference equation is the discrete-time analogue of a differentail equation. We simply use differences ( xn−xn−1 x n x n 1 ) rather than derivatives ( ddtxt t x t ).
  • An important subclass of linear systems consists of those whose input xn x n and output xn x n obey an N N-th order LCCDE:
    ∑K=0N a K yn−K=∑K=0M b K xn−K K 0 N a K y n K K 0 M b K x n K (1)

Example 1: Moving average system

yn=1 M 1 + M 2 +1∑K= M 1 M 2 xn−K y n 1 M 1 M 2 1 K M 1 M 2 x n K where set M 1 =0 M 1 0 and M 2 =M M 2 M , a K = 1ifK=00otherwise a K 1 K 0 0 , b K = 1M+1if0≤K≤M0otherwise b K 1 M 1 0 K M 0

How to Implement?

code / hardware M=2 M 2 : yn=13∑K=02xn−K y n 1 3 K 0 2 x n K

IIR? FIR?

Example 2: Recursive System

yn=∑K=1N α K yn−K+xn y n K 1 N α K y n K x n ∑K=0N a K yn−K=xn K 0 N a K y n K x n Where a K =1ifK=0- α K if1≤K≤N0otherwise a K 1 K 0 α K 1 K N 0

How to Implement?

N=2 N 2 : yn=∑K=12 α K yn−K+xn y n K 1 2 α K y n K x n

IIR? FIR?

FIR ~ MOVING AVERAGE ~ FEED FORWARD

IIR ~ RECURSIVE ~ FEED BACK

  • The SOLUTION of a difference equation is similar to a differential equation.
  • In particular, note that a single input-output pair ( xn x n , y p n y p n ) that solves the DE is not enough to characterize the solution. ∑k=0N a k y p n−k=∑k=0M b k xn−k k N 0 a k y p n k k M 0 b k x n k Add in zero to get the homogenous equation: ∑k=0N a k y h n−k=0 k N 0 a k y h n k 0 . ∑k=0N a k y p n−k+ y h n−k=∑k=0M b k xn−k k N 0 a k y p n k y h n k k M 0 b k x n k where the particular solution ("forced") y p n y p n and homogeneous solution ("unforced") y h n y h n

    General Solution

    yn= y p n+ y h n y n y p n y h n (2)

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