Linear ConstantCoefficient Difference E iEquations

Some physical systems can be modelled using linear difference equations withusing linear difference equations with constant coefficients (LCCDE)

These systems are LTI systems

LCCDELCCDE LCCDEs can be generally represented by theLCCDE s can be generally represented by the

following expressions:

== = Mm mNk k mnxbknya 00 ][][

MN == += Mm mNk k mnxbknyany 01 ][][][

Example of a LCCDEExample of a LCCDE

][][][][ 1 kxnxkxny nk

n

k+== ==

][]1[ 1 kxny nk

= =][]1[][]1[][][nxnyny

nynxny=+=

][]1[][ nxnyny =

Realisation of and LCCDERealisation of and LCCDE

]1[][][ += nynxny+

x[n] y[n]

One sample delay

y[n-1]

S l i LCCDESolving LCCDEs Solution is given by:

[ ] i h ( l t )][][][ nynyny ph +=

yh[n] is homogeneous (or complementary)solution for input x[n]=0, represents natural modes of unforced oscillation

yp[n] is particular solution due to input x[n], represents forced response

SolvingSolving

LCCDEs are solved in two parts.

First solve the complementary equation, set to zero.

Then solve the particular solution.

The Homogenous SolutionThe Homogenous Solution

Assume:][ n

0

][

==

N knn

a

ny

( ) 0...0

11

10

0

=++++ =

NNNNNn

k k

aaaa

a

( )110 44444 344444 21 NNCharacteristic polynomial of the system

The Homogenous SolutionThe Homogenous Solution Factorizing characteristic polynomial we get:

( )( ) ( ) 0:such that = So the homogeneous solution is:

( )( ) ( ) 0...-:such that , 21 = Ni g

nNN

nnh cccny +++= ...][ 2211

Where ci are free to match initial conditions (IC)

Example Homogenous SolutionExample Homogenous Solution][]1[60][

:isequationsticcharacterithe

][]1[6.0][ nxnyny =

06.0

:isequation sticcharacteri the

10 aa ==+ 6.0=

)6.0(][, cnnythus h =

Repeated RootsRepeated Roots

Repeated roots in characteristic polynomial:

( ) ( )( )

.....12

21 L

L ( )...][ 1112210

++++++++=

nn

nLLh

ccncncnccny

...312 +++ +LL cc

P i l l iParticular solution Represents the forced response of the system

Solution depends on the form of the input

W h f f [ ] di th f f [ ] We choose form of yp[n] according the form of x[n]

Substitute y [n] into nonhomogeneous DE andSubstitute yp[n] into non homogeneous DE and compare coefficients on both sides of the equation

LCCDE lLCCDE example Consider following LCCDE:

][2]2[06.0]1[5.0][ =+ nxnynyny

)10(][:isinput where

][][][][ yyy

n

:are conditions initial and)1.0(][ =nx n

0]2[1]1[==

yy

0]2[y

LCCDE example contdLCCDE example cont d

Characteristic polynomial:

( )( ) 02.03.0006.05.02

==+

( )( )2.0,3.0

02.03.0

21 ==

( ) ( )nnh ccny 2.03.0][ 21 +=

LCCDE example contdLCCDE example cont d

Particular solution is in the form:

( )np Bny eq.nonhomog.tosubstitute,1.0][ = ( )( )nppp

p

nynyny

y

1.02]2[06.0]1[5.0][

qg,][

=+( ) ( ) ( ) ( )

( ) ( )nnnn

BBB

BBB

2100601050

1.021.006.01.05.01.021

21

=+=+

( ) ( )

( )nBB

BBB

1,22

21.006.01.05.0

==+

( )np ny 1.0][ =

LCCDE example contd

Total solution:

( ) ( ) ( )102030][][][ nnn( ) ( ) ( )( ) ( ) ( ) ( )1.02]2[06.0]1[5.01.02.03.0:0

1.02.03.0][][][000

20

1

21

yyccn

ccnynyny nnnph

=+++=++=+=

( ) ( ) ( ) ( )4.8,9.9

1.02]1[06.0]0[5.01.02.03.0:1

21

1112

11

cc

yyccn

===+++=

( ) ( ) ( )( ) ][1.02.04.83.09.9][ 21 nuny nnn +=

LCCDE example contdLCCDE example cont d

2.5

2

1.5

1

0

0.5

0 0.5 1 1.5 2 2.5 3 3.5 40

LCCDE solution

Impulse response of LCCDEImpulse response of LCCDE

If input is impulse, and IC are initial rest conditions output is p p , pimpulse response:

][][][][ h ][][][][ nhnynnx ==

Impulse response exampleImpulse response example

Consider the same LCCDE:

nxnynyny ][2]2[06.0]1[5.0][ =+nnx

:are conditions initial and][][ =

( ) ( )nnh ccnynny

2.03.0][

0,0][

21 +=

Impulse response exampleImpulse response example

Find unknown ci][2]2[06.0]1[5.0][ nxnynyny =+

( ) ( )2.03.0][0,0][],[][

21 ccny

nnynnxnn

h +=

Impulse response exampleImpulse response example

Finally, solving system of 2 equations for two unknowns we get:

4,6 21 cc ==( ) ( )( ) ][2.043.06][ 4,6 21 nunh cc nn = ( )