lecture 6
DESCRIPTION
signals and systemsTRANSCRIPT
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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
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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 ][][][
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Example of a LCCDEExample of a LCCDE
][][][][ 1 kxnxkxny nk
n
k+== ==
][]1[ 1 kxny nk
= =][]1[][]1[][][nxnyny
nynxny=+=
][]1[][ nxnyny =
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Realisation of and LCCDERealisation of and LCCDE
]1[][][ += nynxny+
x[n] y[n]
One sample delay
y[n-1]
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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
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SolvingSolving
LCCDEs are solved in two parts.
First solve the complementary equation, set to zero.
Then solve the particular solution.
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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
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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)
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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 =
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Repeated RootsRepeated Roots
Repeated roots in characteristic polynomial:
( ) ( )( )
.....12
21 L
L ( )...][ 1112210
++++++++=
nn
nLLh
ccncncnccny
...312 +++ +LL cc
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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
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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
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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 +=
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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][ =
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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 +=
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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
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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 ==
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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 +=
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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 +=
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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 = ( )