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EECE 301 Note Set 28a Reading on Partial Fractions

EECE 301 Note Set 28a Reading on Partial Fractions - 1...

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1 Partial Fraction Expansion When trying to find the inverse Laplace transform (or inverse z transform) it is helpful to be able to break a complicated ratio of two polynomials into forms that are on the Laplace Transform (or z transform) table. We will illustrate here using Laplace transforms. This can be done using the method of “partial fraction expansion” (PFE), which is the reverse of finding a common denominator and combining fractions. It is possible to do PFE by hand or it is possible to use MATLAB to help. We will illustrate hand computation only for the simplest case when there are no repeated roots and the order of the numerator polynomial is strictly less than the order of the denominator polynomial. First we will show why the “by-hand” method works and then we will show how one actually does it. Then we will illustrate how to use MATLAB to do PFE. Why It Works For that case, suppose we have a LT Y ( s ) that we wish to invert but it is not on our table. Suppose that the denominator of Y ( s ) can be factored as () ( ) ( ) N N N N N N p s p s p s d s N d s d s d s d s N s D s N s Y = + + + + = = " " 2 1 0 1 1 1 ) ( ) ( ) ( ) ( ) (

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2 For the case we are considering this can be expanded into the form () () () N N p s r p s r p s r s Y + + + = " 2 2 1 1 ) ( where the r i are numbers called the “residues” of the expansion. The goal of doing a PFE is to find the residues so you can form the right-hand side of the above equation. So… we need a way to solve this for each r i . Let’s see how to do that for r 1 (the other r i ’s are done the same). We multiply this equation on both sides by ( s p 1 ) and get () () () ( ) () () () () () () () N N N N p s p s r p s p s r r p s p s r p s p s r p s p s r s Y p s + + + = + + + = 1 2 1 2 1 1 2 1 2 1 1 1 1 ) ( " " where in the second line we have canceled the common ( s p 1 ) in the first term on the right hand side (…. this is the whole point of the process!! We have isolated r 1 !! But there are still a bunch of terms left in the way… but we can now get rid of them as follows.).
3 Now note that if we let s = p 1 then all the terms containing the other residues disappear: () [] ( ) () () () 1 1 1 2 1 1 2 1 1 1 ) ( r p s p p r p s p p r r s Y p s N N p s = + + + = = " Now… you might wonder why this doesn’t make the left-hand side

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EECE 301 Note Set 28a Reading on Partial Fractions - 1...

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