# BU EECE 301 - Partial Fraction Expansion (13 pages)

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## Partial Fraction Expansion

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## Partial Fraction Expansion

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Pages:
13
School:
Binghamton University
Course:
Eece 301 - Signals and Systems (DIS)
##### Signals and Systems (DIS) Documents

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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 Y s N s N s D s d N s N d N 1s N 1 d1s d 0 N s d N s p1 s p2 s p N 2 For the case we are considering this can be expanded into the form Y s rN r1 r2 s p1 s p2 s p N where the ri 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 ri Let s see how to do that for r1 the other ri s are done the same We multiply this equation on both sides by s p1 and get rN s p1 r1 s p1 r2 s p1 s p1 Y s s p1 s p2 s p N r1 r s p1 r2 s p1 N s p N s p2 where in the second line we have canceled the common s p1 in the first term on the right hand side this is the whole point of the process We have isolated r1 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 p1 then all the terms containing the other residues disappear s p1 Y s s p 1 r1 r p p1 r2 p1 p1 N 1 r1 s p2 s p N Now you might

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