Why It WorksHow to Do It1Partial 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 ()()( )NNNNNNpspspsdsNdsdsdsdsNsDsNsY−−−=++++==−−!!210111)()()()()(2For the case we are considering this can be expanded into the form ()() ()NNpsrpsrpsrsY−++−+−= !2211)( 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 ()()()()()()()()()()()NNNNpspsrpspsrrpspsrpspsrpspsrsYps−−++−−+=−−++−−+−−=−1212112121111)(!! 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.).3Now note that if we let s = p1 then all the terms containing the other residues disappear: ()[]()()()()1112112111)( rpspprpspprrsYpsNNps=−−++−−+=−=! Now… you might wonder why this doesn’t make the left-hand side disappear since there is now also a (p1 – p1) term on that side!!! Remember that there is an (s – p1) term in the denominator of Y(s) that cancels the multiplied (s – p1) term!!! So this gives us a nice mathematical result on how to find the residues: ()[]ipsiisYpsr=−= )( Remember that this equation only hold for the case when there are no repeated roots and the order of the numerator of Y(s) is strictly less than the order of the denominator of Y(s) How to Do It Illustrate with an example. Find the inverse LT of 2313)(2++−=ssssY First we have to factor the denominator. Using the quadratic formula we get the roots and the write the factored form as ()()1213)(++−=ssssY4Then we use the equation we derived above: ()[]()( )()()()()222111312132)(2−=−=−=+−=++−+=+=ssssssssssYsr or finally r1 = 7. Notice that we MUST cancel the multiplied factor BEFORE we evaluate at the root value. Similarly, ()[]()( )()()()()111221312131)(1−=−=−=+−=++−+=+=ssssssssssYsr or finally r2 = -4.5Partial Fraction Expansion via MATLAB The “residue” function of MATLAB can be used to compute the partial fraction expansion (PFE) of a ratio of two polynomials. This can be used for Laplace transforms or Z transforms, although we will illustrate it with Laplace transforms here. The “residue” command gives three pieces of information: ! the residues are given in output vector r, ! the poles are given in output vector p, ! the so-called direct terms are given in output vector k. When the order of the numerator polynomial: ! is less than the order of the denominator polynomial there will be no direct terms. ! equals the order of the denominator there will be one direct term;. ! is one greater than the order of the denominator there will be two direct terms. ! etc. That is, when the order of the numerator is p greater than the order of the denominator (with p≥0) there will be p+1 direct terms. The “residue” command requires two input vectors: ! one holding the coefficients of the numerator and ! one holding the coefficients of the numerator.6The right-most element in these vectors corresponds to the s0 coefficient, the next element to the left is the s1 coefficient, etc., until you reach the highest power; if a power is not present it has a zero coefficient. It is easiest to explain how to use “residue” by giving examples. Ex. #1: No Direct Terms, No Repeated Roots, No Complex Roots 2313)(2++−=ssssY In this case there will be no direct terms because the numerator order is lower than the denominator order. The numerator vector is [3 –1] The denominator vector is [1 3 2]. The command and its result is: » [r,p,k]=residue([3 -1],[1 3 2]) r = 7 -4 p = -2 -1 k = []7 Note that k is empty (denoted by []) showing that there are no direct terms. The numbers in r are the residues – they are the numbers that go in the numerator of each term in the expansion. The numbers in p are the poles – they are the numbers subtracted from s to form the denominators of the terms. We have a pole at s = -2 with a residue of 7 and…. …a pole at s = -1 with a residue of –4; thus: 14272313)(2+−++=++−=ssssssY and then using the Laplace Transform Table for each of the two terms in the expansion we get: )(4)(7)(2tuetuetytt −−−=8Ex. #2: One Direct Term, No Repeated Roots, No Complex Roots 23132)(22++−+=sssssY » [r,p,k]=residue([2 3 -1],[1 3 2]) r = -1 -2 p = -2 -1 k = 2 Thus, we have: a pole at: -2 w/ residue -1 a pole at: -1 w/ residue -2; …. But in this case we also have a single direct term of 2, which just adds a term of 2 to the expansion: 2122123132)(22++−++−=++−+=sssssssY and then using the Laplace Transform Table for each of the three terms in the expansion we get: )(2)(2)()(2ttuetuetyttδ+−−=−−9Ex. #3: Two Direct Terms, No Repeated Roots, No Complex Roots 231327)(223++−++=ssssssH » [r,p,k]=residue([7 2 3 -1],[1 3 2]) r = 55 -9 p = -2 -1 k = 7 -19 This gives two terms in the vector k