Residue Computation. Let PsQs be a rational function with degP<degQ and suppose s-r is a non-repeated linear factor of Qs . Prove that the portion of the partial fraction expansion of PsQs corresponding to s-r is As-r, where A (called the residue) is given by the formularole="math" localid="1664365632102" A=lims→rs-rPsQs=PrQ'r

Writing Qs=s-rs-r1...s-rn,using partial fractions and then applying on the equation s-rPsQs=A+s-rA1s-r+...+s-rAns-rnwe obtain the wanted result.

38E - Fundamentals Of Differential Equations And Boundary Value Problems

38E

Question

Residue Computation. Let $\frac{P (s)}{Q (s)}$ be a rational function with deg $P < d e g Q$ and suppose $s - r$ is a non-repeated linear factor of $Q (s)$ . Prove that the portion of the partial fraction expansion of $\frac{P (s)}{Q (s)}$ corresponding to $s - r$ is $\frac{A}{s - r}$ , where $A$ (called the residue) is given by the formula

$A = \underset{s \to r}{l i m} \frac{(s - r) P (s)}{Q (s)} = \frac{P (r)}{Q^{'} (r)}$

Step-by-Step Solution

Verified

Answer

Writing $Q (s) = (s - r) (s - r_{1}) . . . (s - r_{n})$ ,using partial fractions and then applying on the equation $\frac{(s - r) P (s)}{Q (s)} = A + \frac{(s - r) A_{1}}{s - r_{_{}}} + . . . + \frac{(s - r) A_{n}}{s - r_{n}}$ we obtain the wanted result.

1Step 1: Define Inverse Laplace transform

Given a function , if there is a function that is continuous on

and satisfies,then we say that is the inverse Laplace transform of and employ the notation

Non-repeated Linear Factors

If can be factored into a product of distinct linear factors,

where the 's are all distinct real numbers, then the partial fraction expansion has the form

where the 's are real numbers. There are various ways of determining the constants . In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If is a factor of and is the highest power of that divides , then the portion of the partial fraction expansion of that corresponds to the term is

where the 's are real numbers.

Quadratic Factors

If is a quadratic factor of that cannot be reduced to linear factors with real coefficients and is the highest power of that divides , then the portion of the partial fraction expansion that corresponds to is

it is more convenient to express in the form when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

2Step 2: Find the factor of the denominator

Since

$Q (s) = (s - r_{1}) . . . (s - r_{n_{}})$

Then, using partial fractions we get

$\frac{P (s)}{Q (s)} = \frac{A_{1}}{s - r_{1}} + . . . + \frac{A_{n}}{s - r_{n}} = \sum_{i = 1}^{n} \frac{A_{i}}{s - r_{i}}$

from where it follows

$L^{- 1} \{\frac{P (s)}{Q (s)}\} (t) = \sum_{i = 1}^{n} L^{- 1} \{\frac{A_{I}}{s - r_{i}}\} = \sum_{i = 1}^{n} A_{i} e^{r_{i} t}$

Multiplying equation 1 with $s - r_{i}$ we get

$\frac{s - r_{i} P (s)}{Q (s)} = A_{i} + \sum_{i \neq j, j = 1}^{n} \frac{(s - r_{i}) A_{j}}{s - r_{j}}$

3Step 3: Applying the limit

Applying " width="9" height="19" style="max-width: none;" > $\lim_{s \to r}$ gives

$\lim_{s \to r} \frac{(s - r) P (s)}{Q (s)} = \lim_{s \to r} (A + \frac{(s - r) A_{1}}{s - r_{1}} + . . . + \frac{(s - r) A_{n}}{s - r_{n}}) = A + 0 + . . . + 0 = A$