Simple Poles

Recall from Chapter 21 that a simple pole is a first-order pole. If $ F(s) $ has only simple poles, then $ D(s) $ becomes a product of factors, so that where $ s=-p_{1},-p_{2}, \ldots,-p_{n} $ are the simple poles, and $ p_{i} \neq p_{j} $ for all $ i \neq j $ (i.e., the poles are distinct). Assuming that the degree of $ N(s) $ is less than the degree of $ D(s) $. we use partial fraction expansion to decompose $ F(s) $ in Eq. $ (1) $ as The expansion coefficients $ k_{1}, k_{2}, \ldots, k_{n} $ are known as the residues of $ F(s) $. There are many ways of finding the expansion coefficients. One way is using the residue method. If we multiply both sides of Eq. (1) by $ \left(s+p_{1}\right) $, we obtain Since $ p_{i} \neq p_{j} $, setting $ s=-p_{1} $ in Eq. (3) leaves only $ k_{1} $ on the right-hand side of Eq. (3). Hence, Thus, in general, This is known as Heaviside's theorem. Once the values of $ k_{i} $ of we proceed to find the inverse of $ F(s) $ using Eq. (2). Since the inverse . $ \quad $ transform of each term in Eq. (2) is $ \mathcal{L}^{-1}[k /(s+a)]=k e^{-a t} u(t) $, then, from Table 1,