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
$$\bbox[10px,border:1px solid grey]{F(s)=\frac{N(s)}{\left(s+p_{1}\right)\left(s+p_{2}\right) \cdots\left(s+p_{n}\right)}} \tag{1}$$
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
$$F(s)=\frac{k_{1}}{s+p_{1}}+\frac{k_{2}}{s+p_{2}}+\cdots+\frac{k_{n}}{s+p_{n}} \tag{2}$$
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
$$\left(s+p_{1}\right) F(s)=k_{1}+\frac{\left(s+p_{1}\right) k_{2}}{s+p_{2}}+\cdots+\frac{\left(s+p_{1}\right) k_{n}}{s+p_{n}} \tag{3}$$
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,
$$\left.\left(s+p_{1}\right) F(s)\right|_{s=-p_{1}}=k_{1}$$
Thus, in general,
$$k_{i}=\left.\left(s+p_{i}\right) F(s)\right|_{s=-p_{i}}$$
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,
$$f(t)=\left(k_{1} e^{-p_{1} t}+k_{2} e^{-p_{2} t}+\cdots+k_{n} e^{-p_{n} t}\right)$$
Historical note: Named after Oliver Heaviside are known, (1850-1925), an English engineer, the pioneer operational calculus.