# 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.