# Repeated Poles

Suppose $F(s)$ has $n$ repeated poles at $s=-p$. Then we may represent $F(s)$ as
\bbox[10px,border:1px solid grey]{\begin{aligned}F(s)=& \frac{k_{n}}{(s+p)^{n}}+\frac{k_{n-1}}{(s+p)^{n-1}}+\cdots+\frac{k_{2}}{(s+p)^{2}} \\&+\frac{k_{1}}{s+p}+F_{1}(s)\end{aligned}} \tag{1}
where $F_{1}(s)$ is the remaining part of $F(s)$ that does not have a pole at $s=-p$. We determine the expansion coefficient $k_{n}$ as
$$k_{n}=\left.(s+p)^{n} F(s)\right|_{s=-p}$$
as we did above. To determine $k_{n-1}$, we multiply each term in Eq. (1) by $(s+p)^{n}$ and differentiate to get rid of $k_{n}$, then evaluate the result at $s=-p$ to get rid of the other coefficients except $k_{n-1}$. Thus, we obtain
$$k_{n-1}=\left.\frac{d}{d s}\left[(s+p)^{n} F(s)\right]\right|_{s=-p}$$
Repeating this gives
$$k_{n-2}=\left.\frac{1}{2 !} \frac{d^{2}}{d s^{2}}\left[(s+p)^{n} F(s)\right]\right|_{s=-p}$$
The $m$ th term becomes
$$k_{n-m}=\left.\frac{1}{m !} \frac{d^{m}}{d s^{m}}\left[(s+p)^{n} F(s)\right]\right|_{s=-p}$$
where $m=1,2, \ldots, n-1$. One can expect the differentiation to be difficult to handle as $m$ increases. Once we obtain the values of $k_{1}$, $k_{2}, \ldots, k_{n}$ by partial fraction expansion, we apply the inverse transform
$$\mathcal{L}^{-1}\left[\frac{1}{(s+a)^{n}}\right]=\frac{t^{n-1} e^{-a t}}{(n-1) !}$$
to each term in the right-hand side of Eq. (1) and obtain
\begin{aligned}f(t)=& k_{1} e^{-p t}+k_{2} t e^{-p t}+\frac{k_{3}}{2 !} t^{2} e^{-p t} \\&+\cdots+\frac{k_{n}}{(n-1) !} t^{n-1} e^{-p t}+f_{1}(t)\end{aligned}
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