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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