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## Mapping the system function from the s-plane to the z-plane in the presence of multiple order poles.

April 22, 2010   Compiled on May 20, 2020 at 5:19pm

Given $$H\left ( s\right )$$ of order $$N$$ with all its poles $$p_{i}$$ being distinct, it can be expressed in terms of partial fraction expansion in the form of $$H\left ( s\right ) ={\sum \limits _{k=1}^{N}}\frac{A_{k}}{s-p_{k}}$$ and the resulting $$H\left ( z\right )$$ can be found to be $${\sum \limits _{k=1}^{N}}\frac{zA_{k}}{z-e^{p_{k}T}}$$ where $$T$$ is the sampling period.

In the case when $$H\left ( s\right )$$ contains a pole $$q$$ of order $$2$$, then $$H\left ( s\right )$$ can be written as $$\left ({\sum \limits _{k=1}^{N-2}}\frac{A_{k}}{s-p_{k}}\right ) +\frac{A_{q}}{\left ( s-q\right ) ^{2}}$$ and the resulting $$H\left ( z\right )$$ can be found to be $$\left ({\sum \limits _{k=1}^{N-2}}\frac{zA_{k}}{z-e^{p_{k}T}}\right ) +\frac{Tze^{qT}}{\left ( e^{qT}-z\right ) ^{2}}$$.

In the case when $$H\left ( s\right )$$ contains a pole $$q$$ of order $$3$$, then $$H\left ( s\right )$$ can be written as$$\left ({\sum \limits _{k=1}^{N-3}}\frac{A_{k}}{s-p_{k}}\right ) +\frac{A_{q}}{\left ( s-q\right ) ^{3}}$$ and the resulting $$H\left ( z\right )$$ can be found to be $$\left ({\sum \limits _{k=1}^{N-3}}\frac{zA_{k}}{z-e^{p_{k}T}}\right ) +\left ( -\frac{e^{2qT}T^{2}z+e^{qT}T^{2}z^{2}}{2\left ( e^{qT}-z\right ) ^{3}}\right )$$.

The following table was generated in order to obtain the general formula. This table below shows only the part of $$H\left ( z\right )$$ due to the multiple order pole.

 $$n$$ pole order $$H\left ( z\right )$$ $$2$$ $$\frac{Tze^{qT}}{\left ( e^{qT}-z\right ) ^{2}}$$ $$3$$ $$-\frac{e^{2qT}T^{2}z+e^{qT}T^{2}z^{2}}{2\left ( e^{qT}-z\right ) ^{3}}$$ $$4$$ $$\frac{e^{3qT}T^{3}z+4e^{2qT}T^{3}z^{2}+e^{qT}T^{3}z^{3}}{6\left ( e^{qT}-z\right ) ^{4}}$$ $$5$$ $$\frac{-e^{4qT}T^{4}z-11e^{3qT}T^{4}z^{2}-11e^{2qT}T^{4}z^{3}-e^{qT}T^{4}z^{4}}{24\left ( e^{qT}-z\right ) ^{5}}$$ $$6$$ $$\frac{e^{5qT}T^{5}z+26e^{4qT}T^{5}z^{2}+66e^{3qT}T^{5}z^{3}+26e^{2qT}T^{5}z^{4}+e^{qT}T^{5}z^{5}}{120\left ( e^{qT}-z\right ) ^{6}}$$

It is easy to see that the denominator of $$H\left ( z\right )$$ has the general form $$\left ( n-1\right ) !\left ( e^{qT}-z\right ) ^{n}$$ where $$n$$ is the pole order, the hard part is to ﬁnd the general formula for the numerator. The following table is a rewrite of the above table, where only the numerator is show, and $$e^{qT}$$ was written as $$A$$ to make it easier to see the general pattern

 $$n$$ pole order numerator of $$H\left ( z\right )$$ $$2$$ $$\left ( -1\right ) ^{n}\left ( AT\right ) z$$ $$3$$ $$\left ( -1\right ) ^{n}\left [ \left ( AT\right ) ^{2}z-A\left ( Tz\right ) ^{2}\right ]$$ $$4$$ $$\left ( -1\right ) ^{n}\left [ \left ( AT\right ) ^{3}z+4A^{2}T^{3}z^{2}+A\left ( Tz\right ) ^{3}\right ]$$ $$5$$ $$\left ( -1\right ) ^{n}\left [ \left ( AT\right ) ^{4}z-11A^{3}T^{4}z^{2}-11A^{2}T^{4}z^{3}-A\left ( Tz\right ) ^{4}\right ]$$ $$6$$ $$\left ( -1\right ) ^{n}\left [ \left ( AT\right ) ^{5}z+26A^{4}T^{5}z^{2}+66A^{3}T^{5}z^{3}+26A^{2}T^{5}z^{4}+A\left ( Tz\right ) ^{5}\right ]$$

I am trying to determine the general formula to generate the above. This seems to involve some combination of binomial coeﬃcient. But so far, I did not ﬁnd the general formula.

### 1 References

1. Digital signal processing, by Oppenheim and Scafer, page 201
2. Mathematica software version 7