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

Nasser M. Abbasi

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