April 22, 2010 Compiled on January 29, 2024 at 2:53am

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.

- Digital signal processing, by Oppenheim and Scafer, page 201
- Mathematica software version 7