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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 page compiled on July 1, 2015 at 2:34am

Given H (s)  of order N  with all its poles pi  being distinct, it can be expressed in terms of partial fraction expansion in the form of         ∑N
H (s) =     -Ak-
            s−pk
        k=1   and the resulting H  (z)  can be found to be ∑N
   --zAkp-T
   z−e k
k=1   where T  is the sampling period.

In the case when H (s)  contains a pole q  of order 2  , then H (s)  can be written as (N − 2    )
  ∑   -Ak-  + --Aq-
      s− pk    (s−q)2
  k=1   and the resulting H (z)  can be found to be (N −2      )
  ∑  --zAk--     -TzeqT--
     z−epkT  +  (eqT−z)2
  k=1   .

In the case when H (s)  contains a pole q  of order 3  , then H (s)  can be written as(        )
 N∑ −3 A        Aq
     s−kpk  +  (s−q)3
  k=1   and the resulting H (z)  can be found to be (          )
  N∑−3           (   2qT 2   qT 2 2)
     z−zAekpkT  +   − e--T2(ezq+Te−z)T3-z-
  k=1 .

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

|--------------|--------------------------------------------|
| n pole order |H (z)                                       |
|--------------|-TzeqT--------------------------------------|
| 2            |(eqT−-z)2                                     |
|--------------|--e2qTT2z+eqTT2z2----------------------------|
|-3------------|−----2(eqT−z)3-------------------------------|
| 4            |e3qTT-3z+4e2qTT-3z2+eqTT3z3                     |
|--------------|--------6(eqT-−z)4-----------------------------|
| 5            |−e4qTT4z−-11e3qT-T4z2q−T11e2q5TT4z3−eqTT4z4-         |
|--------------|-5qT--5----4qT-254(e2--−z3)qT-5-3---2qT-54--qT-5-5-|
| 6            |e--T-z+26e--T-z-+61620e(eqTT−-zz)6+26e---T-z+e--T-z- |
------------------------------------------------------------

It is easy to see that the denominator of H (z)  has the general form (n − 1)!(eqT − z)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 eqT  was written as A  to make it easier to see the general pattern

|-------------|----------------------------------------------------------------|
|n pole order |numerator  of H (z)                                             |
|-------------|-----n----------------------------------------------------------|
|2------------|(−-1)-(AT--)z---------------------------------------------------|
|3            |(− 1)n[(AT  )2 z − A (Tz)2]                                      |
|-------------|------[-----------------------------]---------------------------|
-4-------------(−-1)n-(AT--)3-z +-4A2T-3z2 +-A-(Tz)3----------------------------|
|             |     n[     4         3 4 2       2 4 3          4]              |
|5------------|(−-1)-[(AT--)-z −-11A-T--z--−-11A--T-z--−-A-(T-z)-------------]-|
|6            |(− 1)n (AT  )5 z + 26A4T 5z2 + 66A3T 5z3 + 26A2T 5z4 + A (T z)5  |
-------------------------------------------------------------------------------

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