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IIR digital filter design for a low pass filter based on impulse invariance and bilinear transformation methods using butterworth analog filter

Nasser M. Abbasi. Written May 10, 2010

June 21, 2013

Contents

1 Introduction
 1.1 Filter specifications
2 Analytical derivation of the design steps
 2.1 Impulse invariance method
 2.2 bilinear transformation method
 2.3 Summary of analytical derivation method
3 Numerical design examples
 3.1 Example 1
  3.1.1 using impulse invariance method (using T=1)
  3.1.2 bilinear method
 3.2 Example 2
  3.2.1 using impulse invariance method
  3.2.2 Using bilinear
4 References

1 Introduction

This report derives a symbolic procedure to design a low pass IIR digital filter from an analog Butterworth  filter using 2 methods: impulse invariance and bilinear transformation.  Two numerical examples are then used to illustrate using the symbolic procedure. There are a total of 13 steps.

A Mathematica program with GUI is written to enable one to use this design for different parameters. A Matlab script is being written as well.

1.1 Filter specifications

Filter specifications are 5 parameters. The frequency specifications are analog frequencies, while the attenuations for the passband and the stopband are given in db



Fs  The sampling frequency in Hz


fc  The passband cutoff frequency in Hz


f
 s  The stopband corner frequency in Hz


δp  The attenuation in db at fc


δs  The attenuation in db at fs


This diagram below illustrates these specifications

PIC

The frequency specifications are in Hz  and they must be first converted to digital frequencies ω  where 0 ≤ |ω| ≤ π  before using the attenuation specifications, The sampling frequency Fs  is used to do this conversion since Fs  corresponds to 2π  on the digital frequency scale.

2 Analytical derivation of the design steps

These are the design steps

  1. Convert specifications from analog to digital frequencies
  2. Based on design method (impulse invariance of bilinear) apply the attenuation criteria to determine Ωc  and N  (the filter order)
  3. Using Ωc  and N  find the locations of the poles of H  (s)  , the butterworth analog filter.
  4. Find H  (z)  from H (s)  . The method of doing this depends if we are using impulse invariance or bilinear. This step is much simpler for the bilinear method as it does not require performing partial fractions decomposition on H (s)

Now we begin the analytical design procedure.

2.1 Impulse invariance method

We first convert analog specifications to digital specifications: Fs   fp
2π = ωp   , hence         fp-
ωp = 2π Fs   and ωs = 2π fsFs

Convert the criteria relative to the digital normalized scale:

pict

Therefore

pict

Notice that we added the T  scaling factor. Now, Butterworth analog filter squared magnitude Fourier transform is given by

         2       T2
|Ha (jΩ)| =  ---(---)2N--
             1+  jjΩΩc

hence equations (A) and (B) above are now written in terms of the analog Butterworth amplitude frequency response and become

pict

Therefore the above becomes

pict

Now, for impulse invariance,       ωp-
Ωp =  T   and       ωs
Ωs =  T  .  

pict

Change inequalities to equalities and simplify

pict

Divide the above 2 equations

pict

We need to round the above to then nearest integer using the Ceiling function  i.e. N  =  ⌈N ⌉

Now for impulse invariance method, use equation (1) above to solve for Ωc

pict

Now that we found N  and Ωc  , next find the poles of H (s).  Since the butterworth magnitude square of the transfer function is

               T 2
|Ha (s)|2 = ---(---)2N--
           1+   jΩsc-

Hence H (s)  poles are found by setting the denominator of the above to zero

pict

We only need to find the LHS poles, which are located at i = 0 ⋅⋅⋅N - 1  , because these are the stable poles. Hence the ith  pole is

        j(π(1+2i+N-))
si = Ωc e    2N

For example for i = 0  , N = 6  we get

          (π(1+N))        π7
s0 = Ωc ej  2N   =  Ωc ej(12)

Now we can write the analog filter generated based on the above selected poles, which is, for impulse invariance

        ---T-K-----
Ha (s) = N-∏ 1
            (s- si)
        i=0
(3)

K  is found by solving H  (0) = T
 a   hence

    N∏-1
k =    (- si)
    i=0

Now we need to write poles in non-polar form and plug them into (3)

         j(π(1+2i+N))     (    π-(1-+-2i+-N-)       π-(1-+-2i+-N-))
si = Ωc e    2N     = Ωc  cos     2N       + jsin      2N                       i = 0 ⋅⋅⋅N - 1

Hence,

         ------------------T-K--------------------
Ha (s) = N∏-1 (      (                          ) )
              s- Ωc  cos π(1+22iN+N-)+ jsin π(1+22Ni+N-)
         i=0
(4)

Where

Ωc = --(-----ωp∕(T------))-
         -1 log  10- δ1p0-1
     10  2N  10

and

    ⌊     [    δp-   ]     [    δs-   ]⌋
    | 1log 10- 10 - 1 - log 10- 10 - 1 |
N = || 2-------log(ω-)--log(ω-)-------||
    |              p         s       |

Now that we have found H (s)  we need to convert it to H (z)

We need to make sure that we multiply poles of complex conjugates with each others to make the result simple to see.

Now that we have Ha (s)  , we do the A  → D  conversion. I.e. obtain H  (z)  from the above H (s)  . When using impulse invariance, we need to perform partial fraction decomposition on (4) above in order to write H (s)  in this form

       N -1
        ∑  -Ai---
H (s) = i=0 s- si

For example, to obtain A
  j  , we write

Aj =  lim Ha (s) = ---T-k-----
     s→sj         N∏-1
                     (s - si)
                  ii=⁄=0j

Once we find all the A ′s  , we now write H (z)  as follows

        N- 1
        ∑   ------Ai--------
H  (z ) = i=0 1- exp (si T)z- 1

This completes the design. We can try to convert the above form of H (z)  to a rational expression as         N(z)
H (z) = D(z)

2.2 bilinear transformation method

We first convert analog specifications to digital specifications: F2sπ = fωp
      p   , hence ωp = 2π fpF-
        s   and         fs-
ωs = 2π Fs

Convert the criteria relative to the digital normalized scale:

pict

Hence

pict

Butterworth analog filter squared magnitude Fourier transform is given by

         2        1
|Ha (jΩ)| =  ---(---)2N--
             1+  jjΩΩc

hence equations (A) and (B) above are now written in terms of the analog Butterworth amplitude frequency response and become

pict

Now we assign values for Ωp  and Ωs  as follows,      2-   (ωp)
Ωp =  T tan 2 ,       2-   (ωs)
Ωs =  T tan 2

pict

Change inequalities to equalities and simplify

pict

Divide the above 2 equations

pict

We need to round the above to then nearest integer using the Ceiling function  i.e. N  =  ⌈N ⌉

Now for bilinear transformation we used equation (2) above to find Ωc,  Hence we now solve for Ωc

pict

Now that we found N  and Ωc  we find the poles of H  (s)  . Since for bilinear the magnitude square of the transfer function is

       2        1
|Ha (s)| =  ---(---)2N--
           1+   jΩsc-

Hence H (s)  poles are found by setting the denominator of the above to zero

pict

We only need to find the LHS poles, which are located at i = 0 ⋅⋅⋅N - 1  , because these are the stable poles. Hence the ith  pole is

        j(π(1+2i+N-))
si = Ωc e    2N

For example for i = 0  , N = 6  we get

          (π(1+N))        π7
s0 = Ωc ej  2N   =  Ωc ej(12)

For bilinear, H (s)  is given by

H  (s) = ----K------
 a      N-∏ 1
            (s- si)
        i=0
(3)

K  is found by solving Ha (0) = 1  , hence we obtain

    N-1
k = ∏  (- s )
    i=0    i

We see that the same expression results for k  for both cases.

Now we need to write poles in non-polar form and plug them into (3)

         (        )     (                                     )
         jπ(1+22iN+N)           π-(1-+-2i+-N-)       π-(1-+-2i+-N-)
si = Ωc e           = Ωc  cos     2N       + jsin      2N                       i = 0 ⋅⋅⋅N - 1

then

         --------------------K--------------------
Ha (s) = N∏-1 (      (                          ) )
              s- Ωc  cos π(1+22iN+N-)+ jsin π(1+22Ni+N-)
         i=0
(4)

Where

          2    (ω )
Ω  =  --(-T-tan(-s2---)-)-
  c      21N log10 10δs10- 1
      10

and

    ⌊       (  δ1p0    )       (   δs10-   )⌋
N = || 1log10-1(0--(--1))--log10(10-(--)1)-||
    || 2 log10 tan ωp2  - log10 tan ωs2   ||

Now that we have found H (s)  we need to convert it to H (z)  . After finding H(s)  as shown above, we simply replace s  by      -1
T211-+zz-1   . This is much simpler than the impulse invariance method. Before doing this substitution, make sure to multiply poles which are complex conjugate of each others in the denominator of H (s)  . After this, then do the above substitution

2.3 Summary of analytical derivation method

We will now make a table with the derivation equations to follow to design in either bilinear or impulse invariance. Note that the same steps are used in both designs except for step 5,6,8,13  . This table make it easy to develop a program.

|-----|------------------------|-------------------------------------------|------------------------|
|step-|Impulse--invariance------|common-fpequation--------------------------|bilinear----------------|
|1----|------------------------|ωp =-2πFs----------------------------------|------------------------|
|2    |                        |ωs = 2πfFs                                  |                        |
|3----|------------------------|α--=--1-s----------------------------------|------------------------|
|-----|------------------------|-p---10δp10-----------------------------------|------------------------|
|4    |                        |αs = -1δs-                                  |                        |
|-----|------ωp----------------|-----1010-----------------------------------|-2----(ωp)--------------|
|5----|Ωp-=--T-----------------|-------------------------------------------|-T tan-(2-)-------------|
|6----|Ωs-=--ωTs----------------|----⌈-----------------⌉--------------------|-2T tan-ω2s---------------|
|     |                        |      1log[αp-1]-log[αs-1]                     |                        |
|7----|------------------------|N-=---2-log(Ωp)-log(Ωs)-----------------------|------------------------|
|8    |Ωc =  ---1--Ωp------    |                                           |Ωc =  ---1-Ωs-------    |
|-----|------10(2N-log10[αp-1])-----|----------------------(π(1+2i+N))-----------|------10(2N-log10[αs-1])-----|
|9    |                        |poles of H (S) si = Ωc ej-2N---    i=0⋅⋅⋅N-1 |                        |
|-----|------------------------|----N∏-1------------------------------------|------------------------|
|10   |                        |k =    (- si)                               |                        |
|-----|------------------------|----i=0------------------------------------|------------------------|
|11   |Ha (s) = N-T1K---       |                                           |Ha (s) = N-1K----       |
|     |         ∏  (s-s)       |                                           |         ∏  (s-s)       |
----------------i=0---i--------------------------------------------------------------i=0---i---------
|     |do  partial fractions:    |                                           |                        |
|-----|---------N∑-1------------|-------------------------------------------|------------------------|
|12   |Ha (s) =    -Ai-        |                                           |                        |
|-----|---------i=0s-si--------|-------------------------------------------|------------------------|
|     |        N∑- 1            |                                           |                        |
|13   |H  (z) =     1--expA(Tisi)z-1- |                                           |H  (z) = Ha (s)|s=-21-z--1 |
|-----|--------i=0--------------------------------------------------------------------------T-1+z--1--
|     |

3 Numerical design examples

3.1 Example 1

Sampling frequency Fs = 20khz  , passband frequency fp = 2khz  , stopband frequency fs = 3khz  , with δp ≥ - 1db  and δstop ≤ - 15db

3.1.1 using impulse invariance method (using T=1)
|------|-------------------------------------------------------------------------------------|
| step  |Impulse invariance                                                                   |
|-1----|ω--=-2πfp-→--2π(2000)-→-0.2π-----------------------------------------------------------|
|------|-p-----Ffss----2π2(03000000)------------------------------------------------------------------|
|-2----|ωs-=-2πFs-→---20000-→--0.3π-----------------------------------------------------------|
| 3    |αp = -1δp-→  --1-1 →  1.2589                                                            |
|------|-----10110---10110---------------------------------------------------------------------|
|-4----|αs-=-10δs10-→--10-1105-→-31.623-----------------------------------------------------------|
| 5    |Ωp = ωp →  ωp→  0.2π                                                                 |
|------|-----ωTs----10.3π-----------------------------------------------------------------------|
|-6----|Ωs-=-⌈T-→---1-→--0.3π-⌉--------------------------------------------------------------|
| 7    |N =   12log[αp--1]-log[αs-1]  →  12log10(lo1g.258(90-.21π))--lologg10((301..36π23)-1)→  ⌈5.8859⌉ → 6                   |
|------|---------log(ΩΩpp)-log(Ωs)--------0.2π-10-------10------------------------------------------|
| 8    |Ωc = --( 12N-log10[αp-1]) →--(21×6log10(1.2589-1)) →  0.70321                                    |
|------|-----10---------------1j0(π(1+2i+N))----------------j(π(7+2i))-----------------------------|
|-9----|poles-of H-(S)-si =-Ωc-e--2N-----→-si-=-0.70321-e----12-----i =-0⋅⋅⋅5------------------|
|------|s0 =---0.182+-j0.67925,s1 =-- 0.49724+-j0.49724,s2 =-- 0.67925+-j0.182---------------|
|      |s3 = - 0.67925- j0.182,s4 = - 0.49724- j0.49724,s5 = - 0.182 - j0.67925               |
|------|----N∏-1------------------------------------------------------------------------------|
| 10   |k =    (- si) →  (0.182-  j0.67925)(0.49724-  j0.49724)(0.67925- j0.182)                 |
|------|----i=0------------------------------------------------------------------------------|
|      |(0.67925 + j0.182)(0.49724 + j0.49724) (0.182 + j0.67925) → 0.12092                       |
|-11---|H--(s) =---T-K---→-------------------------------------------------------------------|
|      |  a     N∏-1                                                                         |
|      |            (s-si)                                                                    |
|------|---------i=0----------------------------1×0.12092---------------------------------------|
|------|(s+0.182-j0.67925)(s+0.49724-j0.49724)(s+0.67925-j0.182)(s+0.67925+j0.182)(s+0.49724+j0.49724)(s+0.182+j0.67925)-|
|      |→multiply  complex  conjugates→  -s2+0.364s+0.4945--s2+0.9904.1428s09+20.49450--s2+1.3585s+0.4945-       |
|------|---------------------0.12092------(------------)(--------------)(-------------)-------|
|------|→--s6+2.7170s5+3.6910s4+3.1789s3+1.8252s2+0.66438s+0.12092----------------------------------------|
|      |                        N∑-1-Ai-   0.14354+j0.24861-  1.0714-j1.1668×10-5  -0.92785-j1.6071-     |
| 12   |partial fraction Ha (s) =   s-si → s+0.182-j0.67925 - s+0.49724+j0.49724 + s+0.67925- j0.182     |
|------|--0.92785+j1.6071---1.0714+j1i.1=6068×10-5---0.14354-j0.24861--------------------------------------|
|------|+-s+0.67925+j0.182----s+0.49724-j0.49724-+-s+0.182+j0.67925--------------------------------------|
|      |        N∑-1----Ai-----   ----0.14354+j0.24861----                                        |
| 13   |H (z) =    1-exp(Tsi)z-1 → 1-exp(-0.182+j0.67925)z-1-                                       |
|------|---1.0714i=-0j1.1668×10-5---------0.92785--j1.6071----------------------------------------------|
|------|1-exp(-0.49724-j0.49724)z--1 +-1-exp(-0.67925+j0.182)------------------------------------------|
|------|-------------------------------------------------------------------------------------|
|      |+ ----0.92785+j1.6071------ ----1.0714+j1.1668×10-5---+  ----0.14354-j0.24861----                |
|------|--1-exp(-0.67920.51-43j50.41+82j)0.z2-48161-1-exp(-1.0.04719742-4j+1j0.1.649687×214)0z--51---1--exp(-00..192827-8j5-0.6j17.960257)1z-1----------------|
|      |H (z) = 1-(0.64858+j0.52368)z-1 - 1-(0.53455-j0.29012)z-1 + 1-(0.49862+j9.1765×10-2)z-1             |
|------|-------0.92785+j1.6071---------1.0714+j1.1668×10-5-------0.14354-j0.24861-----------------------|
|------|+-1-(0.49862--j9.1765×10-2)z-1 --1--(0.53455+j0.29012)z-1 +-1-(0.64858-j0.52368)z-1--------------------
|      |

3.1.2 bilinear method

Using the above design table, these are the numerical values:     -1   --1-
T = Fs = 20000

|------|-------------------------------------------------------------------------------|
| step  |Bilinear                                                                        |
|-1----|ω--=-2πfp-→--2π(2000)-→-0.2π-----------------------------------------------------|
|------|-p-----Ffss----22π(03000000)------------------------------------------------------------|
|-2----|ωs-=-2πFs-→---20000--→-0.3π-----------------------------------------------------|
| 3    |αp = --1δp → --1-1 →  1.2589                                                     |
|------|-----10110---10110---------------------------------------------------------------|
|-4----|αs-=-10δ1s0 →-10-1105-→--31.623-----------------------------------------------------|
| 5    |Ωp = -2tan(ωp) →  2× 20000 tan (0.2π) →  12997                                   |
|------|-----T2----(ω2s)----------------(02.3π)--------------------------------------------|
|-6----|Ωs-=-T⌈-tan--2--→-2-×-20⌉000-tan---2---→-20381------------------------------------|
| 7    |N =   1log[αp-1]-log[αs--1]  →  1log10(1.2589-1)-log10(31.623-1)→  ⌈5.3048⌉ → 6             |
|------|------2--log(ΩΩp)-log(Ωs)------2-20l3og8110(12997)-log10(20381)------------------------------|
| 8    |Ωc = --( 1-logs10[αs-1]) →--(21×6log10(31.623-1)) →  15325.6                              |
|------|-----10-2N-------------1j0(π(1+2i+N))---------------j(π(7+2i))------------------------|
|-9----|poles-of H-(S)-si =-Ωc-e--2N-----→-si-=-15325-e---12----i =-0⋅⋅⋅5--------------|
|------|s0 =-- 3966.4+-j14803,s1-=---10836+-j10836.,s2 =---14803-+-j3966.4--------------|
|      |s3 = - 14803 - j3966.4,s4 = - 10836- j10836.,s5 = - 3966.4 - j14803             |
|------|----N∏-1------------------------------------------------------------------------|
| 10   |k =     (- si) → (3966.4 - j14803)(10836- j10836 )(14803 - j3966.4)               |
|------|----i=0------------------------------------------------------------------------|
|      |(14803+ j3966.4)(10836 + j10836)(3966.4 + j14803)                               |
|------|→---multiply-complex-conjugate-terms-------------------------------------------|
|------|--------------8------------8------------8---------------25---------------------|
|------|→--(2.3486-×-10-)(2.3484-×-10-) (2.3486×-10-)-→-1.2954-×-10-----------------------|
| 11   |Ha (s) = N∏-1K----→                                                             |
|      |            (s-si)                                                              |
|------|---------i=0--------------------------------------------------------------------|
|      |-----------------------------------1.2954×1025----------------------------------- |
|------|(s+3966.4-j14803)(s+10836-j10836)(s+14803-j3966.4)(s+14803+j3966.4)(s+10836+j10836)(s+3966.4+j14803)-|
|------|→---multiply-complex-conjugate25------------------------------------------------|
|      |(s2+7932.8s+2.3486×108)(s2+21.16279524s+×21304837792)(s2+29606.s+2.3486×108)                         |
|------|-------------------------------------------------------------------------------|
|------|-----------------8------------------8-------------------8----------------------|
|------|=-s262+5779.432s.-8s1+.325.3584×816×0108---s42+6271026.7s+2.5s.+0620.344×841×0108 +-s24+0424956.s06+.s8+.726.5344×8160×108-→-------------------|
|      |→  -6------5-------9-4-------1.2193534×1025---172--------21---------25-              |
|------|---s+59211.s+1.7530×10s-+3.2902×10-s-+4.1169×10--s+3.2658×10--s+1.2953×10----------------|
|------|-------------------------------------------------------------------------------|
|-12---|-------------------------------------------------------------------------------|
|------|-------------------------------------------------------------------------------|
| 13   |H (z) = Ha (s)|  2-1-z-1-→  TODO                                                 |
|------|--------------s=T-1+z-1---------------------------------------------------------|
|------|⋅⋅⋅----------------------------------------------------------------------------|
|      |

3.2 Example 2

Sampling frequency Fs = 10khz  , passband corner frequency fp = 1khz  , stopband corner frequency fs = 2khz  , with criteria δp ≥ - 3db  and δstop ≤ - 10db

3.2.1 using impulse invariance method

T = 1

|-----|-------------------------------------------------------------------------|
|step-|Impulse-finvaria2πn(c1e000)-----------------------------------------------------|
|1----|ωp-=-2π-Fps-→---10000-→--0.2π----------------------------------------------|
|2    |ωs = 2π fs→  2π(2000)→  0.4π                                               |
|3----|α--=---1Fs→---110000→-1.9953-----------------------------------------------|
|-----|--p---10δ1p0----10-130--------------------------------------------------------|
|4    |αs =  -1δs-→  --1-10-→  10.0                                                |
|-----|------10ω1p0--ωp1010--------------------------------------------------------|
|5----|Ωp-=--T-→---1-→--0.2π-----------------------------------------------------|
|6----|Ωs-=-⌈ωTs→--0.41π-→--0.4π-⌉--------------------------------------------------|
|     |      1 log[αp-1]-log[αs-1]    1log10(1.9953-1)-log10(10.0-1)                       |
|7----|N--=--2--log(Ωp)-log(Ωs)---→--2---log10(0.2π)--log10(0.4π)--→--1.5884-→-2----------|
|8    |Ωc =  ------Ωp------→  -------0.2π-------→  0.62906                       |
|-----|------10(21N-log10[αp-1])----1(0( 12×2-log10(1).9953-1))--------(------)----------------|
|9    |poles of H(S ) si = Ωc ej π(1+22iN+N) → si = 0.62906 ej π(3+42i) i = 0⋅⋅⋅1     |
|-----|-------------------------------------------------------------------------|
|-----|s0-=---0.44481-+-j0.44481,s1 =---0.44481---j0.44481------------------------|
|-----|-------------------------------------------------------------------------|
|     |    N∏-1                                                                 |
|10   |k =     (- si) → (0.44481- j0.44481)(0.44481 + j0.44481) → 0.39571         |
|-----|-----i=0------------------------------------------------------------------|
|-----|-------------------------------------------------------------------------|
|11   |Ha (s) = N∏-T1K--- → (s+0.44481-j0.44048.319)(57s1+0.44481+j0.44481) → s2+0.808.3969257s1+0.39571    |
|     |            (s-si)                                                        |
|-----|---------i=0-------------------------------------------------------------|
|     |                                                                         |
|-----|------------------------N∑-1---------------------------------------------|
|12   |partial fraction Ha (s) =    sA-is-→  s+0.4044.484148+10j.44481j - s+0.440.448441-810.j44481j       |
|-----|-------------------------i=0---i------------------------------------------|
|-----|→--z2-10.1.524752z3+5z0.41081-------------------------------------------------------|
|     |        N∑- 1                                                             |
|13   |H  (z) =    1--expA(Tisi)z-1-→  1-exp(-0.404.4484418-1ji0.44481)z-1-- 1-exp(-0.404.4448148+1ji0.44481)z--1 |
|-----|---------i=0--------------------------------------------------------------|
|-----|-------------------------------------------------------------------------|
|     |

3.2.2 Using bilinear

     -1--
T =  10000

|------|------------------------------------------------------------------|
|step  |Impulse invariance                                                |
|1-----|ω--=-2πfp-→--2π(1000)-→-0.2π----------------------------------------|
|------|-p-----Ffss----2π1(00200000)-----------------------------------------------|
|2-----|ωs =-2πFs-→---10000-→--0.4π----------------------------------------|
|3     |αp = -1δp-→  --1-3 →  1.9953                                         |
|------|-----10110----10110--------------------------------------------------|
|4-----|αs =-10δs10-→--10-1100→--10.0------------------------------------------|
|5     |Ωp = 2-tan(ωp) → 2 × 10000tan (0.2π) → 6498.4                      |
|------|-----T2----(ω2s)----------------(0.24π)-------------------------------|
|6-----|Ωs-=⌈T-tan--2--→-2-×-10⌉000tan---2--→--14531.----------------------|
|7     |N =   1log[αp-1]-log[αs-1] →  1 log10(1.9953-1)--log10(10.0--1)-→  1.3681 →  2   |
|------|------2-log(ΩΩp)-log(Ωs)------2-1lo45g3101(6498.4)-log10(14531.)------------------|
|8     |Ωc = --( 1-logs10[αs-1]) →--(2×12-log10(10.0-1))-→  8389.5                   |
|------|-----10-2N-------------1j(0π(1+2i+N))---------------j(π(3+2i))-----------|
|9-----|poles of-H-(S) si-=-Ωc-e--2N-----→-si =-8389.5-e----4----i-=-0⋅⋅⋅1-|
|------|s0 =---5932.3+-j5932.3,s1 =---5932.3--j5932.3---------------------|
|      |                                                                  |
|------|----N∏-1-----------------------------------------------------------|
|10    |k =    (- si) → (5932.3- j5932.3)(5932.3+ j5932.3) → 7.0384× 107   |
|------|----i=0-----------------------------------------------------------|
|      |                                                                  |
|11----|H--(s) =---K-----→-----------7.0384×107-----------------------------|
|      | a      N∏-1        (s+5932.3- i5932.3)(s+5932.3+i5932.3)                   |
|      |            (s-si)                                                  |
|------|---------i=0---7---------------------------------------------------|
|12----|→--s2+1178.0635.8s+47×.100384×107----------------------------------------------|
|13    |H (z) = Ha (s)|  21-z-1 →  0.0994529z2+0.19892z+0.099459-                  |
|------|--------------s=-T1+z-1-------z--0.93156z+0.32938----------------------|
|------|-------------------------------------------------------------------
|      |

4 References

  1. Alan Oppenheim, Ronald Schafer, Digital Signal Processing,  Prentice-Hall, inc. 1975, Chapter 5.
  2. Mostafa Shiva, Electrical engineering department, California state university, Fullerton, Lecture notes, handout H.
  3. John Proakis, Dimitris Manolakis, digital signal processing, 3rd edition, section 8.3.