inverse z-transform - partial fractions

Started by June 8, 2010
```[IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
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```>Please refer to the below picture for my question.
[IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
>
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```On 06/07/2010 09:09 PM, bos1234 wrote:
> [IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]

Because the z transform tables are in the form z / (z - d), which
transforms to u(n) * d^n.  1 / (z - d) transforms to u(n-1) * d^(n-1),
which is awkward.

So you want to do your partial fraction expansion such that the result
is convenient.

What's the book?  I know that's how I did it in mine.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
```
```On 6/8/2010 12:09 AM, bos1234 wrote:
> [IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]

You need a z in the numerator. To keep the denominator simple (as you
did), you have to put it there explicitly.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
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```On Jun 8, 12:35&#2013266080;am, "bos1234" <suren130@n_o_s_p_a_m.gmail.com> wrote:
> >Please refer to the below picture for my question.
>
> [IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
>
>
>
> - Hide quoted text -
>
> - Show quoted text -

There's an entire theory about how to do partial fraction expansion.
See here:

http://en.wikipedia.org/wiki/Partial_fraction

Just like with integration, you want to break your complicated product
into a sum of simpler pieces where each piece can be handled easily.
I.e., by lookup in a table. Otherwise you will do contour integration,
but even applying the Cauchy Goursat theorem is simplified via partial
fraction decomposition.

IHTH,
Clay

```
```>>Please refer to the below picture for my question.
>[IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
>>
>

I think I know the answer to this,  They want to do partial fraction
expansion in the simplest way possible, so
1) they divide z from both sides so they have H(z)/z = 1/(z-a)(z-b) etc...

2) they do partial fraction expansion on the right side.
3) they multiply both sides by Z. Hope this helps this is how I remember
doing it in class.
```
```On Jun 8, 11:15&#2013266080;am, Clay <c...@claysturner.com> wrote:
> On Jun 8, 12:35&#2013266080;am, "bos1234" <suren130@n_o_s_p_a_m.gmail.com> wrote:
>
> > >Please refer to the below picture for my question.
>
> > [IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
>
> > - Hide quoted text -
>
> > - Show quoted text -
>
> There's an entire theory about how to do partial fraction expansion.
> See here:
>
> http://en.wikipedia.org/wiki/Partial_fraction
>
> Just like with integration, you want to break your complicated product
> into a sum of simpler pieces where each piece can be handled easily.
> I.e., by lookup in a table. Otherwise you will do contour integration,
> but even applying the Cauchy Goursat theorem is simplified via partial
> fraction decomposition.

a wiki-reference:
http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem

Clay, isn't the Heaviside partial fraction expansion technique
applicable?  There is a wonderfully complete wiki article at
http://en.wikipedia.org/wiki/Heaviside_cover-up_method .

r b-j

```
```On Jun 8, 1:33&#2013266080;pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Jun 8, 11:15&#2013266080;am, Clay <c...@claysturner.com> wrote:
>
>
>
>
>
> > On Jun 8, 12:35&#2013266080;am, "bos1234" <suren130@n_o_s_p_a_m.gmail.com> wrote:
>
> > > >Please refer to the below picture for my question.
>
> > > [IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
>
> > > - Hide quoted text -
>
> > > - Show quoted text -
>
> > There's an entire theory about how to do partial fraction expansion.
> > See here:
>
> >http://en.wikipedia.org/wiki/Partial_fraction
>
> > Just like with integration, you want to break your complicated product
> > into a sum of simpler pieces where each piece can be handled easily.
> > I.e., by lookup in a table. Otherwise you will do contour integration,
> > but even applying the Cauchy Goursat theorem is simplified via partial
> > fraction decomposition.
>
> a wiki-reference:
> &#2013266080;http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
>
> Clay, isn't the Heaviside partial fraction expansion technique
> applicable? &#2013266080;There is a wonderfully complete wiki article at
> &#2013266080;http://en.wikipedia.org/wiki/Heaviside_cover-up_method.

Robert,

Yes you may do this. I found that simply multiplying everything by the
denominator (unfactored one) and then substituting the roots one by
one, that the coefs fall right out. This works especially well for the
OP's problem where there are just two factors. This is essentially the
"cover up" method. It gets more complicated whene there are repeated
roots.

The OP may also want to dig out his calculus book, as it likely has a
section on partial fraction expansion as a means to finding some
integrals.

Clay

```
```>On 06/07/2010 09:09 PM, bos1234 wrote:
>> [IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
>
>Because the z transform tables are in the form z / (z - d), which
>transforms to u(n) * d^n.  1 / (z - d) transforms to u(n-1) * d^(n-1),
>which is awkward.
>
>So you want to do your partial fraction expansion such that the result
>is convenient.
>
>What's the book?  I know that's how I did it in mine.
>
>--
>Tim Wescott
>Control system and signal processing consulting
>www.wescottdesign.com
>
thanks for the helps everyone. The books name is Digital SIgnal Processing:
A practical Approach by Ifeachor and Jervis
```
```On Jun 9, 3:34&#2013266080;am, "bos1234" <suren130@n_o_s_p_a_m.gmail.com> wrote:
> >On 06/07/2010 09:09 PM, bos1234 wrote:
> >> [IMG]http://i50.tinypic.com/28sbn6f.png[/IMG]
>
> >Because the z transform tables are in the form z / (z - d), which
> >transforms to u(n) * d^n. &#2013266080;1 / (z - d) transforms to u(n-1) * d^(n-1),
> >which is awkward.
>
> >So you want to do your partial fraction expansion such that the result
> >is convenient.
>
> >What's the book? &#2013266080;I know that's how I did it in mine.
>
> >--
> >Tim Wescott
> >Control system and signal processing consulting
> >www.wescottdesign.com
>
> thanks for the helps everyone. The books name is Digital SIgnal Processing:
> A practical Approach by Ifeachor and Jervis- Hide quoted text -
>
> - Show quoted text -

Here is a simple example of inverse z transformation using partial
fraction expansion.

http://www.claysturner.com/dsp/FibonacciFormulaProof.pdf

Clay

```