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### Re: InverseZTransform numerical issue on some expression, version 8.0.4

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```\$Version

8.0 for Mac OS X x86 (64-bit) (October 5, 2011)

On my Mac, I had to truncate even more to get a result

sol1=InverseZTransform[
1/(0.95122942-1.95122942 z+z^2),z,n]//
Simplify

\[Piecewise]	0.	n<=0
20.5042 -21.5554 E^(-0.05 n)	True

However, if you add an assumption that n > 0

sol2=InverseZTransform[
1/(0.9512294245007143-1.9512294245007142 z+z^2),
z,n,Assumptions->{n>0}]//Simplify//Chop

20.5042 -21.5554 E^(-0.05 n)

Alternatively, you can do it symbolically

sol3[a_,b_]=InverseZTransform[1/(a+b*z+z^2),z,n];

sol3[0.9512294245007143,-1.9512294245007142]//
Simplify//Rationalize

\[Piecewise]	0	n<=0
20.5042 -21.5554 0.951229^n	True

Bob Hanlon

On Thu, Jun 7, 2012 at 5:20 AM, Nasser M. Abbasi <nma@12000.org> wrote:
>
> Using 8.0.4 on windows 7
>
> ---------------------------
> InverseZTransform[1/(0.9512294245007143 -1.9512294245007142  z+z^2),z,n]
> ---------------------------
>
> does NOT generate an answer.  But
>
> ------------------------------
> InverseZTransform[1/(0.9512294245 -1.9512294245 z+z^2),z,n]
> ------------------------------
>
> does. I simply chopped few digits (kept removing one digit
> at a time until it worked)
>
>    0.9512294245007143->0.9512294245
> and
>   1.9512294245007142->1.9512294245
>
> In order to get a result since I had a feeling it is a numerical
> issue.
>
> screen shot
>
> http://12000.org/tmp/060612/invz.png
>
> question is: Is this due to numerical issue? and
> how to better handle this whole thing. Clearly what I did
> above is not what I would do all the time. May be I am
> missing an option or something else. Because in a larger
> program, I would need to make sure InverseZTransform
> worked all the time or a way to check if it did not work.
>
> thanks
> --Nasser
>

```
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Reply hanlonr357 (710) 6/8/2012 7:39:25 AM

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