On Sunday, July 3, 2011 4:57:02 PM UTC-4, tinkerz wrote:
> I should have said that is a small project
> "Laplace transform is derived from the
> continuous-time Fourier transform"
It is possible, but nit-picky, to disagree.
> I was working with FFT, but the data is continuous.
Once you plug data into an FFT, it isn't continuous any more. Digital computers and the numbers they operate on are inherently discrete.
> Therefore laplace and the continuous-time Fourier transform are the
> combination I am after.
You might do that symbolically with a math package like Maple. I doubt that you want a solution of that sort.
> Still looking and posting on liveperson to find someone to hire for this
> small project is quite tedious.
Seriously, talk to Tim Wescott. If you tell him what you want done rather than how to do it, you'll get along fine.
Engineering is the art of making what you want from things you can get.
Reply by Randy Yates●July 4, 20112011-07-04
On 07/03/2011 11:53 PM, robert bristow-johnson wrote:
> On Jul 3, 1:03 pm, Tim Wescott<t...@seemywebsite.com> wrote:
>> On 07/02/2011 07:28 PM, Randy Yates wrote:
>>> On 07/02/2011 05:02 PM, Tim Wescott wrote:
>>>> Randy is probably forgetting that the z transform (which is what is
>>>> used to analyze sampled-time systems) can be derived from the
>>>> Laplace transform.
>>> There may be some convoluted way to do that, but I don't think it's
>>> going to be of any utility. None of Lyons, Proakis, Mitra, or
>>> Oppenheim discuss deriving it from the Laplace transform.
>>> Of course it has the similar function as the Laplace transform in the
>>> digital domain, and as the Laplace transform is derived from the
>>> continuous-time Fourier transform, the z-transform is derived from the
>>> discrete time Fourier transform. However it isn't "derived" from the
>>> Laplace transform in any way I've seen or can imagine to be useful.
>> Houpis& Lamont, "Digital Control". Shows how, if you model the
>> sampling process as multiplying by a chain of impulses then the z
>> transform just falls out of the Laplace transform.
>> IIRC, Oppenheim, Wilsky& Young, "Signals and Systems" models sampling
>> the same way, and derives the Fourier series from the Fourier transform;
>> I'd be surprised if they don't derive the z transform the same way,
>> later in the book.
> it's straight forward. we've done it here in comp.dsp several times
> since the mid 90s.
> the Z transform is nothing other than the Laplace transform of a
> sampled signal (with them weighted dirac impulses) and the
> z = e^(s*T) where 1/T = sampling frequency
> it's nothing more than that.
You're right Robert (and Tim), we have done that here, and I did
forget. It does bring out the mapping between the two domains nicely,
Randy Yates % "Watching all the days go by...
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Reply by tinkerz●July 4, 20112011-07-04
I email Tim I hope he has a simple solution, I am sure its simple solution
Reply by HardySpicer●July 4, 20112011-07-04
On Jul 5, 1:37=A0am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> HardySpicer wrote:
> > On Jul 3, 3:17 pm, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> >>HardySpicer wrote:
> >>>On Jul 3, 3:24 am, Vladimir Vassilevsky <nos...@nowhere.com> wrote:
> >>>>tinkerz wrote:
> >>>>>Hi All
> >>>>>Where is the best place to hire a programmer for a DSP/Laplace contr=
> >>>>Programmer is a tool. What you are looking for is probably a coder. T=
> >>>>get a coder, go to a forest and find some orangutan.
> >>>User interface is fulling functional I see Mr Impaler..
> >>The orangutan Ph.D. ?
> > What? You don't have a Ph.D and you think your are qualified to answer
> > questions on this NG?
> > Shame on you.
> You see, Hardy, I do have Ph.D. However I am not orangutan.