continuous fourier transform c#

Hi,

I have seen lots of links on google and yahoo, to the FFT, DFT etc.

But nothing for CTFT, can anybody please point me in the right directions
to coded examples of CTFT and laplace transforms in a control process.

Thanks

Tinkerz

On 5/27/2011 10:10 AM, tinkerz wrote:
> Hi,
>
> I have seen lots of links on google and yahoo, to the FFT, DFT etc.
>
> But nothing for CTFT, can anybody please point me in the right directions
> to coded examples of CTFT and laplace transforms in a control process.
>
> Thanks
>
> Tinkerz

If it's continuous, then it's not sampled so it's not digital either. 
If it's not sampled then you might look for the Finite Fourier Transform 
which is at least continuous. and then consider looking at the Fourier 
Series which is likely equivalent to what you're asking about.

Either way, it sounds like you're asking about an analytic solution 
which you might get with Maple, Matlab, Derive, etc.

So, "coded examples" sounds a bit strange unless you plan to code things 
up using Matlab or ... whatever.  And whatever "code" means to you.

In the sampled, digital world:
Laplace Transform goes to Z-Transform
Fourier Transform goes to Finite, Discrete Fourier Transform


Fred

On 5/27/2011 10:10 AM, tinkerz wrote:
> Hi,
>
> I have seen lots of links on google and yahoo, to the FFT, DFT etc.
>
> But nothing for CTFT, can anybody please point me in the right directions
> to coded examples of CTFT and laplace transforms in a control process.
>
> Thanks
>
> Tinkerz

Mathematica has analytic CTFT

http://reference.wolfram.com/mathematica/ref/FourierTransform.html

But ss was mentioned, in the real wold, one used the discrete versions.

To go from CTFT to DTFT, here is a demo

http://demonstrations.wolfram.com/FromContinuousToDiscreteTimeFourierTransformBySamplingMethod/

--Nasser

On May 27, 1:10=A0pm, "tinkerz" <nwoodhamuk@n_o_s_p_a_m.googlemail.com>
wrote:
> Hi,
>
> I have seen lots of links on google and yahoo, to the FFT, DFT etc.
>
> But nothing for CTFT, can anybody please point me in the right directions
> to coded examples of CTFT and laplace transforms in a control process.
>
> Thanks
>
> Tinkerz

1/s in laplace is an integral, so I write it out the transform as
integrals and numerically integrate it, trapezoidal rule is good enough

On 5/29/2011 6:30 PM, steve wrote:
> On May 27, 1:10 pm, "tinkerz"<nwoodhamuk@n_o_s_p_a_m.googlemail.com>
> wrote:
>> Hi,
>>
>> I have seen lots of links on google and yahoo, to the FFT, DFT etc.
>>
>> But nothing for CTFT, can anybody please point me in the right directions
>> to coded examples of CTFT and laplace transforms in a control process.
>>
>> Thanks
>>
>> Tinkerz
>
> 1/s in laplace is an integral, so I write it out the transform as
> integrals and numerically integrate it, trapezoidal rule is good enough

It *is* all about context isn't it?

Fred

On May 27, 7:10=A0pm, "tinkerz" <nwoodhamuk@n_o_s_p_a_m.googlemail.com>
wrote:
> Hi,
>
> I have seen lots of links on google and yahoo, to the FFT, DFT etc.
>
> But nothing for CTFT, can anybody please point me in the right directions
> to coded examples of CTFT and laplace transforms in a control process.

Those kinds of things are not done as part of the numerical
computations. They are analysis tools that help decide how
to configure the numerical computations.

Rune

On Sun, 29 May 2011 17:24:13 -0700, "Nasser M. Abbasi" <nma@12000.org>
wrote:

>On 5/27/2011 10:10 AM, tinkerz wrote:
>> Hi,
>>
>> I have seen lots of links on google and yahoo, to the FFT, DFT etc.
>>
>> But nothing for CTFT, can anybody please point me in the right directions
>> to coded examples of CTFT and laplace transforms in a control process.
>>
>> Thanks
>>
>> Tinkerz
>
>Mathematica has analytic CTFT
>
>http://reference.wolfram.com/mathematica/ref/FourierTransform.html
>
>But ss was mentioned, in the real wold, one used the discrete versions.
>
>To go from CTFT to DTFT, here is a demo
>
>http://demonstrations.wolfram.com/FromContinuousToDiscreteTimeFourierTransformBySamplingMethod/
>
>--Nasser

Hello Nasser,
   At that 2nd web site, the text under the 
sub-title 'Details' states that Mathematica's 
built-in function FourierTransform implements 
the CTFT of a continuous x_a(t) signal.

Can you tell us how a User is supposed to apply a 
continuous x_a(t) input signal (perhaps a voltage) 
to a computer running Mathematica software? 

Does Mathematica have a way of performing 
continous (analog) signal processing?  The reason 
I ask is because the frequency variable in the 
CTFT is radians-per-second and is a continuous 
variable.  Likewise the frequency variable in the 
DTFT is radians-per-sample (not 'radians' as stated 
on the web page) and is also a continuous variable.

I'm trying to figure out how Mathematica can compute 
the values of continous (analog) variables.  Wouldn't 
such processing require an infinite number of binary 
bits to represent a single value?

[-Rick-]

On May 31, 8:25&#4294967295;am, Rick Lyons <R.Lyons@_BOGUS_ieee.org> wrote:
 &#4294967295;
I'm trying to figure out how Mathematica can compute
the values of continous (analog) variables.Wouldn't
> such processing require an infinite number of binary
> bits to represent a single value?

that's true for all fourier transforms

Rick Lyons <R.Lyons@_bogus_ieee.org> wrote:

(snip)
> Does Mathematica have a way of performing 
> continous (analog) signal processing?  The reason 
> I ask is because the frequency variable in the 
> CTFT is radians-per-second and is a continuous 
> variable.  Likewise the frequency variable in the 
> DTFT is radians-per-sample (not 'radians' as stated 
> on the web page) and is also a continuous variable.

Mathematica can do symbolic math.  That way you can
do things that can't be done numerically.
 
> I'm trying to figure out how Mathematica can compute 
> the values of continous (analog) variables.  Wouldn't 
> such processing require an infinite number of binary 
> bits to represent a single value?

I have nearby my TI-92 calculator, but not Mathematica.

If I wanted, for example, the integral from 0 to 2pi
of sin(a x) I just put that in ...

\int(sin(a*x),x,0,2\pi)

(using the TeX names, but the keyboard has an integral sign
and a pi sign.)  I get:

-cos(2*a*x)/a + 1/a

If I wanted the Fourier transform of a continuous square
wave, then I could call that the transform.  The result is
accurate to any number of bits.

I can do the integral of 1/sqrt(1-x^2) from -1 to 1, and
the result comes out as one character, accurate to an infinite
number of bits.  You can always ask for approximate answers.

If you put in ln(256) you get out 8*ln(2).

Mathematica can do all these, and more.

-- glen

On 5/31/2011 3:56 PM, steve wrote:
> On May 31, 8:25 am, Rick Lyons<R.Lyons@_BOGUS_ieee.org>  wrote:
>
> I'm trying to figure out how Mathematica can compute
> the values of continous (analog) variables.Wouldn't
>> such processing require an infinite number of binary
>> bits to represent a single value?
>
> that's true for all fourier transforms

Precision and continuity are two different things.

Let us say that you want to know the expression for some integral of 
some other expression - with defined limits, etc. including +/- infinity.

OK - so you get a symbolic result which can be evaluated for any value 
of the independent variable.

Let's just say for the sake of argument that the value selected is a 
suitably small integer - so no precision problems there!
But the result will have some precision depending on how many digits you 
choose to carry it out on paper, with a sliderule, with a calculator, 
with a computer ... except in special cases where the result has nothing 
but trailing zeros..... where the sliderule may not be the best tool to 
get infinite precision.
This has nothing to do with the expression being continuous and 
everything to do with your arithmetic.

Fred