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integration of a continuous function

Started by Alex_001 February 23, 2009
Hi,
it's well known that sampling a continuous function according to the
sampling theorem requirements, you get all the information on the
contunuous function just from the samples. 
now, if you have to get an accurate estimate of the integral of the
continous function from samples satisfying the sampling theorem
requirements, how can you  get a good estimate of such an integral?
I noticed that if you use a sampling frequency close to the Nyquist rate
and apply the definition of integration in the time domain ( sum(Xi*
delta(x))) the value you get for the integral is totally inaccurate...
I hope the problem is clear...
thanks
Alex



On Feb 23, 2:23&#2013266080;pm, "Alex_001" <a.bast...@email.it> wrote:
> Hi, > it's well known that sampling a continuous function according to the > sampling theorem requirements, you get all the information on the > contunuous function just from the samples. > now, if you have to get an accurate estimate of the integral of the > continous function from samples satisfying the sampling theorem > requirements, how can you &#2013266080;get a good estimate of such an integral? > I noticed that if you use a sampling frequency close to the Nyquist rate > and apply the definition of integration in the time domain ( sum(Xi* > delta(x))) the value you get for the integral is totally inaccurate...
You can express the original continuous function as a linear combination of shifted sinc functions weighted by the sample values. This is the reconstruction theorem. Any integral of the original continuous function is therefore a linear combination of integrated shifted sinc functions weighted by the sample values. illywhacker;
On 23 Feb, 14:23, "Alex_001" <a.bast...@email.it> wrote:
> Hi, > it's well known that sampling a continuous function according to the > sampling theorem requirements, you get all the information on the > contunuous function just from the samples. > now, if you have to get an accurate estimate of the integral of the > continous function from samples satisfying the sampling theorem > requirements, how can you &#2013266080;get a good estimate of such an integral? > I noticed that if you use a sampling frequency close to the Nyquist rate > and apply the definition of integration in the time domain ( sum(Xi* > delta(x))) the value you get for the integral is totally inaccurate...
Numerically integrating continuous functions and sampling are two different cups of tea. If you have an analytical expression for the function, you can use integration schemes with different inherent accuracies, as well as adaptive integration schemes. Besides, sampling close to the Nyquist limit is a bad idea anyway; it would do you no favours with integration eiter. Rune


Alex_001 wrote:
> Hi, > it's well known that sampling a continuous function according to the > sampling theorem requirements, you get all the information on the > contunuous function just from the samples.
Cut the bullshit. What are you really trying to do?
> now, if you have to get an accurate estimate of the integral of the > continous function from samples satisfying the sampling theorem > requirements, how can you get a good estimate of such an integral?
Practical advice: represent the function as the polynomial approximation based on the sampled data. Take the integral from there.
> I noticed that if you use a sampling frequency close to the Nyquist rate > and apply the definition of integration in the time domain ( sum(Xi* > delta(x))) the value you get for the integral is totally inaccurate... > I hope the problem is clear...
Yes, the problem is clear: you are clueles, stupid and lazy.
> thanks > Alex
Go to hell. VLV
On Feb 23, 8:46&#2013266080;am, illywhacker <illywac...@gmail.com> wrote:
> On Feb 23, 2:23&#2013266080;pm, "Alex_001" <a.bast...@email.it> wrote: > > > Hi, > > it's well known that sampling a continuous function according to the > > sampling theorem requirements, you get all the information on the > > contunuous function just from the samples. > > now, if you have to get an accurate estimate of the integral of the > > continous function from samples satisfying the sampling theorem > > requirements, how can you &#2013266080;get a good estimate of such an integral? > > I noticed that if you use a sampling frequency close to the Nyquist rate > > and apply the definition of integration in the time domain ( sum(Xi* > > delta(x))) the value you get for the integral is totally inaccurate... > > You can express the original continuous function as a linear > combination of shifted sinc functions weighted by the sample values. > This is the reconstruction theorem. Any integral of the original > continuous function is therefore a linear combination of integrated > shifted sinc functions weighted by the sample values. > > illywhacker;
Is this another way of saying interpolate by a large factor using a sinc filter and then do sum(Xi*deltaX)? John
On Feb 23, 10:15&#2013266080;am, Vladimir Vassilevsky <antispam_bo...@hotmail.com>
wrote:
> Alex_001 wrote: > > Hi, > > it's well known that sampling a continuous function according to the > > sampling theorem requirements, you get all the information on the > > contunuous function just from the samples. > > Cut the bullshit. What are you really trying to do? > > > now, if you have to get an accurate estimate of the integral of the > > continous function from samples satisfying the sampling theorem > > requirements, how can you &#2013266080;get a good estimate of such an integral? > > Practical advice: represent the function as the polynomial approximation > based on the sampled data. Take the integral from there. > > > I noticed that if you use a sampling frequency close to the Nyquist rate > > and apply the definition of integration in the time domain ( sum(Xi* > > delta(x))) the value you get for the integral is totally inaccurate... > > I hope the problem is clear... > > Yes, the problem is clear: you are clueles, stupid and lazy. > > > thanks > > Alex > > Go to hell. > > VLV
Did you sleep well last night? John
>On Feb 23, 10:15=A0am, Vladimir Vassilevsky <antispam_bo...@hotmail.com> >wrote: >> Alex_001 wrote: >> > Hi, >> > it's well known that sampling a continuous function according to the >> > sampling theorem requirements, you get all the information on the >> > contunuous function just from the samples. >> >> Cut the bullshit. What are you really trying to do? >> >> > now, if you have to get an accurate estimate of the integral of the >> > continous function from samples satisfying the sampling theorem >> > requirements, how can you =A0get a good estimate of such an
integral?
>> >> Practical advice: represent the function as the polynomial
approximation
>> based on the sampled data. Take the integral from there. >> >> > I noticed that if you use a sampling frequency close to the Nyquist
rat=
>e >> > and apply the definition of integration in the time domain ( sum(Xi* >> > delta(x))) the value you get for the integral is totally
inaccurate...
>> > I hope the problem is clear... >> >> Yes, the problem is clear: you are clueles, stupid and lazy. >> >> > thanks >> > Alex >> >> Go to hell. >> >> VLV > >Did you sleep well last night? > >John >
Thank you guys, I appreciate your help. for VLV: I know the problem could be pretty trivial, but I do not work 100% of my time on this...and I just wanted to know if there was a esasy solution. Anyway, nobody forced you to read my thread and reply, did someone? Alex
On Feb 23, 4:19&#2013266080;pm, John <sampson...@gmail.com> wrote:
> On Feb 23, 8:46&#2013266080;am, illywhacker <illywac...@gmail.com> wrote: > > > > > On Feb 23, 2:23&#2013266080;pm, "Alex_001" <a.bast...@email.it> wrote: > > > > Hi, > > > it's well known that sampling a continuous function according to the > > > sampling theorem requirements, you get all the information on the > > > contunuous function just from the samples. > > > now, if you have to get an accurate estimate of the integral of the > > > continous function from samples satisfying the sampling theorem > > > requirements, how can you &#2013266080;get a good estimate of such an integral? > > > I noticed that if you use a sampling frequency close to the Nyquist rate > > > and apply the definition of integration in the time domain ( sum(Xi* > > > delta(x))) the value you get for the integral is totally inaccurate... > > > You can express the original continuous function as a linear > > combination of shifted sinc functions weighted by the sample values. > > This is the reconstruction theorem. Any integral of the original > > continuous function is therefore a linear combination of integrated > > shifted sinc functions weighted by the sample values. > > > illywhacker; > > Is this another way of saying interpolate by a large factor using a > sinc filter and then do sum(Xi*deltaX)?
Yes. but the sinc function is determined by the sampling procedure and the maximum frequency in the original signal, which it is, I think, necessary to know. illywhacker;
On Feb 23, 8:23&#2013266080;am, "Alex_001" <a.bast...@email.it> wrote:
> Hi, > it's well known that sampling a continuous function according to the > sampling theorem requirements, you get all the information on the > contunuous function just from the samples. > now, if you have to get an accurate estimate of the integral of the > continous function from samples satisfying the sampling theorem > requirements, how can you &#2013266080;get a good estimate of such an integral? > I noticed that if you use a sampling frequency close to the Nyquist rate > and apply the definition of integration in the time domain ( sum(Xi* > delta(x))) the value you get for the integral is totally inaccurate... > I hope the problem is clear... > thanks > Alex
Hello Alex, There are several ways of looking at this. I assume your sampling rate is uniform. Then if you are assuming the signal was properly bandlimited before sampling, then your intergral is a sum of integrals. I.e., integration is linear. So what is your interpolating function - how do you connect the dots. Well basically for a bandlimited thing you connect the dots with scaled sync functions. So you need to find the integral of a sync function - well that is not too bad. You may find it in a table of integrals if you don't know the trick. Your overall integral is a sum of weighted (by the samples) integrals of sync functions.. You may also connect the dots with polynomials. For equally spaced dots you can used simple things like trapezoidal rule or Simpson's rule. You may even fit higher order polys to the data and then find the integral. Fortunately this has already been done so you just need the weights provided by Newton-Cotes formulae. If you are free to pick your sampling points and don't mind nonuniform sampling then with N points you can come up with an integral formula that is exact of polys up to order 2N-1. Yeap that is true. Look up Gauss Legendre integration. You can also look up Lobatto's rule and Redeau integration - these are all variations on a theme. IHTH, Clay p.s. How long a data span do you need to integrate? How many points?
"Alex_001" <a.bastari@email.it> wrote in message 
news:B5qdnUTCiKLJPj_UnZ2dnUVZ_j6WnZ2d@giganews.com...
> it's well known that sampling a continuous function according to the > sampling theorem requirements, you get all the information on the > contunuous function just from the samples. > now, if you have to get an accurate estimate of the integral of the > continous function from samples satisfying the sampling theorem > requirements, how can you get a good estimate of such an integral? > I noticed that if you use a sampling frequency close to the Nyquist rate > and apply the definition of integration in the time domain ( sum(Xi* > delta(x))) the value you get for the integral is totally inaccurate...
Try T * (sum(Xi*delta(x))) where T is the time between samples. Plot in on paper ..... the sample points are infinitesimallly short but by considering each sample as lasting until the next sample, then your integral will take on a meaningful value, subject to any interpolation that you might like to try, such as linear (over 2 points) or quadratic (over 3 points). Or else try out Simpson's rule.