>qaisar wrote:
>
>> Dear why should not repeat the process for 'N' sample points?
>> secondly in my view we should divide each error term by 'N-1'
>> as follow
>>
>> e[0] = ( x[0] - f(x[1],x[2],...,x[N-1] )/N-1
>> e[1] = ( x[1] - f(x[0],x[2],...,x[N-1] )/N-1
>> e[2] = ( x[2] - f(x[0],x[1],x[3]...,x[N-1] )/N-1
>>
>> are you agreed.
>
>Absolutely not.
>
>The error is the ( x[?] - f() ) term.  The average (mean) of all of
>them will include a 1/N term, which may be why you're thinking you need
>to include it.
>
>Ciao,
>
>Peter K.
>
>How ( x[?] - f() )is a term, look for example if i have non irregularly
sampled data stored in vector 'x' the vector of corresponding time
instants is 't'. Then according to you i chosse 'N' sample points.

let t1 : is samping instant of sample x(1)
    tN : is samping instant of sample x(N)
    td = tN-t1
    ts= td/N
    ti = t1:ts:tN
    xi = interp1(t,x,ti,'linear')     

where 'xi' and 'ti' are regularly placed data points and corresponding
time instants. 

Now how to proceed for the interpolation error? 
 
 


		
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qaisar wrote:

> Dear why should not repeat the process for 'N' sample points?
> secondly in my view we should divide each error term by 'N-1'
> as follow
>
> e[0] = ( x[0] - f(x[1],x[2],...,x[N-1] )/N-1
> e[1] = ( x[1] - f(x[0],x[2],...,x[N-1] )/N-1
> e[2] = ( x[2] - f(x[0],x[1],x[3]...,x[N-1] )/N-1
>
> are you agreed.

Absolutely not.

The error is the ( x[?] - f() ) term.  The average (mean) of all of
them will include a 1/N term, which may be why you're thinking you need
to include it.

Ciao,

Peter K.

>Should have been:
>
>e[0] = ( x[0] - f(x[1],x[2],...,x[N-1] )
>e[1] = ( x[1] - f(x[0],x[2],...,x[N-1] )
>e[2] = ( x[2] - f(x[0],x[1],x[3]...,x[N-1] )
>              ^ Forgot to change these indices. :-( It's pre-coffee.
>
>Ciao,
>
>Peter K.
>
>Dear why should not repeat the process for 'N' sample points? secondly in
my view we should divide each error term by 'N-1' as follow

e[0] = ( x[0] - f(x[1],x[2],...,x[N-1] )/N-1
e[1] = ( x[1] - f(x[0],x[2],...,x[N-1] )/N-1
e[2] = ( x[2] - f(x[0],x[1],x[3]...,x[N-1] )/N-1

are you agreed.

Regards.
		
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Should have been:

e[0] = ( x[0] - f(x[1],x[2],...,x[N-1] )
e[1] = ( x[1] - f(x[0],x[2],...,x[N-1] )
e[2] = ( x[2] - f(x[0],x[1],x[3]...,x[N-1] )
              ^ Forgot to change these indices. :-( It's pre-coffee.

Ciao,

Peter K.

qaisar wrote:

> I want to calculate the interpolation error occured when I resampled
> regularly the irregularly sampled data. Infact there exists the
> resampled points corresponding to whos time of existance there are
> no sample points lies in original irregularly sampled data.

It's a little hard, given that you don't know what the underlying
signal was to begin with.

One attempt to estimate the error would be to take your irregularly
sampled data --- N points, say, x[0], x[1], ..., x[N-1] --- and do the
interpolation with N-1 of them.  Then, look at what the interpolation
says the dropped sample value should be.

Repeat N-1 more times, dropping all N points once each (and keeping the
other N-1 points in the interpolation).

e.g.

e[0] = ( x[0] - f(x[1],x[2],...,x[N-1] )
e[1] = ( x[0] - f(x[0],x[2],...,x[N-1] )
e[2] = ( x[0] - f(x[0],x[1],x[3]...,x[N-1] )

etc.
where f() is the interpolation function you're using for N-1 points.

Then the mean "error" is just mean of all the e[n].

Ciao,

Peter K.

I want to calculate the interpolation error occured when I resampled
regularly the irregularly sampled data. Infact there exists the resampled
points corresponding to whos time of existance there are no sample points
lies in original irregularly sampled data.

Thnks in advance for your ideas.
		
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