FFT accuracy and 'Noise Floor'

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Cem
Cem on 14 Jul 2014
Answered: Daniel kiracofe on 15 Jul 2014

Hi all,

I have a question regarding the accuracy of FFT (I think). I'm trying to reconstruct a function with an exponential tail (hence the function amplitude changes some 25 orders of magnitude). I have the code below to illustrate what I try. For each of the trials I have a similar noise floor of 1e-15, no matter the sampling rate (see figure). I believe the accuracy is gets killed by the fact that the regions with very low amplitude has to be constructed by subtracting harmonics with a relatively higher amplitude (e.g. ~1e-3 - ~1e-3 =~1e-15). I know it is pretty hopeless, but I'd be glad if anyone has any suggestions to help me lower the noise floor.

Cem

a=0:0.001:0.999;
funct=sin(2*pi*a(1:500));
funct=[funct,funct(end)*exp(-1*(1:500)/10)];
semilogy(a,funct)
hold all
semilogy(a,ifft(fft(funct)))

Accepted Answer

Daniel kiracofe
Daniel kiracofe on 15 Jul 2014
Standard double precision floating point arithmetic is good to about 16 significant digits. So trying to do what you are doing to an accuracy of more than approximately 1e-15 is plain and simply impossible.
to illustrate, try funct=single(sin(2*pi*a(1:500))); this will convert your number to a single precision floating point, which is accurate to about 7 significant digits. You should find that your code is good only to about 1e-6.
the only way you can succeed is to use something with more precision than a double . I don't know of a way to do it natively in matlab. A google search turned this up: http://www.advanpix.com/ might be a good place to start. Or you might need to try something like a computer algebra system (e.g. Maple) that has native support for arbitrary precision arithmetic.

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