Timeline for What is the effect of reducing NFFT less than the signal length
Current License: CC BY-SA 3.0
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| Feb 12, 2018 at 14:41 | comment | added | Usman Ashraf | It was to set different L's in the powers of 2. For example, 256, 512, 1024, 2048 etc. | |
| Feb 12, 2018 at 14:05 | comment | added | Gilles |
@UsmanAshraf , why do you have the "+3" in 2^( nextpow2( lenData ) + 3) ?
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| Feb 12, 2018 at 11:24 | comment | added | Usman Ashraf | @Giles I am using a very similar code but have different results for energy. Maybe because your signal is a periodic sinusoid. Will your observation remain valid if signal is not periodic. 'lenData = length( data ); L = 2^( nextpow2( lenData ) + 3); Y = fft(data, L)/L; Y3energy = (Y.*conj(Y)); energy3 = sum(Y3energy); energy1 = sum(abs(data).^2);' | |
| Feb 12, 2018 at 11:16 | history | edited | Gilles | CC BY-SA 3.0 |
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| Feb 12, 2018 at 11:15 | comment | added | applesoup | @UsmanAshraf, it is not directly possible to use a DFT length smaller than the signal length but still take the complete signal into account. If you want a lower resolution of the DFT, it may be possible to 1. compute the DFT with the length of the signal and 2. interpolate the result to find the spectrum in the desired (lower number of) bins. | |
| Feb 12, 2018 at 10:59 | comment | added | Gilles | @UsmanAshraf, please see the edit in my answer. | |
| Feb 12, 2018 at 10:56 | history | edited | Gilles | CC BY-SA 3.0 |
filling up on OP comments
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| Feb 12, 2018 at 10:49 | comment | added | Usman Ashraf | @applesoup, can you suggest a way in which I don't lose information from the signal and still be able to reduce the NFFT below the input sample length. I want to do this for the more optimized implementation on verilog (hardware design language). | |
| Feb 12, 2018 at 10:41 | comment | added | Usman Ashraf | @applesoup, this is exactly what I have observed in matlab code. If I reduce NFFT below the signal length, the energy value is the same as if the last samples of the signal X are dropped from it. | |
| Feb 12, 2018 at 10:14 | comment | added | applesoup | I think it is also important to note that not all signal samples will be used for the spectrum computation if the DFT length is lower than the signal length. Of course, this is related to a reduced frequency resolution, but may have further consequences, e.g., if the signal is nonzero only in the last samples. | |
| Feb 12, 2018 at 9:58 | comment | added | Usman Ashraf | I understand your point that the resolution of the bins would worsen as N is decreased. But when I use the produced fft indexes to compute the energy in the signal, that gives me the wrong answer. Can you explain, why I can't use Y = fft(data, L); Y3energy = (Y.*conj(Y))/L; energy3 = sum(Y3energy); to find the energy of the in the signal. | |
| Feb 12, 2018 at 9:56 | history | edited | Gilles | CC BY-SA 3.0 |
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| Feb 12, 2018 at 9:09 | history | edited | Gilles | CC BY-SA 3.0 |
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| Feb 12, 2018 at 8:53 | history | edited | Gilles | CC BY-SA 3.0 |
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| Feb 12, 2018 at 8:46 | history | answered | Gilles | CC BY-SA 3.0 |