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faq:why_is_there_a_residual_50hz_line-noise_component_after_applying_a_dft_filter [2014/05/02 10:13]
robert [Why is there a residual 50Hz line-noise component after applying a DFT filter?]
faq:why_is_there_a_residual_50hz_line-noise_component_after_applying_a_dft_filter [2017/08/17 11:21] (current)
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 {{:​faq:​dftfilter1.png?​500|}} {{:​faq:​dftfilter1.png?​500|}}
  
-Then imagine subtracting the estimated 5 Hz component. At the begin you subtract too much, causing a negative (sign-flipped) 50Hz signal remaining in the data, and towards the end of the trial you are not subtracting enough, causing a positive (non sign-flipped) 5 Hz signal remaining (black line in third figure). Computed over the whole interval, the 5 Hz amplitude ​are zero. However, for a short window ​at the begin, there is non-zero amplitude at 5 Hz. In the middle the amplitude dips, but towards the end of the trial the amplitude increases and is again non-zero. I.e. the time-varying amplitude is V or U shaped: large at the edges, small in the middle. ​+Then imagine subtracting the estimated 5 Hz component. At the begin you subtract too much, causing a negative (sign-flipped) 50Hz signal remaining in the data, and towards the end of the trial you are not subtracting enough, causing a positive (non sign-flipped) 5 Hz signal remaining (black line in third figure). Computed over the whole time interval, the 5 Hz amplitude ​is zero. However, for a short time interval ​at the begin, there is non-zero amplitude at 5 Hz. In the middle the amplitude dips, but towards the end of the trial the amplitude increases and is again non-zero. I.e. the time-varying amplitude is V-shaped: large at the edges, small in the middle. Similarly, if you were to look at the power, it would be U-shaped.
  
   % subtract the 5 Hz fit   % subtract the 5 Hz fit