# Differences

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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) 2014/05/02 10:15 robert 2014/05/02 10:13 robert [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:37 jmhorschig [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:34 jmhorschig [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:34 jmhorschig [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:25 jmhorschig see bug 25572012/03/02 16:48 [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2011/01/28 15:03 robert added artifact tag2011/01/03 20:53 robert 2010/12/03 13:50 robert 2010/12/03 13:50 robert created 2014/05/02 10:15 robert 2014/05/02 10:13 robert [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:37 jmhorschig [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:34 jmhorschig [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:34 jmhorschig [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2014/05/02 09:25 jmhorschig see bug 25572012/03/02 16:48 [Why is there a residual 50Hz line-noise component after applying a DFT filter?] 2011/01/28 15:03 robert added artifact tag2011/01/03 20:53 robert 2010/12/03 13:50 robert 2010/12/03 13:50 robert created Line 29: Line 29: {{:​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