# Why is it a bad idea to filter by zeroing out FFT bins?

It's very easy to filter a signal by performing an FFT on it, zeroing out some of the bins, and then performing an IFFT. For instance:

t = linspace(0, 1, 256, endpoint=False)
x = sin(2 * pi * 3 * t) + cos(2 * pi * 100 * t)
X = fft(x)
X[64:192] = 0
y = ifft(X)


The high frequency component is completely removed by this "brickwall" FFT filter.

But I've heard this is not a good method to use.

• Why is it generally a bad idea?
• Are there circumstances in which it's an ok or good choice?

Zeroing bins in the frequency domain is the same as multiplying by a rectangular window in the frequency domain. Multiplying by a window in the frequency domain is the same as circular convolution by the transform of that window in the time domain. The transform of a rectangular window is the Sinc function ($\sin(\omega t)/\omega t$). Note that the Sinc function has lots of large ripples and ripples that extend the full width of time domain aperture. If a time-domain filter that can output all those ripples (ringing) is a "bad idea", then so is zeroing bins.

These ripples will be largest for any spectral content that is "between bins" or non-integer-periodic in the FFT aperture width. So if your original FFT input data is a window on any data that is somewhat non-periodic in that window (e.g. most non-synchronously sampled "real world" signals), then those particular artifacts will be produced by zero-ing bins.

Another way to look at it is that each FFT result bin represents a certain frequency of sine wave in the time domain. Thus zeroing a bin will produce the same result as subtracting that sine wave, or, equivalently, adding a sine wave of an exact FFT bin center frequency but with the opposite phase. Note that if the frequency of some content in the time domain is not purely integer periodic in the FFT width, then trying to cancel it by adding the inverse of an exactly integer periodic sine wave will produce, not silence, but something that looks more like a "beat" note (AM modulated sine wave of a different frequency). Again, probably not what is wanted.

Conversely, if your original time domain signal is just a few pure unmodulated sinusoids that are exactly integer periodic in the FFT aperture width, then zero-ing FFT bins will remove the designated ones without artifacts.

• This answer has good stuff, but I would prefer more of a focus on the Gibbs effect. – Jim Clay Dec 10 '12 at 2:04
• An attempt to get an answer to the Gibbs effect was already asked here: dsp.stackexchange.com/questions/1144/… – hotpaw2 Dec 10 '12 at 2:45
• @hotpaw2 This is a good explanation. However, I need a reference to this and am finding some difficulty in identifying one. It is the reason we do time domain filtering rather than work in the frequency domain. (Also, time domain can be real time.) However, nobody seems to start by stating this! – Hugh Feb 17 '18 at 8:55
• How can this be related with the window method for filter design? – Filipe Pinto May 17 '19 at 11:47
• Compare the transform of a Von Hann window (et.al.) with that of any rectangular window. Much better filter response in general, especially between FFT bins in the stop-band. In general, abruptly zeroing bins is worse than not-zeroing near the transitions. – hotpaw2 May 17 '19 at 13:58

This question has also confused me for a long time. @hotpaw2's explanation is good. You may be interested in the simple experiment using matlab.

https://poweidsplearningpath.blogspot.com/2019/04/dftidft.html

updated information.

To verify this fact is simple, we just need to cautiously observe the spectrum of impulse response of an ideal(?) band pass filter which just zeros out FFT bins. Why do I need to add the adverb "cautiously"? If we just use the same size of the FFT to observe the response of the impulse, we will be deceived as shown in Fig 1. Nonetheless, if we add the order of DFT when observing the output of the filter, that is, zero padding the impulse response, we can find the so called Gibbs phenomenon, ripples in frequency domain, as depicted in Fig 2.

The results in fact comes from the windowing effect. If you want to entirely understand the problem, please refer to chapter 7.6 and chapter 10.1-10.2 of the bible of DSP (1). To sum up, three key points are noted here.

1. Size of window and order of DFT(FFT) are totally independent. Don't mix them together.
2. Properties of window (type/size) dominate the shape of DTFT. (ex. wider main lobe lead to wider transient band in frequency response.)
3. DFT is just the sampling of DTFT in frequency domain. Moreover, the higher the order of DFT, the denser the spectrum of DFT is.

So, with the help of denser spectrum in Fig.2, we can see through the mask of ideal(fake) Band pass filter.

Deceitfully Freq. Response.

Gibbs phenomenon in Freq. Response.

(1) Alan V. Oppenheim and Ronald W. Schafer. 2009. Discrete-Time Signal Processing (3rd ed.). Prentice Hall Press, Upper Saddle River, NJ, USA.

fps = 15;

LPF = 1;
HPF = 2;

n = -511:512;
n0 = 0;
imp = (n==n0);

NyquistF = 1/2*fps;

%% Ideal BPF
tmp_N = 512;
tmp_n = 0:1:tmp_N-1;
freq = ( n .* fps) ./ tmp_N;
F = fft(imp, tmp_N);
F_bpf = IdealBandpassFilter(F, fps, LPF, HPF);
imp_rep =[real(ifft(F_bpf))'];

imp_rep2 =[zeros(1,2048) real(ifft(F_bpf))' zeros(1,2048)];

N = 2^nextpow2(length(imp_rep));
F = fft(imp_rep,N);
freq_step = fps/N;
freq = -fps/2:freq_step:fps/2-freq_step;
freq = freq(N/2+1:end)';

figure;
plot(freq,abs(F(1:N/2)));
xlabel('freq(Hz)');
ylabel('mag');

N = 2^nextpow2(length(imp_rep2));
F = fft(imp_rep2,N);
freq_step = fps/N;
freq = -fps/2:freq_step:fps/2-freq_step;
freq = freq(N/2+1:end)';

figure;
plot(freq,abs(F(1:N/2)));
xlabel('freq(Hz)');
ylabel('mag');
title('Zero Padding (DFT) with more points');

%% Function
function filered_signal = IdealBandpassFilter(input_signal, fs, w1, w2)

N = length(input_signal);
n = 0:1:N-1;
freq = ( n .* fs) ./ N;

filered_signal = zeros(N, 1);

for i = 1:N
if freq(i) > w1 & freq(i) < w2
filered_signal(i) = input_signal(i);
end

end
end

• Can this be converted into a comment? – endolith Apr 22 '19 at 16:20
• Sorry I don't have enough reputation. The explanation in the URL is written by my self as well. I just wanna provide a test code which can visualize the bad influence of ripple. – Po-wei Huang Apr 22 '19 at 19:20

FFT gives poor time resolution i.e it doesn't give information at what time that particular frequency exist. It gives information on existing frequency components for given signal duration.

By zeroing bins in FFT gives poor resolution after IFFT in time domain.

• however there are computational difficulties for very long signal to take fft and then ifft. To avoid zitters/ringing filtering of a signal has to be transit smoothly from pass band to stop band. – Itta Gouthami Apr 5 '18 at 5:58
• "FFT gives poor time resolution" FFT gives no time resolution, it's a spectral domain transform and so, like said afterwards, gives only information about frequency components of a signal. – EdParadox Aug 5 '19 at 15:11
• The resolution provided by an FFT is the length of it's window. Anything outside the FFT's window is not resolved as being inside the FFT's window. – hotpaw2 Aug 31 '19 at 5:50