Filter DC offset - apply High-pass filter (in time-domain or frequency domain?)

Hello all,

I want to filter DC offset for a signal by applying a high-pass filter to it (I've been told that to filter the DC offset this is what I should do - I'm not sure why exactly!!)

My question is; do I do that in the time-domain or frequency-domain after applying FFT? And I would appreciate some explanation if possible.


Roronoa Zoro


Gold Member
think about DC offset as a signal with frequency of 0, a high pass filter will attenuate low frequencies, so at frequency of 0, all DC offset becomes attenuated (under steady state conditions).

The easiest way is to stick a capacitor (what we call a coupling capacitor) in the path of the signal, this effective creates an open circuit for DC. And if you choose your capacitance value correctly, its reactance at the frequency of your concerned signal will be negligible
Thank you wukunlin.

It's really hard for me to picture that. I can picture a sin wave that is shifted up/down the y-axis and I assume that this is the effect of a DC offset. Now when I apply the filter, how would the filter know that these are actually two signals, one 0 frequency signal and a sin signal in order to remove the 0 frequency signal? Am I missing something here?

And in programming (like Matlab), do I apply the filter in the time-domain of in the frequency-domain? Or does it not matter?


Science Advisor
A DC voltage offset on a sinewave signal will charge up a series capacitor to the DC voltage and, from then on, the output will be zero volts DC with the sinewave varying the output voltage positively and negatively about this zero volt level.

The filter doesn't "know" anything. What gets through a filter just depends on the properties of the filter and the frequency that is applied to it.

For example, a sinewave only has one frequency, so applying a sinewave to a high-pass filter will produce an output that varies according to the frequency of the sinewave.

If its frequency was below the cutoff frequency, the output would ideally be zero, but at least it should be greatly reduced compared with the input.

If its frequency was above the cutoff frequency, the output should ideally be the same as the input. It will be less attenuated than below the cutoff frequency anyway.

If the frequency was varied, then you could trace out a frequency response of the filter. Filters are rarely perfect, so this is worth doing.

A square wave contains the fundamental frequency plus odd harmonics of that frequency.
So, a 1000 Hz square wave will contain components of 1000 Hz , 3000 Hz , 5000 Hz etc
If this was applied to a high pass filter with a cutoff of 2000 Hz, the result would be a distorted waveform that looked hardly like a square wave at all. It would have a dip in the normally flat top and bottom of the square wave.

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