# Gaussian filter as a low pass filter

Can anyone explain me how is a Gaussian filter a low pass filter? It may be a simple thing but I just can't seem to wrap my head around it. Also while applying a low pass filter for bandlimiting (to prevent aliasing), which of the following two is better: (a) Applying Gaussian filter to the signal. (b) Applying Box filter in the frequency domain and then using IFT on the resultant. I think option (a) is better because we just convolve the signal with a Gaussian while in option (b) we first need to get the FT, then apply the filter and then convert back using IFT. But what about the quality of the resultant signal?

I'm still a newbie so answers with less jargon will be appreciated.

• Do you perhaps mean a Gaussian window, rather then a Gaussian filter? Also, I can't understand what a) has to do with b) (although filtering in the frequency domain is often not a good idea unless it's FFT convolution) ? Maybe you can be more specific about what you need to do. – dsp_user Dec 26 '17 at 18:47
• Yes, a Gaussian window. I'm talking about using a low pass filter to prevent aliasing – user3033326 Dec 26 '17 at 18:51
• You should understand that filtering can be accomplished by using either a time-domain convolution (easy to implement but relatively slow) or FFT convolution (fast but more difficult to implement). Both of these techniques are equivalent and produce the same result. As for the windows used, a Gaussian is just one of the many windows to choose from. It has a relatively narrow main lobe (good frequency resolution) but the noise floor is not that good. I suggest that you start with a simple time domain convolution filter. – dsp_user Dec 26 '17 at 19:10
• It may also seem tempting to just adjust certain bins in the frequency domain. While multiplication in the frequency domain corresponds to convolution in the time domain, the chances are that you'll add enough aliasing (when IFFT) to introduce possibly audible digital artifacts to your signal. – dsp_user Dec 26 '17 at 19:14

it's because all of the filter coefficients (which is the impulse response $h[n]$) are positive and there is no "ringing" in the impulse response.
so, assuming the impulse response is scaled so that all of the $h[n]$ add to 1, the output of the gaussian filter will be a weighted average of the input. averaging is a process that removes high-frequency variation. so what is left are the low frequencies, like DC. that's what a low-pass filter does.