0
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – dsp_user
    Dec 26 '17 at 18:47
  • $\begingroup$ Yes, a Gaussian window. I'm talking about using a low pass filter to prevent aliasing $\endgroup$
    – user3033326
    Dec 26 '17 at 18:51
  • $\begingroup$ 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. $\endgroup$
    – dsp_user
    Dec 26 '17 at 19:10
  • $\begingroup$ 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. $\endgroup$
    – dsp_user
    Dec 26 '17 at 19:14
1
$\begingroup$

Frequency selective filters are defined according to their frequency domain characterstics. And a lowpass filter is the one whose frequency response magnitude is very small for high frequencies that are rejected and is about unity for low frequencies that are passed, hence the name lowpass.

Looking at the frequency response magnitude of a time-domain Gaussian window reveals the fact that it resembles a lowpass filter charactheristic, though not an ideal brickwall type with a steep transition from passband to stopband, but a mild and smooth one instead.

One application of a lowpass filter is in baseband sampling to prevent aliasing. In such an application the type and quality of the utilized lowpass filter is determined by a number of factors such as the bandwidth of the input signal, its spectral characthersitics, the required sampling rate, target SNR etc. When these conditions are mild. You can use a Gaussian filter to bandlimit the input signal for anti-aliasing purposes. However note that a Gaussian filter in continuous time would generally be replaced by a simpler RC lowpass filter as it would be much simpler to implement.

When you are doing discrete-time sample rate conversion, then a Gaussian filter can be applied in time domain. Or as you said you can try a frequency domain masking and inversion back to time: apparently depicting the effect of a nonrealizable (ideal) brickwall filter on the frequency domain. However note that the results will not be identical to a time domain Gausisan filtering and you shall judge it better by testing, before attempting a full analytical description of the difference in between.

Improve this answer
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.