I am having trouble understanding a concept that I have seen in many published papers. For example, in this paper (http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0082748) they wrote in their methods that to "obtain adequate frequency resolution we downsampled from 44.1 kHz to 8 kHz".

My understanding is that the sample rate is how many times (per second) the actual sound wave was "sampled", and that the maximum frequency that can be detected is 1/2 the sampling rate (or the Nyquist frequency). So by downsampling, you will lose the ability to detect higher frequencies, and in a sense downsampling acts as a sort of filter, as it gets rid of the higher frequency sounds from the recording.

This makes sense to do if your signal of interest is at lower frequencies, but I still don't understand how this gives "adequate frequency resolution". Does downsampling have any influence on the ability to detect lower frequencies? What do they mean by "frequency resolution"? Thank you so much in advance!

  • $\begingroup$ My interpretation: When analyzing a DFT, it makes sense to limit the sampling rate to something close to Nyquist. Otherwise, one ends up with a lot of bins that do not have any information. When plotting (either on screen or on paper), it makes sense to use as many pixels or inches of paper as possible to display the frequencies of interest. $\endgroup$
    – MBaz
    Apr 26, 2016 at 14:57

2 Answers 2


Your understanding is correct. I don't have the full text of the paper, but it sounds like they aren't being very precise with their description. As you pointed out, downsampling doesn't improve frequency resolution in and of itself, it merely reduces the amount of bandwidth that can be represented unambiguously.

However, I can take a guess at what they were referring to. Assuming they are using a DFT for spectral analysis (the common choice), note that the fact that for a given DFT size $N$, the output bin spacing is a function of the sample rate:

$$ f_{res} = \frac{f_s}{N} $$

So, if a smaller frequency resolution in your spectral calculation is desired, decreasing the sample rate of the data can help you do that. Of course, you could also increase the DFT size $N$ as well; there are some applications where that's not an option, though (e.g. if you're using hardware that is limited to certain FFT sizes). Also, this method can have advantages when the signal of interest is in a small band of frequencies relative to the original sample rate; in this case, downsampling followed by a smaller DFT can be more computationally advantageous than doing one large DFT (this is known as the zoom FFT technique).

Note that there isn't a free lunch here: to actually realize an increase in resolution, you need $N$ samples at the lower sample rate, which requires a longer observation time. This is the same total observation time that is required by just collecting more data at the higher sample rate and then doing a larger DFT.


It took me a while to visualize this stuff.

In the $z$-domain, the frequencies from 0 to Nyquist are always represented as 0 to $\pi$. This means that if you increase the sampling rate, there are more frequencies that must be represented in your 0 to $\pi$ region (represented by $\omega$). This means as the sample rate increases, a 50Hz signal, say, is represented by a smaller $\omega$.

Some types of digital filter (FIR) and also FFT have a poor resolution at the lower frequencies, so you downsample to effectively increase their frequency resolution at lower frequencies. Downsampling increases the $\omega$ value you need to represent say, 50Hz.

An interesting problem will arise though if you want to increase your FFT resolution. Downsampling will not help, you must increase your sample set and ultimately the number of FFT bins you have.

I've mainly hit this problem designing IIRs where trying to design certain types of IIR of, say, 5Hz at a sampling rate of 192KHz can be more prone to problems than at 48KHz.


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