# How do you simultaneously undersample and oversample?

"Undersampling" and "Oversampling" are common terms when referring to the choice of sampling frequency for analog to digital data conversion. Is it possible to Undersample and Oversample a waveform at the same time? Examples showing why or why not will be helpful.

The system in question has one real passband analog input and one real passband digital output. This is a practical example of a technique used in digital radio receivers.

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• Is that as a building block with 1 input, 2 outputs and real-time operation? I.e is the question about a specific topology or the operations themselves? How does this "thing" look like in the end? – A_A Jul 3 '16 at 8:17
• @A_A I added to the question based on your comment. – Dan Boschen Jul 3 '16 at 11:38
• Thank you. It potentially contains a big hint but I don't get it :) Sampling is a "fancy" modulation so if used in DAB it probably produces an IF somewhere but that would be just bandpass sampling. There is the possibility of course to undersample a very short target bandwidth in such a way that it is still considered oversampling from "its point of view"...but that's taking too far the wrong way maybe :) – A_A Jul 4 '16 at 8:55
• I think you are close! – Dan Boschen Jul 4 '16 at 14:20
• Here is another clue: dsp.stackexchange.com/questions/31843/… – Dan Boschen Jul 8 '16 at 1:16

Undersampling is also known as "bandpass sampling" and "IF sampling". Consider the sampling of an 11 Hz sine wave and a 1 Hz sine wave, both sampled by a 10 Hz sampling clock as in the figure below:

The samples that are produced (the red circles) for both waveforms are identical. Thus the 11 Hz "Intermediate Freuency" or IF has been downconverted in the sampling process to 1 Hz. We can use this technique to undersample a bandpass waveform, and thereby down-convert it at the same time (putting our waveform of interest in the first Nyquist zone: ±$F_s/2$ where $F_s$ is the sampling freqquency). A key requirement that satisfies Nyquist is that the sampling rate used must exceed twice the bandwidth of the signal (note I said bandwidth, not highest frequency!).

That is "Undersampling" as most often described. To oversample a waveform, we sample at a rate significantly higher than the Nyquist criteria, with one motivation to increase the SNR as limited by quantization noise (by sampling hgiher, the same quantization noise is spread over a wider digital frequency, which we can subsequently filter down to our bandwidth of interest and thereby eliminate a significant amount of the noise). Therefore to both undersample and oversample a waveform, we choose a sampling frequency that is significantly larger than the signal bandwidth (thus oversample), and then use a multiple of this sampling clock to down-convert the waveform of interest positioned at a higher IF frequency (undersample).

For example, consider a waveform that occupies 10 KHz of BW that resides at 20.25 MHz. If we sample this with a 10 MSps sampling clock, we would be both undersampling and oversampling at the same time. (The undersampling creates a digital waveform with 10 KHz of BW at 250 KHz, which is oversampled at 10 MSps).

• Thanks for interesting points. Could you please talk more about "by sampling hgiher, the same quantization noise is spread over a wider digital frequency, which we can subsequently filter down to our bandwidth of interest and thereby eliminate a significant amount of the noise". Why ? I always think that quantization noise can be modeled as i.i.d uniform distribution random variables. Because they are i.i.d, the PSD should be white. You could show me some documents if you don't have time to explain it. Thanks again. – AlexTP Apr 14 '17 at 16:21
• The PSD is in fact well approximated by a white noise process, meaning spread over all frequencies. Just in the case of a digital system "All Frequencies" is in the range of $-f_s/2$ to $+f_s/2$. Therefore if you increase $f_s$ the noise density in power/Hz goes down. Does that clear it up? – Dan Boschen Apr 14 '17 at 17:01
• I am sorry but I still don't get it. Are you telling me that the total quantization noise is $[-f_s/2,+f_s/2]$ is the same of $[-f_s,+f_s]$ if we double the sampling rate ? Imagine you have $n$ samples then you have $n$ quantization noise random variables ~ uniform distribution between $[-q/2,+q/2]$. Doubling the sampling rate yields $2n$ random variables with the same distribution. You scale the band $[-f_s/2,+f_s/2]$ but you have more sample with the variance unchanged. Am I missing something ? – AlexTP Apr 14 '17 at 18:45
• @AlexTP Yes that is exactly correct! One way to explain it (and I like this view as I deal with mixed signal analog/digital design) is consider an analog noise distribution. Regardless of what sampling rate you choose, as long as you take enough samples so that your estimate is close to the actual variance, you will end up with the same variance. (Also assuming your sampling clock has no correlation to the signal, if we are concerned with white noise that would indeed be the case). – Dan Boschen Apr 14 '17 at 21:45
• To make it even simpler perhaps, sample a sine wave at different rates with a clock that in not commensurate with the sinewave; and for each case determine the standard deviation of the sine wave-- you will see that the standard deviation (if you take enough samples) is the same every time. – Dan Boschen Apr 14 '17 at 21:46