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I am struggling with understanding the consequences of oversampling on the frequency spectrum of the signal.

If I understand correctly, with an oversampling rate of 8X we insert 7 new values for each value measured. In case we keep this new values at 0 then we are not inserting any new frequencies to the signal, so the frequency spectrum should stay the same.

On the other hand, if we interpolate the new values eg by averaging the adjacent ones, then we MIGHT generate some new frequencies, that might appear in our frequency spectrum. This new inserted freuquencies will be of higher magnitude and should be filtered out if we want to keep the signal as "clean" as possible.

Am I correct with my understanding of oversampling and interpolation? Thanks in advance for any answer!

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  • $\begingroup$ Inserting zeros between samples has the effect of replicating the signal's spectrum. Thus, one would apply a low-pass filter to this zero-interleaved data to remove these copies and complete the upsampling. $\endgroup$ – AnonSubmitter85 Jan 23 at 18:05
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Yes inserting zeros does insert new frequencies in the unique digital spectrum that extends from $0$ to $2\pi$ radians/sample or equivalently $\pm \pi$ radians/sample corresponding to $\pm F_s/2$ where $F_s$ is the sampling rate. The easiest way to see this intuitively is to consider a DC signal represented by a stream of constants, such as:

$x_1 = \begin{bmatrix}1 & 1 & 1 & 1 &1 ...\end{bmatrix}$

This is clearly a sampled DC signal, but insert zeros and we get:

$x_2 = \begin{bmatrix}1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 &1 ...\end{bmatrix}$

Now we have a series of periodic impulses (unit samples).

Consider the same in the continuous time domain if that is easier to see: The Fourier Transform for a series of repeat impulses in the time domain is a series of repeated impulses in the Frequency domain. Each harmonic will be a multiple of the repetition rate (which makes sense).

In short, changing the time response of the signal requires non-zero frequency content. We started with DC so the frequency content was a tone at frequency = 0. If we then make that constant value in time abruptly change in one single sample all the way to zero; such a relatively fast change would require very high frequencies. If we instead changes slowly toward zero over many samples, the frequency content would dominate in lower frequencies (slow change).

Further when we are strictly in the digital domain inserting a zero is NOT the same as the empty undefined space between samples. The periodic spectrum when extended beyond the sampling rate prior to inserting zeros, does become the new spectrum with the zeros inserted, as I explain further in the linked posts that provide a more theoretical insight. However this is indeed a change and with that can certainly be described as a creation of new signals in our digital spectrum from the point of consideration that our unique digital spectrum of interest extends from $0$ to $2\pi$ radians/sample. I can understand from this philosophically how one can say "No new frequencies are inserted" since the periodic spectrum that exists is simply compressed on our frequency axis. Personally, when I am working in the digital domain, I typically normalize the sampling rate to $1$ cycle/sample or $2\pi$ radians/sample and view it from that perspective until actually having to translate to/from the analog world. So if I do a process in the digital domain that changes the spectrum in that range such as zero inserts- I would describe that as inserting new frequencies.

For further details on that and applying this to interpolation see:

Fourier Transform of an Impulse Train

Higher order harmonics during sampling

Interpolation through zero insert and filtering:

Choosing right cut-off frequency for a LP filter in upsampler

What is the impulse response used in an interpolation filter when upsampling?

Ideal Interpolation Filter for zero insert resampling:

Downsample: resample vs antialias fitlering + decimation

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  • $\begingroup$ The problem here is that you're considering the first example as representing a DC value and nothing else, while the second you're considering as having repeated frequency content. However, the first signal is sampled and therefore also has repeated frequency content. In fact, it's exactly the sample frequency content as the second, assume the first represents a 1x sample rate and the second 4x. That is, if you play back the first, without filtering, and the second at 4x that rate, both will look the same in both time and frequency domains. $\endgroup$ – earlevel Jan 24 at 0:02
  • $\begingroup$ @earlevel Yes agreed but in that context I am only referring to the first Nyquist zone- the unique digital frequency span from 0 to $F_s$ and for that this all applies. $\endgroup$ – Dan Boschen Jan 24 at 0:15
  • $\begingroup$ Think it it this way, all ones is the sampled representation of a DC signal. Insert zeros and you will have additional frequencies within your normalized frequency span between 0 and $F_s$. In the links I explain the points you are also making but this is an intuitive way to look at it that is completely valid. $\endgroup$ – Dan Boschen Jan 24 at 0:20
  • $\begingroup$ Yes, I understand the point of view, since frequencies (that were there in the first place) are now in the band of interest (because it has been widened. I did want to comment on it, though, because it conflicts with my answer here, if you boil both of our answer down ("yes it does" vs "no it doesn't" posts of view). $\endgroup$ – earlevel Jan 24 at 0:52
  • $\begingroup$ Ah i see your reason to comment - yes important to clarify what one means by “new frequencies” and then almost gets philosophical. Both good answers I think and helpful. $\endgroup$ – Dan Boschen Jan 24 at 0:57
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Interpolation by avaraging does introduce new frequencies as it doesn't reproduce the signal assumed to be the original one. The correct way to interpolate the new values is the Shannon interpolation. Ps: This method is equally correct in the time and frequency domains.

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What you define as oversampling is actually sequence expansion by zero stuffing in between its samples. Which is an operation performed as a prerequisite of interpolation. And yes; zero stuffing a sequence will alter its spectrum as explained by DanBoschen.

Oversampling implies an ADC operation in which a signal is sampled above its Nyquist rate. This operation does not alter the frequency spectrum of the signal but it effects the amplitude scaling of it. Furthermore in an oversampled signal the spectrum will be zero after the signal bandwidth up to the Nyquist frequency.

Note that consequence of oversampling in the discrete-time frequency is that, the frequency axis is compressed from the frequency $\omega = \pi$ towards $\omega = 0$; so this is also a change in relative positioning of frequencies. (But which can be reverted.)

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The key to understanding what inserting zeros does is to understand two things: what samples represent in the time domain (because we want to insert zeros in the time domain), and what they represent in the frequency domain (because we want to know what it did to the spectrum).

First, sampling is a modulation (PCM—Pulse Code Modulation), equivalent to multiplying an impulse train by our analog signal, and creates images in the frequency domain. Here is an example spectrum represented by our samples; the original analog signal's spectrum is shown in green, and images in red:

Original spectrum

These images are the price we pay to represent the analog signal as samples. As such, the usable bandwidth is from 0 Hz up to (but not including) half the sample rate. Above that is a backwards image of our original spectrum, with images repeated around multiples of the sample rate. It's OK, we remove the images when we convert back to analog, using a DAC's lowpass filter.

In the time domain, samples represent impulses. They are instantaneous values taken at a constant interval. Inserting zeros changes nothing except what we consider the sample rate.

For instance, sample a signal one time per second. That represents an impulse train, so consider playing it back raw, as an impulse train.

Now consider placing a zero-valued sample between each of the original samples. Consider playing it back as an impulse train but at twice the original rate, two times per second.

Zero insertion

Can you see that nothing has changed in the signal, except the sample rate? Likewise, if we look at the spectrum, nothing has changed in the frequency domain—this is obvious, since the time domain signal has not changed.

Upsampled spectrum

However, our usable bandwidth has doubled. The first inverted image now lies in our usable band, now shown in green. It will not be removed by the DAC when played back through a DAC at the new, higher rate. And it will be a problem with any non-linear processing in the digital domain.

That is why we follow the zero-insertion with a lowpass filter (or combine the two steps for efficiency). Here it is again after proper filtering, below half the sample rate:

Upsampled and filtered spectrum

So, the answer is that nothing changes in proper integer sample rate conversion by zero-insertion, and the result is as perfect as the lowpass filter used to clean up the exposed images.

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