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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 '20 at 18:05
  • $\begingroup$ Related $\endgroup$ – OverLordGoldDragon Nov 1 '20 at 16:06
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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 type of modulation (PAM—Pulse Amplitude Modulation), equivalent to multiplying an impulse train by our analog signal, and creates images in the frequency domain. When converted to digital sample values, we have PCM (Pulse Code Modulation), a common term for digital audio. 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 original 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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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. $\endgroup$ – Nigel Redmon Jan 24 '20 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 '20 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 '20 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). $\endgroup$ – Nigel Redmon Jan 24 '20 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 '20 at 0:57
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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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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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Zero stuffing does not insert additional frequencies and the frequencies above the original signal frequency are not present in the original signal; however, because the original signal is a set of samples, there are convolved images in the frequency domain and those higher frequencies will be taken up by the first duplicate image in this case. It simply increases the amount of frequencies that are sampled, i.e. the period of sampling in the frequency domain widens and ends up including the next image of the original signal in the frequency domain, which is known as imaging.

enter image description here

Because the original signal has a finite period in the time domain, it must therefore have discrete frequency samples in the frequency domain (which is shown on row 2; the dotted line indicates an impulse envelope and the ellipses indicate that the period repeats infinitely). The signal has a total bandwidth of Fs or a Fmax of 1/2 Fs, therefore it needs to be sampled at at least Fs as the Nyquist rate, which is whatever the bandwidth happens to be. When this time domain signal is sampled on row 3 at every Ts seconds, it convolves the frequency domain every Fs.

On the 4th row, we upsample a time domain signal 2x which is already a set of samples and hence already has infinite images in the frequency domain. Because it is multiplying a set of impulses with a denser set of impulses, there will be 0s where the impulses don't intersect (the dotted line shows the envelope of the impulses and not a continuous signal, so it is actually 0 between the impulses) (in this case every other sample will be a 0 because we are upsampling 2x. Ts is now half the Ts of the original signal). The frequency domain of the signal will now be convolved every 2Fs. The new Fs is 2x the original. The time domain samples will just be the original samples but with 0s interspersed between the original samples.

The resulting frequency domain is identical, except Fs now covers a 2x larger window of the frequency domain. This means that you need a low pass filter at the frequency of the original Fs to remove the unwanted frequencies to get the resulting Fs window you would have got from sampling the original continuous signal at that sampling rate as opposed to a set of samples.

Sampling the original time domain signal 2x would have yielded:

enter image description here

You don't need a low pass filter here because the images are greater than the Nyquist frequency.

When you filter out the imaging of the upsample in the first scenario, the frequency domain resembles the above scenario. It is called interpolation, because it turns each 0 into an interpolation between the points either side, to match identically the time domain of above scenario. The zero stuffing is the prerequisite part of interpolation.

All in all, upsampling is the process of zero stuffing and interpolating (filtering) a set of samples of signal to give the set of samples a higher sampling rate, as if they had been taken from the original analogue signal at that higher sampling rate. The underlying frequency components in the signal does not change. You are just sampling it at a higher rate.

The DFT of the zero stuffed samples is the original frequency domain samples and another set of samples the same size added to the end which sample an image. If you make the samples of the image all 0 and then perform an IDFT, the 0s in the resulting set of time domain samples now become interpolated points.

The frequency domain is a series of infinite images which has been multiplied by the filter, which is why the rolloff matters, because the filter is not being multiplied by 0, but by images (and the side lobes of the frequency domain sinc impulses). It is a series of infinite images because the time domain was a series of impulses, which were windowed to the signal length (which merely causes a convolution around the impulses in the images) and then the multiplication with the filter removes all of these images and the most possible of the 2 immediately adjacent images either side.

The original signal and the upsampled signal have the same window size and therefore the impulses representing the samples in the frequency domain have the same shape and size. The impulses in the time domain of the upsampled signal are smaller and hence the filter in the frequency domain is wider to match the sampling frequency and so is the ZOH DAC frequency response. The resulting pulse shape filter is wider and this allows for some side lobes of the frequency domain impulse sincs to be included in the output because they are not removed by the filter (this is identical to the scenario where the original signal is sampled at this frequency and results in the same reconstruction). The rolloff of the filter is actually the result of the windowing, because the sinc time domain pulse shape that is used to reconstruct cannot be infinite, and therefore it is not a perfect rect shape brick wall filter in the frequency domain. The rolloff of the upsampled signal is of course identical. The fact that the impulse sinc side lobes in the frequency domain that also get multiplied with it have decreased in magnitude slightly by this point means that the rolloff x images x side lobes is less, so there is less spectral leakage outside of the desired band, though twice as large. This means that aliasing under the rolloff has less effect, as well as it being moved outside the audible range.

The point of upsampling is to move the reconstruction/anti-aliasing filter well above audible frequencies, where it can be more gentle and not affect the audible spectrum. A high-quality filter at 22khz is hard to make in hardware without distorting phase and without cutting into frequencies below 20khz. A gentle filter at 88khz is much easier, and it can distort there all it wants without affecting the audible range. It's a simple engineering solution to a problem that could otherwise become audible... at least to some of us (not to me, not for a while :() [1]

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