20 ms·

Nyquist Frequency

81 points by 1_over_n 3y ago

polalavik 3y ago
If you're interested in learning more about various DSP topics, I run a blog on over at https://signalprocessingjobs.com/ https://signalprocessingjobs.com/ - a signal processing job board and blog! One of the more popular series is the Journal2Matlab blog about translating academic journal papers into easy to read matlab.
gooseyard 3y ago
Dan Worrall made a fantastic video which touches on Nyquist. His youtube channel is a tremendous resource: https://www.youtube.com/watch?v=-jCwIsT0X8M https://www.youtube.com/watch?v=-jCwIsT0X8M
elromulous 3y ago
To add another misconception, the Nyquist frequency is a lower bound, below which you necessarily get aliasing. It doesn't say anything about whether said sampling rate is sufficient for reconstruction or whatever your intended use is. E.g. sampling a 1hz signal at 2hz still doesn't tell you if the signal was a 1hz sin or a 1hz sawtooth (depending on how lucky or unlucky you are).
- kardos 3y ago
  A 1hz sawtooth would not be band limited below 2hz
- Evidlo 3y ago
  A 1Hz sawtooth contains frequencies above 1Hz. It actually has frequency components that go out to infinity, so its impossible to perfectly reconstruct a sawtooth without knowing beforehand that its a sawtooth. This is true for any signal with discontinuities (i.e. not "band-limited").
  - PaulDavisThe1st 3y ago
    This is incorrect, though subtly, and for several different reasons: 1) It is completely possible to create a sawtooth wave that contains only a single frequency. However, you could also consider the wave to be an (infinite) sum of sinusoids at different frequencies. Both views are "correct", and which is more appropriate depends on the context. 2) Related to (1): natural (acoustic) sounds are almost always best considered as a sine series. While there are such sounds which are most easily described as a sawtooth, when you consider the physical/mechanical process by which they are formed, the sine series is a more obvious approach. 3) A digital 1Hz sinusoid can trivially contain no harmonics at all. However, the moment you attempt to convert this into an acoustic pressure wave, the nature of the physical world essentially guarantees that the acoustic pressure wave will have a series of harmonics going out far beyond the base frequency. Once you start actually moving things (like magnetic coils, speaker cones and air), it's more or less impossible to avoid generating harmonics. But since the original signal was genuinely a pure sine tone, it becomes a little tricky to decide what the correct way to describe this is.
    - stagger87 3y ago
      At the "textbook"/"theory" level, the person you are replying to is not wrong. A sawtooth waveform has infinite harmonics. If you were going to be nitpicky (which your response was in that spirit), the best thing to have said (IMO) was that the high frequency harmonics are going to drop off and be below any sort of "noise floor" or sensitivity of the system and not matter anyways. Instead you wrote a bunch of stuff about sounds and pressure waves that I don't think had the effect you intended. I think you lost the plot somewhere along the way.
      - PaulDavisThe1st 3y ago
        > A sawtooth waveform has infinite harmonics This is only true if you consider the waveform to be a sine series. As I indicated, this is a perfectly legitimate way to think about a sawtooth (and indeed, it appears to be fundamentally how the human ear works too). But a sawtooth waveform is also nothing more than a very sharp rise/drop in air pressure followed by a longer drop/rise, repeated over and over again. If you want to synthesize a sawtooth wavefrom with analog equipment, then thinking of it as an (infinite) sine series makes sense, because that's how you will end up approximating the (perfect) sawtooth. However, digital synthesis does not require this sort of conception at all, and can be constructed without any summing of a harmonic series. Also, I find it assuming that in the comments of a post about nyquist, you would write > a bunch of stuff about sounds and pressure waves that I don't think had the effect you intended. I think you lost the plot somewhere along the way. What do you the plot is?
- GeompMankle 3y ago
  What is the point of adding a misconception? The conception of the theorem is that if the signal being sampled is sufficiently integrable AND bandlimited AND the signal is uniformly sampled at at least the Nyquist rate over all time/space THEN then reconstruction of the bandlimited signal is exactly possible using the sinc interpolator. The proof is covered in "Shannon's original proof" in the Wikipedia article and most books on signal analysis such as Gaskill's Linear System book. Most EE people will have to do the proof as an intro course assignment in the first month of a DSP class. OTOH, if you are not able to sample the function over all space or time AND the function happens to be periodic outside the interval you did sample THEN reconstruction of the bandlimited periodic signal is possible using the Dirchlet kernel. If you are not able to sample the function overall space (from the first) AND that function is not periodic, you have small problems which occasionally become big problems if you are no careful. Most DSP books have a chapter about windowing discrete data and dealing with this conundrum. Basically, exact reconstruction is not guaranteed and context-specific techniques need to be employed to ensure desirable fidelity.
- duped 3y ago
  > It doesn't say anything about whether said sampling rate is sufficient for reconstruction or whatever your intended use is. Formally, the Shannon-Nyquist theorem states that if you sample a band limited signal at twice its bandwidth, an ideal reconstruction filter can be used to perfectly reconstruct the input signal. There's some wiggle room over ideal sampling/filtering, but the point is that it tells you exactly what the input was, provided it was band limited. The misconception I think you're having is that band width is not the period of a signal.
- ska 3y ago
  That isn't really what is going on. If the signal doesn't contain any higher frequency information, the Nyquist limit establishes what you need to exactly reconstruct the signal. It is therefore sufficient for any use. So your case, a 1hz sin doesn't contain any higher frequencies, and will be reconstructed perfectly. A 1hz sawtooth contains higher frequencies, and so is not. I think what you are really getting into is that a signal with periodicity of, say 1hz, does not mean that the Nyquist limit is 1hz. Square waves and sawtooths are particularly obvious examples of this, because the sharp edges cannot be achieved without (very many) high frequency contributions. Now you can avoid this by creating a different set of component functions and a different sense of "frequency" but that just pushes the problem around. Also, since you are doing non-standard things you need to explain it, especially if what you are using doens't form a proper basis. Finally, of course this is all in the idea mathematical setting, in real world noise etc. also has to be taken into effect.
- Sesse__ 3y ago
  > To add another misconception, the Nyquist frequency is a lower bound, below which you necessarily get aliasing. It doesn't say anything about whether said sampling rate is sufficient for reconstruction or whatever your intended use is. Yes, it does. The Nyquist criterion gives exactly the (minimum) sampling frequency you need for perfect reconstruction of a bandlimited signal. > E.g. sampling a 1hz signal at 2hz still doesn't tell you if the signal was a 1hz sin or a 1hz sawtooth (depending on how lucky or unlucky you are). A 1 Hz sawtooth is not a bandlimited signal, so the Nyquist theorem does not apply.
- TimTheTinker 3y ago
  A low pass filter at 2hz would filter out the high frequencies contained in a sawtooth waveform, thus rendering a 1hz sine waveform. To accurately sample a 1hz sawtooth waveform, you'd have to filter/sample at a much higher frequency.
- Gordonjcp 3y ago
  Sampling a 1Hz sawtooth at 2Hz will alias.
- femto 3y ago
  If you know the signal is periodic with known frequency/period, you can be clever and sample it at that that frequency +/- a small offset. The frequency spurs then will not fall on top of each other and you can "unwrap" them to give a more complete picture of the signal. In that way you could determine whether a signal with a known frequency of 1Hz is a sawtooth or sine. Nyquist more or less says "If I you know nothing about the signal, by sampling at X Hz, you can determine what the signal looks like over a bandwidth of 0 Hz to X/2 Hz". If you have additional knowledge about the signal (eg. band limited, periodic or other) you can exceed those limits. It can also be looked at from an information viewpoint. Nyquist says "if you sample a signal at a certain rate you will get a certain amount of new information about it". You might "spend" this information by saying something about the signal over the band DC-f/2, or you might choose to say something about the signal over a different band of frequencies. In the example above we chose to say something about a set of discrete harmonic frequencies over a very wide bandwidth, ignoring the frequencies in between the harmonics as the 1Hz constraint told us they will be zero.
_kb 3y ago
Had a great encounter with this recently! In an environment I work there's multichannel audio recordings that are archived. The archival recordings all had a perfect 4kHz tone appearing, seemingly out of nowhere. This was happening on every channel, across every room, but only in one building. Nowhere else. Absolutely nothing of the sort showed up on live monitoring. The systems were all the same and yet this behaviour was consistent across all systems only at one location. The full system was reviewed: from processing, recording, signal distribution, audio capture, and in room. Maybe there was a test gen that had accidentally deployed? Nope. Some odd bug in an echo canceller? Also no. Something weird with interference from lighting or power? Slim chance, but also no. Complete mystery. When looking for acoustic sources there was an odd little blip on the RTA at 20kHz. This was traced back to a test tone emitted from the fire safety system (ultrasonic signal for continuous monitoring). It's inaudible to most people and will be filtered before any voice-to-text processing so no reason for concern. Anyway 20kHz is nowhere near 4kHz though so the search continued. The dissimilarly of 20kHz and 4kHz is true, until you consider what happens in a non-bandwidth limited signal. The initial capture was taking place at a 48kHz sampling rate. It turns out the archival was downsampling to 24kHz, without applying an anti-aliasing filter. Without filtering, any frequency content above the Nyquist 'folds' back over the reproducible range. So in this case a clean 24kHz bandwidth signal with a little bit of inaudible ultrasonic background noise was being folded at 12kHz to create a very audible 4kHz tone. It was essentially a capture the flag for signals nerds and a whole lot of fun to trace.
- spacechild1 3y ago
  > It turns out the archival was downsampling to 24kHz But... why?
  - deleted 3y ago
    [deleted]
  - InitialLastName 3y ago
    In situations where you don't need the archival to be at "perfect reproduction" quality (including things like broadcast archives or recordings of voice comms) you can get by with a 12kHz maximum frequency without losing the essentials (especially clarity of voices). Many adults can't hear much past 12kHz anyway and most music and voice content doesn't have content past 10khz. You don't lose much, but you save half your file size by x2 downsampling.
lumb63 3y ago
Signals and systems was a tough course for me. It was what crushed my 4.0 GPA. Nyquist frequency was a concept I could not wrap my head around. I’ve improved, but it still doesn’t click as I’d like it to. When I took the course, it made no sense to me that you could sample at twice the frequency of the signal and reconstruct it. Consider a sine wave at 1 Hz. If you sample at 2 Hz, you’d get readings of 0, 1, 0, -1, etc. If you graph that, it’s a perfect triangle wave, not a sine wave! That’s what I couldn’t not get past. I thought you’d need an infinite sampling rate to accurately capture the sine wave. As I type this out, I’m realizing that a critical component of this that I wasn’t taught (or I didn’t grasp) is the need for the signal to be bandlimited. Returning to my sine example from above, what bothered me was, if I don’t sample more points, how do I know that it’s only a sine wave, and nothing more? That only works if you pretend there are no higher frequencies (or filter them out, though an ideal filter is impossible in practice). If there aren’t higher frequencies, there can’t be anything you “can’t capture” by sampling at the Nyquist frequency.
- tomjakubowski 3y ago
  A triangle wave at 1Hz would have many higher frequency components. If you know a priori that the highest frequency of the signal is 1Hz, sampling at 2Hz is enough to infer 0, 1, 0, -1, ... came from a sine wave.
- jancsika 3y ago
  I've had an open GSoC project for some years to create a library that makes a handful of these audio misconceptions true. So the student would design an oscillator or oscillator bank where the closer you get to Nyquist, the more some "bad thing" happens to the corresponding output. Morphing into a triangle would be one way to do it.
- rnpk 3y ago
  You get the original sine wave back from 0, 1, 0, -1 not by plotting it linearly (which gives you the triangle) but by using a sinc interpolation function.
- Sesse__ 3y ago
  What you are saying is generally correct, but: If you sample a 1 Hz sine at 2 Hz, you wouldn't get readings of 0, 1, 0, -1, etc.; you would get readings of 1, -1, 1, -1, etc., or if you're very unlucky, 0, 0, 0, 0, …! The _exact_ case is of Fs/2 is, well, an edge case.
abhaynayar 3y ago
Soothing.
coolandsmartrr 3y ago
I saw the Nyquist Frequency mentioned in the American Cinematographer Magazine. The article illustrate how detailed patterns, like sweaters, can produce a fuzzy jagged artifact called moire. This is because there is too much information for camera's sensor to interpret and summarize the details into pixels (ie. surpassing the Nyquist Frequency). Their suggested solutions were to 1) get a wide-angle lens to reduce detail beamed into the sensor 2) use a larger image sensor or 3) remove the object causing moire artifacts.
- regularfry 3y ago
  Yep. Strictly speaking what's happening is that the pattern has a higher spatial frequency than the sensor, and the light detection acts as a non-linear interaction which aliases the higher frequencies down into the bandwidth of the sensor. A wide-angle lens would change the effective bandwidth of the system, as would a larger sensor: all either would do is change the apparent size of the moire pattern (possibly so it's less annoying). What you really want is something that would act as a spatial low-pass filter in front of the sensor; something like a very slightly frosted piece of glass which would prevent any feature size smaller than two sensor pixels from being resolved on the far side. I imagine if that wasn't a completely stupid idea for some other reason that you could buy them.
  - regularfry 3y ago
    Of course, it occurs that the non-stupid version of this is to defocus the lens slightly, so the point spread function does the low-pass filtering job. That also tells you why the advice isn't "put a low-pass filter in front of the sensor": it'll be because for the non-moire case you need the resolution.
  - kimburgess 3y ago
    You can buy them: https://rawlite.com/olpf-for-blackmagic-pocket-cinema-camera-4k/ https://rawlite.com/olpf-for-blackmagic-pocket-cinema-camera..., https://www.grassvalley.com/products/optical_low_pass_filter https://www.grassvalley.com/products/optical_low_pass_filter
  - 3y ago
monkeycantype 3y ago
The coolest nyquist frequency application I've every come across, if you look up how modulation of nerve impulses works in the optic never you can figure up the fastest rate of blinking your eye can perceive, and it checks out in reality.
xchip 3y ago
Beware, there are lots of misconceptions in the comments.
kurokawad 3y ago
[flagged]
EarthIsHome 3y ago
One misconception that many make regarding the Nyquist frequency is thinking that the sampling rate needs to be twice the highest frequency. Your sampling should really really be twice the bandwidth. e.g. your bandwidth is 100 MHz centered at 1 GHz (it needs to actually be bandlimited to 100 MHz**). You do not need to sample at 2.2 GHz. You sample at 200 MSPS (really, you should sample a little more than that, say 210 MSPS, so that the bandwidth of interest doesn't butt up against the Nyquist zone edges.)
- mikepavone 3y ago
  Is this assuming you have some analog hardware that's demodulating the signal in front of your ADC? How do you demodulate a signal from a 1GHz carrier with 200 MSPS?
  - klodolph 3y ago
    No, this assumption is incorrect. You can ADC first and then demodulate afterwards. The spectrum of your high-frequency (near 1 GHz) signal will be aliased at frequencies below the Nyquist frequency, but it’s easy to calculate the original frequency, if you know that the signal is band-limited.
  - labcomputer 3y ago
    As the sibling comment mentioned, you don’t need to demodulate first, because that is actually what the sampling process of your ADC does. You can think of it as multiplying the original signal by a comb (in the time domain) of delta functions, which folds everything (in the frequency domain) back into the nyquist frequency of your ADC. Each delta function corresponds to one sample. If your original signal was truly band-limited to 100MHz, then what comes out is a replica of the band limited signal. One catch (which is actually fairly easy to do in practice) is that the sampling window needs to correspond to around 1/f of the carrier frequency. This is what YakBizzaro is talking about (ADC analog bandwidth) in their sibling post.
    - mikepavone 3y ago

81 points by 1_over_n 3y ago

If you're interested in learning more about various DSP topics, I run a blog on over at https://signalprocessingjobs.com/ https://signalprocessingjobs.com/ - a signal processing job board and blog! One of the more popular series is the Journal2Matlab blog about translating academic journal papers into easy to read matlab.

gooseyard 3y ago

Dan Worrall made a fantastic video which touches on Nyquist. His youtube channel is a tremendous resource: https://www.youtube.com/watch?v=-jCwIsT0X8M https://www.youtube.com/watch?v=-jCwIsT0X8M

elromulous 3y ago

To add another misconception, the Nyquist frequency is a lower bound, below which you necessarily get aliasing. It doesn't say anything about whether said sampling rate is sufficient for reconstruction or whatever your intended use is. E.g. sampling a 1hz signal at 2hz still doesn't tell you if the signal was a 1hz sin or a 1hz sawtooth (depending on how lucky or unlucky you are).

kardos 3y ago
A 1hz sawtooth would not be band limited below 2hz
Evidlo 3y ago
A 1Hz sawtooth contains frequencies above 1Hz. It actually has frequency components that go out to infinity, so its impossible to perfectly reconstruct a sawtooth without knowing beforehand that its a sawtooth. This is true for any signal with discontinuities (i.e. not "band-limited").
- PaulDavisThe1st 3y ago
  This is incorrect, though subtly, and for several different reasons: 1) It is completely possible to create a sawtooth wave that contains only a single frequency. However, you could also consider the wave to be an (infinite) sum of sinusoids at different frequencies. Both views are "correct", and which is more appropriate depends on the context. 2) Related to (1): natural (acoustic) sounds are almost always best considered as a sine series. While there are such sounds which are most easily described as a sawtooth, when you consider the physical/mechanical process by which they are formed, the sine series is a more obvious approach. 3) A digital 1Hz sinusoid can trivially contain no harmonics at all. However, the moment you attempt to convert this into an acoustic pressure wave, the nature of the physical world essentially guarantees that the acoustic pressure wave will have a series of harmonics going out far beyond the base frequency. Once you start actually moving things (like magnetic coils, speaker cones and air), it's more or less impossible to avoid generating harmonics. But since the original signal was genuinely a pure sine tone, it becomes a little tricky to decide what the correct way to describe this is.
  - stagger87 3y ago
    At the "textbook"/"theory" level, the person you are replying to is not wrong. A sawtooth waveform has infinite harmonics. If you were going to be nitpicky (which your response was in that spirit), the best thing to have said (IMO) was that the high frequency harmonics are going to drop off and be below any sort of "noise floor" or sensitivity of the system and not matter anyways. Instead you wrote a bunch of stuff about sounds and pressure waves that I don't think had the effect you intended. I think you lost the plot somewhere along the way.
    - PaulDavisThe1st 3y ago
      > A sawtooth waveform has infinite harmonics This is only true if you consider the waveform to be a sine series. As I indicated, this is a perfectly legitimate way to think about a sawtooth (and indeed, it appears to be fundamentally how the human ear works too). But a sawtooth waveform is also nothing more than a very sharp rise/drop in air pressure followed by a longer drop/rise, repeated over and over again. If you want to synthesize a sawtooth wavefrom with analog equipment, then thinking of it as an (infinite) sine series makes sense, because that's how you will end up approximating the (perfect) sawtooth. However, digital synthesis does not require this sort of conception at all, and can be constructed without any summing of a harmonic series. Also, I find it assuming that in the comments of a post about nyquist, you would write > a bunch of stuff about sounds and pressure waves that I don't think had the effect you intended. I think you lost the plot somewhere along the way. What do you the plot is?
GeompMankle 3y ago
What is the point of adding a misconception? The conception of the theorem is that if the signal being sampled is sufficiently integrable AND bandlimited AND the signal is uniformly sampled at at least the Nyquist rate over all time/space THEN then reconstruction of the bandlimited signal is exactly possible using the sinc interpolator. The proof is covered in "Shannon's original proof" in the Wikipedia article and most books on signal analysis such as Gaskill's Linear System book. Most EE people will have to do the proof as an intro course assignment in the first month of a DSP class. OTOH, if you are not able to sample the function over all space or time AND the function happens to be periodic outside the interval you did sample THEN reconstruction of the bandlimited periodic signal is possible using the Dirchlet kernel. If you are not able to sample the function overall space (from the first) AND that function is not periodic, you have small problems which occasionally become big problems if you are no careful. Most DSP books have a chapter about windowing discrete data and dealing with this conundrum. Basically, exact reconstruction is not guaranteed and context-specific techniques need to be employed to ensure desirable fidelity.
duped 3y ago
> It doesn't say anything about whether said sampling rate is sufficient for reconstruction or whatever your intended use is. Formally, the Shannon-Nyquist theorem states that if you sample a band limited signal at twice its bandwidth, an ideal reconstruction filter can be used to perfectly reconstruct the input signal. There's some wiggle room over ideal sampling/filtering, but the point is that it tells you exactly what the input was, provided it was band limited. The misconception I think you're having is that band width is not the period of a signal.
ska 3y ago
That isn't really what is going on. If the signal doesn't contain any higher frequency information, the Nyquist limit establishes what you need to exactly reconstruct the signal. It is therefore sufficient for any use. So your case, a 1hz sin doesn't contain any higher frequencies, and will be reconstructed perfectly. A 1hz sawtooth contains higher frequencies, and so is not. I think what you are really getting into is that a signal with periodicity of, say 1hz, does not mean that the Nyquist limit is 1hz. Square waves and sawtooths are particularly obvious examples of this, because the sharp edges cannot be achieved without (very many) high frequency contributions. Now you can avoid this by creating a different set of component functions and a different sense of "frequency" but that just pushes the problem around. Also, since you are doing non-standard things you need to explain it, especially if what you are using doens't form a proper basis. Finally, of course this is all in the idea mathematical setting, in real world noise etc. also has to be taken into effect.
Sesse__ 3y ago
> To add another misconception, the Nyquist frequency is a lower bound, below which you necessarily get aliasing. It doesn't say anything about whether said sampling rate is sufficient for reconstruction or whatever your intended use is. Yes, it does. The Nyquist criterion gives exactly the (minimum) sampling frequency you need for perfect reconstruction of a bandlimited signal. > E.g. sampling a 1hz signal at 2hz still doesn't tell you if the signal was a 1hz sin or a 1hz sawtooth (depending on how lucky or unlucky you are). A 1 Hz sawtooth is not a bandlimited signal, so the Nyquist theorem does not apply.
TimTheTinker 3y ago
A low pass filter at 2hz would filter out the high frequencies contained in a sawtooth waveform, thus rendering a 1hz sine waveform. To accurately sample a 1hz sawtooth waveform, you'd have to filter/sample at a much higher frequency.
Gordonjcp 3y ago
Sampling a 1Hz sawtooth at 2Hz will alias.
femto 3y ago
If you know the signal is periodic with known frequency/period, you can be clever and sample it at that that frequency +/- a small offset. The frequency spurs then will not fall on top of each other and you can "unwrap" them to give a more complete picture of the signal. In that way you could determine whether a signal with a known frequency of 1Hz is a sawtooth or sine. Nyquist more or less says "If I you know nothing about the signal, by sampling at X Hz, you can determine what the signal looks like over a bandwidth of 0 Hz to X/2 Hz". If you have additional knowledge about the signal (eg. band limited, periodic or other) you can exceed those limits. It can also be looked at from an information viewpoint. Nyquist says "if you sample a signal at a certain rate you will get a certain amount of new information about it". You might "spend" this information by saying something about the signal over the band DC-f/2, or you might choose to say something about the signal over a different band of frequencies. In the example above we chose to say something about a set of discrete harmonic frequencies over a very wide bandwidth, ignoring the frequencies in between the harmonics as the 1Hz constraint told us they will be zero.

_kb 3y ago

Had a great encounter with this recently! In an environment I work there's multichannel audio recordings that are archived. The archival recordings all had a perfect 4kHz tone appearing, seemingly out of nowhere. This was happening on every channel, across every room, but only in one building. Nowhere else. Absolutely nothing of the sort showed up on live monitoring. The systems were all the same and yet this behaviour was consistent across all systems only at one location. The full system was reviewed: from processing, recording, signal distribution, audio capture, and in room. Maybe there was a test gen that had accidentally deployed? Nope. Some odd bug in an echo canceller? Also no. Something weird with interference from lighting or power? Slim chance, but also no. Complete mystery. When looking for acoustic sources there was an odd little blip on the RTA at 20kHz. This was traced back to a test tone emitted from the fire safety system (ultrasonic signal for continuous monitoring). It's inaudible to most people and will be filtered before any voice-to-text processing so no reason for concern. Anyway 20kHz is nowhere near 4kHz though so the search continued. The dissimilarly of 20kHz and 4kHz is true, until you consider what happens in a non-bandwidth limited signal. The initial capture was taking place at a 48kHz sampling rate. It turns out the archival was downsampling to 24kHz, without applying an anti-aliasing filter. Without filtering, any frequency content above the Nyquist 'folds' back over the reproducible range. So in this case a clean 24kHz bandwidth signal with a little bit of inaudible ultrasonic background noise was being folded at 12kHz to create a very audible 4kHz tone. It was essentially a capture the flag for signals nerds and a whole lot of fun to trace.

spacechild1 3y ago
> It turns out the archival was downsampling to 24kHz But... why?
- deleted 3y ago
  [deleted]
- InitialLastName 3y ago
  In situations where you don't need the archival to be at "perfect reproduction" quality (including things like broadcast archives or recordings of voice comms) you can get by with a 12kHz maximum frequency without losing the essentials (especially clarity of voices). Many adults can't hear much past 12kHz anyway and most music and voice content doesn't have content past 10khz. You don't lose much, but you save half your file size by x2 downsampling.

lumb63 3y ago

Signals and systems was a tough course for me. It was what crushed my 4.0 GPA. Nyquist frequency was a concept I could not wrap my head around. I’ve improved, but it still doesn’t click as I’d like it to. When I took the course, it made no sense to me that you could sample at twice the frequency of the signal and reconstruct it. Consider a sine wave at 1 Hz. If you sample at 2 Hz, you’d get readings of 0, 1, 0, -1, etc. If you graph that, it’s a perfect triangle wave, not a sine wave! That’s what I couldn’t not get past. I thought you’d need an infinite sampling rate to accurately capture the sine wave. As I type this out, I’m realizing that a critical component of this that I wasn’t taught (or I didn’t grasp) is the need for the signal to be bandlimited. Returning to my sine example from above, what bothered me was, if I don’t sample more points, how do I know that it’s only a sine wave, and nothing more? That only works if you pretend there are no higher frequencies (or filter them out, though an ideal filter is impossible in practice). If there aren’t higher frequencies, there can’t be anything you “can’t capture” by sampling at the Nyquist frequency.

tomjakubowski 3y ago
A triangle wave at 1Hz would have many higher frequency components. If you know a priori that the highest frequency of the signal is 1Hz, sampling at 2Hz is enough to infer 0, 1, 0, -1, ... came from a sine wave.
jancsika 3y ago
I've had an open GSoC project for some years to create a library that makes a handful of these audio misconceptions true. So the student would design an oscillator or oscillator bank where the closer you get to Nyquist, the more some "bad thing" happens to the corresponding output. Morphing into a triangle would be one way to do it.
rnpk 3y ago
You get the original sine wave back from 0, 1, 0, -1 not by plotting it linearly (which gives you the triangle) but by using a sinc interpolation function.
Sesse__ 3y ago
What you are saying is generally correct, but: If you sample a 1 Hz sine at 2 Hz, you wouldn't get readings of 0, 1, 0, -1, etc.; you would get readings of 1, -1, 1, -1, etc., or if you're very unlucky, 0, 0, 0, 0, …! The _exact_ case is of Fs/2 is, well, an edge case.

abhaynayar 3y ago

Soothing.

coolandsmartrr 3y ago

I saw the Nyquist Frequency mentioned in the American Cinematographer Magazine. The article illustrate how detailed patterns, like sweaters, can produce a fuzzy jagged artifact called moire. This is because there is too much information for camera's sensor to interpret and summarize the details into pixels (ie. surpassing the Nyquist Frequency). Their suggested solutions were to 1) get a wide-angle lens to reduce detail beamed into the sensor 2) use a larger image sensor or 3) remove the object causing moire artifacts.

regularfry 3y ago
Yep. Strictly speaking what's happening is that the pattern has a higher spatial frequency than the sensor, and the light detection acts as a non-linear interaction which aliases the higher frequencies down into the bandwidth of the sensor. A wide-angle lens would change the effective bandwidth of the system, as would a larger sensor: all either would do is change the apparent size of the moire pattern (possibly so it's less annoying). What you really want is something that would act as a spatial low-pass filter in front of the sensor; something like a very slightly frosted piece of glass which would prevent any feature size smaller than two sensor pixels from being resolved on the far side. I imagine if that wasn't a completely stupid idea for some other reason that you could buy them.
- regularfry 3y ago
  Of course, it occurs that the non-stupid version of this is to defocus the lens slightly, so the point spread function does the low-pass filtering job. That also tells you why the advice isn't "put a low-pass filter in front of the sensor": it'll be because for the non-moire case you need the resolution.
- kimburgess 3y ago
  You can buy them: https://rawlite.com/olpf-for-blackmagic-pocket-cinema-camera-4k/ https://rawlite.com/olpf-for-blackmagic-pocket-cinema-camera..., https://www.grassvalley.com/products/optical_low_pass_filter https://www.grassvalley.com/products/optical_low_pass_filter
- 3y ago

monkeycantype 3y ago

The coolest nyquist frequency application I've every come across, if you look up how modulation of nerve impulses works in the optic never you can figure up the fastest rate of blinking your eye can perceive, and it checks out in reality.

xchip 3y ago

Beware, there are lots of misconceptions in the comments.

kurokawad 3y ago

[flagged]

EarthIsHome 3y ago

One misconception that many make regarding the Nyquist frequency is thinking that the sampling rate needs to be twice the highest frequency. Your sampling should really really be twice the bandwidth. e.g. your bandwidth is 100 MHz centered at 1 GHz (it needs to actually be bandlimited to 100 MHz**). You do not need to sample at 2.2 GHz. You sample at 200 MSPS (really, you should sample a little more than that, say 210 MSPS, so that the bandwidth of interest doesn't butt up against the Nyquist zone edges.)

mikepavone 3y ago
Is this assuming you have some analog hardware that's demodulating the signal in front of your ADC? How do you demodulate a signal from a 1GHz carrier with 200 MSPS?
- klodolph 3y ago
  No, this assumption is incorrect. You can ADC first and then demodulate afterwards. The spectrum of your high-frequency (near 1 GHz) signal will be aliased at frequencies below the Nyquist frequency, but it’s easy to calculate the original frequency, if you know that the signal is band-limited.
- labcomputer 3y ago
  As the sibling comment mentioned, you don’t need to demodulate first, because that is actually what the sampling process of your ADC does. You can think of it as multiplying the original signal by a comb (in the time domain) of delta functions, which folds everything (in the frequency domain) back into the nyquist frequency of your ADC. Each delta function corresponds to one sample. If your original signal was truly band-limited to 100MHz, then what comes out is a replica of the band limited signal. One catch (which is actually fairly easy to do in practice) is that the sampling window needs to correspond to around 1/f of the carrier frequency. This is what YakBizzaro is talking about (ADC analog bandwidth) in their sibling post.
  - mikepavone 3y ago