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One misconception that many make regarding the Nyquist frequency is thinking that the sampling rate needs to be twice the highest frequency. Your sampling shou

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
    Thanks for the explanation! Between your comment and the Undersampling wiki page diydsp linked to I think I am on the path to enlightenment. > If your original signal was truly band-limited to 100MHz In practice, this means you need to band pass before the ADC, right? i.e. "signal" in this case is the entire input to the ADC and not just the particular modulated signal you care about
    - labcomputer 3y ago
      > In practice, this means you need to band pass before the ADC, right? i.e. "signal" in this case is the entire input to the ADC and not just the particular modulated signal you care about Right and right. And, you’d normally want that to be a contiguous 100 MHz band of frequencies (you could in principle have multiple discontiguous bands that add up to 100 MHz if they are spaced right (they don’t fold down to the same base frequencies), but that would be quite an unusual application).
- cushychicken 3y ago
  To quote a meme: “That’s the neat part. You don’t.” If you bandlimit your input, aliasing effectively strips out the carrier tone and leaves the modulated signal. In a way, you’re relying on aliasing / frequency folding to do it for you. https://ars.els-cdn.com/content/image/3-s2.0-B9780124158931000020-f02-06-9780124158931.jpg https://ars.els-cdn.com/content/image/3-s2.0-B97801241589310... You can even improve information transfer in these scenarios by using a synchronizer, which allows you to phase shift your sampling to be at the ideal transition point in your information stream.
IIAOPSW 3y ago
I'm mentally filling in the gaps here and assuming MSPS is MegaSamplesPerSecond?
- BenjiWiebe 3y ago
  Yes
YakBizzarro 3y ago
That's true, but there are a couple of things more. First, your DAC or ADC need to have such analog bandwidth. Working in a higher Nyquist zone also require higher amplification since the signal would be considerably weaker and more complex filtering to remove the signal from the other zones
wittenbunk 3y ago
Only true for continuous RF sources. For transient signals you need at least Nyquist frequency.
- azalemeth 3y ago
  Or use the traditional "lock-in" amplifier technique of mixing with a known reference at the frequency mid-point of the range you care about? (That's how NMR spectrometers / MRI scanners worked for decades
  - muffles 3y ago
    Isnt the lock-in amplifier technique used to improve the SNR ratio of a signal by filtering out noise at frequencies outside a specific range of interest? High-speed sampling would still be required to accurately measure transient signals.
- mhh__ 3y ago
  In that sense isn't the bandwidth 0-Max anyway though?
kayson 3y ago
I know what you're getting at, but your statement, as others have pointed out, is incorrect. Your sampling rate always always has to be twice the highest frequency of the signal you are sampling. If you are sampling an RF-modulated signal with a center frequency of 1GHz and 100MHz of baseband bandwidth, then yes, you do need to sample at 2.2GHz+. And some applications do exactly that. If you're taking the RF signal, mixing it down to baseband, and filtering it to bandlimit, then you have a signal with maximum frequency component of 100MHz, and in that case, yes, your sampling rate can be 200MHz+
- abstrakraft 3y ago
  From an information theoretic perspective (which is the perspective Nyquist was originally coming from, though it didn't yet have that name), you don't need to mix the signal down. Assuming it is truly band-limited, you can sample the signal directly at RF, and reproduce it from those samples. Additionally, you will need to modulate the reproduced signal into the original band, which means you need to know where that band is - perhaps this is the detail you're pointing out? Another way of looking at it is that sampling inherently does the mixing down to baseband. Although it may not be exactly the baseband you want if the spectrum isn't cleanly symmetric about a multiple of the sample frequency.
  - Sesse__ 3y ago
    I've worked on ultrasound systems that definitely worked this way, not just in theory but also in practice. Bandpass filter 20–40 kHz, sample directly at 40 kHz (giving 20 kHz bandwidth). No mixer step involved, but your spectrum becomes inverted (e.g. if you do an FFT, a 22 kHz tone will be in the 18 kHz bin, not the 2 kHz bin as you would perhaps expect).
    - abstrakraft 3y ago
      Aliasing makes more sense (to me, anyway) if you think about the spectrum of complex signals, in which signals of real samples are modeled as the sum of positive and negative frequencies. In the sampling operation, all sinusoids are shifted down to the "natural baseband" by adding or subtracting some multiple of the sampling frequency that places the resulting frequency within +/- half of the sampling frequency. So for your example of 22kHz, that real frequency has two components: +22kHz that gets shifted down to -18kHz=22kHz-40kHz, and -22kHz that gets shifted up to +18kHz=-22kHz+40kHz. Note that this "natural baseband" is an abstraction of our own invention. You can just as easily think of the spectrum as ranging from 0Hz to the sampling frequency f_s, rather than -f_s/2 to f_s/2. The fact that some prefer one over the other is precisely why fftshift exists.
gaze 3y ago
Yeah but you also need the bandwidth of the sampler to exceed the highest frequency of the sample. Most samplers are limited by some kind of RC time and not their sinc envelope. Most.
cushychicken 3y ago
The folks who are telling you you’re wrong don’t understand Nyquist’s criterion very well. Curse those undergrad courses for only effectively teaching about Nyquist at baseband frequencies. You can sample 100MHz of bandwidth at 1GHz just as you describe at 210MSPS. You’ll get everything in the 950-1050MHz band. Trouble is, without an antialiasing filter, you’ll get every other band that’s a multiple of that sampling rate. The Nyquist criterion works at every multiple of the sampling frequency. Bandpass filter your analog input appropriately from 950-1050MHz and you’re golden. This is the way nearly every commodity Wi-Fi chip downsamples 2.4/5GHz raw RF. Sigma-delta ADCs are cheap, fast, and space efficient for die area using this method.
- femto 3y ago
  The most fiendish application of this effect that I've seen is polyphase filtering. I can't remember the details, but at the time I can remember the wonder of understanding (in a lecture by fred harris) how most the logic was running at a low sampling rate yet the input was at a high rate. The mixing was done by aliasing. Details here: https://www.dsprelated.com/thread/7758/understanding-the-concept-of-polyphase-filters https://www.dsprelated.com/thread/7758/understanding-the-con... https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_book_ch_6_text_5_61615.pdf https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_bo... https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_ch_6_supplement_3_934.pdf https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_ch...
  - Sesse__ 3y ago
    Polyphase filtering is less crazy than it initially sounds. Conceptually, you can think of it as: I have this signal in frequency f. I want to resample it to frequency (b/a)*f, where a and b are integers. (You can also do polyphase filtering to resample of non-rational or varying ratios, by essentially approximating towards a rational, but let's ignore that for the moment.) a and b can be pretty large if you want, e.g. a=160,b=147 will downsample from 48 kHz to 44100 Hz. So what you do to resample a signal (again conceptually), is: 1. Add <a> zeros between every input sample (which repeats the spectrum <a> times), 2. Apply a suitable (long!) FIR lowpass filter so that the signal is bandlimited, 3. Take every <b>-th sample (which doesn't cause any aliasing due to #2). Now the core of the polyphase filtering idea: We don't need to actually calculate the FIR filter for the samples we don't want in #3. And most of the input values to the filter will be zero due to #1. So instead of storing all the zeros and stuff, we simply pick out every <a>-th tap of the FIR filter and use that on the input signal directly. But since a and b don't line up perfectly, this means we get a different subset of the FIR filter for every output sample; we have a time-varying filter (or a filterbank, if you want). You get <b> different such filters before you're back where you started.
Chinjut 3y ago
Consider a signal whose value at x seconds is f(2x) - 2 f(3x) + f(4x), where f(x) = sin(2πx)/x. Considering that the absolute frequencies of f(x) are uniformly distributed from 0 to 1 Hz, the absolute frequencies of this total signal should be constrained to between 2 and 4 Hz. Thus, a bandwidth of 2 Hz. But if we sample at 6 Hz (three times the bandwidth!) including x = 0, we'll get all zeros. Granted, we might say that from the perspective of the complex Fourier transform using signed frequencies, the frequencies of this signal actually range over [-4 Hz, -2 Hz] U [+2 Hz, +4 Hz]. But I'm not sure that's the interpretation you had in mind. Let me know if I've screwed anything up here!
- Chinjut 3y ago
  That is, it's not quite as simple as saying you just need to sample at any frequency at least twice the bandwidth. Rather, it's the more complicated behavior described by this graph: https://en.wikipedia.org/wiki/Undersampling#/media/File:Samplerates.svg https://en.wikipedia.org/wiki/Undersampling#/media/File:Samp.... That is, the general rule is that the ratio of the highest frequency in the signal to half the sample rate, and the ratio of the lowest frequency in the signal to half the sample rate, have to lie within an interval of consecutive natural numbers. When the lowest frequency is zero, this is the familiar rule that the sample rate has to be at least twice the highest frequency in the signal. But more generally, it's more complicated.
- Chinjut 3y ago
  Whoops, I should've pulled the division by x out of the definition of f. The example I had in mind was [sin(4πx) - 2 sin(6πx) + sin(8πx)]/x. [Another good example is [sin(6πx) - 2 sin(8πx) + sin(10πx)]/x, whose frequencies are between 3 Hz and 5 Hz, thus a bandwidth of 2 Hz, but sampling at 4 Hz or even 8Hz gets all zeroes.] Anyway, the details on that example don't matter, the Wikipedia graph and article makes things more clear.
paulsutter 3y ago
Thank you I came here to post exactly this. Suggestion, you might want to correct the wikipedia page
- stagger87 3y ago
  You do not want to "correct" the wiki because the wiki is not wrong. The person you are replying to is clearly thinking about some sort of RF system (given the frequencies mentioned) where it's important to have a baseband filter to eliminate aliasing, and that filter will have some sort of roll off region, resulting in a higher sample rate than available bandwidth. That's all great, but the Nyquist theorem isn't talking about an RF system. It's referring to sampling. When the wiki uses the word "bandwidth", they mean the frequencies that don't alias given a specific sample rate.
- eternauta3k 3y ago
  Is the wikipedia page really wrong though? Highest frequency is what the mathematicians care about. EEs care about bandwidth because they're always modulating stuff and thinking in terms of carrier and baseband. Strictly speaking, what the EE grandparent suggested is using aliasing to mix the signal down to baseband.

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
  Thanks for the explanation! Between your comment and the Undersampling wiki page diydsp linked to I think I am on the path to enlightenment. > If your original signal was truly band-limited to 100MHz In practice, this means you need to band pass before the ADC, right? i.e. "signal" in this case is the entire input to the ADC and not just the particular modulated signal you care about
  - labcomputer 3y ago
    > In practice, this means you need to band pass before the ADC, right? i.e. "signal" in this case is the entire input to the ADC and not just the particular modulated signal you care about Right and right. And, you’d normally want that to be a contiguous 100 MHz band of frequencies (you could in principle have multiple discontiguous bands that add up to 100 MHz if they are spaced right (they don’t fold down to the same base frequencies), but that would be quite an unusual application).
cushychicken 3y ago
To quote a meme: “That’s the neat part. You don’t.” If you bandlimit your input, aliasing effectively strips out the carrier tone and leaves the modulated signal. In a way, you’re relying on aliasing / frequency folding to do it for you. https://ars.els-cdn.com/content/image/3-s2.0-B9780124158931000020-f02-06-9780124158931.jpg https://ars.els-cdn.com/content/image/3-s2.0-B97801241589310... You can even improve information transfer in these scenarios by using a synchronizer, which allows you to phase shift your sampling to be at the ideal transition point in your information stream.

IIAOPSW 3y ago

I'm mentally filling in the gaps here and assuming MSPS is MegaSamplesPerSecond?

BenjiWiebe 3y ago
Yes

YakBizzarro 3y ago

That's true, but there are a couple of things more. First, your DAC or ADC need to have such analog bandwidth. Working in a higher Nyquist zone also require higher amplification since the signal would be considerably weaker and more complex filtering to remove the signal from the other zones

wittenbunk 3y ago

Only true for continuous RF sources. For transient signals you need at least Nyquist frequency.

azalemeth 3y ago
Or use the traditional "lock-in" amplifier technique of mixing with a known reference at the frequency mid-point of the range you care about? (That's how NMR spectrometers / MRI scanners worked for decades
- muffles 3y ago
  Isnt the lock-in amplifier technique used to improve the SNR ratio of a signal by filtering out noise at frequencies outside a specific range of interest? High-speed sampling would still be required to accurately measure transient signals.
mhh__ 3y ago
In that sense isn't the bandwidth 0-Max anyway though?

kayson 3y ago

I know what you're getting at, but your statement, as others have pointed out, is incorrect. Your sampling rate always always has to be twice the highest frequency of the signal you are sampling. If you are sampling an RF-modulated signal with a center frequency of 1GHz and 100MHz of baseband bandwidth, then yes, you do need to sample at 2.2GHz+. And some applications do exactly that. If you're taking the RF signal, mixing it down to baseband, and filtering it to bandlimit, then you have a signal with maximum frequency component of 100MHz, and in that case, yes, your sampling rate can be 200MHz+

abstrakraft 3y ago
From an information theoretic perspective (which is the perspective Nyquist was originally coming from, though it didn't yet have that name), you don't need to mix the signal down. Assuming it is truly band-limited, you can sample the signal directly at RF, and reproduce it from those samples. Additionally, you will need to modulate the reproduced signal into the original band, which means you need to know where that band is - perhaps this is the detail you're pointing out? Another way of looking at it is that sampling inherently does the mixing down to baseband. Although it may not be exactly the baseband you want if the spectrum isn't cleanly symmetric about a multiple of the sample frequency.
- Sesse__ 3y ago
  I've worked on ultrasound systems that definitely worked this way, not just in theory but also in practice. Bandpass filter 20–40 kHz, sample directly at 40 kHz (giving 20 kHz bandwidth). No mixer step involved, but your spectrum becomes inverted (e.g. if you do an FFT, a 22 kHz tone will be in the 18 kHz bin, not the 2 kHz bin as you would perhaps expect).
  - abstrakraft 3y ago
    Aliasing makes more sense (to me, anyway) if you think about the spectrum of complex signals, in which signals of real samples are modeled as the sum of positive and negative frequencies. In the sampling operation, all sinusoids are shifted down to the "natural baseband" by adding or subtracting some multiple of the sampling frequency that places the resulting frequency within +/- half of the sampling frequency. So for your example of 22kHz, that real frequency has two components: +22kHz that gets shifted down to -18kHz=22kHz-40kHz, and -22kHz that gets shifted up to +18kHz=-22kHz+40kHz. Note that this "natural baseband" is an abstraction of our own invention. You can just as easily think of the spectrum as ranging from 0Hz to the sampling frequency f_s, rather than -f_s/2 to f_s/2. The fact that some prefer one over the other is precisely why fftshift exists.

gaze 3y ago

Yeah but you also need the bandwidth of the sampler to exceed the highest frequency of the sample. Most samplers are limited by some kind of RC time and not their sinc envelope. Most.

cushychicken 3y ago

The folks who are telling you you’re wrong don’t understand Nyquist’s criterion very well. Curse those undergrad courses for only effectively teaching about Nyquist at baseband frequencies. You can sample 100MHz of bandwidth at 1GHz just as you describe at 210MSPS. You’ll get everything in the 950-1050MHz band. Trouble is, without an antialiasing filter, you’ll get every other band that’s a multiple of that sampling rate. The Nyquist criterion works at every multiple of the sampling frequency. Bandpass filter your analog input appropriately from 950-1050MHz and you’re golden. This is the way nearly every commodity Wi-Fi chip downsamples 2.4/5GHz raw RF. Sigma-delta ADCs are cheap, fast, and space efficient for die area using this method.

femto 3y ago
The most fiendish application of this effect that I've seen is polyphase filtering. I can't remember the details, but at the time I can remember the wonder of understanding (in a lecture by fred harris) how most the logic was running at a low sampling rate yet the input was at a high rate. The mixing was done by aliasing. Details here: https://www.dsprelated.com/thread/7758/understanding-the-concept-of-polyphase-filters https://www.dsprelated.com/thread/7758/understanding-the-con... https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_book_ch_6_text_5_61615.pdf https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_bo... https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_ch_6_supplement_3_934.pdf https://s3.amazonaws.com/embeddedrelated/user/124841/fbmc_ch...
- Sesse__ 3y ago
  Polyphase filtering is less crazy than it initially sounds. Conceptually, you can think of it as: I have this signal in frequency f. I want to resample it to frequency (b/a)*f, where a and b are integers. (You can also do polyphase filtering to resample of non-rational or varying ratios, by essentially approximating towards a rational, but let's ignore that for the moment.) a and b can be pretty large if you want, e.g. a=160,b=147 will downsample from 48 kHz to 44100 Hz. So what you do to resample a signal (again conceptually), is: 1. Add <a> zeros between every input sample (which repeats the spectrum <a> times), 2. Apply a suitable (long!) FIR lowpass filter so that the signal is bandlimited, 3. Take every <b>-th sample (which doesn't cause any aliasing due to #2). Now the core of the polyphase filtering idea: We don't need to actually calculate the FIR filter for the samples we don't want in #3. And most of the input values to the filter will be zero due to #1. So instead of storing all the zeros and stuff, we simply pick out every <a>-th tap of the FIR filter and use that on the input signal directly. But since a and b don't line up perfectly, this means we get a different subset of the FIR filter for every output sample; we have a time-varying filter (or a filterbank, if you want). You get <b> different such filters before you're back where you started.

Chinjut 3y ago

Consider a signal whose value at x seconds is f(2x) - 2 f(3x) + f(4x), where f(x) = sin(2πx)/x. Considering that the absolute frequencies of f(x) are uniformly distributed from 0 to 1 Hz, the absolute frequencies of this total signal should be constrained to between 2 and 4 Hz. Thus, a bandwidth of 2 Hz. But if we sample at 6 Hz (three times the bandwidth!) including x = 0, we'll get all zeros. Granted, we might say that from the perspective of the complex Fourier transform using signed frequencies, the frequencies of this signal actually range over [-4 Hz, -2 Hz] U [+2 Hz, +4 Hz]. But I'm not sure that's the interpretation you had in mind. Let me know if I've screwed anything up here!

Chinjut 3y ago
That is, it's not quite as simple as saying you just need to sample at any frequency at least twice the bandwidth. Rather, it's the more complicated behavior described by this graph: https://en.wikipedia.org/wiki/Undersampling#/media/File:Samplerates.svg https://en.wikipedia.org/wiki/Undersampling#/media/File:Samp.... That is, the general rule is that the ratio of the highest frequency in the signal to half the sample rate, and the ratio of the lowest frequency in the signal to half the sample rate, have to lie within an interval of consecutive natural numbers. When the lowest frequency is zero, this is the familiar rule that the sample rate has to be at least twice the highest frequency in the signal. But more generally, it's more complicated.
Chinjut 3y ago
Whoops, I should've pulled the division by x out of the definition of f. The example I had in mind was [sin(4πx) - 2 sin(6πx) + sin(8πx)]/x. [Another good example is [sin(6πx) - 2 sin(8πx) + sin(10πx)]/x, whose frequencies are between 3 Hz and 5 Hz, thus a bandwidth of 2 Hz, but sampling at 4 Hz or even 8Hz gets all zeroes.] Anyway, the details on that example don't matter, the Wikipedia graph and article makes things more clear.

paulsutter 3y ago

Thank you I came here to post exactly this. Suggestion, you might want to correct the wikipedia page

stagger87 3y ago
You do not want to "correct" the wiki because the wiki is not wrong. The person you are replying to is clearly thinking about some sort of RF system (given the frequencies mentioned) where it's important to have a baseband filter to eliminate aliasing, and that filter will have some sort of roll off region, resulting in a higher sample rate than available bandwidth. That's all great, but the Nyquist theorem isn't talking about an RF system. It's referring to sampling. When the wiki uses the word "bandwidth", they mean the frequencies that don't alias given a specific sample rate.
eternauta3k 3y ago
Is the wikipedia page really wrong though? Highest frequency is what the mathematicians care about. EEs care about bandwidth because they're always modulating stuff and thinking in terms of carrier and baseband. Strictly speaking, what the EE grandparent suggested is using aliasing to mix the signal down to baseband.