Sampling Theorem and reconstruction
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I do not understand a concept about the Nyquist - Shannon sampling theorem.
It says that it is possibile to perfectly get the original analog signal from the signal obtained by sampling if and only if the sampling frequency is higher than twice the maximum frequency of the initial signal.
I can understand it if I think at what happens in the frequency domain, in which the sampling produces replicas of the initial spectrum and therefore a low pass filter reconstructor can delete them and keep the original spectrum.
But in time domain sampling simply means to extract values of the original signal at instants separated by the sampling time T.
Once I have extracted these values, I have lost all the informations about the points between two consecutive instants of sampling. How can the reconstructor device perfectly obtain the original signal? It does not know how to connect the sampled points (they can be connected by infinite mathematical curves and all the information inside T time are lost). For example, it can connect them as in figure 1 (the correct original signal), or as in figure 2.
figure 1
figure 2
This makes me think that a very high sampling frequency is surely a good thing, since the points are very close together, but there is not a frequency that if overcome, allows a 100% perfect reconstruction, since sampling implies losing information.
analog signal signal-processing sampling signal-theory
asked 2 days ago
Kinka-ByoKinka-Byo
612
$endgroup$
add a comment |
$begingroup$
I do not understand a concept about the Nyquist - Shannon sampling theorem.
It says that it is possibile to perfectly get the original analog signal from the signal obtained by sampling if and only if the sampling frequency is higher than twice the maximum frequency of the initial signal.
I can understand it if I think at what happens in the frequency domain, in which the sampling produces replicas of the initial spectrum and therefore a low pass filter reconstructor can delete them and keep the original spectrum.
But in time domain sampling simply means to extract values of the original signal at instants separated by the sampling time T.
Once I have extracted these values, I have lost all the informations about the points between two consecutive instants of sampling. How can the reconstructor device perfectly obtain the original signal? It does not know how to connect the sampled points (they can be connected by infinite mathematical curves and all the information inside T time are lost). For example, it can connect them as in figure 1 (the correct original signal), or as in figure 2.
figure 1
figure 2
This makes me think that a very high sampling frequency is surely a good thing, since the points are very close together, but there is not a frequency that if overcome, allows a 100% perfect reconstruction, since sampling implies losing information.
analog signal signal-processing sampling signal-theory
asked 2 days ago
Kinka-ByoKinka-Byo
612
$endgroup$
7
$begingroup$
Your bottom signal has some much higher frequency components than the other ones here.
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– Hearth
2 days ago
5
$begingroup$
There is exactly ONE curve that passes thru all those points AND is band limited to strictly less then Fs/2.
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– Dan Mills
2 days ago
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it's important to note that unique reconstruction is only possible if the original signal is strictly bandlimited. Or to put it another way, given the samples, the assumption of strict bandlimiting allows a single signal to be reconstructed. To the extent that the bandlimited assumption is untrue, then the reconstructed signal will not match the original - this is called aliasing.
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– Neil_UK
2 days ago
$begingroup$
Also note that in practice it may require higher sampling frequency to provide acceptable reconstruction since perfect band-limiting are not practical. For example Audio CDs use 44.1kHz sampling to provide 0-20kHz output. Oscilloscopes generally use 5-10 times the required minimum sampling frequency to provide acceptable waveform integrity as a sharp cutoff filter would tend to create waveform artifacts such as ringing.
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– Kevin White
2 days ago
$begingroup$
It says that...if the sampling frequency is higher than twice the maximum frequency of the initial signalIt most certainly does not. Don't confuse the frequency and bandwidth.
$endgroup$
– AndrejaKo
yesterday
add a comment |
$begingroup$
I do not understand a concept about the Nyquist - Shannon sampling theorem.
It says that it is possibile to perfectly get the original analog signal from the signal obtained by sampling if and only if the sampling frequency is higher than twice the maximum frequency of the initial signal.
I can understand it if I think at what happens in the frequency domain, in which the sampling produces replicas of the initial spectrum and therefore a low pass filter reconstructor can delete them and keep the original spectrum.
But in time domain sampling simply means to extract values of the original signal at instants separated by the sampling time T.
Once I have extracted these values, I have lost all the informations about the points between two consecutive instants of sampling. How can the reconstructor device perfectly obtain the original signal? It does not know how to connect the sampled points (they can be connected by infinite mathematical curves and all the information inside T time are lost). For example, it can connect them as in figure 1 (the correct original signal), or as in figure 2.
figure 1
figure 2
This makes me think that a very high sampling frequency is surely a good thing, since the points are very close together, but there is not a frequency that if overcome, allows a 100% perfect reconstruction, since sampling implies losing information.
analog signal signal-processing sampling signal-theory
asked 2 days ago
Kinka-ByoKinka-Byo
612
$endgroup$
I do not understand a concept about the Nyquist - Shannon sampling theorem.
It says that it is possibile to perfectly get the original analog signal from the signal obtained by sampling if and only if the sampling frequency is higher than twice the maximum frequency of the initial signal.
I can understand it if I think at what happens in the frequency domain, in which the sampling produces replicas of the initial spectrum and therefore a low pass filter reconstructor can delete them and keep the original spectrum.
But in time domain sampling simply means to extract values of the original signal at instants separated by the sampling time T.
Once I have extracted these values, I have lost all the informations about the points between two consecutive instants of sampling. How can the reconstructor device perfectly obtain the original signal? It does not know how to connect the sampled points (they can be connected by infinite mathematical curves and all the information inside T time are lost). For example, it can connect them as in figure 1 (the correct original signal), or as in figure 2.
figure 1
figure 2
This makes me think that a very high sampling frequency is surely a good thing, since the points are very close together, but there is not a frequency that if overcome, allows a 100% perfect reconstruction, since sampling implies losing information.
analog signal signal-processing sampling signal-theory
analog signal signal-processing sampling signal-theory
asked 2 days ago
Kinka-ByoKinka-Byo
612
asked 2 days ago
Kinka-ByoKinka-Byo
612
asked 2 days ago
Kinka-ByoKinka-Byo
612
asked 2 days ago
Kinka-ByoKinka-Byo
612
asked 2 days ago
Kinka-ByoKinka-Byo
612
612
7
$begingroup$
Your bottom signal has some much higher frequency components than the other ones here.
$endgroup$
– Hearth
2 days ago
5
$begingroup$
There is exactly ONE curve that passes thru all those points AND is band limited to strictly less then Fs/2.
$endgroup$
– Dan Mills
2 days ago
$begingroup$
it's important to note that unique reconstruction is only possible if the original signal is strictly bandlimited. Or to put it another way, given the samples, the assumption of strict bandlimiting allows a single signal to be reconstructed. To the extent that the bandlimited assumption is untrue, then the reconstructed signal will not match the original - this is called aliasing.
$endgroup$
– Neil_UK
2 days ago
$begingroup$
Also note that in practice it may require higher sampling frequency to provide acceptable reconstruction since perfect band-limiting are not practical. For example Audio CDs use 44.1kHz sampling to provide 0-20kHz output. Oscilloscopes generally use 5-10 times the required minimum sampling frequency to provide acceptable waveform integrity as a sharp cutoff filter would tend to create waveform artifacts such as ringing.
$endgroup$
– Kevin White
2 days ago
$begingroup$
It says that...if the sampling frequency is higher than twice the maximum frequency of the initial signalIt most certainly does not. Don't confuse the frequency and bandwidth.
$endgroup$
– AndrejaKo
yesterday
add a comment |
7
$begingroup$
Your bottom signal has some much higher frequency components than the other ones here.
$endgroup$
– Hearth
2 days ago
5
$begingroup$
There is exactly ONE curve that passes thru all those points AND is band limited to strictly less then Fs/2.
$endgroup$
– Dan Mills
2 days ago
$begingroup$
it's important to note that unique reconstruction is only possible if the original signal is strictly bandlimited. Or to put it another way, given the samples, the assumption of strict bandlimiting allows a single signal to be reconstructed. To the extent that the bandlimited assumption is untrue, then the reconstructed signal will not match the original - this is called aliasing.
$endgroup$
– Neil_UK
2 days ago
$begingroup$
Also note that in practice it may require higher sampling frequency to provide acceptable reconstruction since perfect band-limiting are not practical. For example Audio CDs use 44.1kHz sampling to provide 0-20kHz output. Oscilloscopes generally use 5-10 times the required minimum sampling frequency to provide acceptable waveform integrity as a sharp cutoff filter would tend to create waveform artifacts such as ringing.
$endgroup$
– Kevin White
2 days ago
$begingroup$
It says that...if the sampling frequency is higher than twice the maximum frequency of the initial signalIt most certainly does not. Don't confuse the frequency and bandwidth.
$endgroup$
– AndrejaKo
yesterday
7
7
$begingroup$
Your bottom signal has some much higher frequency components than the other ones here.
$endgroup$
– Hearth
2 days ago
$begingroup$
Your bottom signal has some much higher frequency components than the other ones here.
$endgroup$
– Hearth
2 days ago
5
5
$begingroup$
There is exactly ONE curve that passes thru all those points AND is band limited to strictly less then Fs/2.
$endgroup$
– Dan Mills
2 days ago
$begingroup$
There is exactly ONE curve that passes thru all those points AND is band limited to strictly less then Fs/2.
$endgroup$
– Dan Mills
2 days ago
$begingroup$
it's important to note that unique reconstruction is only possible if the original signal is strictly bandlimited. Or to put it another way, given the samples, the assumption of strict bandlimiting allows a single signal to be reconstructed. To the extent that the bandlimited assumption is untrue, then the reconstructed signal will not match the original - this is called aliasing.
$endgroup$
– Neil_UK
2 days ago
$begingroup$
it's important to note that unique reconstruction is only possible if the original signal is strictly bandlimited. Or to put it another way, given the samples, the assumption of strict bandlimiting allows a single signal to be reconstructed. To the extent that the bandlimited assumption is untrue, then the reconstructed signal will not match the original - this is called aliasing.
$endgroup$
– Neil_UK
2 days ago
$begingroup$
Also note that in practice it may require higher sampling frequency to provide acceptable reconstruction since perfect band-limiting are not practical. For example Audio CDs use 44.1kHz sampling to provide 0-20kHz output. Oscilloscopes generally use 5-10 times the required minimum sampling frequency to provide acceptable waveform integrity as a sharp cutoff filter would tend to create waveform artifacts such as ringing.
$endgroup$
– Kevin White
2 days ago
$begingroup$
Also note that in practice it may require higher sampling frequency to provide acceptable reconstruction since perfect band-limiting are not practical. For example Audio CDs use 44.1kHz sampling to provide 0-20kHz output. Oscilloscopes generally use 5-10 times the required minimum sampling frequency to provide acceptable waveform integrity as a sharp cutoff filter would tend to create waveform artifacts such as ringing.
$endgroup$
– Kevin White
2 days ago
$begingroup$
It says that...if the sampling frequency is higher than twice the maximum frequency of the initial signal It most certainly does not. Don't confuse the frequency and bandwidth.$endgroup$
– AndrejaKo
yesterday
$begingroup$
It says that...if the sampling frequency is higher than twice the maximum frequency of the initial signal It most certainly does not. Don't confuse the frequency and bandwidth.$endgroup$
– AndrejaKo
yesterday
add a comment |
4 Answers
4
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oldest
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You can think of any perfectly bandlimited signal as the superposition of a set of $frac{sin(t)}{t} = text{sinc}(t)$ curves, with their peaks positioned uniformly along the time axis. Their spacing is $frac{2}{BW}$.
sinc(x) also happens to be the time-domain response of a perfect low-pass filter, and it explains how the continuous-time reconstruction (interpolation) is accomplished from a series of discrete samples.
When we uniformly sample a signal, each sample is a direct measurement of the amplitude of one of those sinc() waves. This works because it is a property of the sinc() function that it is zero at every sampling point, except at its own peak. In other words, when you take a measurement, you're not getting any "interference" from any of the other sinc() functions. Therefore, the set of N discrete measurements contains all of the information in the continuous-time signal represented by that collection of sinc() waves.
Now, it gets even weirder than what TimWestcott was alluding to — the samples do not even have to be uniformly spaced in time! It turns out that ANY N unique samples taken within a window of time (with certain limitations) of a perfectly bandlimited signal can be used to reconstruct that signal. It takes a lot more math to do it, though!
With nonuniform sampling, you are no longer getting a clean measurement of just one of the sinc() amplitudes. Instead, you're getting a mix of many, if not all of them. However, since you know exactly where you are on each one (obviously, each sample must be time-stamped), it is possible to solve the large system of linear equations to find the actual amplitudes and therefore reconstruct the original signal. Of course, this process is very sensitive to small perturbations (noise and math errors, for example), and I'm hand-waving away some details about constraints on the set of samples, but the general principle holds.
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
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add a comment |
$begingroup$
If the signal is perfectly bandlimited, then there is no additional information to be gotten out of it by sampling faster than twice the bandwidth. So perfect reconstruction must be possible. It's as @DanMills said: there's one and only one curve that'll pass through the sampled points and be correct, and that's the curve that you'd get from a perfect reconstruction filter.
(Note that it gets weirder -- at least in theory, if the bandwidth is $B$, then you don't need to sample $x(t)$ at $2B$ -- you can sample $x(t)$ and $frac{d x(t)}{dt}$ simultaneously at $B$, or sample out to the third derivative (i.e., collect four samples) at $frac{B}{2}$, or commit various other crimes to the signal before sampling an $N$ wide vector at $frac{2B}{N}$. Most such schemes (definitely the derivatives that I mention) would be horribly impractical, but in theory they'll work, and you do occasionally stumble across schemes that are actually used in reality.)
answered 2 days ago
TimWescottTimWescott
6,4231416
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Why band - limiting means there is one e only one curves that will pass through the sampled points?
$endgroup$
– Kinka-Byo
2 days ago
1
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One of the things that you just have to accept with linear systems theory is that there are some things that are much easier to describe using frequency domain analysis, and there are other things that are much easier to describe using time domain analysis, and they are equivalent. So it is true in the time domain because it is true in the frequency domain. There is probably some mostly-time-domain explanation, but it'll take pages and pages more mathematics to describe it -- and there's still going to be frequency domain arguments in the proof, because of the way it's stated.
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– TimWescott
yesterday
add a comment |
$begingroup$
One way to represent a signal is by its evolution in time (the time domain). Another equivalent way is by its frequency components (the frequency domain). The "steeper" or "sharper" the variation in a time domain signal, the higher that signal's frequency content.
The signal you show in Figure 1 may have a frequency spectrum like the one below:
There is a maximum frequency fmax1 beyond which the frequency content is zero, meaning the signal is bandlimited. To reconstruct the signal without error, sample at a minimum rate of fs1 = 2*fmax1 (the Nyquist rate).
The signal you show in Figure 2 has higher frequency content since you placed sharp variations between several samples. Its frequency spectrum may resemble this:
The highest frequency component in the Figure 2 signal (fmax2) is greater than the highest frequency component in the Figure 1 signal (fmax1). This means you must sample at a higher rate to reconstruct the signal, fs2 = 2*fmax2 > fs1.
In summary, you can add as many small variations to the signals as you want, but in doing so you introduce higher frequency content, and you must increase the sampling rate according to the Nyquist theorem.
You may argue that a non-bandlimited signal has frequency content extending to infinity, and for such a signal we can't define a minimum sampling rate for error-free reconstruction. In practical situations, we bandlimit signals with antialiasing filters to ensure the frequency content above a certain frequency is zero.
answered yesterday
w_hilew_hile
1011
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w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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There are already good answers here. Maybe here a very short one which might be useful. I assume you now something about FFT, spectrum or at least Fourier rows.
Look at the second picture you posted. What frequencies appear in its Fourier decomposition? Obviously it has much higher frequencies in it than the curve in picture one. But with the distance between the sampling points (and therefor the sampling frequency) you are limited to something which highest frequency has to be lower than 1/2 the sampling frequency. Or the other way: $$f_{highest} < 2*f_s$$ to be able to reconstruct it.
The points will reconstruct something with the least amount of sinusoids needed to hit the points.
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can think of any perfectly bandlimited signal as the superposition of a set of $frac{sin(t)}{t} = text{sinc}(t)$ curves, with their peaks positioned uniformly along the time axis. Their spacing is $frac{2}{BW}$.
sinc(x) also happens to be the time-domain response of a perfect low-pass filter, and it explains how the continuous-time reconstruction (interpolation) is accomplished from a series of discrete samples.
When we uniformly sample a signal, each sample is a direct measurement of the amplitude of one of those sinc() waves. This works because it is a property of the sinc() function that it is zero at every sampling point, except at its own peak. In other words, when you take a measurement, you're not getting any "interference" from any of the other sinc() functions. Therefore, the set of N discrete measurements contains all of the information in the continuous-time signal represented by that collection of sinc() waves.
Now, it gets even weirder than what TimWestcott was alluding to — the samples do not even have to be uniformly spaced in time! It turns out that ANY N unique samples taken within a window of time (with certain limitations) of a perfectly bandlimited signal can be used to reconstruct that signal. It takes a lot more math to do it, though!
With nonuniform sampling, you are no longer getting a clean measurement of just one of the sinc() amplitudes. Instead, you're getting a mix of many, if not all of them. However, since you know exactly where you are on each one (obviously, each sample must be time-stamped), it is possible to solve the large system of linear equations to find the actual amplitudes and therefore reconstruct the original signal. Of course, this process is very sensitive to small perturbations (noise and math errors, for example), and I'm hand-waving away some details about constraints on the set of samples, but the general principle holds.
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
$endgroup$
add a comment |
$begingroup$
You can think of any perfectly bandlimited signal as the superposition of a set of $frac{sin(t)}{t} = text{sinc}(t)$ curves, with their peaks positioned uniformly along the time axis. Their spacing is $frac{2}{BW}$.
sinc(x) also happens to be the time-domain response of a perfect low-pass filter, and it explains how the continuous-time reconstruction (interpolation) is accomplished from a series of discrete samples.
When we uniformly sample a signal, each sample is a direct measurement of the amplitude of one of those sinc() waves. This works because it is a property of the sinc() function that it is zero at every sampling point, except at its own peak. In other words, when you take a measurement, you're not getting any "interference" from any of the other sinc() functions. Therefore, the set of N discrete measurements contains all of the information in the continuous-time signal represented by that collection of sinc() waves.
Now, it gets even weirder than what TimWestcott was alluding to — the samples do not even have to be uniformly spaced in time! It turns out that ANY N unique samples taken within a window of time (with certain limitations) of a perfectly bandlimited signal can be used to reconstruct that signal. It takes a lot more math to do it, though!
With nonuniform sampling, you are no longer getting a clean measurement of just one of the sinc() amplitudes. Instead, you're getting a mix of many, if not all of them. However, since you know exactly where you are on each one (obviously, each sample must be time-stamped), it is possible to solve the large system of linear equations to find the actual amplitudes and therefore reconstruct the original signal. Of course, this process is very sensitive to small perturbations (noise and math errors, for example), and I'm hand-waving away some details about constraints on the set of samples, but the general principle holds.
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
$endgroup$
add a comment |
$begingroup$
You can think of any perfectly bandlimited signal as the superposition of a set of $frac{sin(t)}{t} = text{sinc}(t)$ curves, with their peaks positioned uniformly along the time axis. Their spacing is $frac{2}{BW}$.
sinc(x) also happens to be the time-domain response of a perfect low-pass filter, and it explains how the continuous-time reconstruction (interpolation) is accomplished from a series of discrete samples.
When we uniformly sample a signal, each sample is a direct measurement of the amplitude of one of those sinc() waves. This works because it is a property of the sinc() function that it is zero at every sampling point, except at its own peak. In other words, when you take a measurement, you're not getting any "interference" from any of the other sinc() functions. Therefore, the set of N discrete measurements contains all of the information in the continuous-time signal represented by that collection of sinc() waves.
Now, it gets even weirder than what TimWestcott was alluding to — the samples do not even have to be uniformly spaced in time! It turns out that ANY N unique samples taken within a window of time (with certain limitations) of a perfectly bandlimited signal can be used to reconstruct that signal. It takes a lot more math to do it, though!
With nonuniform sampling, you are no longer getting a clean measurement of just one of the sinc() amplitudes. Instead, you're getting a mix of many, if not all of them. However, since you know exactly where you are on each one (obviously, each sample must be time-stamped), it is possible to solve the large system of linear equations to find the actual amplitudes and therefore reconstruct the original signal. Of course, this process is very sensitive to small perturbations (noise and math errors, for example), and I'm hand-waving away some details about constraints on the set of samples, but the general principle holds.
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
$endgroup$
You can think of any perfectly bandlimited signal as the superposition of a set of $frac{sin(t)}{t} = text{sinc}(t)$ curves, with their peaks positioned uniformly along the time axis. Their spacing is $frac{2}{BW}$.
sinc(x) also happens to be the time-domain response of a perfect low-pass filter, and it explains how the continuous-time reconstruction (interpolation) is accomplished from a series of discrete samples.
When we uniformly sample a signal, each sample is a direct measurement of the amplitude of one of those sinc() waves. This works because it is a property of the sinc() function that it is zero at every sampling point, except at its own peak. In other words, when you take a measurement, you're not getting any "interference" from any of the other sinc() functions. Therefore, the set of N discrete measurements contains all of the information in the continuous-time signal represented by that collection of sinc() waves.
Now, it gets even weirder than what TimWestcott was alluding to — the samples do not even have to be uniformly spaced in time! It turns out that ANY N unique samples taken within a window of time (with certain limitations) of a perfectly bandlimited signal can be used to reconstruct that signal. It takes a lot more math to do it, though!
With nonuniform sampling, you are no longer getting a clean measurement of just one of the sinc() amplitudes. Instead, you're getting a mix of many, if not all of them. However, since you know exactly where you are on each one (obviously, each sample must be time-stamped), it is possible to solve the large system of linear equations to find the actual amplitudes and therefore reconstruct the original signal. Of course, this process is very sensitive to small perturbations (noise and math errors, for example), and I'm hand-waving away some details about constraints on the set of samples, but the general principle holds.
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
answered 2 days ago
Dave Tweed♦Dave Tweed
122k9152264
122k9152264
add a comment |
add a comment |
$begingroup$
If the signal is perfectly bandlimited, then there is no additional information to be gotten out of it by sampling faster than twice the bandwidth. So perfect reconstruction must be possible. It's as @DanMills said: there's one and only one curve that'll pass through the sampled points and be correct, and that's the curve that you'd get from a perfect reconstruction filter.
(Note that it gets weirder -- at least in theory, if the bandwidth is $B$, then you don't need to sample $x(t)$ at $2B$ -- you can sample $x(t)$ and $frac{d x(t)}{dt}$ simultaneously at $B$, or sample out to the third derivative (i.e., collect four samples) at $frac{B}{2}$, or commit various other crimes to the signal before sampling an $N$ wide vector at $frac{2B}{N}$. Most such schemes (definitely the derivatives that I mention) would be horribly impractical, but in theory they'll work, and you do occasionally stumble across schemes that are actually used in reality.)
answered 2 days ago
TimWescottTimWescott
6,4231416
$endgroup$
$begingroup$
Why band - limiting means there is one e only one curves that will pass through the sampled points?
$endgroup$
– Kinka-Byo
2 days ago
1
$begingroup$
One of the things that you just have to accept with linear systems theory is that there are some things that are much easier to describe using frequency domain analysis, and there are other things that are much easier to describe using time domain analysis, and they are equivalent. So it is true in the time domain because it is true in the frequency domain. There is probably some mostly-time-domain explanation, but it'll take pages and pages more mathematics to describe it -- and there's still going to be frequency domain arguments in the proof, because of the way it's stated.
$endgroup$
– TimWescott
yesterday
add a comment |
$begingroup$
If the signal is perfectly bandlimited, then there is no additional information to be gotten out of it by sampling faster than twice the bandwidth. So perfect reconstruction must be possible. It's as @DanMills said: there's one and only one curve that'll pass through the sampled points and be correct, and that's the curve that you'd get from a perfect reconstruction filter.
(Note that it gets weirder -- at least in theory, if the bandwidth is $B$, then you don't need to sample $x(t)$ at $2B$ -- you can sample $x(t)$ and $frac{d x(t)}{dt}$ simultaneously at $B$, or sample out to the third derivative (i.e., collect four samples) at $frac{B}{2}$, or commit various other crimes to the signal before sampling an $N$ wide vector at $frac{2B}{N}$. Most such schemes (definitely the derivatives that I mention) would be horribly impractical, but in theory they'll work, and you do occasionally stumble across schemes that are actually used in reality.)
answered 2 days ago
TimWescottTimWescott
6,4231416
$endgroup$
$begingroup$
Why band - limiting means there is one e only one curves that will pass through the sampled points?
$endgroup$
– Kinka-Byo
2 days ago
1
$begingroup$
One of the things that you just have to accept with linear systems theory is that there are some things that are much easier to describe using frequency domain analysis, and there are other things that are much easier to describe using time domain analysis, and they are equivalent. So it is true in the time domain because it is true in the frequency domain. There is probably some mostly-time-domain explanation, but it'll take pages and pages more mathematics to describe it -- and there's still going to be frequency domain arguments in the proof, because of the way it's stated.
$endgroup$
– TimWescott
yesterday
add a comment |
$begingroup$
If the signal is perfectly bandlimited, then there is no additional information to be gotten out of it by sampling faster than twice the bandwidth. So perfect reconstruction must be possible. It's as @DanMills said: there's one and only one curve that'll pass through the sampled points and be correct, and that's the curve that you'd get from a perfect reconstruction filter.
(Note that it gets weirder -- at least in theory, if the bandwidth is $B$, then you don't need to sample $x(t)$ at $2B$ -- you can sample $x(t)$ and $frac{d x(t)}{dt}$ simultaneously at $B$, or sample out to the third derivative (i.e., collect four samples) at $frac{B}{2}$, or commit various other crimes to the signal before sampling an $N$ wide vector at $frac{2B}{N}$. Most such schemes (definitely the derivatives that I mention) would be horribly impractical, but in theory they'll work, and you do occasionally stumble across schemes that are actually used in reality.)
answered 2 days ago
TimWescottTimWescott
6,4231416
$endgroup$
If the signal is perfectly bandlimited, then there is no additional information to be gotten out of it by sampling faster than twice the bandwidth. So perfect reconstruction must be possible. It's as @DanMills said: there's one and only one curve that'll pass through the sampled points and be correct, and that's the curve that you'd get from a perfect reconstruction filter.
(Note that it gets weirder -- at least in theory, if the bandwidth is $B$, then you don't need to sample $x(t)$ at $2B$ -- you can sample $x(t)$ and $frac{d x(t)}{dt}$ simultaneously at $B$, or sample out to the third derivative (i.e., collect four samples) at $frac{B}{2}$, or commit various other crimes to the signal before sampling an $N$ wide vector at $frac{2B}{N}$. Most such schemes (definitely the derivatives that I mention) would be horribly impractical, but in theory they'll work, and you do occasionally stumble across schemes that are actually used in reality.)
answered 2 days ago
TimWescottTimWescott
6,4231416
answered 2 days ago
TimWescottTimWescott
6,4231416
answered 2 days ago
TimWescottTimWescott
6,4231416
answered 2 days ago
TimWescottTimWescott
6,4231416
6,4231416
$begingroup$
Why band - limiting means there is one e only one curves that will pass through the sampled points?
$endgroup$
– Kinka-Byo
2 days ago
1
$begingroup$
One of the things that you just have to accept with linear systems theory is that there are some things that are much easier to describe using frequency domain analysis, and there are other things that are much easier to describe using time domain analysis, and they are equivalent. So it is true in the time domain because it is true in the frequency domain. There is probably some mostly-time-domain explanation, but it'll take pages and pages more mathematics to describe it -- and there's still going to be frequency domain arguments in the proof, because of the way it's stated.
$endgroup$
– TimWescott
yesterday
add a comment |
$begingroup$
Why band - limiting means there is one e only one curves that will pass through the sampled points?
$endgroup$
– Kinka-Byo
2 days ago
1
$begingroup$
One of the things that you just have to accept with linear systems theory is that there are some things that are much easier to describe using frequency domain analysis, and there are other things that are much easier to describe using time domain analysis, and they are equivalent. So it is true in the time domain because it is true in the frequency domain. There is probably some mostly-time-domain explanation, but it'll take pages and pages more mathematics to describe it -- and there's still going to be frequency domain arguments in the proof, because of the way it's stated.
$endgroup$
– TimWescott
yesterday
$begingroup$
Why band - limiting means there is one e only one curves that will pass through the sampled points?
$endgroup$
– Kinka-Byo
2 days ago
$begingroup$
Why band - limiting means there is one e only one curves that will pass through the sampled points?
$endgroup$
– Kinka-Byo
2 days ago
1
1
$begingroup$
One of the things that you just have to accept with linear systems theory is that there are some things that are much easier to describe using frequency domain analysis, and there are other things that are much easier to describe using time domain analysis, and they are equivalent. So it is true in the time domain because it is true in the frequency domain. There is probably some mostly-time-domain explanation, but it'll take pages and pages more mathematics to describe it -- and there's still going to be frequency domain arguments in the proof, because of the way it's stated.
$endgroup$
– TimWescott
yesterday
$begingroup$
One of the things that you just have to accept with linear systems theory is that there are some things that are much easier to describe using frequency domain analysis, and there are other things that are much easier to describe using time domain analysis, and they are equivalent. So it is true in the time domain because it is true in the frequency domain. There is probably some mostly-time-domain explanation, but it'll take pages and pages more mathematics to describe it -- and there's still going to be frequency domain arguments in the proof, because of the way it's stated.
$endgroup$
– TimWescott
yesterday
add a comment |
$begingroup$
One way to represent a signal is by its evolution in time (the time domain). Another equivalent way is by its frequency components (the frequency domain). The "steeper" or "sharper" the variation in a time domain signal, the higher that signal's frequency content.
The signal you show in Figure 1 may have a frequency spectrum like the one below:
There is a maximum frequency fmax1 beyond which the frequency content is zero, meaning the signal is bandlimited. To reconstruct the signal without error, sample at a minimum rate of fs1 = 2*fmax1 (the Nyquist rate).
The signal you show in Figure 2 has higher frequency content since you placed sharp variations between several samples. Its frequency spectrum may resemble this:
The highest frequency component in the Figure 2 signal (fmax2) is greater than the highest frequency component in the Figure 1 signal (fmax1). This means you must sample at a higher rate to reconstruct the signal, fs2 = 2*fmax2 > fs1.
In summary, you can add as many small variations to the signals as you want, but in doing so you introduce higher frequency content, and you must increase the sampling rate according to the Nyquist theorem.
You may argue that a non-bandlimited signal has frequency content extending to infinity, and for such a signal we can't define a minimum sampling rate for error-free reconstruction. In practical situations, we bandlimit signals with antialiasing filters to ensure the frequency content above a certain frequency is zero.
answered yesterday
w_hilew_hile
1011
New contributor
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
One way to represent a signal is by its evolution in time (the time domain). Another equivalent way is by its frequency components (the frequency domain). The "steeper" or "sharper" the variation in a time domain signal, the higher that signal's frequency content.
The signal you show in Figure 1 may have a frequency spectrum like the one below:
There is a maximum frequency fmax1 beyond which the frequency content is zero, meaning the signal is bandlimited. To reconstruct the signal without error, sample at a minimum rate of fs1 = 2*fmax1 (the Nyquist rate).
The signal you show in Figure 2 has higher frequency content since you placed sharp variations between several samples. Its frequency spectrum may resemble this:
The highest frequency component in the Figure 2 signal (fmax2) is greater than the highest frequency component in the Figure 1 signal (fmax1). This means you must sample at a higher rate to reconstruct the signal, fs2 = 2*fmax2 > fs1.
In summary, you can add as many small variations to the signals as you want, but in doing so you introduce higher frequency content, and you must increase the sampling rate according to the Nyquist theorem.
You may argue that a non-bandlimited signal has frequency content extending to infinity, and for such a signal we can't define a minimum sampling rate for error-free reconstruction. In practical situations, we bandlimit signals with antialiasing filters to ensure the frequency content above a certain frequency is zero.
answered yesterday
w_hilew_hile
1011
New contributor
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
One way to represent a signal is by its evolution in time (the time domain). Another equivalent way is by its frequency components (the frequency domain). The "steeper" or "sharper" the variation in a time domain signal, the higher that signal's frequency content.
The signal you show in Figure 1 may have a frequency spectrum like the one below:
There is a maximum frequency fmax1 beyond which the frequency content is zero, meaning the signal is bandlimited. To reconstruct the signal without error, sample at a minimum rate of fs1 = 2*fmax1 (the Nyquist rate).
The signal you show in Figure 2 has higher frequency content since you placed sharp variations between several samples. Its frequency spectrum may resemble this:
The highest frequency component in the Figure 2 signal (fmax2) is greater than the highest frequency component in the Figure 1 signal (fmax1). This means you must sample at a higher rate to reconstruct the signal, fs2 = 2*fmax2 > fs1.
In summary, you can add as many small variations to the signals as you want, but in doing so you introduce higher frequency content, and you must increase the sampling rate according to the Nyquist theorem.
You may argue that a non-bandlimited signal has frequency content extending to infinity, and for such a signal we can't define a minimum sampling rate for error-free reconstruction. In practical situations, we bandlimit signals with antialiasing filters to ensure the frequency content above a certain frequency is zero.
answered yesterday
w_hilew_hile
1011
New contributor
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
One way to represent a signal is by its evolution in time (the time domain). Another equivalent way is by its frequency components (the frequency domain). The "steeper" or "sharper" the variation in a time domain signal, the higher that signal's frequency content.
The signal you show in Figure 1 may have a frequency spectrum like the one below:
There is a maximum frequency fmax1 beyond which the frequency content is zero, meaning the signal is bandlimited. To reconstruct the signal without error, sample at a minimum rate of fs1 = 2*fmax1 (the Nyquist rate).
The signal you show in Figure 2 has higher frequency content since you placed sharp variations between several samples. Its frequency spectrum may resemble this:
The highest frequency component in the Figure 2 signal (fmax2) is greater than the highest frequency component in the Figure 1 signal (fmax1). This means you must sample at a higher rate to reconstruct the signal, fs2 = 2*fmax2 > fs1.
In summary, you can add as many small variations to the signals as you want, but in doing so you introduce higher frequency content, and you must increase the sampling rate according to the Nyquist theorem.
You may argue that a non-bandlimited signal has frequency content extending to infinity, and for such a signal we can't define a minimum sampling rate for error-free reconstruction. In practical situations, we bandlimit signals with antialiasing filters to ensure the frequency content above a certain frequency is zero.
answered yesterday
w_hilew_hile
1011
New contributor
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
w_hilew_hile
1011
New contributor
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
w_hilew_hile
1011
answered yesterday
w_hilew_hile
1011
1011
New contributor
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
w_hile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
There are already good answers here. Maybe here a very short one which might be useful. I assume you now something about FFT, spectrum or at least Fourier rows.
Look at the second picture you posted. What frequencies appear in its Fourier decomposition? Obviously it has much higher frequencies in it than the curve in picture one. But with the distance between the sampling points (and therefor the sampling frequency) you are limited to something which highest frequency has to be lower than 1/2 the sampling frequency. Or the other way: $$f_{highest} < 2*f_s$$ to be able to reconstruct it.
The points will reconstruct something with the least amount of sinusoids needed to hit the points.
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
$endgroup$
add a comment |
$begingroup$
There are already good answers here. Maybe here a very short one which might be useful. I assume you now something about FFT, spectrum or at least Fourier rows.
Look at the second picture you posted. What frequencies appear in its Fourier decomposition? Obviously it has much higher frequencies in it than the curve in picture one. But with the distance between the sampling points (and therefor the sampling frequency) you are limited to something which highest frequency has to be lower than 1/2 the sampling frequency. Or the other way: $$f_{highest} < 2*f_s$$ to be able to reconstruct it.
The points will reconstruct something with the least amount of sinusoids needed to hit the points.
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
$endgroup$
add a comment |
$begingroup$
There are already good answers here. Maybe here a very short one which might be useful. I assume you now something about FFT, spectrum or at least Fourier rows.
Look at the second picture you posted. What frequencies appear in its Fourier decomposition? Obviously it has much higher frequencies in it than the curve in picture one. But with the distance between the sampling points (and therefor the sampling frequency) you are limited to something which highest frequency has to be lower than 1/2 the sampling frequency. Or the other way: $$f_{highest} < 2*f_s$$ to be able to reconstruct it.
The points will reconstruct something with the least amount of sinusoids needed to hit the points.
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
$endgroup$
There are already good answers here. Maybe here a very short one which might be useful. I assume you now something about FFT, spectrum or at least Fourier rows.
Look at the second picture you posted. What frequencies appear in its Fourier decomposition? Obviously it has much higher frequencies in it than the curve in picture one. But with the distance between the sampling points (and therefor the sampling frequency) you are limited to something which highest frequency has to be lower than 1/2 the sampling frequency. Or the other way: $$f_{highest} < 2*f_s$$ to be able to reconstruct it.
The points will reconstruct something with the least amount of sinusoids needed to hit the points.
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
answered yesterday
Mr.Sh4nnonMr.Sh4nnon
1339
1339
add a comment |
add a comment |
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$begingroup$
Your bottom signal has some much higher frequency components than the other ones here.
$endgroup$
– Hearth
2 days ago
5
$begingroup$
There is exactly ONE curve that passes thru all those points AND is band limited to strictly less then Fs/2.
$endgroup$
– Dan Mills
2 days ago
$begingroup$
it's important to note that unique reconstruction is only possible if the original signal is strictly bandlimited. Or to put it another way, given the samples, the assumption of strict bandlimiting allows a single signal to be reconstructed. To the extent that the bandlimited assumption is untrue, then the reconstructed signal will not match the original - this is called aliasing.
$endgroup$
– Neil_UK
2 days ago
$begingroup$
Also note that in practice it may require higher sampling frequency to provide acceptable reconstruction since perfect band-limiting are not practical. For example Audio CDs use 44.1kHz sampling to provide 0-20kHz output. Oscilloscopes generally use 5-10 times the required minimum sampling frequency to provide acceptable waveform integrity as a sharp cutoff filter would tend to create waveform artifacts such as ringing.
$endgroup$
– Kevin White
2 days ago
$begingroup$
It says that...if the sampling frequency is higher than twice the maximum frequency of the initial signalIt most certainly does not. Don't confuse the frequency and bandwidth.$endgroup$
– AndrejaKo
yesterday