Why doesn't sampling a periodic continuous-time signal yield a periodic discrete-time signal?Discretizing a continuous time signalMain differences to take into account between continuous and discrete time signalsDoes the hold part in sample and hold turns the signal from discrete to continuous?$2pi$ periodicity of discrete-time Fourier transformWhy do we use discrete-time sampling?What is the difference between continuous, discrete, analog and digital signal?Why DTFT coefficients are periodic and why continuous Fourier transform coefficients are not periodic?Minimum possible sampling frequency for continuous time signalFourier Series Representation of Continuous-Time Periodic SignalsContinuous-time RNN and Shannon sampling theorem
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Why doesn't sampling a periodic continuous-time signal yield a periodic discrete-time signal?
Discretizing a continuous time signalMain differences to take into account between continuous and discrete time signalsDoes the hold part in sample and hold turns the signal from discrete to continuous?$2pi$ periodicity of discrete-time Fourier transformWhy do we use discrete-time sampling?What is the difference between continuous, discrete, analog and digital signal?Why DTFT coefficients are periodic and why continuous Fourier transform coefficients are not periodic?Minimum possible sampling frequency for continuous time signalFourier Series Representation of Continuous-Time Periodic SignalsContinuous-time RNN and Shannon sampling theorem
$begingroup$
I have been studying signals and systems lately and I have came across the following claim:
The uniform sampling of a periodic continuous-time signal may not be periodic!
Can someone please explain why this statement is true?
discrete-signals signal-analysis sampling continuous-signals
edited yesterday
Rodrigo de Azevedo
578314
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
New contributor
Ahmed Wessam 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 |
$begingroup$
I have been studying signals and systems lately and I have came across the following claim:
The uniform sampling of a periodic continuous-time signal may not be periodic!
Can someone please explain why this statement is true?
discrete-signals signal-analysis sampling continuous-signals
edited yesterday
Rodrigo de Azevedo
578314
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
New contributor
Ahmed Wessam 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 |
$begingroup$
I have been studying signals and systems lately and I have came across the following claim:
The uniform sampling of a periodic continuous-time signal may not be periodic!
Can someone please explain why this statement is true?
discrete-signals signal-analysis sampling continuous-signals
edited yesterday
Rodrigo de Azevedo
578314
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
New contributor
Ahmed Wessam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have been studying signals and systems lately and I have came across the following claim:
The uniform sampling of a periodic continuous-time signal may not be periodic!
Can someone please explain why this statement is true?
discrete-signals signal-analysis sampling continuous-signals
discrete-signals signal-analysis sampling continuous-signals
edited yesterday
Rodrigo de Azevedo
578314
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
New contributor
Ahmed Wessam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Rodrigo de Azevedo
578314
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
New contributor
Ahmed Wessam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Rodrigo de Azevedo
578314
edited yesterday
Rodrigo de Azevedo
578314
edited yesterday
Rodrigo de Azevedo
578314
578314
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
New contributor
Ahmed Wessam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
asked May 21 at 10:46
Ahmed WessamAhmed Wessam
1466
1466
New contributor
Ahmed Wessam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ahmed Wessam 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 |
1 Answer
1
active
oldest
votes
$begingroup$
If the ratio between your sampling frequency and the frequency of your signal is irrational, you will not have a periodic discrete signal.
Assuming you have a 1-kHz sine wave and you sample at 3000*sqrt(2) Hz. You will have approximately 4.2 samples per period. However you will not be able to sample the sine wave exactly at the same place. Hence your digital signal will not be periodic.
However, if you sampled the same 1-kHz signal at 4 kHz, you would get a periodic discrete signal. The period would be 4 samples.
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
answered May 21 at 10:54
BenBen
878411
$endgroup$
2
$begingroup$
And quite interestingly (correct me if I am wrong), since the measure of the rationals is zero, if you sample a continuous periodic signal discretely without knowing its frequency, the probability of getting a periodic discrete signal is zero (theoretically speaking, however in practice due to quantization things won't be so bad).
$endgroup$
– Apollys
May 21 at 22:18
4
$begingroup$
@Apollys On the other hand, the rationals are dense in the reals and the lifetime of the Universe is perhaps and ours is certainly bounded, hence getting something close-enough to periodic (though perhaps with a long period) is more than likely - in particular, when the signal and sample are not generated by controlled processes in zero-gravity and near the absolute zero temperatur and whatnot ...
$endgroup$
– Hagen von Eitzen
May 22 at 4:26
$begingroup$
Correct me if i am wrong: But when the input singal is1kHzand you sample with3.5kHz, you get a periodic signal with a period time of2ms. To get a periodic signal,f_sdoes not need to ben*f_inbut can ben*f_in/m
$endgroup$
– 12431234123412341234123
May 22 at 10:44
$begingroup$
Yes, the ratio between 3.5 kHz and 1 kHz is rational number, 2/7 i.e not irrational.
$endgroup$
– Ben
May 22 at 11:59
1
$begingroup$
@Apollys : Yes but in some systems they implement a control loop to adjust the sampling frequency to a multiple of the signal of interest frequency. For example in power systems, where the sampling frequency is tracking the grid frequency. This makes some calculations easier, calculating the mean, RMS and harmonics for example.
$endgroup$
– Ben
May 22 at 12:05
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
If the ratio between your sampling frequency and the frequency of your signal is irrational, you will not have a periodic discrete signal.
Assuming you have a 1-kHz sine wave and you sample at 3000*sqrt(2) Hz. You will have approximately 4.2 samples per period. However you will not be able to sample the sine wave exactly at the same place. Hence your digital signal will not be periodic.
However, if you sampled the same 1-kHz signal at 4 kHz, you would get a periodic discrete signal. The period would be 4 samples.
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
answered May 21 at 10:54
BenBen
878411
$endgroup$
2
$begingroup$
And quite interestingly (correct me if I am wrong), since the measure of the rationals is zero, if you sample a continuous periodic signal discretely without knowing its frequency, the probability of getting a periodic discrete signal is zero (theoretically speaking, however in practice due to quantization things won't be so bad).
$endgroup$
– Apollys
May 21 at 22:18
4
$begingroup$
@Apollys On the other hand, the rationals are dense in the reals and the lifetime of the Universe is perhaps and ours is certainly bounded, hence getting something close-enough to periodic (though perhaps with a long period) is more than likely - in particular, when the signal and sample are not generated by controlled processes in zero-gravity and near the absolute zero temperatur and whatnot ...
$endgroup$
– Hagen von Eitzen
May 22 at 4:26
$begingroup$
Correct me if i am wrong: But when the input singal is1kHzand you sample with3.5kHz, you get a periodic signal with a period time of2ms. To get a periodic signal,f_sdoes not need to ben*f_inbut can ben*f_in/m
$endgroup$
– 12431234123412341234123
May 22 at 10:44
$begingroup$
Yes, the ratio between 3.5 kHz and 1 kHz is rational number, 2/7 i.e not irrational.
$endgroup$
– Ben
May 22 at 11:59
1
$begingroup$
@Apollys : Yes but in some systems they implement a control loop to adjust the sampling frequency to a multiple of the signal of interest frequency. For example in power systems, where the sampling frequency is tracking the grid frequency. This makes some calculations easier, calculating the mean, RMS and harmonics for example.
$endgroup$
– Ben
May 22 at 12:05
add a comment |
$begingroup$
If the ratio between your sampling frequency and the frequency of your signal is irrational, you will not have a periodic discrete signal.
Assuming you have a 1-kHz sine wave and you sample at 3000*sqrt(2) Hz. You will have approximately 4.2 samples per period. However you will not be able to sample the sine wave exactly at the same place. Hence your digital signal will not be periodic.
However, if you sampled the same 1-kHz signal at 4 kHz, you would get a periodic discrete signal. The period would be 4 samples.
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
answered May 21 at 10:54
BenBen
878411
$endgroup$
2
$begingroup$
And quite interestingly (correct me if I am wrong), since the measure of the rationals is zero, if you sample a continuous periodic signal discretely without knowing its frequency, the probability of getting a periodic discrete signal is zero (theoretically speaking, however in practice due to quantization things won't be so bad).
$endgroup$
– Apollys
May 21 at 22:18
4
$begingroup$
@Apollys On the other hand, the rationals are dense in the reals and the lifetime of the Universe is perhaps and ours is certainly bounded, hence getting something close-enough to periodic (though perhaps with a long period) is more than likely - in particular, when the signal and sample are not generated by controlled processes in zero-gravity and near the absolute zero temperatur and whatnot ...
$endgroup$
– Hagen von Eitzen
May 22 at 4:26
$begingroup$
Correct me if i am wrong: But when the input singal is1kHzand you sample with3.5kHz, you get a periodic signal with a period time of2ms. To get a periodic signal,f_sdoes not need to ben*f_inbut can ben*f_in/m
$endgroup$
– 12431234123412341234123
May 22 at 10:44
$begingroup$
Yes, the ratio between 3.5 kHz and 1 kHz is rational number, 2/7 i.e not irrational.
$endgroup$
– Ben
May 22 at 11:59
1
$begingroup$
@Apollys : Yes but in some systems they implement a control loop to adjust the sampling frequency to a multiple of the signal of interest frequency. For example in power systems, where the sampling frequency is tracking the grid frequency. This makes some calculations easier, calculating the mean, RMS and harmonics for example.
$endgroup$
– Ben
May 22 at 12:05
add a comment |
$begingroup$
If the ratio between your sampling frequency and the frequency of your signal is irrational, you will not have a periodic discrete signal.
Assuming you have a 1-kHz sine wave and you sample at 3000*sqrt(2) Hz. You will have approximately 4.2 samples per period. However you will not be able to sample the sine wave exactly at the same place. Hence your digital signal will not be periodic.
However, if you sampled the same 1-kHz signal at 4 kHz, you would get a periodic discrete signal. The period would be 4 samples.
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
answered May 21 at 10:54
BenBen
878411
$endgroup$
If the ratio between your sampling frequency and the frequency of your signal is irrational, you will not have a periodic discrete signal.
Assuming you have a 1-kHz sine wave and you sample at 3000*sqrt(2) Hz. You will have approximately 4.2 samples per period. However you will not be able to sample the sine wave exactly at the same place. Hence your digital signal will not be periodic.
However, if you sampled the same 1-kHz signal at 4 kHz, you would get a periodic discrete signal. The period would be 4 samples.
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
answered May 21 at 10:54
BenBen
878411
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
edited May 21 at 16:23
Dilip Sarwate
13.6k12563
13.6k12563
answered May 21 at 10:54
BenBen
878411
answered May 21 at 10:54
BenBen
878411
answered May 21 at 10:54
BenBen
878411
878411
2
$begingroup$
And quite interestingly (correct me if I am wrong), since the measure of the rationals is zero, if you sample a continuous periodic signal discretely without knowing its frequency, the probability of getting a periodic discrete signal is zero (theoretically speaking, however in practice due to quantization things won't be so bad).
$endgroup$
– Apollys
May 21 at 22:18
4
$begingroup$
@Apollys On the other hand, the rationals are dense in the reals and the lifetime of the Universe is perhaps and ours is certainly bounded, hence getting something close-enough to periodic (though perhaps with a long period) is more than likely - in particular, when the signal and sample are not generated by controlled processes in zero-gravity and near the absolute zero temperatur and whatnot ...
$endgroup$
– Hagen von Eitzen
May 22 at 4:26
$begingroup$
Correct me if i am wrong: But when the input singal is1kHzand you sample with3.5kHz, you get a periodic signal with a period time of2ms. To get a periodic signal,f_sdoes not need to ben*f_inbut can ben*f_in/m
$endgroup$
– 12431234123412341234123
May 22 at 10:44
$begingroup$
Yes, the ratio between 3.5 kHz and 1 kHz is rational number, 2/7 i.e not irrational.
$endgroup$
– Ben
May 22 at 11:59
1
$begingroup$
@Apollys : Yes but in some systems they implement a control loop to adjust the sampling frequency to a multiple of the signal of interest frequency. For example in power systems, where the sampling frequency is tracking the grid frequency. This makes some calculations easier, calculating the mean, RMS and harmonics for example.
$endgroup$
– Ben
May 22 at 12:05
add a comment |
2
$begingroup$
And quite interestingly (correct me if I am wrong), since the measure of the rationals is zero, if you sample a continuous periodic signal discretely without knowing its frequency, the probability of getting a periodic discrete signal is zero (theoretically speaking, however in practice due to quantization things won't be so bad).
$endgroup$
– Apollys
May 21 at 22:18
4
$begingroup$
@Apollys On the other hand, the rationals are dense in the reals and the lifetime of the Universe is perhaps and ours is certainly bounded, hence getting something close-enough to periodic (though perhaps with a long period) is more than likely - in particular, when the signal and sample are not generated by controlled processes in zero-gravity and near the absolute zero temperatur and whatnot ...
$endgroup$
– Hagen von Eitzen
May 22 at 4:26
$begingroup$
Correct me if i am wrong: But when the input singal is1kHzand you sample with3.5kHz, you get a periodic signal with a period time of2ms. To get a periodic signal,f_sdoes not need to ben*f_inbut can ben*f_in/m
$endgroup$
– 12431234123412341234123
May 22 at 10:44
$begingroup$
Yes, the ratio between 3.5 kHz and 1 kHz is rational number, 2/7 i.e not irrational.
$endgroup$
– Ben
May 22 at 11:59
1
$begingroup$
@Apollys : Yes but in some systems they implement a control loop to adjust the sampling frequency to a multiple of the signal of interest frequency. For example in power systems, where the sampling frequency is tracking the grid frequency. This makes some calculations easier, calculating the mean, RMS and harmonics for example.
$endgroup$
– Ben
May 22 at 12:05
2
2
$begingroup$
And quite interestingly (correct me if I am wrong), since the measure of the rationals is zero, if you sample a continuous periodic signal discretely without knowing its frequency, the probability of getting a periodic discrete signal is zero (theoretically speaking, however in practice due to quantization things won't be so bad).
$endgroup$
– Apollys
May 21 at 22:18
$begingroup$
And quite interestingly (correct me if I am wrong), since the measure of the rationals is zero, if you sample a continuous periodic signal discretely without knowing its frequency, the probability of getting a periodic discrete signal is zero (theoretically speaking, however in practice due to quantization things won't be so bad).
$endgroup$
– Apollys
May 21 at 22:18
4
4
$begingroup$
@Apollys On the other hand, the rationals are dense in the reals and the lifetime of the Universe is perhaps and ours is certainly bounded, hence getting something close-enough to periodic (though perhaps with a long period) is more than likely - in particular, when the signal and sample are not generated by controlled processes in zero-gravity and near the absolute zero temperatur and whatnot ...
$endgroup$
– Hagen von Eitzen
May 22 at 4:26
$begingroup$
@Apollys On the other hand, the rationals are dense in the reals and the lifetime of the Universe is perhaps and ours is certainly bounded, hence getting something close-enough to periodic (though perhaps with a long period) is more than likely - in particular, when the signal and sample are not generated by controlled processes in zero-gravity and near the absolute zero temperatur and whatnot ...
$endgroup$
– Hagen von Eitzen
May 22 at 4:26
$begingroup$
Correct me if i am wrong: But when the input singal is
1kHz and you sample with 3.5kHz, you get a periodic signal with a period time of 2ms. To get a periodic signal, f_s does not need to be n*f_in but can be n*f_in/m$endgroup$
– 12431234123412341234123
May 22 at 10:44
$begingroup$
Correct me if i am wrong: But when the input singal is
1kHz and you sample with 3.5kHz, you get a periodic signal with a period time of 2ms. To get a periodic signal, f_s does not need to be n*f_in but can be n*f_in/m$endgroup$
– 12431234123412341234123
May 22 at 10:44
$begingroup$
Yes, the ratio between 3.5 kHz and 1 kHz is rational number, 2/7 i.e not irrational.
$endgroup$
– Ben
May 22 at 11:59
$begingroup$
Yes, the ratio between 3.5 kHz and 1 kHz is rational number, 2/7 i.e not irrational.
$endgroup$
– Ben
May 22 at 11:59
1
1
$begingroup$
@Apollys : Yes but in some systems they implement a control loop to adjust the sampling frequency to a multiple of the signal of interest frequency. For example in power systems, where the sampling frequency is tracking the grid frequency. This makes some calculations easier, calculating the mean, RMS and harmonics for example.
$endgroup$
– Ben
May 22 at 12:05
$begingroup$
@Apollys : Yes but in some systems they implement a control loop to adjust the sampling frequency to a multiple of the signal of interest frequency. For example in power systems, where the sampling frequency is tracking the grid frequency. This makes some calculations easier, calculating the mean, RMS and harmonics for example.
$endgroup$
– Ben
May 22 at 12:05
add a comment |
Ahmed Wessam is a new contributor. Be nice, and check out our Code of Conduct.
Ahmed Wessam is a new contributor. Be nice, and check out our Code of Conduct.
Ahmed Wessam is a new contributor. Be nice, and check out our Code of Conduct.
Ahmed Wessam is a new contributor. Be nice, and check out our Code of Conduct.
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