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Can sampling rate be a floating point number?

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5

$begingroup$

Suppose we have a sampling frequency for a signal of 15.5 samples/sec and we take samples for a period of 7 seconds. This means total samples are 108.5, does this make any sense?

Shouldn't the number of samples taken be an integer like 108 or 109? Or can the particular points in time from 0 second to 7 seconds on which to take the samples be determined in this case? How would one do that?

frequency signal-processing sampling

share|improve this question

edited 18 hours ago

jusaca

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Ron HowardRon Howard

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Ron Howard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What is the sampling rate if you take 217 samples over 14 seconds?
  $endgroup$
  – Harry Svensson
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes, video, 29.97fps
  $endgroup$
  – user3528438
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I plan on using a sample rate of exactly 45000000/11 Hz in a future project. Can't even use a float to represent that.
  $endgroup$
  – pipe
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasWeller Maybe OP understands it even if you don't? As you can see from the answers, it can be any real value, including arbitrary rationals.
  $endgroup$
  – pipe
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasWeller, I pretty sure pipe's point is that sample rates and the binary representation of floating point numbers are 2.0 unrelated concepts.
  $endgroup$
  – Glen Yates
  6 hours ago
  

  
  
  
  

 | 
show 3 more comments

5

$begingroup$

Suppose we have a sampling frequency for a signal of 15.5 samples/sec and we take samples for a period of 7 seconds. This means total samples are 108.5, does this make any sense?

Shouldn't the number of samples taken be an integer like 108 or 109? Or can the particular points in time from 0 second to 7 seconds on which to take the samples be determined in this case? How would one do that?

frequency signal-processing sampling

share|improve this question

edited 18 hours ago

jusaca

2,52688 silver badges3030 bronze badges

asked 19 hours ago

Ron HowardRon Howard

2822 bronze badges

New contributor

Ron Howard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What is the sampling rate if you take 217 samples over 14 seconds?
  $endgroup$
  – Harry Svensson
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes, video, 29.97fps
  $endgroup$
  – user3528438
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I plan on using a sample rate of exactly 45000000/11 Hz in a future project. Can't even use a float to represent that.
  $endgroup$
  – pipe
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasWeller Maybe OP understands it even if you don't? As you can see from the answers, it can be any real value, including arbitrary rationals.
  $endgroup$
  – pipe
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasWeller, I pretty sure pipe's point is that sample rates and the binary representation of floating point numbers are 2.0 unrelated concepts.
  $endgroup$
  – Glen Yates
  6 hours ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Suppose we have a sampling frequency for a signal of 15.5 samples/sec and we take samples for a period of 7 seconds. This means total samples are 108.5, does this make any sense?

Shouldn't the number of samples taken be an integer like 108 or 109? Or can the particular points in time from 0 second to 7 seconds on which to take the samples be determined in this case? How would one do that?

frequency signal-processing sampling

share|improve this question

edited 18 hours ago

jusaca

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asked 19 hours ago

Ron HowardRon Howard

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Ron Howard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 18 hours ago

jusaca

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asked 19 hours ago

Ron HowardRon Howard

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New contributor

Ron Howard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 18 hours ago

jusaca

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asked 19 hours ago

Ron HowardRon Howard

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New contributor

Ron Howard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 18 hours ago

jusaca

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edited 18 hours ago

jusaca

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asked 19 hours ago

Ron HowardRon Howard

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Ron Howard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 19 hours ago

Ron HowardRon Howard

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3 Answers
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14

$begingroup$

Forget sampling rate for a few seconds... Think about sampling period for a second, which is the time interval between two consecutive samples. This time can be an integer or any real number (as long as it’s positive, of course).

Sampling rate is simply the inverse of sampling period. Does it make more sense this way?

share|improve this answer

edited 6 hours ago

answered 18 hours ago

joribamajoribama

1,21122 silver badges1212 bronze badges

$endgroup$

add a comment | 

9

$begingroup$

Some things are always an integer. Samples are always integer. You can take 108 or 109 samples.

Sample rate can be a floating point number, or more generally a rational, or even a real.

You calculate the sample rate by dividing the number of samples (less one to get the number of periods between samples) by the time it takes to obtain those samples.

Generally a floating point number is an approximation to the real number you want. With double precision, it's a very good approximation, but it's usually inexact.

If you're given a sample rate, and a time, the product might be an exact integer, if the numbers are chosen carefully, but it probably won't be. It might be in error a small amount, due to the approximation of floating point representation. It might be in error a lot, because the source of your information chose very approximate numbers, or even made up the numbers to start with.

share|improve this answer

answered 18 hours ago

Neil_UKNeil_UK

85.5k22 gold badges8686 silver badges197197 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

Yes, the sampling rate can be any number you want.

But you obviously would not get partial samples in the end, you just have to round down.

In your example the first sample is taken at $ frac{1}{15.5}s $ = 64.5 ms and then at every multiple from that. This means you get your last sample at 6,966 s. That is the 108's sample. So at 7 s you still have taken only 108 samples. And then at 7,0305 s you get the next sample.

You can imagine the samples beeing taken in a way like this dirac comb:

If you stop sampling between 3T and 4T you do not have partial samples. You just round down. T is the inverse of the sample frequency, or in your case 64.5 ms.

share|improve this answer

edited 18 hours ago

answered 18 hours ago

jusacajusaca

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$endgroup$

add a comment | 

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14

$begingroup$

Forget sampling rate for a few seconds... Think about sampling period for a second, which is the time interval between two consecutive samples. This time can be an integer or any real number (as long as it’s positive, of course).

Sampling rate is simply the inverse of sampling period. Does it make more sense this way?

share|improve this answer

edited 6 hours ago

answered 18 hours ago

joribamajoribama

1,21122 silver badges1212 bronze badges

$endgroup$

add a comment | 

14

$begingroup$

Forget sampling rate for a few seconds... Think about sampling period for a second, which is the time interval between two consecutive samples. This time can be an integer or any real number (as long as it’s positive, of course).

Sampling rate is simply the inverse of sampling period. Does it make more sense this way?

share|improve this answer

edited 6 hours ago

answered 18 hours ago

joribamajoribama

1,21122 silver badges1212 bronze badges

$endgroup$

add a comment | 

$begingroup$

Forget sampling rate for a few seconds... Think about sampling period for a second, which is the time interval between two consecutive samples. This time can be an integer or any real number (as long as it’s positive, of course).

Sampling rate is simply the inverse of sampling period. Does it make more sense this way?

share|improve this answer

edited 6 hours ago

answered 18 hours ago

joribamajoribama

1,21122 silver badges1212 bronze badges

$endgroup$

share|improve this answer

edited 6 hours ago

answered 18 hours ago

joribamajoribama

1,21122 silver badges1212 bronze badges

answered 18 hours ago

joribamajoribama

1,21122 silver badges1212 bronze badges

answered 18 hours ago

joribamajoribama

1,21122 silver badges1212 bronze badges

9

$begingroup$

Some things are always an integer. Samples are always integer. You can take 108 or 109 samples.

Sample rate can be a floating point number, or more generally a rational, or even a real.

You calculate the sample rate by dividing the number of samples (less one to get the number of periods between samples) by the time it takes to obtain those samples.

Generally a floating point number is an approximation to the real number you want. With double precision, it's a very good approximation, but it's usually inexact.

If you're given a sample rate, and a time, the product might be an exact integer, if the numbers are chosen carefully, but it probably won't be. It might be in error a small amount, due to the approximation of floating point representation. It might be in error a lot, because the source of your information chose very approximate numbers, or even made up the numbers to start with.

share|improve this answer

answered 18 hours ago

Neil_UKNeil_UK

85.5k22 gold badges8686 silver badges197197 bronze badges

$endgroup$

add a comment | 

9

$begingroup$

Some things are always an integer. Samples are always integer. You can take 108 or 109 samples.

Sample rate can be a floating point number, or more generally a rational, or even a real.

You calculate the sample rate by dividing the number of samples (less one to get the number of periods between samples) by the time it takes to obtain those samples.

Generally a floating point number is an approximation to the real number you want. With double precision, it's a very good approximation, but it's usually inexact.

If you're given a sample rate, and a time, the product might be an exact integer, if the numbers are chosen carefully, but it probably won't be. It might be in error a small amount, due to the approximation of floating point representation. It might be in error a lot, because the source of your information chose very approximate numbers, or even made up the numbers to start with.

share|improve this answer

answered 18 hours ago

Neil_UKNeil_UK

85.5k22 gold badges8686 silver badges197197 bronze badges

$endgroup$

add a comment | 

$begingroup$

Some things are always an integer. Samples are always integer. You can take 108 or 109 samples.

Sample rate can be a floating point number, or more generally a rational, or even a real.

You calculate the sample rate by dividing the number of samples (less one to get the number of periods between samples) by the time it takes to obtain those samples.

Generally a floating point number is an approximation to the real number you want. With double precision, it's a very good approximation, but it's usually inexact.

If you're given a sample rate, and a time, the product might be an exact integer, if the numbers are chosen carefully, but it probably won't be. It might be in error a small amount, due to the approximation of floating point representation. It might be in error a lot, because the source of your information chose very approximate numbers, or even made up the numbers to start with.

share|improve this answer

answered 18 hours ago

Neil_UKNeil_UK

85.5k22 gold badges8686 silver badges197197 bronze badges

$endgroup$

share|improve this answer

answered 18 hours ago

Neil_UKNeil_UK

85.5k22 gold badges8686 silver badges197197 bronze badges

answered 18 hours ago

Neil_UKNeil_UK

85.5k22 gold badges8686 silver badges197197 bronze badges

answered 18 hours ago

Neil_UKNeil_UK

85.5k22 gold badges8686 silver badges197197 bronze badges

6

$begingroup$

Yes, the sampling rate can be any number you want.

But you obviously would not get partial samples in the end, you just have to round down.

In your example the first sample is taken at $ frac{1}{15.5}s $ = 64.5 ms and then at every multiple from that. This means you get your last sample at 6,966 s. That is the 108's sample. So at 7 s you still have taken only 108 samples. And then at 7,0305 s you get the next sample.

You can imagine the samples beeing taken in a way like this dirac comb:

If you stop sampling between 3T and 4T you do not have partial samples. You just round down. T is the inverse of the sample frequency, or in your case 64.5 ms.

share|improve this answer

edited 18 hours ago

answered 18 hours ago

jusacajusaca

2,52688 silver badges3030 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

Yes, the sampling rate can be any number you want.

But you obviously would not get partial samples in the end, you just have to round down.

In your example the first sample is taken at $ frac{1}{15.5}s $ = 64.5 ms and then at every multiple from that. This means you get your last sample at 6,966 s. That is the 108's sample. So at 7 s you still have taken only 108 samples. And then at 7,0305 s you get the next sample.

You can imagine the samples beeing taken in a way like this dirac comb:

If you stop sampling between 3T and 4T you do not have partial samples. You just round down. T is the inverse of the sample frequency, or in your case 64.5 ms.

share|improve this answer

edited 18 hours ago

answered 18 hours ago

jusacajusaca

2,52688 silver badges3030 bronze badges

$endgroup$

add a comment | 

$begingroup$

Yes, the sampling rate can be any number you want.

But you obviously would not get partial samples in the end, you just have to round down.

In your example the first sample is taken at $ frac{1}{15.5}s $ = 64.5 ms and then at every multiple from that. This means you get your last sample at 6,966 s. That is the 108's sample. So at 7 s you still have taken only 108 samples. And then at 7,0305 s you get the next sample.

You can imagine the samples beeing taken in a way like this dirac comb:

If you stop sampling between 3T and 4T you do not have partial samples. You just round down. T is the inverse of the sample frequency, or in your case 64.5 ms.

share|improve this answer

edited 18 hours ago

answered 18 hours ago

jusacajusaca

2,52688 silver badges3030 bronze badges

$endgroup$

share|improve this answer

edited 18 hours ago

answered 18 hours ago

jusacajusaca

2,52688 silver badges3030 bronze badges

answered 18 hours ago

jusacajusaca

2,52688 silver badges3030 bronze badges

answered 18 hours ago

jusacajusaca

2,52688 silver badges3030 bronze badges