use the oversamplling followed by '' decimation method ''to increasee the ADC resolution and not normal averagingLTC2485 I2C Data FormatAny tricks to generate triangle wave to add to analog signal for oversampling?About the meaning of “oversampling”Delta Sigma Converters: ResolutionSampling rate analog to digital converterQuestion about the registers of an ADCOptimal tradeoff between ADC bit depth and sampling rateWhy can't you just average ADC samples to get more resolution from an ADC?Resolution enhancement for a 2wire PT100 setup with ADC and no external amplificationA question about 16 bit ADC representation
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use the oversamplling followed by '' decimation method ''to increasee the ADC resolution and not normal averaging
LTC2485 I2C Data FormatAny tricks to generate triangle wave to add to analog signal for oversampling?About the meaning of “oversampling”Delta Sigma Converters: ResolutionSampling rate analog to digital converterQuestion about the registers of an ADCOptimal tradeoff between ADC bit depth and sampling rateWhy can't you just average ADC samples to get more resolution from an ADC?Resolution enhancement for a 2wire PT100 setup with ADC and no external amplificationA question about 16 bit ADC representation
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
To increase the 12-bit resolution of ADC from 12 bit to 14 bit , this can be done through 'oversampling and decimation method'
An Atmel Application note says that
The higher the number of samples averaged is, the more selective the low-pass filter will be, and the better the interpolation. The extra samples, m, achieved by oversampling the signal are added, just as in normal averaging, but the result are not divided by m as in normal averaging. Instead the result is right shifted by n, where n is the desired extra bit of resolution, to scale the answer correctly. Right shifting a binary number once is equal to dividing the binary number by a factor of 2.
It is important to remember that normal averaging does not increase the resolution of the conversion. Decimation, or Interpolation, is the averaging method, which combined with oversampling, which increases the resolution
This reference clearly says that in decimation method the result is right shifted by the desired extra bit of resolution, and not divided by m as in normal average.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
microcontroller adc
edited 2 hours ago
Dave Tweed♦
126k10156272
asked 2 hours ago
AliAli
61
New contributor
Ali 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 |
$begingroup$
To increase the 12-bit resolution of ADC from 12 bit to 14 bit , this can be done through 'oversampling and decimation method'
An Atmel Application note says that
The higher the number of samples averaged is, the more selective the low-pass filter will be, and the better the interpolation. The extra samples, m, achieved by oversampling the signal are added, just as in normal averaging, but the result are not divided by m as in normal averaging. Instead the result is right shifted by n, where n is the desired extra bit of resolution, to scale the answer correctly. Right shifting a binary number once is equal to dividing the binary number by a factor of 2.
It is important to remember that normal averaging does not increase the resolution of the conversion. Decimation, or Interpolation, is the averaging method, which combined with oversampling, which increases the resolution
This reference clearly says that in decimation method the result is right shifted by the desired extra bit of resolution, and not divided by m as in normal average.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
microcontroller adc
edited 2 hours ago
Dave Tweed♦
126k10156272
asked 2 hours ago
AliAli
61
New contributor
Ali is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
How do you define "normal averaging"?
$endgroup$
– TimWescott
2 hours ago
$begingroup$
"but what if we don't use a binary number" - all numbers are binary numbers in a microcontroller.
$endgroup$
– brhans
2 hours ago
$begingroup$
@TimWescott: It's "defined" (sort of) in the paper.
$endgroup$
– Dave Tweed♦
2 hours ago
add a comment |
$begingroup$
To increase the 12-bit resolution of ADC from 12 bit to 14 bit , this can be done through 'oversampling and decimation method'
An Atmel Application note says that
The higher the number of samples averaged is, the more selective the low-pass filter will be, and the better the interpolation. The extra samples, m, achieved by oversampling the signal are added, just as in normal averaging, but the result are not divided by m as in normal averaging. Instead the result is right shifted by n, where n is the desired extra bit of resolution, to scale the answer correctly. Right shifting a binary number once is equal to dividing the binary number by a factor of 2.
It is important to remember that normal averaging does not increase the resolution of the conversion. Decimation, or Interpolation, is the averaging method, which combined with oversampling, which increases the resolution
This reference clearly says that in decimation method the result is right shifted by the desired extra bit of resolution, and not divided by m as in normal average.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
microcontroller adc
edited 2 hours ago
Dave Tweed♦
126k10156272
asked 2 hours ago
AliAli
61
New contributor
Ali is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
To increase the 12-bit resolution of ADC from 12 bit to 14 bit , this can be done through 'oversampling and decimation method'
An Atmel Application note says that
The higher the number of samples averaged is, the more selective the low-pass filter will be, and the better the interpolation. The extra samples, m, achieved by oversampling the signal are added, just as in normal averaging, but the result are not divided by m as in normal averaging. Instead the result is right shifted by n, where n is the desired extra bit of resolution, to scale the answer correctly. Right shifting a binary number once is equal to dividing the binary number by a factor of 2.
It is important to remember that normal averaging does not increase the resolution of the conversion. Decimation, or Interpolation, is the averaging method, which combined with oversampling, which increases the resolution
This reference clearly says that in decimation method the result is right shifted by the desired extra bit of resolution, and not divided by m as in normal average.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
microcontroller adc
microcontroller adc
edited 2 hours ago
Dave Tweed♦
126k10156272
asked 2 hours ago
AliAli
61
New contributor
Ali is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Dave Tweed♦
126k10156272
asked 2 hours ago
AliAli
61
New contributor
Ali is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Dave Tweed♦
126k10156272
edited 2 hours ago
Dave Tweed♦
126k10156272
edited 2 hours ago
Dave Tweed♦
126k10156272
126k10156272
asked 2 hours ago
AliAli
61
New contributor
Ali is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
AliAli
61
asked 2 hours ago
AliAli
61
61
New contributor
Ali is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ali is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
How do you define "normal averaging"?
$endgroup$
– TimWescott
2 hours ago
$begingroup$
"but what if we don't use a binary number" - all numbers are binary numbers in a microcontroller.
$endgroup$
– brhans
2 hours ago
$begingroup$
@TimWescott: It's "defined" (sort of) in the paper.
$endgroup$
– Dave Tweed♦
2 hours ago
add a comment |
$begingroup$
How do you define "normal averaging"?
$endgroup$
– TimWescott
2 hours ago
$begingroup$
"but what if we don't use a binary number" - all numbers are binary numbers in a microcontroller.
$endgroup$
– brhans
2 hours ago
$begingroup$
@TimWescott: It's "defined" (sort of) in the paper.
$endgroup$
– Dave Tweed♦
2 hours ago
$begingroup$
How do you define "normal averaging"?
$endgroup$
– TimWescott
2 hours ago
$begingroup$
How do you define "normal averaging"?
$endgroup$
– TimWescott
2 hours ago
$begingroup$
"but what if we don't use a binary number" - all numbers are binary numbers in a microcontroller.
$endgroup$
– brhans
2 hours ago
$begingroup$
"but what if we don't use a binary number" - all numbers are binary numbers in a microcontroller.
$endgroup$
– brhans
2 hours ago
$begingroup$
@TimWescott: It's "defined" (sort of) in the paper.
$endgroup$
– Dave Tweed♦
2 hours ago
$begingroup$
@TimWescott: It's "defined" (sort of) in the paper.
$endgroup$
– Dave Tweed♦
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I wouldn't take that application note too seriously — it contains many errors, both conceptual1 and typographical.
Adding up a bunch of samples and then scaling the sum by some factor, no matter what you call it, IS averaging. It's also filtering. It is, in fact, just one special case of a finite impulse response (FIR) filter, in which every sample gets its own scale factor and then they get added together to create the result.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It's all the same thing in the end.
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
Just use ordinary division if the divisor isn't a power of 2.
1 For example, "white" noise is NOT equivalent to "gaussian" noise, although many natural noise sources are both gaussian AND white.
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
$endgroup$
3
$begingroup$
Although, the distinction that the paper is making (however badly) is that whatever method you use for averaging (or filtering), you need to save the least significant bits. If you average 16 samples, you (roughly) reduce the RMS noise by a factor of 4. If you don't keep the two additional "good" bits one way or another, you lose the advantage. Whether you do that by shifting, or multiplying by floats, or whatever -- it still needs to be done.
$endgroup$
– TimWescott
1 hour ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
I wouldn't take that application note too seriously — it contains many errors, both conceptual1 and typographical.
Adding up a bunch of samples and then scaling the sum by some factor, no matter what you call it, IS averaging. It's also filtering. It is, in fact, just one special case of a finite impulse response (FIR) filter, in which every sample gets its own scale factor and then they get added together to create the result.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It's all the same thing in the end.
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
Just use ordinary division if the divisor isn't a power of 2.
1 For example, "white" noise is NOT equivalent to "gaussian" noise, although many natural noise sources are both gaussian AND white.
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
$endgroup$
3
$begingroup$
Although, the distinction that the paper is making (however badly) is that whatever method you use for averaging (or filtering), you need to save the least significant bits. If you average 16 samples, you (roughly) reduce the RMS noise by a factor of 4. If you don't keep the two additional "good" bits one way or another, you lose the advantage. Whether you do that by shifting, or multiplying by floats, or whatever -- it still needs to be done.
$endgroup$
– TimWescott
1 hour ago
add a comment |
$begingroup$
I wouldn't take that application note too seriously — it contains many errors, both conceptual1 and typographical.
Adding up a bunch of samples and then scaling the sum by some factor, no matter what you call it, IS averaging. It's also filtering. It is, in fact, just one special case of a finite impulse response (FIR) filter, in which every sample gets its own scale factor and then they get added together to create the result.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It's all the same thing in the end.
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
Just use ordinary division if the divisor isn't a power of 2.
1 For example, "white" noise is NOT equivalent to "gaussian" noise, although many natural noise sources are both gaussian AND white.
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
$endgroup$
3
$begingroup$
Although, the distinction that the paper is making (however badly) is that whatever method you use for averaging (or filtering), you need to save the least significant bits. If you average 16 samples, you (roughly) reduce the RMS noise by a factor of 4. If you don't keep the two additional "good" bits one way or another, you lose the advantage. Whether you do that by shifting, or multiplying by floats, or whatever -- it still needs to be done.
$endgroup$
– TimWescott
1 hour ago
add a comment |
$begingroup$
I wouldn't take that application note too seriously — it contains many errors, both conceptual1 and typographical.
Adding up a bunch of samples and then scaling the sum by some factor, no matter what you call it, IS averaging. It's also filtering. It is, in fact, just one special case of a finite impulse response (FIR) filter, in which every sample gets its own scale factor and then they get added together to create the result.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It's all the same thing in the end.
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
Just use ordinary division if the divisor isn't a power of 2.
1 For example, "white" noise is NOT equivalent to "gaussian" noise, although many natural noise sources are both gaussian AND white.
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
$endgroup$
I wouldn't take that application note too seriously — it contains many errors, both conceptual1 and typographical.
Adding up a bunch of samples and then scaling the sum by some factor, no matter what you call it, IS averaging. It's also filtering. It is, in fact, just one special case of a finite impulse response (FIR) filter, in which every sample gets its own scale factor and then they get added together to create the result.
So, the question is, why do we need to use decimation method instead of the normal averaging after the oversampling to increase the ADC resolution?
It's all the same thing in the end.
It says above "Right shifting a binary number once is equal to dividing the binary number by a factor of 2", but what if we don't use a binary number, how do we use the decimation method in this case?
Just use ordinary division if the divisor isn't a power of 2.
1 For example, "white" noise is NOT equivalent to "gaussian" noise, although many natural noise sources are both gaussian AND white.
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
edited 2 hours ago
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
answered 2 hours ago
Dave Tweed♦Dave Tweed
126k10156272
126k10156272
3
$begingroup$
Although, the distinction that the paper is making (however badly) is that whatever method you use for averaging (or filtering), you need to save the least significant bits. If you average 16 samples, you (roughly) reduce the RMS noise by a factor of 4. If you don't keep the two additional "good" bits one way or another, you lose the advantage. Whether you do that by shifting, or multiplying by floats, or whatever -- it still needs to be done.
$endgroup$
– TimWescott
1 hour ago
add a comment |
3
$begingroup$
Although, the distinction that the paper is making (however badly) is that whatever method you use for averaging (or filtering), you need to save the least significant bits. If you average 16 samples, you (roughly) reduce the RMS noise by a factor of 4. If you don't keep the two additional "good" bits one way or another, you lose the advantage. Whether you do that by shifting, or multiplying by floats, or whatever -- it still needs to be done.
$endgroup$
– TimWescott
1 hour ago
3
3
$begingroup$
Although, the distinction that the paper is making (however badly) is that whatever method you use for averaging (or filtering), you need to save the least significant bits. If you average 16 samples, you (roughly) reduce the RMS noise by a factor of 4. If you don't keep the two additional "good" bits one way or another, you lose the advantage. Whether you do that by shifting, or multiplying by floats, or whatever -- it still needs to be done.
$endgroup$
– TimWescott
1 hour ago
$begingroup$
Although, the distinction that the paper is making (however badly) is that whatever method you use for averaging (or filtering), you need to save the least significant bits. If you average 16 samples, you (roughly) reduce the RMS noise by a factor of 4. If you don't keep the two additional "good" bits one way or another, you lose the advantage. Whether you do that by shifting, or multiplying by floats, or whatever -- it still needs to be done.
$endgroup$
– TimWescott
1 hour ago
add a comment |
Ali is a new contributor. Be nice, and check out our Code of Conduct.
Ali is a new contributor. Be nice, and check out our Code of Conduct.
Ali is a new contributor. Be nice, and check out our Code of Conduct.
Ali is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
How do you define "normal averaging"?
$endgroup$
– TimWescott
2 hours ago
$begingroup$
"but what if we don't use a binary number" - all numbers are binary numbers in a microcontroller.
$endgroup$
– brhans
2 hours ago
$begingroup$
@TimWescott: It's "defined" (sort of) in the paper.
$endgroup$
– Dave Tweed♦
2 hours ago