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How to find an ADC circuit's bandwidth experimentally?
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$begingroup$
I'm trying to measure the bandwidth of a circuit that I've built. It's an analogue to digital converter with some input protection and biasing. The ADC samples at a constant/fixed 50kSa/s.
My approach is to inject a 2V sine wave of variable frequency $f_{in}$ and take maximum ($Vmax$) and minimum ($Vmin$) samples. That gives me $Vpp = Vmax - Vmin$ and I sweep $fin$ until $ Vpp = 2 sqrt{frac{1}{2}}V approx 1.41V $.
I've read the Wiki articles Bandwidth (signal processing) and Half-power point but remain confused.
When I find the -3dB point, what is the corresponding official bandwidth I've found? Is it $fin$ or $ frac{fin}{2} $ with consideration of the Nyquist rate?
circuit-analysis bandwidth
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm trying to measure the bandwidth of a circuit that I've built. It's an analogue to digital converter with some input protection and biasing. The ADC samples at a constant/fixed 50kSa/s.
My approach is to inject a 2V sine wave of variable frequency $f_{in}$ and take maximum ($Vmax$) and minimum ($Vmin$) samples. That gives me $Vpp = Vmax - Vmin$ and I sweep $fin$ until $ Vpp = 2 sqrt{frac{1}{2}}V approx 1.41V $.
I've read the Wiki articles Bandwidth (signal processing) and Half-power point but remain confused.
When I find the -3dB point, what is the corresponding official bandwidth I've found? Is it $fin$ or $ frac{fin}{2} $ with consideration of the Nyquist rate?
circuit-analysis bandwidth
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
$endgroup$
2
$begingroup$
It's not really part of your question, but the RMS voltage (or just mean-squared voltage) is a much better indication of signal strength as you approach the sample rate. Even at $f_s / 4$, using $V_{pp}$ gets problematical.
$endgroup$
– TimWescott
8 hours ago
1
$begingroup$
@TimWescott That's interesting. Is it appropriate to ask here why Vpp gets problematical, or should I break it out into a new question?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
It would be good as a new question. The answer is semi-long.
$endgroup$
– TimWescott
7 hours ago
$begingroup$
to find the bandwidth, input 1 microsecond pulses with 99 microseconds at the opposite level. Then, if your ADC accurately quantizes the voltage difference, reduce the 1uS to 100 nanosecond, and retest.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
$begingroup$
I'm trying to measure the bandwidth of a circuit that I've built. It's an analogue to digital converter with some input protection and biasing. The ADC samples at a constant/fixed 50kSa/s.
My approach is to inject a 2V sine wave of variable frequency $f_{in}$ and take maximum ($Vmax$) and minimum ($Vmin$) samples. That gives me $Vpp = Vmax - Vmin$ and I sweep $fin$ until $ Vpp = 2 sqrt{frac{1}{2}}V approx 1.41V $.
I've read the Wiki articles Bandwidth (signal processing) and Half-power point but remain confused.
When I find the -3dB point, what is the corresponding official bandwidth I've found? Is it $fin$ or $ frac{fin}{2} $ with consideration of the Nyquist rate?
circuit-analysis bandwidth
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
$endgroup$
I'm trying to measure the bandwidth of a circuit that I've built. It's an analogue to digital converter with some input protection and biasing. The ADC samples at a constant/fixed 50kSa/s.
My approach is to inject a 2V sine wave of variable frequency $f_{in}$ and take maximum ($Vmax$) and minimum ($Vmin$) samples. That gives me $Vpp = Vmax - Vmin$ and I sweep $fin$ until $ Vpp = 2 sqrt{frac{1}{2}}V approx 1.41V $.
I've read the Wiki articles Bandwidth (signal processing) and Half-power point but remain confused.
When I find the -3dB point, what is the corresponding official bandwidth I've found? Is it $fin$ or $ frac{fin}{2} $ with consideration of the Nyquist rate?
circuit-analysis bandwidth
circuit-analysis bandwidth
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
asked 8 hours ago
Paul UszakPaul Uszak
3,6502 gold badges20 silver badges44 bronze badges
3,6502 gold badges20 silver badges44 bronze badges
2
$begingroup$
It's not really part of your question, but the RMS voltage (or just mean-squared voltage) is a much better indication of signal strength as you approach the sample rate. Even at $f_s / 4$, using $V_{pp}$ gets problematical.
$endgroup$
– TimWescott
8 hours ago
1
$begingroup$
@TimWescott That's interesting. Is it appropriate to ask here why Vpp gets problematical, or should I break it out into a new question?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
It would be good as a new question. The answer is semi-long.
$endgroup$
– TimWescott
7 hours ago
$begingroup$
to find the bandwidth, input 1 microsecond pulses with 99 microseconds at the opposite level. Then, if your ADC accurately quantizes the voltage difference, reduce the 1uS to 100 nanosecond, and retest.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
2
$begingroup$
It's not really part of your question, but the RMS voltage (or just mean-squared voltage) is a much better indication of signal strength as you approach the sample rate. Even at $f_s / 4$, using $V_{pp}$ gets problematical.
$endgroup$
– TimWescott
8 hours ago
1
$begingroup$
@TimWescott That's interesting. Is it appropriate to ask here why Vpp gets problematical, or should I break it out into a new question?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
It would be good as a new question. The answer is semi-long.
$endgroup$
– TimWescott
7 hours ago
$begingroup$
to find the bandwidth, input 1 microsecond pulses with 99 microseconds at the opposite level. Then, if your ADC accurately quantizes the voltage difference, reduce the 1uS to 100 nanosecond, and retest.
$endgroup$
– analogsystemsrf
2 hours ago
2
2
$begingroup$
It's not really part of your question, but the RMS voltage (or just mean-squared voltage) is a much better indication of signal strength as you approach the sample rate. Even at $f_s / 4$, using $V_{pp}$ gets problematical.
$endgroup$
– TimWescott
8 hours ago
$begingroup$
It's not really part of your question, but the RMS voltage (or just mean-squared voltage) is a much better indication of signal strength as you approach the sample rate. Even at $f_s / 4$, using $V_{pp}$ gets problematical.
$endgroup$
– TimWescott
8 hours ago
1
1
$begingroup$
@TimWescott That's interesting. Is it appropriate to ask here why Vpp gets problematical, or should I break it out into a new question?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
@TimWescott That's interesting. Is it appropriate to ask here why Vpp gets problematical, or should I break it out into a new question?
$endgroup$
– Paul Uszak
8 hours ago
1
1
$begingroup$
It would be good as a new question. The answer is semi-long.
$endgroup$
– TimWescott
7 hours ago
$begingroup$
It would be good as a new question. The answer is semi-long.
$endgroup$
– TimWescott
7 hours ago
$begingroup$
to find the bandwidth, input 1 microsecond pulses with 99 microseconds at the opposite level. Then, if your ADC accurately quantizes the voltage difference, reduce the 1uS to 100 nanosecond, and retest.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
to find the bandwidth, input 1 microsecond pulses with 99 microseconds at the opposite level. Then, if your ADC accurately quantizes the voltage difference, reduce the 1uS to 100 nanosecond, and retest.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
$endgroup$
$begingroup$
That simple...?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
Yes. The bandwidth does not depend on sampling rates and Nyquist rates.
$endgroup$
– Barry
7 hours ago
$begingroup$
It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency.
$endgroup$
– Caleb Reister
5 hours ago
add a comment |
$begingroup$
I think you are confused about what happens when the input frequency exceeds the Nyquist limit. The peak-to-peak voltage doesn't change...assuming you collect enough samples...but the apparent frequency of the signal changes. If there is a "bandwidth" beyond which the signal is attenuated, then it is due to the analog electronics of your ADC, not to the effect of sampling.
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
$endgroup$
$begingroup$
Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal?
$endgroup$
– Paul Uszak
7 hours ago
$begingroup$
Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC.
$endgroup$
– Elliot Alderson
7 hours ago
add a comment |
$begingroup$
The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic.
The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way tell what the actual input frequency was. Despite this, the ADC is still producing the same peak amplitude for a frequency way above its 'bandwidth'.
Note also that while the peak amplitudes are identical, neither of them actually reach the peak amplitude of the input signal. This may be a problem because the sampling points depend on the relative phase of the input and sampling frequencies. At certain frequencies and phases there will be 'nulls' of lower peak amplitude, going down to zero amplitude when the sampling rate exactly equals the input frequency. However, provided the input frequency and sampling rate are not locked in phase you will (eventually) see the true peak values.
When I find the -3dB point, what is the corresponding official
bandwidth I've found? Is it fin or fin/2 with consideration of the
Nyquist rate?
I'm not sure that an 'official' Nyquist related bandwidth exists, but if it did I imagine it would be based on the rms level of many samples with random phases, not the peak sample values. Usually the aliased output above half the sampling rate is considered to be anomalous, so the first dip to -3dB below that (if it existed) would be considered the bandwidth.
You may not detect any dip in peak values due to the sampling rate, but you should see any attenuation in the analog circuitry before or inside the ADC (antialiasing filter, sample-and-hold circuit etc.). Depending on the sample rate applied, that analog bandwidth may be much higher than the Nyquist frequency or sampling rate.
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
$endgroup$
$begingroup$
That simple...?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
Yes. The bandwidth does not depend on sampling rates and Nyquist rates.
$endgroup$
– Barry
7 hours ago
$begingroup$
It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency.
$endgroup$
– Caleb Reister
5 hours ago
add a comment |
$begingroup$
The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
$endgroup$
$begingroup$
That simple...?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
Yes. The bandwidth does not depend on sampling rates and Nyquist rates.
$endgroup$
– Barry
7 hours ago
$begingroup$
It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency.
$endgroup$
– Caleb Reister
5 hours ago
add a comment |
$begingroup$
The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
$endgroup$
The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
answered 8 hours ago
BarryBarry
10.7k1 gold badge17 silver badges17 bronze badges
10.7k1 gold badge17 silver badges17 bronze badges
$begingroup$
That simple...?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
Yes. The bandwidth does not depend on sampling rates and Nyquist rates.
$endgroup$
– Barry
7 hours ago
$begingroup$
It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency.
$endgroup$
– Caleb Reister
5 hours ago
add a comment |
$begingroup$
That simple...?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
Yes. The bandwidth does not depend on sampling rates and Nyquist rates.
$endgroup$
– Barry
7 hours ago
$begingroup$
It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency.
$endgroup$
– Caleb Reister
5 hours ago
$begingroup$
That simple...?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
That simple...?
$endgroup$
– Paul Uszak
8 hours ago
1
1
$begingroup$
Yes. The bandwidth does not depend on sampling rates and Nyquist rates.
$endgroup$
– Barry
7 hours ago
$begingroup$
Yes. The bandwidth does not depend on sampling rates and Nyquist rates.
$endgroup$
– Barry
7 hours ago
$begingroup$
It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency.
$endgroup$
– Caleb Reister
5 hours ago
$begingroup$
It is also worth noting that many ADC datasheets define this as analog bandwidth, and it is typically an order of magnitude higher than the sampling frequency.
$endgroup$
– Caleb Reister
5 hours ago
add a comment |
$begingroup$
I think you are confused about what happens when the input frequency exceeds the Nyquist limit. The peak-to-peak voltage doesn't change...assuming you collect enough samples...but the apparent frequency of the signal changes. If there is a "bandwidth" beyond which the signal is attenuated, then it is due to the analog electronics of your ADC, not to the effect of sampling.
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
$endgroup$
$begingroup$
Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal?
$endgroup$
– Paul Uszak
7 hours ago
$begingroup$
Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC.
$endgroup$
– Elliot Alderson
7 hours ago
add a comment |
$begingroup$
I think you are confused about what happens when the input frequency exceeds the Nyquist limit. The peak-to-peak voltage doesn't change...assuming you collect enough samples...but the apparent frequency of the signal changes. If there is a "bandwidth" beyond which the signal is attenuated, then it is due to the analog electronics of your ADC, not to the effect of sampling.
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
$endgroup$
$begingroup$
Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal?
$endgroup$
– Paul Uszak
7 hours ago
$begingroup$
Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC.
$endgroup$
– Elliot Alderson
7 hours ago
add a comment |
$begingroup$
I think you are confused about what happens when the input frequency exceeds the Nyquist limit. The peak-to-peak voltage doesn't change...assuming you collect enough samples...but the apparent frequency of the signal changes. If there is a "bandwidth" beyond which the signal is attenuated, then it is due to the analog electronics of your ADC, not to the effect of sampling.
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
$endgroup$
I think you are confused about what happens when the input frequency exceeds the Nyquist limit. The peak-to-peak voltage doesn't change...assuming you collect enough samples...but the apparent frequency of the signal changes. If there is a "bandwidth" beyond which the signal is attenuated, then it is due to the analog electronics of your ADC, not to the effect of sampling.
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
answered 8 hours ago
Elliot AldersonElliot Alderson
12.2k2 gold badges12 silver badges25 bronze badges
12.2k2 gold badges12 silver badges25 bronze badges
$begingroup$
Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal?
$endgroup$
– Paul Uszak
7 hours ago
$begingroup$
Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC.
$endgroup$
– Elliot Alderson
7 hours ago
add a comment |
$begingroup$
Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal?
$endgroup$
– Paul Uszak
7 hours ago
$begingroup$
Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
Thanks. Yes I understand that the frequency changes. That's my problem. Thus is it fin or fin/2?
$endgroup$
– Paul Uszak
8 hours ago
$begingroup$
I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
I don't understand why you think it would be fin/2. The Nyquist limit and the sampling rate do not determine or factor into the determination of bandwidth.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal?
$endgroup$
– Paul Uszak
7 hours ago
$begingroup$
I guess I'm confused... Hence the question. So to be crystal clear as glass, bandwidth is only related to signal level? It is not related at all to the ability to reproduce said signal?
$endgroup$
– Paul Uszak
7 hours ago
$begingroup$
Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC.
$endgroup$
– Elliot Alderson
7 hours ago
$begingroup$
Yes, of course it is related to the ability to reproduce the signal...the signal level is part of that ability. But the sampling frequency of an ideal ADC has nothing to do with the signal level of the sampled data. The sampling frequency is related to the apparent frequency of the sampled data, and this effect is independent of the bandwidth of the analog part of the ADC.
$endgroup$
– Elliot Alderson
7 hours ago
add a comment |
$begingroup$
The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic.
The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way tell what the actual input frequency was. Despite this, the ADC is still producing the same peak amplitude for a frequency way above its 'bandwidth'.
Note also that while the peak amplitudes are identical, neither of them actually reach the peak amplitude of the input signal. This may be a problem because the sampling points depend on the relative phase of the input and sampling frequencies. At certain frequencies and phases there will be 'nulls' of lower peak amplitude, going down to zero amplitude when the sampling rate exactly equals the input frequency. However, provided the input frequency and sampling rate are not locked in phase you will (eventually) see the true peak values.
When I find the -3dB point, what is the corresponding official
bandwidth I've found? Is it fin or fin/2 with consideration of the
Nyquist rate?
I'm not sure that an 'official' Nyquist related bandwidth exists, but if it did I imagine it would be based on the rms level of many samples with random phases, not the peak sample values. Usually the aliased output above half the sampling rate is considered to be anomalous, so the first dip to -3dB below that (if it existed) would be considered the bandwidth.
You may not detect any dip in peak values due to the sampling rate, but you should see any attenuation in the analog circuitry before or inside the ADC (antialiasing filter, sample-and-hold circuit etc.). Depending on the sample rate applied, that analog bandwidth may be much higher than the Nyquist frequency or sampling rate.
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
$endgroup$
add a comment |
$begingroup$
The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic.
The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way tell what the actual input frequency was. Despite this, the ADC is still producing the same peak amplitude for a frequency way above its 'bandwidth'.
Note also that while the peak amplitudes are identical, neither of them actually reach the peak amplitude of the input signal. This may be a problem because the sampling points depend on the relative phase of the input and sampling frequencies. At certain frequencies and phases there will be 'nulls' of lower peak amplitude, going down to zero amplitude when the sampling rate exactly equals the input frequency. However, provided the input frequency and sampling rate are not locked in phase you will (eventually) see the true peak values.
When I find the -3dB point, what is the corresponding official
bandwidth I've found? Is it fin or fin/2 with consideration of the
Nyquist rate?
I'm not sure that an 'official' Nyquist related bandwidth exists, but if it did I imagine it would be based on the rms level of many samples with random phases, not the peak sample values. Usually the aliased output above half the sampling rate is considered to be anomalous, so the first dip to -3dB below that (if it existed) would be considered the bandwidth.
You may not detect any dip in peak values due to the sampling rate, but you should see any attenuation in the analog circuitry before or inside the ADC (antialiasing filter, sample-and-hold circuit etc.). Depending on the sample rate applied, that analog bandwidth may be much higher than the Nyquist frequency or sampling rate.
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
$endgroup$
add a comment |
$begingroup$
The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic.
The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way tell what the actual input frequency was. Despite this, the ADC is still producing the same peak amplitude for a frequency way above its 'bandwidth'.
Note also that while the peak amplitudes are identical, neither of them actually reach the peak amplitude of the input signal. This may be a problem because the sampling points depend on the relative phase of the input and sampling frequencies. At certain frequencies and phases there will be 'nulls' of lower peak amplitude, going down to zero amplitude when the sampling rate exactly equals the input frequency. However, provided the input frequency and sampling rate are not locked in phase you will (eventually) see the true peak values.
When I find the -3dB point, what is the corresponding official
bandwidth I've found? Is it fin or fin/2 with consideration of the
Nyquist rate?
I'm not sure that an 'official' Nyquist related bandwidth exists, but if it did I imagine it would be based on the rms level of many samples with random phases, not the peak sample values. Usually the aliased output above half the sampling rate is considered to be anomalous, so the first dip to -3dB below that (if it existed) would be considered the bandwidth.
You may not detect any dip in peak values due to the sampling rate, but you should see any attenuation in the analog circuitry before or inside the ADC (antialiasing filter, sample-and-hold circuit etc.). Depending on the sample rate applied, that analog bandwidth may be much higher than the Nyquist frequency or sampling rate.
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
$endgroup$
The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic.
The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way tell what the actual input frequency was. Despite this, the ADC is still producing the same peak amplitude for a frequency way above its 'bandwidth'.
Note also that while the peak amplitudes are identical, neither of them actually reach the peak amplitude of the input signal. This may be a problem because the sampling points depend on the relative phase of the input and sampling frequencies. At certain frequencies and phases there will be 'nulls' of lower peak amplitude, going down to zero amplitude when the sampling rate exactly equals the input frequency. However, provided the input frequency and sampling rate are not locked in phase you will (eventually) see the true peak values.
When I find the -3dB point, what is the corresponding official
bandwidth I've found? Is it fin or fin/2 with consideration of the
Nyquist rate?
I'm not sure that an 'official' Nyquist related bandwidth exists, but if it did I imagine it would be based on the rms level of many samples with random phases, not the peak sample values. Usually the aliased output above half the sampling rate is considered to be anomalous, so the first dip to -3dB below that (if it existed) would be considered the bandwidth.
You may not detect any dip in peak values due to the sampling rate, but you should see any attenuation in the analog circuitry before or inside the ADC (antialiasing filter, sample-and-hold circuit etc.). Depending on the sample rate applied, that analog bandwidth may be much higher than the Nyquist frequency or sampling rate.
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
answered 4 hours ago
Bruce AbbottBruce Abbott
29k1 gold badge24 silver badges40 bronze badges
29k1 gold badge24 silver badges40 bronze badges
add a comment |
add a comment |
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2
$begingroup$
It's not really part of your question, but the RMS voltage (or just mean-squared voltage) is a much better indication of signal strength as you approach the sample rate. Even at $f_s / 4$, using $V_{pp}$ gets problematical.
$endgroup$
– TimWescott
8 hours ago
1
$begingroup$
@TimWescott That's interesting. Is it appropriate to ask here why Vpp gets problematical, or should I break it out into a new question?
$endgroup$
– Paul Uszak
8 hours ago
1
$begingroup$
It would be good as a new question. The answer is semi-long.
$endgroup$
– TimWescott
7 hours ago
$begingroup$
to find the bandwidth, input 1 microsecond pulses with 99 microseconds at the opposite level. Then, if your ADC accurately quantizes the voltage difference, reduce the 1uS to 100 nanosecond, and retest.
$endgroup$
– analogsystemsrf
2 hours ago