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What is the “Nyquist†rate for sampling the derivative of a signal?
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Background: I'm sampling the current through a capacitor. The signal of interest is the voltage across the capacitor. I will digitally integrate the current measurement to obtain the voltage.
Question: Given that the voltage across the capacitor is bandwidth limited, and I am sampling the derivative of this voltage, what is the minimum sample rate required to perfectly reconstruct the voltage signal from the current samples?
If there is no canned answer to this question, anything that could point me in the right direction would be helpful. Thank you in advance for any help!!
dsp sampling data-acquisition
asked Aug 28 at 22:03
Dweeberkitty
1216
 |Â
show 1 more comment
up vote
12
down vote
favorite
Background: I'm sampling the current through a capacitor. The signal of interest is the voltage across the capacitor. I will digitally integrate the current measurement to obtain the voltage.
Question: Given that the voltage across the capacitor is bandwidth limited, and I am sampling the derivative of this voltage, what is the minimum sample rate required to perfectly reconstruct the voltage signal from the current samples?
If there is no canned answer to this question, anything that could point me in the right direction would be helpful. Thank you in advance for any help!!
dsp sampling data-acquisition
asked Aug 28 at 22:03
Dweeberkitty
1216
1
You want to "perfectly reconstruct" the original signal from the samples? What do you mean by that?
– Elliot Alderson
Aug 28 at 22:10
1
The Nyquist rate is twice the highest frequency in the original signal.
– Peter Karlsen
Aug 29 at 8:25
@Dweerberkitty as Dave mentioned, signal is just a signal :). On a serious note, if you are using real-measurement systems, then there could be delays which will have impact on your derivative operation. So, if you account for them (with some luck, if the system is simple), you could analytically derive the necessary sampling period.
– Raaja
Aug 29 at 10:58
"The voltage across the capacitor is bandwidth limited". Why?
– Rodrigo de Azevedo
Aug 30 at 3:25
@RodrigodeAzevedo, this is just an assumption to simplify the problem statement. In reality, it's not bandwidth limited, but the frequency range of interest is well-defined in this problem. Thanks!
– Dweeberkitty
Aug 31 at 16:23
 |Â
show 1 more comment
up vote
12
down vote
favorite
up vote
12
down vote
favorite
Background: I'm sampling the current through a capacitor. The signal of interest is the voltage across the capacitor. I will digitally integrate the current measurement to obtain the voltage.
Question: Given that the voltage across the capacitor is bandwidth limited, and I am sampling the derivative of this voltage, what is the minimum sample rate required to perfectly reconstruct the voltage signal from the current samples?
If there is no canned answer to this question, anything that could point me in the right direction would be helpful. Thank you in advance for any help!!
dsp sampling data-acquisition
asked Aug 28 at 22:03
Dweeberkitty
1216
Background: I'm sampling the current through a capacitor. The signal of interest is the voltage across the capacitor. I will digitally integrate the current measurement to obtain the voltage.
Question: Given that the voltage across the capacitor is bandwidth limited, and I am sampling the derivative of this voltage, what is the minimum sample rate required to perfectly reconstruct the voltage signal from the current samples?
If there is no canned answer to this question, anything that could point me in the right direction would be helpful. Thank you in advance for any help!!
dsp sampling data-acquisition
asked Aug 28 at 22:03
Dweeberkitty
1216
asked Aug 28 at 22:03
Dweeberkitty
1216
asked Aug 28 at 22:03
Dweeberkitty
1216
asked Aug 28 at 22:03
Dweeberkitty
1216
1216
1
You want to "perfectly reconstruct" the original signal from the samples? What do you mean by that?
– Elliot Alderson
Aug 28 at 22:10
1
The Nyquist rate is twice the highest frequency in the original signal.
– Peter Karlsen
Aug 29 at 8:25
@Dweerberkitty as Dave mentioned, signal is just a signal :). On a serious note, if you are using real-measurement systems, then there could be delays which will have impact on your derivative operation. So, if you account for them (with some luck, if the system is simple), you could analytically derive the necessary sampling period.
– Raaja
Aug 29 at 10:58
"The voltage across the capacitor is bandwidth limited". Why?
– Rodrigo de Azevedo
Aug 30 at 3:25
@RodrigodeAzevedo, this is just an assumption to simplify the problem statement. In reality, it's not bandwidth limited, but the frequency range of interest is well-defined in this problem. Thanks!
– Dweeberkitty
Aug 31 at 16:23
 |Â
show 1 more comment
1
You want to "perfectly reconstruct" the original signal from the samples? What do you mean by that?
– Elliot Alderson
Aug 28 at 22:10
1
The Nyquist rate is twice the highest frequency in the original signal.
– Peter Karlsen
Aug 29 at 8:25
@Dweerberkitty as Dave mentioned, signal is just a signal :). On a serious note, if you are using real-measurement systems, then there could be delays which will have impact on your derivative operation. So, if you account for them (with some luck, if the system is simple), you could analytically derive the necessary sampling period.
– Raaja
Aug 29 at 10:58
"The voltage across the capacitor is bandwidth limited". Why?
– Rodrigo de Azevedo
Aug 30 at 3:25
@RodrigodeAzevedo, this is just an assumption to simplify the problem statement. In reality, it's not bandwidth limited, but the frequency range of interest is well-defined in this problem. Thanks!
– Dweeberkitty
Aug 31 at 16:23
1
1
You want to "perfectly reconstruct" the original signal from the samples? What do you mean by that?
– Elliot Alderson
Aug 28 at 22:10
You want to "perfectly reconstruct" the original signal from the samples? What do you mean by that?
– Elliot Alderson
Aug 28 at 22:10
1
1
The Nyquist rate is twice the highest frequency in the original signal.
– Peter Karlsen
Aug 29 at 8:25
The Nyquist rate is twice the highest frequency in the original signal.
– Peter Karlsen
Aug 29 at 8:25
@Dweerberkitty as Dave mentioned, signal is just a signal :). On a serious note, if you are using real-measurement systems, then there could be delays which will have impact on your derivative operation. So, if you account for them (with some luck, if the system is simple), you could analytically derive the necessary sampling period.
– Raaja
Aug 29 at 10:58
@Dweerberkitty as Dave mentioned, signal is just a signal :). On a serious note, if you are using real-measurement systems, then there could be delays which will have impact on your derivative operation. So, if you account for them (with some luck, if the system is simple), you could analytically derive the necessary sampling period.
– Raaja
Aug 29 at 10:58
"The voltage across the capacitor is bandwidth limited". Why?
– Rodrigo de Azevedo
Aug 30 at 3:25
"The voltage across the capacitor is bandwidth limited". Why?
– Rodrigo de Azevedo
Aug 30 at 3:25
@RodrigodeAzevedo, this is just an assumption to simplify the problem statement. In reality, it's not bandwidth limited, but the frequency range of interest is well-defined in this problem. Thanks!
– Dweeberkitty
Aug 31 at 16:23
@RodrigodeAzevedo, this is just an assumption to simplify the problem statement. In reality, it's not bandwidth limited, but the frequency range of interest is well-defined in this problem. Thanks!
– Dweeberkitty
Aug 31 at 16:23
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
18
down vote
accepted
Taking a derivative (or an integral) is a linear operation — it doesn't create any frequencies that weren't in the original signal (or remove any), it just changes their relative levels.
So the Nyquist rate for the derivative is the same as that for the original signal.
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
3
True in an ideal world in which there are perfectly bandlimited signals, ideal lowpass filters and no thermal noise at all.
– Rodrigo de Azevedo
Aug 29 at 5:10
The whole SNR balance changes. A small high-frequency component, which might alias, but not do much because of it's size, can become a sizable, sure-to-cause-big-low-frequency-components-on-sampling monster.
– Scott Seidman
Aug 30 at 12:33
add a comment |Â
up vote
-1
down vote
Taking the derivative multiplies the transform by s, which effectively rotates the magnitude graph counterclockwise. Thus, may well be higher frequency components in the derivative. A more succinct way to put this is that derivation amplifies the high frequency content.
The Laplace Transform $ frac1s+1 $ (which would be the step response of a single pole high-pass filter)
bode(tf(1, [ 1 1 ]))
The Laplace Transform of it's derivative, $ fracss+1 $
bode(tf([1 0], [ 1 1 ]))
The derivative in this case clearly has higher frequency components. Perhaps more correctly, it has much larger high frequency components than the non-derivative. One might choose to sample the first signal at 200 rads/s with some confidence, as the energy is very small at the nyquist rate, but aliasing would be substantial if you sampled the derivative at the same rate.
Thus, it depends on the nature of the signal. The derivative of a sinusoid will be a sinusoid of the same frequency, but the derivative of band limited noise will have higher frequency components than the noise.
EDIT: In response to the downvote, I'll hammer this home with a concrete example. Let me take a sine wave, and add some random normal noise to it (one tenth the magnitude of the sine wave)
The fft of this signal is:
Now, let me take the derivative of the signal:
and the fft of the derivative
Undersampling will, of course, alias either the signal or the derivative. The effects of the undersampling will be modest for the signal, and the result of undersampling the derivative will be absolutely useless.
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
2
I'm not sure what you think you're plotting here, but it isn't band-limited signals.
– Dave Tweed♦
Aug 29 at 13:40
The Fourier Transform of a signal, and the Fourier Transform of it's derivative.
– Scott Seidman
Aug 29 at 13:46
What language is that, anyway?
– Dave Tweed♦
Aug 29 at 13:47
1
Ah. In that case,tf()does not represent a signal, it represents a transfer function. Definitely not band-limited.
– Dave Tweed♦
Aug 29 at 13:49
1
You are still missing the point that the signal is bandlimited. You're adding non-bandlimited noise to the signal to make your point, which is outside the scope of the question. Yes, that's a practical consideration, but the question (as I see it) is theoretical.
– Dave Tweed♦
Aug 31 at 13:39
 |Â
show 8 more comments
up vote
-2
down vote
You can't.
Integration will only tell you about how the voltage changes during the time you're sampling.
The capacitor will always start with some charge present though, so there will be some initial voltage. Your calculation cannot know that voltage, so it cannot know the actual voltage across the capacitor during your measurement time. This should be familiar from maths classes - you always integrate between two points.
You also have a problem that although your current measurement samples are Nyquist-limited, the actual current through the capacitor may not be. Unless you can guarantee that the current through the capacitor has a hard low-pass filter somewhere below the Nyquist limit, you can never measure the current accurately enough to reproduce the voltage. I need to be clear that this is actually mathematically impossible, because it would require a sample rate of infinity.
But if you know the starting voltage and if the actual current through the capacitor is suitably low-pass-filtered, then DaveTweed is correct that the Nyquist limit for the integral is the same as for the sampled data.
answered Aug 29 at 12:28
Graham
2,347411
I don't see why you need to make a difference between the actual current through the capacitor and the band-limited measured value. What's so magic with this situation that the well known linearity of derivatives, filters, and integration no longer applies?
– pipe
Aug 29 at 18:23
@pipe In a word, sampling. Suppose we're sampling at 1kHz. Now suppose we have a 0.5ms long current spike. The sampled version will never see the spike, but the actual capacitor voltage certainly will. Then you have the residual errors between every form of digital integration and the actual value. And I haven't even started on anything related to resolution, which is yet another can of worms.
– Graham
Aug 29 at 18:44
But the energy in that pulse will spread out into bands that the sampler will see. For example: a pulse train with very short pulses will, after band-limiting, amount to a slightly elevated DC level. The area of your pulse will still be the same, and integrating the band-limited version ends up with the same result.
– pipe
Aug 29 at 19:00
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
accepted
Taking a derivative (or an integral) is a linear operation — it doesn't create any frequencies that weren't in the original signal (or remove any), it just changes their relative levels.
So the Nyquist rate for the derivative is the same as that for the original signal.
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
3
True in an ideal world in which there are perfectly bandlimited signals, ideal lowpass filters and no thermal noise at all.
– Rodrigo de Azevedo
Aug 29 at 5:10
The whole SNR balance changes. A small high-frequency component, which might alias, but not do much because of it's size, can become a sizable, sure-to-cause-big-low-frequency-components-on-sampling monster.
– Scott Seidman
Aug 30 at 12:33
add a comment |Â
up vote
18
down vote
accepted
Taking a derivative (or an integral) is a linear operation — it doesn't create any frequencies that weren't in the original signal (or remove any), it just changes their relative levels.
So the Nyquist rate for the derivative is the same as that for the original signal.
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
3
True in an ideal world in which there are perfectly bandlimited signals, ideal lowpass filters and no thermal noise at all.
– Rodrigo de Azevedo
Aug 29 at 5:10
The whole SNR balance changes. A small high-frequency component, which might alias, but not do much because of it's size, can become a sizable, sure-to-cause-big-low-frequency-components-on-sampling monster.
– Scott Seidman
Aug 30 at 12:33
add a comment |Â
up vote
18
down vote
accepted
up vote
18
down vote
accepted
Taking a derivative (or an integral) is a linear operation — it doesn't create any frequencies that weren't in the original signal (or remove any), it just changes their relative levels.
So the Nyquist rate for the derivative is the same as that for the original signal.
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
Taking a derivative (or an integral) is a linear operation — it doesn't create any frequencies that weren't in the original signal (or remove any), it just changes their relative levels.
So the Nyquist rate for the derivative is the same as that for the original signal.
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
answered Aug 28 at 22:09
Dave Tweed♦
107k9127227
107k9127227
3
True in an ideal world in which there are perfectly bandlimited signals, ideal lowpass filters and no thermal noise at all.
– Rodrigo de Azevedo
Aug 29 at 5:10
The whole SNR balance changes. A small high-frequency component, which might alias, but not do much because of it's size, can become a sizable, sure-to-cause-big-low-frequency-components-on-sampling monster.
– Scott Seidman
Aug 30 at 12:33
add a comment |Â
3
True in an ideal world in which there are perfectly bandlimited signals, ideal lowpass filters and no thermal noise at all.
– Rodrigo de Azevedo
Aug 29 at 5:10
The whole SNR balance changes. A small high-frequency component, which might alias, but not do much because of it's size, can become a sizable, sure-to-cause-big-low-frequency-components-on-sampling monster.
– Scott Seidman
Aug 30 at 12:33
3
3
True in an ideal world in which there are perfectly bandlimited signals, ideal lowpass filters and no thermal noise at all.
– Rodrigo de Azevedo
Aug 29 at 5:10
True in an ideal world in which there are perfectly bandlimited signals, ideal lowpass filters and no thermal noise at all.
– Rodrigo de Azevedo
Aug 29 at 5:10
The whole SNR balance changes. A small high-frequency component, which might alias, but not do much because of it's size, can become a sizable, sure-to-cause-big-low-frequency-components-on-sampling monster.
– Scott Seidman
Aug 30 at 12:33
The whole SNR balance changes. A small high-frequency component, which might alias, but not do much because of it's size, can become a sizable, sure-to-cause-big-low-frequency-components-on-sampling monster.
– Scott Seidman
Aug 30 at 12:33
add a comment |Â
up vote
-1
down vote
Taking the derivative multiplies the transform by s, which effectively rotates the magnitude graph counterclockwise. Thus, may well be higher frequency components in the derivative. A more succinct way to put this is that derivation amplifies the high frequency content.
The Laplace Transform $ frac1s+1 $ (which would be the step response of a single pole high-pass filter)
bode(tf(1, [ 1 1 ]))
The Laplace Transform of it's derivative, $ fracss+1 $
bode(tf([1 0], [ 1 1 ]))
The derivative in this case clearly has higher frequency components. Perhaps more correctly, it has much larger high frequency components than the non-derivative. One might choose to sample the first signal at 200 rads/s with some confidence, as the energy is very small at the nyquist rate, but aliasing would be substantial if you sampled the derivative at the same rate.
Thus, it depends on the nature of the signal. The derivative of a sinusoid will be a sinusoid of the same frequency, but the derivative of band limited noise will have higher frequency components than the noise.
EDIT: In response to the downvote, I'll hammer this home with a concrete example. Let me take a sine wave, and add some random normal noise to it (one tenth the magnitude of the sine wave)
The fft of this signal is:
Now, let me take the derivative of the signal:
and the fft of the derivative
Undersampling will, of course, alias either the signal or the derivative. The effects of the undersampling will be modest for the signal, and the result of undersampling the derivative will be absolutely useless.
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
2
I'm not sure what you think you're plotting here, but it isn't band-limited signals.
– Dave Tweed♦
Aug 29 at 13:40
The Fourier Transform of a signal, and the Fourier Transform of it's derivative.
– Scott Seidman
Aug 29 at 13:46
What language is that, anyway?
– Dave Tweed♦
Aug 29 at 13:47
1
Ah. In that case,tf()does not represent a signal, it represents a transfer function. Definitely not band-limited.
– Dave Tweed♦
Aug 29 at 13:49
1
You are still missing the point that the signal is bandlimited. You're adding non-bandlimited noise to the signal to make your point, which is outside the scope of the question. Yes, that's a practical consideration, but the question (as I see it) is theoretical.
– Dave Tweed♦
Aug 31 at 13:39
 |Â
show 8 more comments
up vote
-1
down vote
Taking the derivative multiplies the transform by s, which effectively rotates the magnitude graph counterclockwise. Thus, may well be higher frequency components in the derivative. A more succinct way to put this is that derivation amplifies the high frequency content.
The Laplace Transform $ frac1s+1 $ (which would be the step response of a single pole high-pass filter)
bode(tf(1, [ 1 1 ]))
The Laplace Transform of it's derivative, $ fracss+1 $
bode(tf([1 0], [ 1 1 ]))
The derivative in this case clearly has higher frequency components. Perhaps more correctly, it has much larger high frequency components than the non-derivative. One might choose to sample the first signal at 200 rads/s with some confidence, as the energy is very small at the nyquist rate, but aliasing would be substantial if you sampled the derivative at the same rate.
Thus, it depends on the nature of the signal. The derivative of a sinusoid will be a sinusoid of the same frequency, but the derivative of band limited noise will have higher frequency components than the noise.
EDIT: In response to the downvote, I'll hammer this home with a concrete example. Let me take a sine wave, and add some random normal noise to it (one tenth the magnitude of the sine wave)
The fft of this signal is:
Now, let me take the derivative of the signal:
and the fft of the derivative
Undersampling will, of course, alias either the signal or the derivative. The effects of the undersampling will be modest for the signal, and the result of undersampling the derivative will be absolutely useless.
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
2
I'm not sure what you think you're plotting here, but it isn't band-limited signals.
– Dave Tweed♦
Aug 29 at 13:40
The Fourier Transform of a signal, and the Fourier Transform of it's derivative.
– Scott Seidman
Aug 29 at 13:46
What language is that, anyway?
– Dave Tweed♦
Aug 29 at 13:47
1
Ah. In that case,tf()does not represent a signal, it represents a transfer function. Definitely not band-limited.
– Dave Tweed♦
Aug 29 at 13:49
1
You are still missing the point that the signal is bandlimited. You're adding non-bandlimited noise to the signal to make your point, which is outside the scope of the question. Yes, that's a practical consideration, but the question (as I see it) is theoretical.
– Dave Tweed♦
Aug 31 at 13:39
 |Â
show 8 more comments
up vote
-1
down vote
up vote
-1
down vote
Taking the derivative multiplies the transform by s, which effectively rotates the magnitude graph counterclockwise. Thus, may well be higher frequency components in the derivative. A more succinct way to put this is that derivation amplifies the high frequency content.
The Laplace Transform $ frac1s+1 $ (which would be the step response of a single pole high-pass filter)
bode(tf(1, [ 1 1 ]))
The Laplace Transform of it's derivative, $ fracss+1 $
bode(tf([1 0], [ 1 1 ]))
The derivative in this case clearly has higher frequency components. Perhaps more correctly, it has much larger high frequency components than the non-derivative. One might choose to sample the first signal at 200 rads/s with some confidence, as the energy is very small at the nyquist rate, but aliasing would be substantial if you sampled the derivative at the same rate.
Thus, it depends on the nature of the signal. The derivative of a sinusoid will be a sinusoid of the same frequency, but the derivative of band limited noise will have higher frequency components than the noise.
EDIT: In response to the downvote, I'll hammer this home with a concrete example. Let me take a sine wave, and add some random normal noise to it (one tenth the magnitude of the sine wave)
The fft of this signal is:
Now, let me take the derivative of the signal:
and the fft of the derivative
Undersampling will, of course, alias either the signal or the derivative. The effects of the undersampling will be modest for the signal, and the result of undersampling the derivative will be absolutely useless.
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
Taking the derivative multiplies the transform by s, which effectively rotates the magnitude graph counterclockwise. Thus, may well be higher frequency components in the derivative. A more succinct way to put this is that derivation amplifies the high frequency content.
The Laplace Transform $ frac1s+1 $ (which would be the step response of a single pole high-pass filter)
bode(tf(1, [ 1 1 ]))
The Laplace Transform of it's derivative, $ fracss+1 $
bode(tf([1 0], [ 1 1 ]))
The derivative in this case clearly has higher frequency components. Perhaps more correctly, it has much larger high frequency components than the non-derivative. One might choose to sample the first signal at 200 rads/s with some confidence, as the energy is very small at the nyquist rate, but aliasing would be substantial if you sampled the derivative at the same rate.
Thus, it depends on the nature of the signal. The derivative of a sinusoid will be a sinusoid of the same frequency, but the derivative of band limited noise will have higher frequency components than the noise.
EDIT: In response to the downvote, I'll hammer this home with a concrete example. Let me take a sine wave, and add some random normal noise to it (one tenth the magnitude of the sine wave)
The fft of this signal is:
Now, let me take the derivative of the signal:
and the fft of the derivative
Undersampling will, of course, alias either the signal or the derivative. The effects of the undersampling will be modest for the signal, and the result of undersampling the derivative will be absolutely useless.
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
edited Aug 31 at 13:30
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
answered Aug 29 at 13:35
Scott Seidman
21.5k43180
21.5k43180
2
I'm not sure what you think you're plotting here, but it isn't band-limited signals.
– Dave Tweed♦
Aug 29 at 13:40
The Fourier Transform of a signal, and the Fourier Transform of it's derivative.
– Scott Seidman
Aug 29 at 13:46
What language is that, anyway?
– Dave Tweed♦
Aug 29 at 13:47
1
Ah. In that case,tf()does not represent a signal, it represents a transfer function. Definitely not band-limited.
– Dave Tweed♦
Aug 29 at 13:49
1
You are still missing the point that the signal is bandlimited. You're adding non-bandlimited noise to the signal to make your point, which is outside the scope of the question. Yes, that's a practical consideration, but the question (as I see it) is theoretical.
– Dave Tweed♦
Aug 31 at 13:39
 |Â
show 8 more comments
2
I'm not sure what you think you're plotting here, but it isn't band-limited signals.
– Dave Tweed♦
Aug 29 at 13:40
The Fourier Transform of a signal, and the Fourier Transform of it's derivative.
– Scott Seidman
Aug 29 at 13:46
What language is that, anyway?
– Dave Tweed♦
Aug 29 at 13:47
1
Ah. In that case,tf()does not represent a signal, it represents a transfer function. Definitely not band-limited.
– Dave Tweed♦
Aug 29 at 13:49
1
You are still missing the point that the signal is bandlimited. You're adding non-bandlimited noise to the signal to make your point, which is outside the scope of the question. Yes, that's a practical consideration, but the question (as I see it) is theoretical.
– Dave Tweed♦
Aug 31 at 13:39
2
2
I'm not sure what you think you're plotting here, but it isn't band-limited signals.
– Dave Tweed♦
Aug 29 at 13:40
I'm not sure what you think you're plotting here, but it isn't band-limited signals.
– Dave Tweed♦
Aug 29 at 13:40
The Fourier Transform of a signal, and the Fourier Transform of it's derivative.
– Scott Seidman
Aug 29 at 13:46
The Fourier Transform of a signal, and the Fourier Transform of it's derivative.
– Scott Seidman
Aug 29 at 13:46
What language is that, anyway?
– Dave Tweed♦
Aug 29 at 13:47
What language is that, anyway?
– Dave Tweed♦
Aug 29 at 13:47
1
1
Ah. In that case,
tf() does not represent a signal, it represents a transfer function. Definitely not band-limited.– Dave Tweed♦
Aug 29 at 13:49
Ah. In that case,
tf() does not represent a signal, it represents a transfer function. Definitely not band-limited.– Dave Tweed♦
Aug 29 at 13:49
1
1
You are still missing the point that the signal is bandlimited. You're adding non-bandlimited noise to the signal to make your point, which is outside the scope of the question. Yes, that's a practical consideration, but the question (as I see it) is theoretical.
– Dave Tweed♦
Aug 31 at 13:39
You are still missing the point that the signal is bandlimited. You're adding non-bandlimited noise to the signal to make your point, which is outside the scope of the question. Yes, that's a practical consideration, but the question (as I see it) is theoretical.
– Dave Tweed♦
Aug 31 at 13:39
 |Â
show 8 more comments
up vote
-2
down vote
You can't.
Integration will only tell you about how the voltage changes during the time you're sampling.
The capacitor will always start with some charge present though, so there will be some initial voltage. Your calculation cannot know that voltage, so it cannot know the actual voltage across the capacitor during your measurement time. This should be familiar from maths classes - you always integrate between two points.
You also have a problem that although your current measurement samples are Nyquist-limited, the actual current through the capacitor may not be. Unless you can guarantee that the current through the capacitor has a hard low-pass filter somewhere below the Nyquist limit, you can never measure the current accurately enough to reproduce the voltage. I need to be clear that this is actually mathematically impossible, because it would require a sample rate of infinity.
But if you know the starting voltage and if the actual current through the capacitor is suitably low-pass-filtered, then DaveTweed is correct that the Nyquist limit for the integral is the same as for the sampled data.
answered Aug 29 at 12:28
Graham
2,347411
I don't see why you need to make a difference between the actual current through the capacitor and the band-limited measured value. What's so magic with this situation that the well known linearity of derivatives, filters, and integration no longer applies?
– pipe
Aug 29 at 18:23
@pipe In a word, sampling. Suppose we're sampling at 1kHz. Now suppose we have a 0.5ms long current spike. The sampled version will never see the spike, but the actual capacitor voltage certainly will. Then you have the residual errors between every form of digital integration and the actual value. And I haven't even started on anything related to resolution, which is yet another can of worms.
– Graham
Aug 29 at 18:44
But the energy in that pulse will spread out into bands that the sampler will see. For example: a pulse train with very short pulses will, after band-limiting, amount to a slightly elevated DC level. The area of your pulse will still be the same, and integrating the band-limited version ends up with the same result.
– pipe
Aug 29 at 19:00
add a comment |Â
up vote
-2
down vote
You can't.
Integration will only tell you about how the voltage changes during the time you're sampling.
The capacitor will always start with some charge present though, so there will be some initial voltage. Your calculation cannot know that voltage, so it cannot know the actual voltage across the capacitor during your measurement time. This should be familiar from maths classes - you always integrate between two points.
You also have a problem that although your current measurement samples are Nyquist-limited, the actual current through the capacitor may not be. Unless you can guarantee that the current through the capacitor has a hard low-pass filter somewhere below the Nyquist limit, you can never measure the current accurately enough to reproduce the voltage. I need to be clear that this is actually mathematically impossible, because it would require a sample rate of infinity.
But if you know the starting voltage and if the actual current through the capacitor is suitably low-pass-filtered, then DaveTweed is correct that the Nyquist limit for the integral is the same as for the sampled data.
answered Aug 29 at 12:28
Graham
2,347411
I don't see why you need to make a difference between the actual current through the capacitor and the band-limited measured value. What's so magic with this situation that the well known linearity of derivatives, filters, and integration no longer applies?
– pipe
Aug 29 at 18:23
@pipe In a word, sampling. Suppose we're sampling at 1kHz. Now suppose we have a 0.5ms long current spike. The sampled version will never see the spike, but the actual capacitor voltage certainly will. Then you have the residual errors between every form of digital integration and the actual value. And I haven't even started on anything related to resolution, which is yet another can of worms.
– Graham
Aug 29 at 18:44
But the energy in that pulse will spread out into bands that the sampler will see. For example: a pulse train with very short pulses will, after band-limiting, amount to a slightly elevated DC level. The area of your pulse will still be the same, and integrating the band-limited version ends up with the same result.
– pipe
Aug 29 at 19:00
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
You can't.
Integration will only tell you about how the voltage changes during the time you're sampling.
The capacitor will always start with some charge present though, so there will be some initial voltage. Your calculation cannot know that voltage, so it cannot know the actual voltage across the capacitor during your measurement time. This should be familiar from maths classes - you always integrate between two points.
You also have a problem that although your current measurement samples are Nyquist-limited, the actual current through the capacitor may not be. Unless you can guarantee that the current through the capacitor has a hard low-pass filter somewhere below the Nyquist limit, you can never measure the current accurately enough to reproduce the voltage. I need to be clear that this is actually mathematically impossible, because it would require a sample rate of infinity.
But if you know the starting voltage and if the actual current through the capacitor is suitably low-pass-filtered, then DaveTweed is correct that the Nyquist limit for the integral is the same as for the sampled data.
answered Aug 29 at 12:28
Graham
2,347411
You can't.
Integration will only tell you about how the voltage changes during the time you're sampling.
The capacitor will always start with some charge present though, so there will be some initial voltage. Your calculation cannot know that voltage, so it cannot know the actual voltage across the capacitor during your measurement time. This should be familiar from maths classes - you always integrate between two points.
You also have a problem that although your current measurement samples are Nyquist-limited, the actual current through the capacitor may not be. Unless you can guarantee that the current through the capacitor has a hard low-pass filter somewhere below the Nyquist limit, you can never measure the current accurately enough to reproduce the voltage. I need to be clear that this is actually mathematically impossible, because it would require a sample rate of infinity.
But if you know the starting voltage and if the actual current through the capacitor is suitably low-pass-filtered, then DaveTweed is correct that the Nyquist limit for the integral is the same as for the sampled data.
answered Aug 29 at 12:28
Graham
2,347411
answered Aug 29 at 12:28
Graham
2,347411
answered Aug 29 at 12:28
Graham
2,347411
answered Aug 29 at 12:28
Graham
2,347411
2,347411
I don't see why you need to make a difference between the actual current through the capacitor and the band-limited measured value. What's so magic with this situation that the well known linearity of derivatives, filters, and integration no longer applies?
– pipe
Aug 29 at 18:23
@pipe In a word, sampling. Suppose we're sampling at 1kHz. Now suppose we have a 0.5ms long current spike. The sampled version will never see the spike, but the actual capacitor voltage certainly will. Then you have the residual errors between every form of digital integration and the actual value. And I haven't even started on anything related to resolution, which is yet another can of worms.
– Graham
Aug 29 at 18:44
But the energy in that pulse will spread out into bands that the sampler will see. For example: a pulse train with very short pulses will, after band-limiting, amount to a slightly elevated DC level. The area of your pulse will still be the same, and integrating the band-limited version ends up with the same result.
– pipe
Aug 29 at 19:00
add a comment |Â
I don't see why you need to make a difference between the actual current through the capacitor and the band-limited measured value. What's so magic with this situation that the well known linearity of derivatives, filters, and integration no longer applies?
– pipe
Aug 29 at 18:23
@pipe In a word, sampling. Suppose we're sampling at 1kHz. Now suppose we have a 0.5ms long current spike. The sampled version will never see the spike, but the actual capacitor voltage certainly will. Then you have the residual errors between every form of digital integration and the actual value. And I haven't even started on anything related to resolution, which is yet another can of worms.
– Graham
Aug 29 at 18:44
But the energy in that pulse will spread out into bands that the sampler will see. For example: a pulse train with very short pulses will, after band-limiting, amount to a slightly elevated DC level. The area of your pulse will still be the same, and integrating the band-limited version ends up with the same result.
– pipe
Aug 29 at 19:00
I don't see why you need to make a difference between the actual current through the capacitor and the band-limited measured value. What's so magic with this situation that the well known linearity of derivatives, filters, and integration no longer applies?
– pipe
Aug 29 at 18:23
I don't see why you need to make a difference between the actual current through the capacitor and the band-limited measured value. What's so magic with this situation that the well known linearity of derivatives, filters, and integration no longer applies?
– pipe
Aug 29 at 18:23
@pipe In a word, sampling. Suppose we're sampling at 1kHz. Now suppose we have a 0.5ms long current spike. The sampled version will never see the spike, but the actual capacitor voltage certainly will. Then you have the residual errors between every form of digital integration and the actual value. And I haven't even started on anything related to resolution, which is yet another can of worms.
– Graham
Aug 29 at 18:44
@pipe In a word, sampling. Suppose we're sampling at 1kHz. Now suppose we have a 0.5ms long current spike. The sampled version will never see the spike, but the actual capacitor voltage certainly will. Then you have the residual errors between every form of digital integration and the actual value. And I haven't even started on anything related to resolution, which is yet another can of worms.
– Graham
Aug 29 at 18:44
But the energy in that pulse will spread out into bands that the sampler will see. For example: a pulse train with very short pulses will, after band-limiting, amount to a slightly elevated DC level. The area of your pulse will still be the same, and integrating the band-limited version ends up with the same result.
– pipe
Aug 29 at 19:00
But the energy in that pulse will spread out into bands that the sampler will see. For example: a pulse train with very short pulses will, after band-limiting, amount to a slightly elevated DC level. The area of your pulse will still be the same, and integrating the band-limited version ends up with the same result.
– pipe
Aug 29 at 19:00
add a comment |Â
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1
You want to "perfectly reconstruct" the original signal from the samples? What do you mean by that?
– Elliot Alderson
Aug 28 at 22:10
1
The Nyquist rate is twice the highest frequency in the original signal.
– Peter Karlsen
Aug 29 at 8:25
@Dweerberkitty as Dave mentioned, signal is just a signal :). On a serious note, if you are using real-measurement systems, then there could be delays which will have impact on your derivative operation. So, if you account for them (with some luck, if the system is simple), you could analytically derive the necessary sampling period.
– Raaja
Aug 29 at 10:58
"The voltage across the capacitor is bandwidth limited". Why?
– Rodrigo de Azevedo
Aug 30 at 3:25
@RodrigodeAzevedo, this is just an assumption to simplify the problem statement. In reality, it's not bandwidth limited, but the frequency range of interest is well-defined in this problem. Thanks!
– Dweeberkitty
Aug 31 at 16:23