What is the "Nyquist" rate for sampling the derivative of a signal?

Question

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!!

Answer 1

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.

Answer 2

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 \$ \frac{1}{s+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, \$ \frac{s}{s+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.

Answer 3

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.