Sample Frequency does not matter if it is sufficiently large

Question

I'm not able to convince a colleague on the following topic.

Statement: If, and only if, the sample frequency is sufficiently large and above the Nyquist frequency it does not matter what fs is as long as the signal length (in time domain) is constant.

Let's say we have a sinusoid with a frequency of 214 kHz. Let's say we take an analog measurement that is a 100 µs long. If we would sample at 10 MHz we would have 1000 samples ($100u*10M=1000$). If we would sample at 5 MHz we would have 500 samples. Regardless $ bin_{size} = \frac{fs}{n_{samples}} = \frac{1}{t_{length}} $.

In the example above the bin size would be 10 kHz, therefore one would find two strong magnitudes at freq 210 and 220, where 210 is slightly larger. With interpolation one can find the 214.

Now it seems counter intuitive that a higher sampling frequency does not improve things.

Is there any reason/benefit to use a higher sampling frequency as the bin size remains constant?

EDIT Thanks for the replies so far, let's add some more! Using a sinusoid and sweeping over it's frequency from 50 to 1000 kHz we can see that we can determine it's frequency quite accurately (I do a parabolic interpolation on the bin, it's quick sue me😂).

Sinusoid sampled at 2.5, 5 and 10 MHz (yellow, red, blue):

Zooming in we can indeed see a marginal improvement due to quantization (note, obviously this also plays in a role in my interpolation!)

Let's say, rather than a 'regular' sinusoid we have a damped sinusoid. We do the same thing and keep the decay-rate constant. Now something interesting happens that I currently do not understand. The accuracy decreases when the sample rate increases! It is now also strongly influenced by the phase, which makes sense as there's "less signal" to work with with phase 0 or $\pi$ vs $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ (due to damping).

Any thoughts on why a higher fs harms the frequency estimation?

Answer 1

As long as fs > 2x the highest frequency of the signal, the signal can in theory be perfectly reconstructed.

For a practical system with practical limitations, one might think of reasons why it would work better for more oversampling rather than less, but that would be tied to the practicalities of that system.

Now, there is a delay associated with the sampling rate. I imagine that some latency sensitive control applications might like to over sample just to reduce that delay? Or does the limitation in bandwidth somehow remove this advantage as well?

Answer 2

It's because of this trig identity

$$ \cos(\omega) = 1 - 2 \sin^2\left(\tfrac{\omega}{2} \right) $$

even if using floating point arithmetic, your cosine term gets so close to 1 at low frequencies relative to the sample rate that the information that differentiates it from 1, those bits just fall offa the edge.

Answer 3

In terms of frequency resolution, you are correct. Sampling rate does not affect the resolution in the frequency domain, rather signal duration does (as you noted). Time duration is to frequency resolution what sampling rate is to frequency duration (bandwidth).

As you're probably aware, one way to interpolate the spectrum is to zero pad it before transforming it, though it's important to note that zero padding does NOT fundamentally change the resolution (it only upsamples the spectrum, without adding any information).

I can't think of many benefits to significantly oversampling a signal, beyond a sufficient amount to account for filter rolloff. I suppose it might make the spectrum estimation problem more over-determined, but I think that would be a pretty marginal benefit in most cases. In addition, there is of course a significant cost to oversampling in terms of memory and processing speed for practical applications.

Edit:

@AlexTP made a good point in the comments that an additional benefit of oversampling is reduction in quantization error.