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  1. Can we sample the Dirac function?

$$ x(t) = \delta(t) $$

  1. Can we sample a DC signal?

$$ x(t) = 1$$

I think that we can't sample $x(t) = 1$ because the Fourier Transform of $x(t) = 1$ is $2\pi \delta(\omega)$ (the Dirac function)

and we can't sample the Dirac function because it extends to $\infty$ and we cant find $f_\mathrm{max}$.

I find this solution but I am not sure. Can anyone help?

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  • $\begingroup$ We can sample a portion of the DC signal over a finite time duration; or if you allow for the mythical option of sampling for infinite time then we can sample the entire DC signal. Also see this: dsp.stackexchange.com/questions/46833/… $\endgroup$ – Dan Boschen Sep 28 '20 at 2:44
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Can we sample the Dirac function?

Strictly speaking: "sampling" would be taking the instantaneous value. Since the Dirac Delta doesn't have a value at $t=0$ (it is not really a function!), NO.

Realistically speaking: an ADC can't measure instantaneous values. That's impossible, because it would require infinite bandwidth of the conversion system (not only of the digital signal), and physical reality doesn't allow for that. So, what a device with limited bandwidth does is measure the (weighted) integral over some (short) time interval. And then, the Dirac isn't any problematic: unlike a value at every point, we know its integral over every interval.

However, when we take physical reality of ADCs into account: there can't be a Dirac Delta in real world, that would require infinite bandwidth... we hence can't measure something that doesn't exist, so yet another: no.

and we can't sample the Dirac function because it extends to ∞ and we cant find fmax.

aaaah! that's not the reason. The signal not being observable is not the same as not being truthfully reconstructable from the digital representation.

Can we sample a DC signal?

yes of course.

I think that we can't sample x(t)=1 because the Fourier Transform of x(t)=1 is 2πδ(ω) (the Dirac function)

No problem with that. Sampling happens in time domain.

It's also not a problem for reconstruction. Nyquist demands limited bandwidth, and your signal very much has limited bandwidth (it has zero bandwidth, to be exact).

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