This is what I know about the basic sampling process enter image description here

I am told that fbar(t) (i.e the sampled signal) is defined for all t. I don't understand that. If n is a finite integer (..,-3, -2, -1, 0, 1...), then clearly it won't be defined n=1.5. So how is it defined for all t? Also I thought that sampled signals are always discrete and I understand that continious signals are defined for all t, it makes no sense that this is continious.

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    $\begingroup$ $\bar{f}(t)$ is not a discrete-time signal; it's still continuous. However, it is equal to zero for all $t \neq nT$. $\endgroup$ – MBaz Apr 18 '18 at 17:37
  • $\begingroup$ I thought the same thing but say if n=1.5 and since fbar_t is equal to a summation that relies on n=1,2,3, then shouldn't it be undefined for n=1.5? I just don't see how you can use a summation sigma symbol that relies on discrete n values to repersent something continuous. $\endgroup$ – AlfroJang80 Apr 18 '18 at 17:42
  • $\begingroup$ @MBaz, Slight correction, it is not continuous although it is defined on the entire domain. $\endgroup$ – Cedron Dawg Apr 18 '18 at 18:21
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    $\begingroup$ The delta function is zero when the argument is not zero, not undefined. $\endgroup$ – Cedron Dawg Apr 18 '18 at 18:24
  • $\begingroup$ @MBaz, careful with terminology. $\bar{f}(t)$ is a continuous-time signal, that is it is a function of continuous $t$, but, unless $f(nT)=0 \quad \forall n\in\mathbb{Z}$, then $\bar{f}(t)$ is not continuous. $\endgroup$ – robert bristow-johnson Apr 18 '18 at 18:36

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