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In the below screenshot I was unable to understand how the equation 4.3 is derived from the above equation.. How $x(nT)$ has become $x(n)$ ? Please clear my doubt and suggest some link to understand DFT. I am sorry for the poor image quality.

enter image description here

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It looks like the author implicitly makes the substitution $x(nT) \rightarrow x(n)$, but even if they were making this explicit, there are more clear ways to get to Eq. 4.3.

One way to do this is to denote the continuous (analog) and discrete time-domain signals as $x_a$ and $x$, respectively. Your text, using this convention, would then look like this (starting from Eq. 4.2):

$$ x_a(t) \xrightarrow{sampling} x_a(nT) $$

and

$$ X(e^{j\omega})\ =\ \sum_{n = -\infty}^{\infty} x_a(nT) e^{-j\Omega nT}. $$

(Note the right-hand-side exponent: it is $-j\Omega nT$, not $-j\Omega nt$ as it appears in your text; this looks like an error to me.) Then if you take into account the following equality (assuming uniform sampling):

$$ x(n)\ =\ x_a(nT), \quad -\infty < n < \infty, $$

you end up with:

$$ X(e^{j\omega})\ =\ \sum_{n = -\infty}^{\infty} x(n) e^{-j(\Omega T)n}, $$

or (noting that $\omega = \Omega T$):

$$ X(e^{j\omega})\ =\ \sum_{n = -\infty}^{\infty} x(n) e^{-j\omega n}. $$

By the way, this is a derivation from CFT to DTFT operation, not CFT to DFT one (as you say in the title of your question).

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That is most likely a typo. In my opinion it should also read x(nT) in Eq. 4.3

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  • $\begingroup$ I wonder if the author / editor forgot to delete one of the options. If you make the substitution $x(n) \rightarrow x(nT)$ in equation (4.3), then the unnumbered equation above it is precisely the same (apart from the grouping of $\Omega t$). $\endgroup$ – Peter K. Feb 10 '14 at 19:52
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It's just a definition of a sequence of numbers representing $x(t)$: $x[0]=x(0), x[1]=x(T), x[2]=x(2T), ... , x[n]=x(nT)$

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  • $\begingroup$ well, they are not using square brackets in Eq 4.3. IMO Eq. 4,3 is formally incorrect. Clearly, Eq. 4.2 and 4.3 refer to the same "x", which is a function of continuous time. I agree, if there would be square brackets situation is different. $\endgroup$ – Andreas H. Feb 10 '14 at 18:09

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