Continuous-time vs Discrete-time Fourier transform

Question

case 1) to calculate the Fourier transform of discrete-time signal(sampled signal) we use Discrete-time Fourier transform.

but my question is:

case 2) if I consider that discrete-time signal as continuous-time signal which is multiplied by Dirac comb(impulse train) then we produced a sampled signal but in continuous mode now we apply continuous Fourier transform to this signal.

notice that in both case we have the same sampled signal(the only difference: in first case signal is in discrete form and in second signal is in continuous form)

the result of "case 1" is equal to result of "case 2"? if not,what does cause it?

Answer 1

This is pretty straightforward to show. Let $x[n]$ be a sampled version of a continuous-time signal $x_c(t)$:

$$x[n]=x_c(nT)\tag{1}$$

The DTFT of $x[n]$ is defined by

$$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{2}$$

The CTFT of the signal

$$x_c(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT)=\sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT)\tag{3}$$

is given by

$$\begin{align}\mathscr{F}\left\{\sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT)\right\}&=\sum_{n=-\infty}^{\infty}x_c(nT)\mathscr{F}\left\{\delta(t-nT)\right\}\\&=\sum_{n=-\infty}^{\infty}x_c(nT)e^{-jn\Omega T}\tag{4}\end{align}$$

which equals $(2)$ with $x[n]=x_c(nT)$ and $\omega=\Omega T$ being the normalized angular frequency.

And now I found this question, which is pretty similar but also asks about some background.