# Continuous-time vs Discrete-time Fourier transform

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?

• it is equal if you cross your i's and dot your t's and make sure your scaling is correct. perhaps someone has asked this question before and there is an answer. i feel like i have been here before. Jul 29 '19 at 21:30
• here is half of your answer. this shows what happens to your spectrum when you uniformly sample with a Dirac comb function (a.k.a. "impulse train"). you will see that the spectrum becomes periodic forever, just like going around the unit circle on the z-plane forever. Jul 29 '19 at 21:34

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.