$2\pi$ periodicity of discrete-time Fourier transform

Question

In my signals and systems course, we have learned that the discrete-time Fourier transform is $2\pi$ periodic, but the continuous-time Fourier transform is not periodic in general. For reference, we are using the following definitions of each transform:

Continuous-Time:

$$ x(t) = \int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} \, d\omega \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \, dt $$

Discrete-Time:

$$ x[n] = \int_{\langle 2\pi \rangle} X(e^{j\omega}) e^{j\omega n} \, d\omega \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} $$

I'm searching for some intuition as for why the DTFT is periodic, but the CTFT is not. In class, my instructor presented the following argument: for a discrete-time signal,

$$ e^{j\omega n} = e^{j\omega(n + 2\pi)} $$

and thus any $x[n]$ can be expressed as a sum of individually $2\pi$ periodic functions. However, I don't see why that argument only applies to discrete-time signals - I feel as if it also works for continuous-time signals.

Any someone explain?

Answer 1

The argument does not work in continuous time. In discrete time the argument is that

$$e^{j\omega n}=e^{j(\omega+2\pi)n},\qquad n\in\mathbb{Z}\tag{1}$$

This is true because by definition $n$ is an integer. In continuous time we generally have

$$e^{j\omega t}\neq e^{j(\omega+2\pi)t},\qquad t\in\mathbb{R}\tag{2}$$

because $t$ is a real variable. Eq. $(2)$ only holds with equality for integer $t$, but not for all other values of $t$ (which would be necessary for $e^{j\omega t}$ to be $2\pi$-periodic).

But the fundamental reason why the DTFT is periodic is the fact that we deal with sequences, and if we assume that the sequence is the result of sampling a continuous-time signal, then it becomes obvious that different continuous-time signals can result in the same sampled sequence. E.g., take a continuous-time cosine

$$x(t)=\cos(2\pi f_0 t)\tag{3}$$

Now we sample $x(t)$ at instances $nT=n/f_s$, where $f_s$ is the sampling frequency:

$$x[n]=\cos(2\pi f_0 nT)=\cos(2\pi n f_0/f_s)\tag{4}$$

I leave it as an exercise to you to show that the frequencies $f_0=f_s-\Delta f$ and $f_0=f_s+\Delta f$ (with some $\Delta f>0$) result in the same sequence $x[n]$. This phenomenon is called aliasing. This means that frequencies above $f_s/2$ (also called the Nyquist frequency) cannot be distinguished from the corresponding frequencies below the Nyquist frequency. That's why the DTFT is periodic. There cannot be any new spectral information outside the frequency interval $[-f_s/2,f_s/2]$. If you haven't learned it yet, check out the sampling theorem, which will also help you understand the nature of spectra of discrete-time signals.

Also have a look at this answer to a related question.