Periodicity of the discrete-time Fourier Transform

Question

The DTFT of a sequence $x[n]$ can be written as $$X(e^{j\omega}) = \sum_{n = -\infty}^{\infty} x[n] e^{-j\omega n}.$$ Is the smallest (fundamental) period in frequency of the DTFT always $2\pi$? Or can it be smaller than $2\pi$?

In addition, I was wondering why we cannot use that same notation for the argument with the CTFT. We typically denote the continuous-time Fourier Transform by $X(j\omega)$. But what's wrong with using $X(e^{j\omega})$ where $$X(e^{j\omega}) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt?$$

I know for a fact that this is wrong because the notation inherently implies the periodicity of $X(e^{j\omega})$, since $X(e^{j(\omega + 2\pi)}) = X(e^{j\omega})$, which is obbiously not true in this case. But I don't understand what mathematically does not permit us to use $e^{j\omega}$ as an argument in the continuous-time case.

Answer 1

The DTFT is always $2\pi$-periodic. However, it can also have a smaller period, namely a fraction of $2\pi$. Take any sequence $x[n]$ for which the DTFT exists and insert $L-1$ zeros between the samples. The DTFT of the new sequency $\hat{x}[n]$ can then be written as

$$\hat{X}(e^{j\omega})=\sum_{n=-\infty}^{\infty}\hat{x}[n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}\hat{x}[nL]e^{-jnL\omega}\tag{1}$$

$\hat{X}(e^{j\omega})$ as given by $(1)$ clearly has a period of $2\pi/L$.

Concerning your second question, the reason why a continuous-time Fourier transform $X(j\omega)$ cannot be written as $X(e^{j\omega})$ is that

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\neq \int_{-\infty}^{\infty}x(t)(e^{j\omega})^{-t}dt\tag{2}$$

simply because generally

$$e^{-j\omega t}\neq (e^{j\omega})^{-t}\tag{3}$$

unless $t$ is an integer (as is the case with the DTFT).

Note that if $(3)$ were not true, i.e., if $e^{-j\omega t}= (e^{j\omega})^{-t}$ were true, we could easily show that $e^{-j\omega t}=1$ for all $t$. Just write $\omega=2\pi/T$ and you get

$$e^{-j\omega t}=e^{-j2\pi t/T}=\left(e^{-j2\pi}\right)^{t/T}=1^{t/T}=1$$

which is of course absurd.

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Proof of Eq. $(3)$:

Let $z$ be a complex number with $|z|=1$ (the magnitude is irrelevant here):

$$z=e^{j\theta}\tag{4}$$

Let $a$ and $b$ be real numbers. Then

$$z^{ab}=e^{j\theta a b}\tag{5}$$

And with $z^a=u$

$$\left(z^a\right)^b=u^b=e^{j\arg\{u\}b}\tag{6}$$

With $\arg\{u\}=\text{pv}\{\theta a\}\in (-\pi,\pi]$, where $\text{pv}$ denotes the principal value, it is clear that generally

$$z^{ab}=e^{j\theta a b}\neq \left(z^a\right)^b=e^{j\,\text{pv}\{\theta a\}b}\tag{7}$$

There are two cases where $(5)$ and $(6)$ are equal:

1. if $\text{pv}\{\theta a\}=\theta a$, which is the case if $\theta a\in (-\pi,\pi]$.

2. if $b$ is integer, since

$$\text{pv}\{\theta a\}=\theta a+2\pi k\tag{8}$$

with some appropriately chosen integer $k$ (such that the result is in the interval $(-\pi,\pi]$), then

$$\left(z^a\right)^b=e^{j\,\text{pv}\{\theta a\}b}=e^{j(\theta a+2\pi k)b}=e^{j\theta ab}=z^{ab},\qquad b\in\mathbb{Z}\tag{9}$$

[Note that $(9)$ would also hold for rational $b$ as long as $kb$ is integer.]

Answer 2

You could define a DTFT that has a fundamental period that is less than $2\pi$, and try to find its IDTFT. We say that it's periodic in $2\pi$ because that's true for all DTFTs, but there might be cases where the period is smaller (even though they might not have an easy-to-find inverse). A trivial case would be the transform of the Dirac delta, whose DTFT is a constant and so has any period you want.

Regarding the notation, it's just a way to distinguish between continuous and discrete signals when working on the frequency domain. When you have ADCs and DACs, things may became a bit confusing if we use the same notation for all the transforms.

Note that the only thing that matters is that they are functions of $\omega$, so one could just write $X(\omega)$ in any case and it would be valid.

The fact that we use $j\omega$ in the continuous case comes from the fact that the Fourier transform is (in general) the Laplace transform evaluated when $s=j\omega$. So if we have some Laplace transform $X_L(s)$, we can evaluate that in the imaginary axis to get the Fourier transform, and thus the convention.

In the discrete case, something similar happens. We usually link the z-transform and the DTFT evaluating the former at $z=e^{j\omega}$, and the same as in the continuous case happens.

Answer 3

to answer your last question, it turns out (this is ultra-fundamental) that an eigenfunction for a Linear Time-Invariant (LTI) System is the exponential input. this is true whether it's Continuous time:

$$ x(t) = e^{st} $$

or if it's Discrete time

$$ x[n] = e^{\log(z) n} = z^n $$

This means if the eigenfunction goes into the LTI, the same function comes out but scaled by a constant. I think they call that constant an "eigenvalue". Turns out that the output of the two cases above are:

$$ y(t) = H(s) \, x(t) $$

and

$$ y[n] = H(z) \, x[n] $$

So, specifically for the complex sinusoid:

$$ x(t) = e^{j \Omega t} $$

or, if it's Discrete time

$$ x[n] = e^{j \omega n} $$

then what comes out is:

$$ y(t) = H(j \Omega) \, x(t) $$

or

$$ y[n] = H(e^{j \omega}) \, x[n] $$