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.


3 Answers 3


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


$\hat{X}(e^{j\omega})$ 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.

Proof of Eq. $(3)$:

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


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

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

And with $z^a=u$


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.]

  • $\begingroup$ Why is $e^{-j\omega t} = (e^{j\omega})^{-t}$ invalid? Is it because we are not allowed to raise a complex number to irrational and/or transcendental powers? $\endgroup$
    – user33568
    Commented Feb 20, 2018 at 16:38
  • $\begingroup$ @0MW: You can raise a complex number to any other complex number, but the two complex numbers under consideration are simply not equal. $(e^{j\omega})^{-t}$ is $2\pi$-periodic in $\omega$ and $e^{-j\omega t}$ isn't. With $e^{j\omega}$ you lose information because it gives the same value for $\omega+2\pi k$, which isn't the case for the other function. $\endgroup$
    – Matt L.
    Commented Feb 20, 2018 at 17:24
  • $\begingroup$ I understand that the DTFT can have a period less than $2\pi$, but I was recently reading up on the duality between the DTFT and the continuous-time fourier series: namely, the DTFT can be thought of as a continuous-time fourier series in $\omega$ with period $2\pi$. This last bit is important, since the factor of $2\pi$ appears in the synthesis equation. Does this duality break down if the DTFT is periodic with a period less than $2\pi$? $\endgroup$
    – user33568
    Commented Feb 23, 2018 at 7:44
  • $\begingroup$ @0MW: The analogy always holds; if the period is smaller, then you have zeros in between the coefficients in both cases. $\endgroup$
    – Matt L.
    Commented Feb 23, 2018 at 9:40

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.

  • $\begingroup$ The question is why a CTFT cannot be written as $X(e^{j\omega})$, and it looks like you discussed the difference between the notations $X(\omega)$ and $X(j\omega)$. $\endgroup$
    – Matt L.
    Commented Feb 20, 2018 at 8:45
  • $\begingroup$ @MattL. You are right, I read the question shallowly and misinterpreted it. I'll leave it here though, as I think it can help someone in the future, maybe. +1 to your answer. $\endgroup$
    – Tendero
    Commented Feb 20, 2018 at 12:25

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:

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


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

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:

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


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

  • $\begingroup$ This is all true and important, but I don't see how it answers the OP's second question. $\endgroup$
    – Matt L.
    Commented Feb 20, 2018 at 8:43
  • $\begingroup$ it answers why we use $X(s)\bigg|_{s=j\omega}$ in the continuous-time case and $X(z)\bigg|_{z=e^{j\omega}}$ in the discrete-time case. $\endgroup$ Commented Feb 20, 2018 at 17:51
  • $\begingroup$ That's right, but the question was - as far as I understood - why we don't (and cannot) use $X(e^{j\omega})$ in the continuous domain. $\endgroup$
    – Matt L.
    Commented Feb 20, 2018 at 17:54
  • $\begingroup$ because we have to use $X(s)\bigg|_{s=j\omega}$ instead. we can't use both. $\endgroup$ Commented Feb 21, 2018 at 1:59
  • $\begingroup$ Yes, but I think that the most important reason is that, unlike the DTFT, the CTFT is no function of $e^{j\omega}$. $\endgroup$
    – Matt L.
    Commented Feb 21, 2018 at 6:33

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