I have started learning DSP on my own and I have this doubt. I have done some googling but haven't found an answer. I hope that someone here would give the answer. It will be of great help.


1 Answer 1


Note that the sequence


is in $\ell^2(\mathbb Z)$ because


but it is not in $\ell^1(\mathbb Z)$ since


We know that for sequences in $\ell^1(\mathbb{Z})$ the DTFT $X(e^{j\omega})$ exists and it is a continuous function of $\omega$. However, it is not necessary for a sequence to be in $\ell^1(\mathbb{Z})$ for the DTFT to exist. We can extend the definition of the DTFT to sequences in $\ell^2(\mathbb{Z})$, which implies that the DTFT converges in a mean square sense (and not uniformly), and that $X(e^{j\omega})$ may have discontinuities.

In that extended sense, the DTFT of $(1)$ exists, just as it exists for the ideal discrete-time differentiator or the discrete-time Hilbert transformer, the impulse responses of which also only decay as $1/n$.

The DTFT of $x[n]$ is


The series $(4)$ is the complex Mercator series

$$\sum_{n=1}^{\infty}\frac{z^{n}}{n}=-\log(1-z),\qquad |z|\le 1,\quad z\neq 1\tag{5}$$

with $z=e^{-j\omega}$. It is the Taylor series for $-\log(1-z)$ and it converges for all $|z|\le 1$, $z\neq 1$. From $(5)$ we can immediately write down the $\mathcal{Z}$-transform of $x[n]$:

$$X(z)=\sum_{n=1}^{\infty}\frac{z^{-n}}{n}=-\log(1-z^{-1}),\qquad |z|\ge 1,\quad z\neq 1\tag{6}$$

Since $(6)$ also converges for $|z|=1$ except for $z=1$, we can obtain the DTFT directly from $(6)$ by setting $z=e^{j\omega}$. For $e^{j\omega}=1$, i.e., $\omega=0$, the DTFT has a singularity.


Equation $(7)$ agrees with WolframAlpha's result. Also take a look at this related answer.

Let's visually check if it is plausible that the series $(4)$ really converges to $(7)$. I've computed the sum over the first $100$ terms of $(4)$, and I've plotted the real and imaginary parts of that partial sum together with the real and imaginary parts of $(7)$: enter image description here

From the figure, it appears that the series $(4)$ indeed converges to $(7)$ in a mean square sense. This is of course no proof but a simple sanity check.

  • $\begingroup$ WolframAlpha thinks!?!?! We've reached the singularity, then! :D $\endgroup$
    – Peter K.
    Jan 3, 2018 at 16:43
  • 1
    $\begingroup$ @PeterK.: yeah, it's when those funny dots appear, then it's thinking hard ... :) $\endgroup$
    – Matt L.
    Jan 3, 2018 at 16:47
  • 2
    $\begingroup$ The question is about the existance of the FT. It does not converge since the region of convergence of the Z transform excludes the unit circle $\endgroup$
    – Juancho
    Jan 4, 2018 at 1:35
  • 1
    $\begingroup$ @Juancho: Thanks, I misread and worked out the Z-transform. However, note that the FT does exist as I'm going to show as soon as I have more time. $\endgroup$
    – Matt L.
    Jan 4, 2018 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.