Define $y[n]:=\displaystyle\sum_{m=-\infty}^{n}x[m]$. The DTFT is found as follows: \begin{align*} y[n]&=\sum_{m=-\infty}^{n}x[m]\\ \\ &=\sum_{m=-\infty}^{n-1}x[m]+x[n]\\ \\ &=y[n-1]+x[n]\\ \\ \implies x[n]&=y[n]-y[n-1] \end{align*} Therefore, applying DTFT : \begin{align*} \mathcal{F}\{x\}[n]=X(e^{j\omega})&=\mathcal{F}\{y[n]-y[n-1]\}\\ \\ &=(1-e^{-j\omega})Y(e^{j\omega}) \end{align*} Hence : $$ Y(e^{j\omega})=\frac{1}{1-e^{-j\omega}}X(e^{j\omega}) $$ However, the correct answer is : $$ \frac{1}{1-e^{-j\omega}}X(e^{j\omega})+\color{red}{\pi X(e^{j0})\sum_{k=-\infty}^{\infty}\delta(\omega-2k\pi)} $$ My question is where did the DC term come from and why didn't it work in my derivation?

  • 1
    $\begingroup$ Dividing by $1-e^{-j\omega}$ is not valid for $\omega=0$ (and integer multiples of $2\pi$). $\endgroup$
    – Matt L.
    Commented Apr 9, 2021 at 9:50
  • $\begingroup$ You are right sir, but how did they fix this when they used the impulse train? @MattL. $\endgroup$
    – SPARSE
    Commented Apr 9, 2021 at 9:52
  • $\begingroup$ I might write up an answer when I have the time to do so, but in the meantime take a look at the answers to this question. $\endgroup$
    – Matt L.
    Commented Apr 9, 2021 at 9:58
  • $\begingroup$ ... and this one. $\endgroup$
    – Matt L.
    Commented Apr 9, 2021 at 9:59
  • $\begingroup$ Alright will do, thank you very much sir! $\endgroup$
    – SPARSE
    Commented Apr 9, 2021 at 10:00

1 Answer 1


Note that the equation


doesn't uniquely determine the sequence $y[n]$. If some $y[n]$ satisfies $(1)$, so does $y[n]+c$ with some constant $c$. Consequently, Eq. $(1)$ determines $y[n]$ only up to a constant, which corresponds to a DC term in the frequency domain.

Hence, the equation


is only valid for $\omega\neq 0$, or, more generally, for $\omega\neq 2\pi k$, $k\in\mathbb{Z}$.

The most straightforward way to derive the DTFT of


is to realize that $y[n]$ is the convolution of $x[n]$ with the unit step sequence $u[n]$, and, consequently, the DTFT of $y[n]$ is given by


where $U(e^{j\omega})$ is the DTFT of the unit step $u[n]$:


Several ways to derive the result $(5)$ are discussed in the answers to this and this question.

Note that in $(5)$ I use $\delta(\omega)$ instead of $\sum_k\delta(\omega-2\pi k)$ to avoid cluttered notation. The fact that the DTFT is always $2\pi$-periodic is understood.

  • $\begingroup$ Thank you very much sir now I have fully understood it. $\endgroup$
    – SPARSE
    Commented Apr 9, 2021 at 11:22
  • 1
    $\begingroup$ "The fact that the DTFT is always $2\pi$-periodic is understood". I have also encountered more mathematically rigorous folks who insist (to the point of engaging in flame wars) that the DTFT is only defined on some interval $\omega \in [\omega_0, \omega_0 + 2 \pi)$ or $\omega \in (\omega_0, \omega_0 + 2 \pi]$. Either definition works equally well for casual engineering use. $\endgroup$
    – TimWescott
    Commented Apr 9, 2021 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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