# Where did we get the DC term of the Accumulator from DTFT?

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?

• Dividing by $1-e^{-j\omega}$ is not valid for $\omega=0$ (and integer multiples of $2\pi$). – Matt L. Apr 9 at 9:50
• You are right sir, but how did they fix this when they used the impulse train? @MattL. – Read my bio pls Apr 9 at 9:52
• 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. – Matt L. Apr 9 at 9:58
• ... and this one. – Matt L. Apr 9 at 9:59
• Alright will do, thank you very much sir! – Read my bio pls Apr 9 at 10:00

Note that the equation

$$x[n]=y[n]-y[n-1]\tag{1}$$

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

$$Y(e^{j\omega})=\frac{X(e^{j\omega})}{1-e^{-j\omega}}\tag{2}$$

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

$$y[n]=\sum_{m=-\infty}^nx[m]\tag{3}$$

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

$$Y(e^{j\omega})=X(e^{j\omega})U(e^{j\omega})\tag{4}$$

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

$$U(e^{j\omega})=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}}\tag{5}$$

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.

• Thank you very much sir now I have fully understood it. – Read my bio pls Apr 9 at 11:22
• "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. – TimWescott Apr 9 at 14:52