If i apply the DTFT on unit step function, then i get follow: $$DTFT\{u[n]\}=\sum_{n=-\infty}^{\infty}u[n]e^{-j\omega n}=\sum_{n=0}^{\infty}e^{-j\omega n} = \frac{1}{1-e^{-j\omega}}$$.

Now i have the problem, if $|e^{-j\omega}|$ = 1, the sum diverges.To handle this case, i know that $e^{-j\omega}$ is $2\pi$ periodic, i get $$\frac{1}{1-e^{-j\omega}}+\underbrace{e^{-j0}}_1 \sum_{k=-\infty}^{\infty}\delta(\omega+2\pi k)$$.

In books i found that the DTFT of the unit step is $$\frac{1}{1-e^{-j\omega}}+\pi \sum_{k=-\infty}^{\infty}\delta(\omega+2\pi k)$$.

Can me anyone explain why get the $\pi$ in the DTFT of the unit step?

  • $\begingroup$ that the sum diverges is a problem, that's why there is the Fourier inversion theorem, the extension of the Fourier transform to $L^p$ functions and then to distributions in term of bounded operators of Banach space, etc. and I don't get what you did. so you can compute the Fourier series of $\frac{1}{1-e^{- i \omega}}$ and check that the coefficients are $u[n]-1/2$. or you can consider the DTFT of $\frac{u[n-1]}{n}$ and prove that it is $-\ln(1-e^{-i \omega})$ with the Taylor series of $-\ln(1-z)$ at $z=0$ (and $(-\ln(1-z))' = \frac{1}{1-z}$) $\endgroup$
    – reuns
    May 5, 2016 at 5:53

2 Answers 2


Hi guys I think this way is better than others,What's Ur idea?

$u[n]=f[n] + g[n] $


$f[n]= {1\over2}$ for $-\infty<n<\infty $


$g[n]=\left\{ \begin{array}{c} {1\over2} \text{ for } n\ge 0 \\ {-1\over2} \text{ for } n<0 \end{array} \right.$


$ \delta [n] = g[n] - g[n-1]$

U Know DTFT of $\delta[n]$ is $1$ and DTFT of $g[n] - g[n-1]\to G(e^{j\omega})-e^{-j\omega}G(e^{j\omega})$ so: $1=G(e^{j\omega})-e^{-j\omega}G(e^{j\omega})$


$G(e^{j\omega})={1\over 1-e^{-j\omega}}$

and we know that the DTFT of $f[n]\to F(e^{j\omega})=\pi\sum_{k=-\infty}^\infty\delta(\omega -2\pi k)$

finally :

$u[n] = f[n]+g[n] \to U(e^{j\omega})=F(e^{j\omega})+G(e^{j\omega})$

$U(e^{j\omega})={1\over 1-e^{-j\omega}}+\pi\sum_{k=-\infty}^\infty\delta(\omega -2\pi k)$


Firstly, you need to be aware that

$$u[n]=\sum_{n=0}^{\infty} \delta[n]$$

And according to Accumulation property of DTFT which implies that: $$y[n]=\sum_{m=-\infty}^{n}x[m]$$ $$\sum_{m=-\infty}^{n}x[m]{\iff}\frac{1}{1-e^{-j\omega}}X(e^{j\omega}) + \pi X(e^{j0})\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)\ \ \ \ *$$ Now assume that: $$x[n] =\sum_{m=-\infty}^{n} g[m]$$ $$g[n] =\delta[n]$$ $$G({e^{j\omega}}) = 1$$

Substitute in *: $$\frac{1}{1-e^{-j\omega}}G(e^{j\omega}) + \pi G(e^{j0})\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$ Now refine the last formula:

HINT: $G(e^{j\omega}) = 1\\ G(e^{j0}) = 1$ $$\frac{1}{1-e^{-j\omega}} + \pi \sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$


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