The unit step signal defined as

$$ u[n]= \lbrace 1; n>=0; \\ \qquad0; n<0 \rbrace $$

has three possible solutions for its Fourier domain representation depending on the type of approach. These are as follows -

  1. The widely followed approach (Oppenheim Textbook)- calculating the Fourier transform of the unit step function from the Fourier transform of the signum function.

$$ F(u[n])=U(j\omega) = \pi\delta(\omega)+\frac{1}{j\omega} $$

  1. Fourier Transform calculated from the Z transform of the unit step function (Refer Proakis Textbook, Digital Signal Processing Algorithms and applications, pages 267,268 section 4.2.8 )

$$ U(j\omega) = \frac {e^{\frac{j\omega}{2}}}{2j\sin \frac{\omega}{2}}; \omega \neq 2\pi k; k=0,1,2,3... $$

  1. Fourier transform calculated by splitting into even and odd functions - followed in Proakis Textbook (Refer Proakis Textbook, Digital Signal Processing Algorithms and applications, page 618 section 8.1 ) $$ U(j\omega) = \pi\delta(\omega)+\frac{1}{1-e^{-j\omega}} $$

The 2nd representation can be ignored since it is not a well-behaved function. But the approaches followed by Proakis and Oppenheim are equally valid (they extend the Fourier transform to include impulses in the frequency domain) But the confusion is that they provide different solutions.

Is there any mistake in my understanding? or am I missing any crucial point?? Kindly help me understand this and the correct form that can be used in all applications. (I found that the Oppenheim approach is used in deriving the Kramers-Kronig Relations and the Proakis approach used in the derivation of the Hilbert transform)


3 Answers 3


Note that the first expression is the Fourier transform of the continuous unit step $u(t)$, so it's not applicable to the discrete-time step sequence $u[n]$. Furthermore, the second and the third expressions are both correct, and they are identical if you take into account that the second expression does not claim validity at integer multiples of $2\pi$.

If we leave out angular frequencies at multiples of $2\pi$, the third expression becomes

$$U(j\omega)=\frac{1}{1-e^{-j\omega}}=\frac{1}{e^{-j\omega /2}(e^{j\omega /2}-e^{-j\omega /2})}=\frac{e^{j\omega /2}}{2j\sin(\omega/2)},\quad \omega\neq 2k\pi$$

which is identical to the second expression.

  • $\begingroup$ Thanks a lot! Yea the second and third are equivalent but in the third they have composition by including the impulse at the poles. Thank you for the clarification $\endgroup$
    – Injitea
    Nov 15, 2015 at 17:16

As Matt said, the second and third definition are the same except for the part with impulse. The impulse ($\pi\delta(\omega)$) account for the DC value of $u[n]$. Without that term (i.e., the second definition) it is actually the FT of $v[n]=\frac{1}{2}\operatorname{sgn}[n]$. We have $u[n]=v[n]+\frac{1}{2}$. And hence the FT of $u[n]$ has the additional term to account for the addition of $\frac{1}{2}$. Also, the discrete time FT (or DTFT) of $u[n]$ is properly written as $U(e^{j\omega})$.

The first definition, $U(j\omega)$ is the "continuous time" FT (or CTFT) of $u(t)$ (not $u[n]$) and hence different from the other two definitions.


The following derivation illustrates the full picture of why this extra $\delta(\omega)$ term appears.

The real axis integral representation in the complex <span class=$\omega = \omega^{\prime}+j\omega^{\prime\prime}$ domain. Note that $ |e^{j\omega t}| = e^{-\omega^{\prime\prime}} $ diminishes at infinite radius in the upper half." />

Considering the Fourier transform pairs

\begin{equation} \begin{split} X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \\ x(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega \\ \end{split} \end{equation}

The integral property is described by

\begin{equation} \int_{-\infty}^{t} x(\tau) d\tau = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) \Big[\int_{-\infty}^{t} e^{j\omega\tau} d\tau\Big] d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega \end{equation}

This integral can be spitted into three parts in the complex $ \omega $ domain because of the singularity at $ \omega=0 $ as shown in Figure \ref{figure1}.

\begin{equation} \begin{split} \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega = \frac{1}{2\pi} \lim_{r\to0} \Bigg[ & \underbrace{\int_{-\infty}^{re^{-j\pi}} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega}_{I} + \underbrace{\int_{re^{j0}}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega}_{II} \\ & + \underbrace{\int_{-\pi}^{0} X(re^{j\theta})e^{j re^{j\theta}t} d\theta}_{III} \Bigg] \end{split} \end{equation}

In third integral above, we substituted $ \omega =re^{j\theta} $ in order to implement the detour illustrated in Figure \ref{figure1}. After taking the limit, this results in

\begin{equation} \begin{split} \int_{-\infty}^{t} x(\tau) d\tau & = \frac{1}{2\pi} \Bigg[\mathcal{P.V.} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega + \pi X(0)\Bigg] \\ & = \frac{1}{2\pi} \Bigg[\mathcal{P.V.} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega + \int_{-\infty}^{\infty} \pi\delta(\omega) X(\omega) e^{j\omega t} d\omega \Bigg] \\ & = \frac{1}{2\pi} \int_{-\infty}^{\infty} \Big(\frac{1}{j\omega}+ \pi \delta(\omega) \Big) X(\omega) e^{j\omega t} d\omega \\ \end{split} \end{equation}

Hence, the effect of the integral operator on $ x(t) $ in $ \omega $ domain becomes

\begin{equation} \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \big[1/j\omega+ \pi \delta(\omega)\big] X(\omega) \end{equation}

It is easy to show that the effect of the derivative operator on $ x(t) $ in $ \omega $ domain is

\begin{equation} \frac{d}{dt}x(t) \Longleftrightarrow j\omega X(\omega) \end{equation}

Taking the derivative of the integral of $ x(t) $ (\textit{i.e.,} $ \frac{d}{dt} \int_{-\infty}^{t} x(\tau) d\tau = x(t)$) results in

\begin{equation} j\omega \big[1/j\omega+ \pi \delta(\omega)\big] X(\omega) = X(\omega) \end{equation}

which illustrates the reversibility of the integral operator in the $ \omega $ domain after applying the derivative operator. A typical example would be the unit step function $ u(t) $

\begin{equation} u(t)= \int_{-\infty}^{t} \delta(\tau) d\tau \end{equation}

From the previous, the Fourier transform is $ U(\omega) = 1/j\omega + \pi\delta(\omega) $.


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