The inverse Fourier transform of sgn w is a fundamental result that connects the signum function in the frequency domain to a specific time‑domain signal. Understanding this transformation is essential for anyone working with signal processing, control theory, or any field where frequency‑domain analysis is performed.
Introduction
In many engineering and physics problems, we analyze systems by examining how they respond to different frequencies. Still, the Fourier transform converts a time‑domain function into its frequency‑domain representation, while the inverse Fourier transform does the opposite. When the frequency‑domain function is the signum function, denoted by (\operatorname{sgn}(w)), the inverse transform yields a well‑known time‑domain expression that has both theoretical importance and practical utility.
Definition of the Signum Function
The signum function, often written as (\operatorname{sgn}(w)) or simply (\text{sgn}(w)), is defined as
[ \operatorname{sgn}(w) = \begin{cases} 1, & w > 0,\ 0, & w = 0,\ -1, & w < 0. \end{cases} ]
It extracts the sign of its argument, ignoring the magnitude. In the context of Fourier analysis, (\operatorname{sgn}(w)) appears as a simple yet non‑trivial frequency‑domain function because it is not absolutely integrable, which requires careful handling using distribution theory or principal‑value integrals.
Fourier Transform Basics
The Fourier transform of a function (x(t)) is given by
[ X(w) = \int_{-\infty}^{\infty} x(t) e^{-j w t}, dt, ]
and its inverse is
[ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(w) e^{j w t}, dw. ]
These relations assume that the integrals converge in the ordinary sense. When they do not, as with (\operatorname{sgn}(w)), we interpret the integrals in the sense of distributions or use a limiting procedure.
Inverse Fourier Transform of (\operatorname{sgn}(w))
To compute the inverse transform, we start with the definition and insert (X(w) = \operatorname{sgn}(w)):
[ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \operatorname{sgn}(w) e^{j w t}, dw. ]
Because (\operatorname{sgn}(w)) is odd, the integral can be rewritten as
[ x(t) = \frac{1}{\pi} \int_{0}^{\infty} \sin(w t), dw, ]
using the identity (\operatorname{sgn}(w) e^{j w t} = \operatorname{sgn}(w) \cos(w t) + j \operatorname{sgn}(w) \sin(w t)) and noting that the cosine part integrates to zero by
