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Fourier Transform Pairs of Conjugation and Time Reversal
November 17, 2021

Table of Contents

Introduction

Too often I find myself asking the questions “what is the Fourier transform of x(-t)”? “What is the inverse Fourier transform of X^*(-f)?” Finding them in a book or on Wikipedia often takes too long so to save myself the time and hopefully be useful to others, I’ve provided the derivations for the Fourier transforms of x(t), x^*(t), x(-t) and x^*(-t) below.

More posts on DSP Math:

Fourier Transform

The Fourier transform of x(t) is given by

(1)   \begin{equation*}\mathcal{F}\{ x(t) \} = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} ~ dt = X(f).\end{equation*}

Fourier Transform of Conjugate

The Fourier transform of x^*(t), the conjugate of x(t), is given by

(2)   \begin{equation*}\mathcal{F}\{ x^*(t) \} = \int_{-\infty}^{\infty} x^*(t) e^{-j2\pi f t} ~ dt.\end{equation*}

A double-conjugate is applied to (2)

(3)   \begin{equation*}\left( \left( \int_{-\infty}^{\infty} x^*(t) e^{-j2\pi f t} ~ dt \right)^* \right)^*\end{equation*}

and the inner conjugate is simplified as

(4)   \begin{equation*}\left( \int_{-\infty}^{\infty} x(t) e^{j2\pi f t} ~ dt \right)^*.\end{equation*}

A variable substitution is made for the frequency \psi,

(5)   \begin{equation*}f = -\psi\end{equation*}

such that (4) is written as

(6)   \begin{equation*}\left( \int_{-\infty}^{\infty} x(t) e^{-j2\pi \psi t} ~ dt \right)^*.\end{equation*}

Using (1) the integral in (6) is simplified to

(7)   \begin{equation*}\left( X(\psi) \right)^* = \left( X(-f) \right)^*\end{equation*}

such that

(8)   \begin{equation*}\mathcal{F} \{x^*(t) \} = X^*(-f).\end{equation*}

Fourier Transform of Time Reversal

The Fourier transform of x(-t), which is time reversed x(t), is given by

(9)   \begin{equation*}\mathcal{F}\{ x(-t) \} = \int_{-\infty}^{\infty} x(-t) e^{-j2\pi f t} ~ dt.\end{equation*}

Substitute \tau = -t such that (9) is written as

(10)   \begin{equation*}-\int_{\infty}^{-\infty} x(t) e^{j2\pi f \tau} ~ d\tau\end{equation*}

where

(11)   \begin{equation*}dt = -d\tau.\end{equation*}

Swapping the integration limits in (10) cancels the negation so the integral is written as

(12)   \begin{equation*}\int_{-\infty}^{\infty} x(\tau) e^{j2\pi f \tau} ~ d\tau.\end{equation*}

Substitute for the frequency f,

(13)   \begin{equation*}f = -\psi\end{equation*}

such that (12) is written as

(14)   \begin{equation*}\int_{-\infty}^{\infty} x(\tau) e^{-j2\pi \psi \tau} ~ d\tau.\end{equation*}

Using (1) the Fourier transform of x(-t) is written as

(15)   \begin{equation*}\mathcal{F} \{ x(-t) \} = X(\psi)\end{equation*}

which is simplified as

(16)   \begin{equation*}\mathcal{F} \{ x(-t) \} = X(-f).\end{equation*}

The blog on negative frequency showed that positive frequencies rotate in a counter-clockwise direction around the unit circle. My interpretation of (16) is that reversing the time, x(-t), also reverses the direction rotation around the unit circle for positive frequencies, which is why the frequencies in X(-f) are reversed as well.

Fourier Transform of Conjugate and Time Reversal

The Fourier transform of x^*(-t), the conjugated and time reversal of x(t), is given by

(17)   \begin{equation*}\mathcal{F}\{ x^*(-t) \} = \int_{-\infty}^{\infty} x^*(-t) e^{-j2\pi f t} ~ dt.\end{equation*}

A double conjugate is applied over (17),

(18)   \begin{equation*}\left( \left( \int_{-\infty}^{\infty} x^*(-t) e^{-j2\pi f t} ~ dt \right)^* \right)^*\end{equation*}

and evaluating the inner conjugate leads to

(19)   \begin{equation*}\left( \int_{-\infty}^{\infty} x(-t) e^{j2\pi f t} ~ dt \right)^*.\end{equation*}

A variable substitution is made for the time,

(20)   \begin{equation*}\tau = -t\end{equation*}

where

(21)   \begin{equation*}d\tau = -dt\end{equation*}

such that (19) is written as

(22)   \begin{equation*}\left( -\int_{\infty}^{-\infty} x(\tau) e^{-j2\pi f \tau} ~ d\tau \right)^*.\end{equation*}

Swapping the integration limits cancels the negation such that (22) is written as

(23)   \begin{equation*}\left( \int_{-\infty}^{\infty} x(\tau) e^{-j2\pi f \tau} ~ d\tau \right)^*.\end{equation*}

The integral in (23) is evaluated as

(24)   \begin{equation*}\left( X(f) \right)^*\end{equation*}

such that

(25)   \begin{equation*}\mathcal{F} \{ x^*(-t) \} = X^*(f).\end{equation*}

Takeaway

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