# Duality property of the Fourier Transform

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This property states that if $F(\omega)$ is the Fourier transform of $f(t)$, then the Fourier transform of $F(t)$ is $2 \pi f(-\omega)$; we write
$$\mathcal{F}[f(t)]=F(\omega) \quad \Longrightarrow \quad \mathcal{F}[F(t)]=2 \pi f(-\omega)$$
This expresses the symmetry property of the Fourier transform. To derive this property, we recall that
$$f(t)=\mathcal{F}^{-1}[F(\omega)]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega$$
or
$$2 \pi f(t)=\int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega$$
Replacing $t$ by $-t$ gives
$$2 \pi f(-t)=\int_{-\infty}^{\infty} F(\omega) e^{-j \omega t} d \omega$$
If we interchange $t$ and $\omega$, we obtain
$$2 \pi f(-\omega)=\int_{-\infty}^{\infty} F(t) e^{-j \omega t} d t=\mathcal{F}[F(t)]$$
as expected.
For example, if $f(t)=e^{-|t|}$, then $$F(\omega)=\frac{2}{\omega^{2}+1}$$ By the duality property, the Fourier transform of $F(t)=2 /\left(t^{2}+1\right)$ is $$2 \pi f(\omega)=2 \pi e^{-|\omega|}$$ Figure $1$ shows another example of the duality property. It illustrates the fact that if $f(t)=\delta(t)$ so that $F(\omega)=1$, as in Fig. 1(a), then the Fourier transform of $F(t)=1$ is $2 \pi f(\omega)=2 \pi \delta(\omega)$ as shown in Fig. 1(b).
Fig. 1: A typical illustration of the duality property of the Fourier transform: (a) transform of impulse, (b) transform of unit dc level.

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