Editing Hartley transform (section)

==Relation to Fourier transform==
This transform differs from the classic Fourier transform 
<math>F(\omega) = \mathcal{F} \{ f(t) \}(\omega)</math> in the choice of the kernel. In the Fourier transform, we have the exponential kernel,
{{nowrap|1=<math>\exp\left({-\mathrm{i}\omega t}\right) = \cos(\omega t) - \mathrm{i} \sin(\omega t)</math>,}}
where <math>\mathrm{i}</math> is the [[imaginary unit]].

The two transforms are closely related, however, and the Fourier transform (assuming it uses the same <math>1/\sqrt{2\pi}</math> normalization convention) can be computed from the Hartley transform via:

<math display=block>F(\omega) = \frac{H(\omega) + H(-\omega)}{2} - \mathrm{i} \frac{H(\omega) - H(-\omega)}{2}\,.</math>

That is, the real and imaginary parts of the Fourier transform are simply given by the [[even and odd functions|even and odd]] parts of the Hartley transform, respectively.

Conversely, for real-valued functions {{nowrap|1=<math>f(t)</math>,}} the Hartley transform is given from the Fourier transform's real and imaginary parts:

<math display=block>\{ \mathcal{H} f \} = \Re \{ \mathcal{F}f \} - \Im \{ \mathcal{F}f \} = \Re \{ \mathcal{F}f \cdot (1+\mathrm{i}) \}\,,</math>

where <math>\Re</math> and <math>\Im</math> denote the real and imaginary parts.