The Hartley_transform is defined as $$ H(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t) \, \mbox{cas}(\omega t) \mathrm{d}t, $$ with $\mbox{cas}(\omega t) = \cos(\omega t) + \sin(\omega t)$.

The Fourier transform on the other hand is defined very similar as $$ F(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t) \, \mbox{exp}(i \omega t) \mathrm{d}t, $$ with $\mbox{exp}(i \omega t) = \cos(\omega t) + i \sin(\omega t)$.

But although the Fourier transform requires complex numbers it is much more widespread than the Hartley transform. Why is that? Are their any properties that make the Fourier transformation much more useful than the Hartley transformation? Or what is the advantage of the Fourier transformation over the Hartley transformation?

  • $\begingroup$ Since $\operatorname{cas}(\omega t) = \sqrt{2} \cos(\omega t + \pi/4)$, is this really any different than a cosine transform? $\endgroup$ – AnonSubmitter85 Dec 8 '15 at 16:52
  • 1
    $\begingroup$ I don't think so. A cosine transform is simply the real part of a Fourier transform. But the real part of the Fourier transform can be computed from a Hartley transform like $Re[F(\omega)] = (H(\omega) + H(-\omega))/2$. So only the even part of the Hartley transform is equivalent to a cosine transform. This is because the Hartley transform kernel is a shifted cosine function, which is not symmetric around the origin. $\endgroup$ – asmaier Dec 9 '15 at 20:54

Homomorphism property

A big advantage of the kernel $e^{i\omega t}$ over $\operatorname{cas}(\omega t)$ is that the former is a homomorphism of the group $(\mathbb{R},+)$ into the multiplicative group of unimodular complex numbers: $e^{i\omega (t+s)} = e^{i\omega t}e^{i\omega s}$. This identity leads to

  • Shift formula: the transform of $f(t-\tau)$ is $e^{i\tau}$ times the transform of $f$
  • Convolution formula: the transform of convolution is the product of transforms.

Both of those can be expressed in terms of the Hartley transform, but in a messy way: essentially one ends up recovering Fourier transform, applying the simple formula for Fourier transform, then going back and ending up with an unintuitive sum of several terms.

Amplitude-phase distinction

Avoiding complex numbers is a questionable benefit. Harmonics have amplitude and phase. Complex numbers have magnitude and argument, which work perfectly for representing amplitude and phase. Thus, the interpretation of the complex number $F(\omega)$ is pretty straightforward: it tells us the amplitude and phase at frequency $\omega$.

In contrast, $H(\omega)$ tells us neither: it's some combination of amplitude and phase from which neither can be recovered without also looking at $H(-\omega)$. So we end up combining $H(\omega)$ and $H(-\omega)$ over and over again, which is not any easier than working with real and imaginary parts of $F$, and is less transparent.


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