Why is the Fourier transform more useful than the Hartley transform in physics?

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 used much more in physics than the Hartley transform. Why is that? Are their any properties that make the Fourier transformation more "physical"? Or what is the advantage of the Fourier transformation over the Hartley transformation?

• This is an unusual question for this site, and I'm not totally sure whether it is (or really, should be) considered on topic, but it's definitely an interesting question. – David Z Jan 9 '16 at 10:04
• @DavidZ I say definite on topic, and interesting enough that I'm writing an answer. – DanielSank Jan 9 '16 at 10:05
• some reasons that come to my mind (to be further elaborated): 1) the residue theorem is way easier using $\mathrm{cis}$ that with $\mathrm{cas}$ (because it decays very fast in the lower/upper half plane). 2) $\mathrm{cis}$ is more easily differentiated (1st order ODE compared to 2nd for $\mathrm{cas}$). 3) $\mathrm{cis}$ is an eigenfunction for the momentum operator (generator of translations) 4) I'm not sure Riemann-Lebesgue works for Hartley (and we love the stationary-phase approximation) 5) I'm not sure the HT is defined for distributions (very useful in QFT)... – AccidentalFourierTransform Jan 9 '16 at 10:46
• @AccidentalFourierTransform All true. I think really this all comes down to a choice of basis and Fourier is nice because $D|\omega \rangle = i\omega|\omega\rangle$, i.e. the Fourier basis diagonalizes the derivative. – DanielSank Jan 9 '16 at 10:52
• @AccidentalFourierTransform I think your points are very good. Why not post an answer? – DanielSank Jan 9 '16 at 11:00

First of all, why are Fourier transforms useful?

The Fourier transform is special because the complex exponential functions are eigenfunctions of he translation. In any linear problem with translation invariance, the Fourier transform turns a differential equation into an algebraic one. A simple example is a driven damped harmonic oscillator. The equation of motion is $$\ddot{\phi}(t) + 2\beta\dot{\phi}(t) + \omega_0^2 \phi(t) = J(t)$$ where $\beta$ is the damping parameter, $\omega_0$ is the natural resonance frequency, and $J(t)$ is the drive. This problem is invariant under translation in time.

This is a differential equation, which means it's pretty complicated; it involves limits and all kinds of complex calculus stuff. If we use $$\phi(t) = \int \frac{d\omega}{2\pi} \tilde{\phi}(\omega) e^{i \omega t}$$ we get a new version of the equation of motion in the frequency basis: \begin{align} (-\omega^2 + i 2 \beta \omega + \omega_0^2)\tilde{\phi}(\omega) =& \tilde{J}(\omega) \\ \tilde{\phi}(\omega) =& \frac{-\tilde{J}(\omega)}{\omega^2 - i 2 \beta \omega - \omega_0^2} \, . \end{align} This is way simpler: we now have an algebraic equation instead of a differential equation. In particular, when $J(t)$ is sinusoidal then $\tilde{J}(\omega)$ is a delta function and we can actually just solve for $\phi(t)$. For example, if $J(t) = A \cos(\Omega t)$, then $\tilde{J}(\omega) = (1/2)(\delta(\Omega - \omega) + \delta(\Omega + \omega))$ and we find $$\phi(t) = \text{Re} \left[ \frac{-A e^{i \Omega t}}{\Omega^2 - i 2 \beta \Omega - \omega_0^2} \right] \, .$$

The mathematical reason this was all so easy is that the exponential functions $\exp(i \omega t)$ are eigenvectors of the derivative: $(d/dt)\exp(i \omega t) = i \omega \exp(i \omega t)$. We could have written all of this stuff in vector notation and it would make a few conceptual issues super-duper clear, but that's a topic for another post (on Math.SE, probably).

Fourier is too much info

The Fourier transform actually contains redundant information if the the original function is purely real valued. If $\phi(t) \in \mathbb{R}$, then it turns out that the negative frequency components in the Fourier transform are just the complex conjugates of the positive parts, i.e. $\tilde{\phi}(-\omega) = \tilde{\phi}(\omega)^*$. Because of this, we can rewrite a Fourier representation of a real signal in terms of only positive frequency parts: \begin{align} \phi(t) =& \int_{-\infty}^\infty \frac{d\omega}{2\pi} \tilde{\phi}(\omega)e^{i \omega t} \\ =& \int_0^\infty \frac{d\omega}{2\pi} \tilde{\phi}(\omega)e^{i \omega t} + \int_{-\infty}^0 \frac{d\omega}{2\pi} \tilde{\phi}(\omega)e^{i \omega t} \\ =& \int_0^\infty \frac{d\omega}{2\pi} \tilde{\phi}(\omega)e^{i \omega t} + \int_0^\infty \frac{d\omega}{2\pi} \tilde{\phi}(-\omega)e^{-i \omega t} \\ =& 2 \text{Re} \left[ \int_0^\infty \frac{d\omega}{2\pi} \tilde{\phi}(\omega) e^{i \omega t}\right] \, . \end{align}

So what's up with Hartley?

An arbitrary signal has an amplitude and phase at each frequency. The Fourier transform encapsulates this in the amplitude and phase of $\tilde{\phi}(\omega)$. To get that in a cosine transform, we have to add a parameter $\theta(\omega)$ to handle the phase, i.e.$^{[a]}$ $$\phi(t) = \int_0^\infty \frac{d\omega}{2\pi} M(\omega) \cos(\omega t + \theta(\omega)) \, .$$ The point is that at each frequency you need two numbers. Hartley deals with this by putting some of the information in the negative frequencies. You can see this in the fact that the Fourier ($F$) and Hartley ($H$) transforms are related by $$F(\omega) = \frac{H(\omega) + H(-\omega)}{2} - i \frac{H(\omega) - H(-\omega)}{2} \, .$$

So... why not use Hartley?

It's just a bit more of a pain in the butt. Fourier turns derivatives into factors of $i \omega$ because $$(d/dt)\exp(i\omega t) = i \omega \exp(i \omega t) \, .$$ The derivative of the Hartley kernel is \begin{align} (d/dt) \text{cas}(\omega t) =& (d/dt) (\cos(\omega t) + \sin(\omega t)) \\ =& \omega (-\sin(\omega t) + \cos(\omega t)) \\ =& \omega \, \text{cas}(-\omega t) \, . \end{align} As you see, the Hartley transform turns derivatives into an inversion in the direction of time. We could definitely still use this, it's just a little less familiar because it loses the resemblance to the eigenvalue problem you get with Fourier.

Is there a physical difference?

No and yes. The Fourier and Hartley transforms are both examples of basis transformations. A function $f(t)$ can be viewed as the components of a vector $\left \lvert f \right \rangle$ expressed in a particular basis. The Fourier or Hartley transform $\tilde{f}(\omega)$ provides the components of the same vector in a different basis. Expressing a single physical object in different bases (or coordinates, or languages) doesn't change the nature of the object, so in a sense neither Fourier nor Hartley is more physical than the other. Fourier is just almost always more convenient mathematically.

However, the reason Fourier is more convenient is that the Fourier transform is an eigenvector of the operation of translations in time. Define $$V'(t) = V(t + t') \, .$$ Then $$\tilde{V'}(\omega) = \int dt \, V(t+t') \exp[-i \omega t] = e^{i \omega t'} \tilde{V}(\omega) \,.$$ This is intimately related to the fact that the kernel of Fourier is the eigenfunction of the derivative. You see, the derivative generates translations. The operator $\exp(t' (d/dt))$ shifts a function by $t'$. One way to see this is via Taylor expansion: $$\left(e^{t' (d/dt)}V \right)(t) \equiv \underbrace{\left( \sum_{n=0}^\infty \frac{t'^n}{n!} \left(\frac{d}{dt}\right)^n V\right)(t)}_\text{Taylor series about t} = V(t + t') \, .$$

Putting it all together: \begin{align} (e^{t' (d/dt)}V)(t) =& e^{t' (d/dt)} \int\frac{d\omega}{2\pi} \tilde{V}(\omega) e^{i \omega t} \\ =& \int\frac{d\omega}{2\pi} \tilde{V}(\omega) \underbrace{\sum_{n=0}^\infty \frac{t'^n}{n!} \left(\frac{d}{dt}\right)^n e^{i \omega t}}_\text{translation op. act on Fourier kernel} \\ =& \int\frac{d\omega}{2\pi} \tilde{V}(\omega) e^{i \omega t'} e^{i \omega t} \\ =& \int\frac{d\omega}{2\pi} \tilde{V}(\omega) e^{i \omega (t+t')} \\ =& V(t+t') \, . \end{align}

So yeah, the Fourier transform is nice because the kernel has a special property with respect to translations. Problems with translational invariance are enormously simplified if you express them in the Fourier basis. The example at the top of this post is such a problem: the homogeneous part of the differential equation is time translation invariant.

$[a]$: You can also write this as a superposition of a sine and cosine transform: $$\phi(t) = \int_0^\infty \frac{d\omega}{2\pi} \left( A(\omega) \cos(\omega t) + B(\omega) \sin(\omega t) \right) \, .$$

• (+1) IMHO, one of the main advantages is mathematical, and not physical: integration in the complex plane is way easier using complex exponentials. Every time we have to integrate $\cos$/$\sin$ in the complex plane we turn them into exponentials, so in practice we use the FT to calculate the HT. Why would we want to introduce the HT if the FT contains the same information and is mathematically easier to handle? – AccidentalFourierTransform Jan 9 '16 at 10:59
• @AccidentalFourierTransform Yeah, that's a good point for sure. Mind if I add a few extra points based on your comments, or were you going to post an answer? – DanielSank Jan 9 '16 at 11:01
• I think it is interesting to note that when looking at second derivatives both kernels show again the same behaviour: $(d^2/dt^2) \text{cas}(\omega t) = -\omega^2 \text{cas}(\omega t)$ and $(d^2/dt^2)\exp(i\omega t) = - \omega^2 \exp(i \omega t)$ – asmaier Jan 9 '16 at 12:13
• I would love if someone could address this part of the question: "Are their any properties that make the Fourier transformation more "physical"?" . – asmaier Jan 12 '16 at 16:52
• @DanielSank oh god I just cant believe I missed the chance to say "Fourier Transform is not an accident"!!! – AccidentalFourierTransform Jan 13 '16 at 20:04
1. On the set ${\cal L}_1(\mathbb{R})$ of integrable functions $f:\mathbb{R}\to \mathbb{C}$, we can define the sine and cosine transforms $$\tag{1}\left\{\begin{array}{ccc} ({\cal C}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~\cos(\omega t) f(t),\cr ({\cal S}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~\sin(\omega t) f(t). \end{array} \right.$$

2. Next, consider the reflection $$\tag{2} ({\cal R}f)(t)~:=~f(-t),$$ and note that $$\tag{3}\left\{\begin{array}{ccc} {\cal C}{\cal R}f &=& {\cal C}f, \cr {\cal S}{\cal R}f &=& -{\cal S} f.\end{array} \right.$$

3. Similarly, we can define the complex Fourier transform $$\tag{4}\left\{\begin{array}{ccccc} ({\cal F}f)(\omega) &:=& \int_{\mathbb{R}}\! dt~ {\rm cis}(-\omega t) f(t) &=& ({\cal C}f)(\omega)-i({\cal S}f)(\omega) ,\cr ({\cal F}{\cal R}f)(\omega) &=& \int_{\mathbb{R}}\! dt ~{\rm cis}(\omega t) f(t)&=& ({\cal C}f)(\omega)+i({\cal S}f)(\omega), \end{array} \right.$$ or the Hartley transform as $$\tag{5}\left\{\begin{array}{ccccc} ({\cal H}f)(\omega) &:=& \int_{\mathbb{R}}\! dt ~{\rm cas}(\omega t) f(t)&=& ({\cal C}f)(\omega)+({\cal S}f)(\omega) ,\cr ({\cal H}{\cal R}f)(\omega) &=& \int_{\mathbb{R}}\! dt ~{\rm cas}(-\omega t) f(t) &=& ({\cal C}f)(\omega)-({\cal S}f)(\omega). \end{array} \right.$$

4. Clearly, the three integral transforms (1), (4) & (5) are bijectively related on ${\cal L}_1(\mathbb{R})$, and it becomes just a matter of convenience which one to use. DanielSank's answer already lists some of these conveniences.

Short answer: The Hartley transform is a subset of the results given by Fourier transforms, which is only the real part (assuming your signal is real, which is almost always the case in physics).

Long answer: Practically, you need the amplitude and phase of the signal, and the Fourier transform gives you both amplitude and phase by taking the

magnitude: $A=\sqrt{x^2+y^2}$ and

the phase: $\phi=\arctan{\frac{y}{x}}$

This all is information contained in the signal. If you use the Hartley transformation, then you only have the $x$ component of your signal samples. This is only useful for very special cases, while I imagine that the computational cost is probably the same.