# Connection between Parseval's Theorem to Fourier Transform is Unitary

In YouTube Parseval's Theorem, Professor Steve Brunton indicated there is a connection between Parseval's Theorem and Fourier Transform is Unitary.

But how a Unitary (preserving inner products, preserving orthonormality of basis) relates to or contributes to Parseval's Theorem?

Useful property (of Parseval's Theorem) when using Fourier Transform to approximate a function f(x) is that there are small Fourier Coefficient that I can ignore and I still do a good job approximating the function f(x).

This is related with the fact that the Fourier Transform is a Unitary Operator which preserves distances between functions. If I have two functions F and G and Fourier Transform them from x coordinates, they are going to have the same inner products in the Fourier coordinates.

That is another subtle property of Parseval's Theorem.

• Parseval’s theorem (and Plancheral’s) shows that the Fourier transform is energy preserving, or more generally, the Fourier transform is an isometry in $L^{2}$. A sort of generalized Parseval’s theorem is that a mapping of a signal to its orthobasis coefficients is inner product preserving. The Fourier transform is an orthobasis expansion, and the energy of a signal is the inner product with itself, so inner product preservation of orthobasis expansions implies energy preservation of the Fourier transform. Commented Jul 19 at 8:13

I didn't have time to write out an answer last night, so I'll expand on my comments here. The goal is to show that orthobasis expansions are geometry preserving.

Let $$\{\phi_{\gamma}\}_{\gamma \in \Gamma}$$ be an orthobasis for a signal in space $$G$$. For any two signals $$x,y \in G$$ we have the synthesis operator $$$$\Phi[\{\alpha(\gamma)\}_{\gamma \in \Gamma}] = \sum_{\gamma \in \Gamma}\alpha(\gamma)\phi_{\gamma}(t)$$$$ and the analysis operator $$$$\Phi^{*}[x(t)] = \{\langle x(t),\phi_{\gamma}(t) \rangle \}_{\gamma \in \Gamma} = \{\alpha(\gamma)\}_{\gamma \in \Gamma}$$$$

We wish to show that for any orthobasis $$$$\langle x(t),y(t) \rangle = \sum_{\gamma \in \Gamma}\alpha(\gamma)\bar{\beta}(\gamma)$$$$ where $$$$\alpha(\gamma) = \langle x(t),\phi_{\gamma}(t) \rangle, \; \beta(\gamma) = \langle y(t),\phi_{\gamma}(t) \rangle$$$$

Explicitly writing this out, we get \begin{aligned} \langle x(t),y(t) \rangle &= \langle \sum_{\gamma \in \Gamma}\alpha(\gamma)\phi_{\gamma}(t),\sum_{\gamma ' \in \Gamma}\beta(\gamma ')\phi_{\gamma '}(t) \rangle \\ &= \sum_{\gamma}\sum_{\gamma '}\alpha(\gamma)\bar{\beta}(\gamma ') \langle \phi_{\gamma}(t),\phi_{\gamma '}(t)\rangle \end{aligned}

Using the definition of an orthobasis $$$$\langle \phi_{\gamma}(t),\phi_{\gamma '}(t) \rangle = \begin{cases} 1, & \gamma = \gamma ' \\ 0, & \gamma \neq \gamma ' \end{cases}$$$$ we get $$$$\langle x(t),y(t) \rangle = \sum_{\gamma}\alpha(\gamma)\bar{\beta}(\gamma)$$$$

The energy in a signal x(t) is then \begin{aligned} \lvert\lvert x(t)\rvert\rvert_{2} &= \int \lvert x(t) \rvert^{2}dt \\ &= \int x(t) \bar{x}(t) dt \\ &= \langle x(t),x(t) \rangle \\ &= \sum_{\gamma}\alpha(\gamma)\bar{\alpha}(\gamma) \\ &= \sum_{\gamma} \lvert \alpha(\gamma) \rvert^{2} \end{aligned}

For the Fourier transform, we get $$$$\{\alpha(\omega)\}_{\omega \in \Omega} = \{\langle x(t),\phi_{\omega}(t)\rangle\}_{\omega \in \Omega}$$$$ where $$$$\phi_{\omega}(t) = e^{-j\omega t}$$$$

Thus, Parseval's theorem with respect to the Fourier transform being energy preserving is a consequence of the more general conclusion that orthobasis expansions are geometry preserving.

1 Justin Romberg's lecture notes on compressed sensing - Lecture 1

As Baddioes points out, the inner product of a signal with itself gives the energy of that signal – and due to Parseval's theorem (for the unitary Fourier transform), that energy is the same in both domains.

So, if you subtract a signal in the frequency domain (by setting small Fourier coefficients to zero, you're effectively subtracting the signal that is made up solely of these), then you subtract only the same energy (so, very little) in the time domain, and your error can't be large.

Note that defining the Fourier transform as unitary is a choice of definition. Not everyone does that. It's often more practical to include a $$1/\sqrt{2\pi}$$ scaling factor in both directions of one's definition, or no such factor at all, and live with the fact that energies aren't identical, but just proportional; it depends on what you want to do with the transform. If the absolute energy of something coming out of a transform matters, you hence need to check whether the material you're working with sticks to unitary Fourier transforms. That's relatively likely for physics literature, but not as inherent for communications engineering literature, where absolutes don't matter as much, and usually, you only care about the ratio of powers (esp. between noise and signal power) – which are preserved, no matter whether you define the Fourier transform to be unitary or not.

• In my opinion what is better than the $1/\sqrt{2 \pi}$ scaling factor is using the "ordinary frequency" definition of the Fourier Transform: $$\mathscr{F}\Big\{x(t)\Big\} \triangleq X(f) \triangleq \int_{-\infty}^{+\infty} x(t) \, e^{-i 2 \pi f t} \, \mathrm{d}t$$ and inverse $$\mathscr{F}^{-1}\Big\{X(f)\Big\} \triangleq x(t) = \int_{-\infty}^{+\infty} X(f) \, e^{i 2 \pi f t} \, \mathrm{d}f$$ $f$ is "ordinary" frequency, not angular frequency. Commented Jul 19 at 21:49
• @robertbristow-johnson 100% this. I mean, engineers here, right. I rarely see a spectrum allocation, a quartz oscillator, or a speaker being spec'ed in radian per second. Commented Jul 20 at 9:35
• But it makes Pasevals Theorem, the Duality Theorem, the relationship between DC and the area under the curve, all that is trivial, at least about remembering the scaling constants (which are gone). Commented Jul 20 at 18:07