1

I think that I almost understood Fourier analysis, but I am stuck with the normalization factor.

Based on the analogy with a change of basis of standard vectors, I would expect that the role of this factor is dividing by the modulus of the analyzing function (this would be the way to convert the dot product into a sort of scalar projection of the signal over the analyzing function, right?).

I would also expect that this modulus is $\sqrt2π$, because in the next example you divide by $2π$, but the correlation 1 seems to be pointing at the equivalent of the cosine of the angle, which would require division by the modulus squared…

$${\rm{if n = m }} \to {\rm{ }}\frac{1}{{2\pi }}\int_{ - \pi }^\pi {{e^{in\frac{{2\pi }}{T}t}}{e^{ - im\frac{{2\pi }}{T}t}}} dt = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {{e^0}} dt = \frac{1}{{2\pi }}\int_{ - \pi }^\pi 1 dt = \frac{{2\pi }}{{2\pi }} = 1 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeyAaiaabA % gacaqGGaGaaeOBaiaab2dacaqGTbGaaeiiaiabgkziUkaabccadaWc % aaqaaiaaigdaaeaacaaIYaGaeqiWdahaamaapedabaGaamyzamaaCa % aaleqabaGaamyAaiaad6gadaWcaaqaaiaaikdacqaHapaCaeaacaWG % ubaaaiaadshaaaGccaWGLbWaaWbaaSqabeaacqGHsislcaWGPbGaam % yBamaalaaabaGaaGOmaiabec8aWbqaaiaadsfaaaGaamiDaaaaaeaa % cqGHsislcqaHapaCaeaacqaHapaCa0Gaey4kIipakiaadsgacaWG0b % Gaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaiabec8aWbaadaWdXaqa % aiaadwgadaahaaWcbeqaaiaaicdaaaaabaGaeyOeI0IaeqiWdahaba % GaeqiWdahaniabgUIiYdGccaWGKbGaamiDaiabg2da9maalaaabaGa % aGymaaqaaiaaikdacqaHapaCaaWaa8qmaeaacaaIXaaaleaacqGHsi % slcqaHapaCaeaacqaHapaCa0Gaey4kIipakiaadsgacaWG0bGaeyyp % a0ZaaSaaaeaacaaIYaGaeqiWdahabaGaaGOmaiabec8aWbaacqGH9a % qpcaaIXaaaaa!7D69! $$

In fact, there is a “symmetrical” convention where the pre-factor is $\sqrt2π$ for both the direct and the inverse Fourier transform.

However, I have plenty of questions:

  1. In Fourier series, what I usually see is the pre-factor $1/π$. Why so? In Fourier series you face a periodic function, which is decomposed into its fundamental frequency plus the discrete harmonics; in Fourier transform, you face a non-periodic signal being decomposed into a continuous set of frequencies… but is that relevant for the purpose of choosing the pre-factor?
  2. Sometimes I see a pre-factor of $1/T$, but only in Fourier series examples... Why that? In order to normalize, do you divide by chance by seconds? Does it have to do with the "weight" of the integral?
  3. In one article I read that if you write $ω$ in the exponent of the analyzing exponential function, instead of $2πf$ or $2π/T$, the pre-factor changes. But aren’t the three expressions equivalent?
  4. Another convention for Fourier transform is having no pre-factor in the direct transform and $1/2π$ in the inverse transform… I understand that everything related to units is characterized by conventionality… but is there an analogue situation at the level of standard vectors…?
Sierra
  • 247
  • For the Fourier series the point is that ${ \frac{1}{\sqrt{T}}e^{2i \pi n ./T} }_{n \in \mathbb{Z}}$ is an orthonormal basis of $L^2([a,a+T])$ so $g= \sum_n \langle g,\frac{1}{\sqrt{T}}e^{2i \pi n ./T} \rangle \frac{1}{\sqrt{T}}e^{2i \pi n./T}$ where the convergence is in $L^2([a,a+T])$ norm – reuns Mar 06 '19 at 01:19
  • @reuns Good, you seem to build the factor into the analyzing functions, make them thus unitary and hence make the basis orthonormal from start. The choice of dividing by $\sqrt{T}$ also looks logical, since a pure cosine correlation would require dividing by the two involved moduli, i.e. a factor of $1/T$, but in Fourier context we need to divide only by one of them. I am also happy dividing by time, which would look like a way of eliminating the weight of the integral… But then why do others use $1/T$ or $1/ π$ in the series or $1/\sqrt2π$ or $1/2π$ in the transform? – Sierra Mar 06 '19 at 16:46

0 Answers0