To obtain the magnitude of a particular frequency contained in a periodic signal, you take the inner product of the signal’s function with the (analyzing) basis function corresponding to such frequency, which in exponential form looks as follows:

$${C_n} = \frac{1}{T}\int_0^T {f(t){e^{ - in\frac{{2\pi }}{T}t}}} dt % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeeaaaaaa6di % eB1vgapeGaam4qamaaBaaaleaacaWGUbaabeaak8aacqGH9aqpdaWc % aaqaaiaaigdaaeaacaWGubaaamaapedabaGaamOzaiaacIcacaWG0b % GaaiykaiaadwgadaahaaWcbeqaaiabgkHiTiaadMgacaWGUbWaaSaa % aeaacaaIYaGaeqiWdahabaGaamivaaaacaWG0baaaaqaaiaaicdaae % aacaWGubaaniabgUIiYdGccaWGKbGaamiDaaaa!4E51! $$

That is to say, you choose the complex conjugate of the exponential function (negative sign in the exponent).

As to the reason for that, most answers that I have read say that the choice is conventional: it could be a positive sign as well, as long as the inverse (synthesizing) transform takes the opposite sign.

On another note, to check if two complex functions are orthogonal by taking their inner product, you also have to use the complex conjugate of one of them. Otherwise you would not get the expected result, which is as follows:

$$\frac{1}{T}\int_0^T {{e^{in\frac{{2\pi }}{T}t}}{e^{ - im\frac{{2\pi }}{T}t}}} dt = \left\{ \begin{array}{l}1{\rm{ if n = m}}\\{\rm{0 if n}} \ne {\rm{m}}\end{array} \right. % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaamivaaaadaWdXaqaaiaadwgadaahaaWcbeqaaiaadMga % caWGUbWaaSaaaeaacaaIYaGaeqiWdahabaGaamivaaaacaWG0baaaO % GaamyzamaaCaaaleqabaGaeyOeI0IaamyAaiaad2gadaWcaaqaaiaa % ikdacqaHapaCaeaacaWGubaaaiaadshaaaaabaGaaGimaaqaaiaads % faa0Gaey4kIipakiaadsgacaWG0bGaeyypa0Zaaiqaaqaabeqaaiaa % igdacaqGGaGaaeiiaiaabccacaqGPbGaaeOzaiaabccacaqGUbGaae % ypaiaab2gaaeaacaqGWaGaaeiiaiaabccacaqGGaGaaeyAaiaabAga % caqGGaGaaeOBaiabgcMi5kaab2gaaaGaay5Eaaaaaa!5F73! $$

I wonder if there is no connection at all between the two things, which would entail (it seems) that the choice of the negative sign in the Fourier formula is not so arbitrary…

This is the farthest that I have arrived in an attempt at answering my own question: what you try to do in the orthogonality test is to find the correlation between the two functions of the same basis (frequencies) and there you need to introduce one conjugate; in the Fourier case you are also testing the correlation, though here between different basis, one signal expressed in the time domain versus one of its frequency components, so shouldn’t the same need (one complex conjugate) arise?


I made some progress in understanding this, but would need feedback.

First, in Fourier case, when you take the inner product, you have already projected the basis function of the frequency basis ($e$) onto the time basis, so $f(t)$ and $e$ are on the same basis.

Second, in the second case, you are testing the outcome of using the complex conjugates for an inner product with the extreme cases of identical functions or orthogonal functions, which give you the extreme results of total or null correlation, respectively; that is fine, that works; hence it looks only logical that the same method (inner product using complex conjugates) works for testing partial correlation; in particular, it seems to me that the reason why the complex conjugates approach works is that this way you are taking out of each product the angle of the analyzing function, so that it is not added and thus the product reduces to a pure multiplication of the moduli (?).

I wonder if anyone could validate or refute this tentative understanding.

  • $\begingroup$ You might want to look up the term "Hilbert basis" $\endgroup$ – Severin Schraven Oct 20 '18 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.