There's an intuitive interpretation of the inverse Fourier Transform with a system of connected wheels spinning in the complex plane. Then the signal (a real one, like usual sine for example) is the the real part of position of a point attached to the system in time. This is pretty easily explained in below article.


I think I understood the concept quite well, however I puzzle over some symmetric intuitive interpretion solely for Fourier Transform (FT) not the inverse. In other words: I know how to reproduce a signal knowing it's FT, but I struggling to imagine how a particular signal can be decomposed into its compononets.

To further explain I will stick to the below convention:

$$f(t) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} F(\omega) e^{i \omega t} \mathrm{d}\omega $$ $$F(\omega) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} f(t) e^{-i \omega t} \mathrm{d}t$$ where $f(t)$ is a signal and $F(\omega)$ its FT.

If we take a look at the equations we can see, that the only difference is the sign in the exponent. It suggests, that one should think of it as spinning backwards through the wheels system with each point of a particular signal. However, now we integrate over frequency. Choosing an arbitrary $\omega$ we can think of moving points with the angular speed of $\omega$. My idea is that if chosen $\omega$ matches a frequency of some compononent of the considered signal, then all the contribution to the integral value comes from that compononent, the not-mathing frequencies somehow cancel.

Let now consider a sine wave. Following the above idea we choose the matching frequency. We know the integral does not exist as it oscilates between -1 and 1 as sine itself. At least if we don't divide it into its positive and negative part. Then we get two delta functions as we should - a negative and a positive one.

There's where my questions arise.

Why is it like that? Why the rest of a signal cancels when integrating at an arbitrary frequency? What about for example constant wave then? Maybe I should look at such a sine wave as a corresponding complex pulse? Should we look differently at odd and even functions or decompose into positive and negative parts?

That bothers me for so long I eventually decided to search here for help.

I would appreciate some deeper explanation as it stays unclear to me. It would be great if it would follow my presented idea.

I've also asked this question one of my professors. He gave me a different approach with matrix diagonalization and spectral theorem, if that's a better way to understand it I would be also grateful for hints or explanation.

Also, maybe you can point me to an appropriate book or paper regarding this topic?

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    $\begingroup$ I think you're missing the fact that sinusoids of different frequencies are orthogonal: $$ \int^{\infty}_{-\infty} \exp(i \omega_1 t)\exp(i \omega_2 t) \mathrm{d}t=0 $$ if $\omega_1 \ne \omega_2$. $\endgroup$ – T L Davis May 1 '17 at 16:57
  • $\begingroup$ Thanks, in fact I've missed it. That's a fair point. $\endgroup$ – piwox May 1 '17 at 17:48
  • $\begingroup$ Also, I guess in this case it should be $\int^{\infty}_{-\infty} \exp(i \omega_1 t)\overline{\exp(i \omega_2 t)} \mathrm{d}t=0$ as that's inner product of complex-valued functions. $\endgroup$ – piwox May 2 '17 at 8:25
  • $\begingroup$ You're right of course. $\endgroup$ – T L Davis May 3 '17 at 16:28

Imagine your function $f(t)$ is a pure tone signal, $\textit{i.e.}$ it is of the form: $$f(t) = \sin{t}$$ with $\omega$ arbitrary.

Because of the fact that the Fourier transform comes from some limit we can define it as a sum: $$\mathcal{F}(f(t)) =\lim_{T\to \infty} a_0(T) + \sum_{k=1}^{\infty}a_k(T)\cos{(2\pi k \,t/T )} + b_k(T) \sin{(2\pi k\,t/T )}$$ Where the coefficients are defined as $$a_0(T) = \frac{1}{T}\int_{-T/2}^{T/2}{f(t)\,dt}$$ $$a_k(T) = \frac{2}{T}\int_{-T/2}^{T/2}{f(t)\cos{(2\pi k \,t/T )}\,dt} \qquad b_k(T) = \frac{2}{T}\int_{-T/2}^{T/2}{f(t)\sin{(2\pi k \,t/T )}\,dt}$$

Take for visualisation $T$ finite, let's say for simplicity $T=2\pi$.

Ii is easy to see that all coefficients give 0 but $b_1$, that equals $1$. That is so because the integrals over one period of product of sinusoidal functions with distinct frequency cancels out as T L Davis pointed out.

Therefore we expect that the "best fit" to the proposed function $f(t)$ is a sine wave with amplitude and frequency unity.

For the continuum case, the delta functions have complex amplitude because of the definition of the transform in terms of complex exponentials. Now there are no basis functions like $\sin{nt}$ or $\cos{nt}$ but $\exp{int}$ (in some sense) that group both. Recall Euler's formula: $$\exp{int}=\cos{nt}+i\sin{nt}$$ You may note that the amplitudes of the waves associated with the sine part must be purely complex to recover the physical significance.

Hope this helps.

  • $\begingroup$ Thanks, with some additional research I finally got it. The thing that made me wonder when reading the answer was why are the Fourier coefficients defined like that. That's simply what all the fuss is about. The coefficents when taking the limit are what we're looking for i.e. the FT. The key here is in fact orthogonality. I came up with a explanation that works for me so I'll extend my thoughs in a separate answer. $\endgroup$ – piwox May 2 '17 at 8:40

I highly, highly recommend "The Fast Fourier Transform" by E. Oran Brigham for developing a deep understanding of the Fourier transform. It's emphasis is on the FFT, but as the cover illustrates, it covers the Fourier transform, the discrete Fourier transform, the FFT and it's applications. It will give you a feel for the Fourier transform by coupling the math with outstanding illustrations of transform pairs, transform operations, etc. One of the best technical books ever. (See the review by Richard N).


I made up my mind and here's what I came up with taking into consideration the orthogonality concept. Orthogonality turns out to be a key to understand what's happening when applying Fourier transforms. Then both matrix interpretation suggested by my professor and analyzing integrals depict the idea well.

Let only consider complex pure tone waves of form $e^{2\pi i \xi t}$. It's simpler and more compact than thinking of both sines and cosines.

Recall (as T L Davis kindly pointed out) that the inner product of complex functions $f$ and $g$ from real line is defined as $$\left<f,g\right>=\int_{-\infty}^{\infty}f(t)\overline{g(t)}\mathrm{d}t$$ It follows that the norm is $$||f||=\left(\int_{-\infty}^{\infty}|f(t)|^2\mathrm{d}t\right)^{{1}/{2}}$$ which are $L^2$ norm and inner product, that can probably be used to make some formal assumptions for existance of FT for particular functions. Having the above in mind let us step back to Fourier series i.e. $f(t)=\sum_\limits{n=-\infty}^{\infty}c_n(T)e^{2\pi i\frac{n}{T} t}$

As HBR pointed out within his answer, the coefficients of Fourier series are given by $$c_n(T)=\frac{1}{T}\int_{-T/2}^{T/2}f(t)e^{-2\pi i \frac{n}{T}t}\mathrm{d}t,$$ but hey, these $c_n(T)$ are exactly what we're asking about. We assume that a periodic function can be decomposed into countably-infinite number of sine waves of given form and we need to know their 'quantitities'. Fourier coefficients are the answer to that problem and Fourier transform is its continuous analogon. So $c_n(T)$ themselves are just a discrete form of FT. By letting $T\to\infty$ and noticing Riemannian sum we get the desired convergence: $$f(t)=\lim_{T\to\infty}\sum_\limits{n=-\infty}^{\infty}e^{2\pi i\frac{n}{T} t}\int_{-T/2}^{T/2}f(t)e^{-2\pi i \frac{n}{T}t}\mathrm{d}t\frac{1}{T}=$$ $$\int_{-\infty}^{\infty}e^{2\pi i\xi t}\int_{-\infty}^{\infty}f(t)e^{-2\pi i\xi t}\mathrm{d}t\mathrm{d}\xi=\int_{-\infty}^{\infty}F(\xi)e^{2\pi i\xi t}\mathrm{d}\xi.$$ where $F(\xi)$ is the FT. We arrived with the continuous form. Letting $T\to\infty$ can be thought of as considering periodic function with an infinite period, so non-periodic actually.

The only question left is why are the coefficients given by such a formulae. That is where the orthogonality comes in. We know that pure tone waves of different frequencies are orthogonal (from inner product and T L Davis comment). So we can form an orthognal linear basis from solely pure tone waves. It follows that there exists a unique representation of any function in such a basis and the 'quantity' coefficients are given by inner product of considered function with each sine. That's the general idea standing behind the whole thing. Now we have to step back to periodic functions once again. If some function $f$ is periodic on interval $[-\frac{T}{2},\frac{T}{2}]$ then it's sufficient to consider inner product over that interval only. $$\left<f(t),e^{2\pi i \xi_n t}\right>=\int_{-T/2}^{T/2}f(t)\overline{e^{2\pi i\xi_n t}}\mathrm{d}t=\int_{-T/2}^{T/2}f(t)e^{-2\pi i\xi_n t}\mathrm{d}t,$$ which is nearly exactly what we had for $c_n(T)$ and $\xi_n$ denotes $\frac{n}{T}$. There's still $\frac{1}{T}$ left. Notice that norm of pure tone wave is $$||e^{2\pi i \xi_n t}||^2=\int_{-T/2}^{T/2}e^{2\pi i \xi_n t}e^{-2\pi i \xi_n t}\mathrm{d}t=\int_{-T/2}^{T/2}1\mathrm{d}t=T.$$

We can see now, that $$c_n(T) = \frac{\left<f(t),e^{2\pi i \xi_n t}\right>}{||e^{2\pi i \xi_n t}||^2}=\cos(\vartheta_n)\frac{||f(t)||}{||e^{2\pi i \xi_n t}||}$$ Extension to continuous form is then more obvious as $\cos(\cdot)$ is bounded, so we get rid of the infinities when $T\to\infty$.

Knowing that we can understand the integral for Fourier transform as taking inner product of a particular signal with pure tone waves frequency by frequency. Note that FT as defined without division by norm, it informally allows to take infinite values and occurence of delta functions. That explains it enough for me.

With the above we're ready to answer questions stated in my post. Terms cancelletion is now obvious from orthogonality. I've asked about a constant wave. It's orthogonal to all non-cosntant basis function. So $$ \mathcal{F}\{C\}(\xi)= \begin{cases} \infty &\mathrm{when} \quad \xi=0 \\ 0 &\mathrm{elsewhere} \\ \end{cases} $$ That goes straightaway from inner product of constant. However, we have to watch out because of limit. We have to distinct between $\delta(k)$ and $C\delta(k)$ they're not the same when though of as limits. The above is $C\delta(k)$ then. What's interesting, if there's a constant component in any singal we can extract it by just integrating this function over the real line.

Focus now on even and odd functions. We notice, that even ones have pure real FT and odd ones pure imaginary. Every function can be easily decomposed into its even and odd part, then we could look at their transforms separately. For example let $f(t)$ be ractantgular pulse on interval $[0,1]$ with amplitude $1$. It's neither odd nor even. Notice that $f(t)=f_e(t)+f_o(t)$, where $f_e(t)$ is recantagular pulse of amplitude $1$ on interval $[-1,1]$ and $f_o(t)$ is $-1$ on $[-1,0]$ and $1$ on $[0,1]$. It can be shown that that first term FT is $\mathrm{sinc}(\xi)$ and second $2i\sin(\xi/2)\mathrm{sinc}(\xi/2)$. They're in fact pure real and pure imaginary, what's more they're even and odd respectively again.

Matrix interpretation is now also easily understable with a bit of linear algebra, but as I see know it's more suitable for discrete Fourier transform, as it regards finite representations. From spectral theorem we know which matrices can be diagonolized. We have $A=U^{-1}DU$, where U is matrix of transition to an orthogonal basis and D is diagonal. D stands for discrete FT. Most probably it has some interesting properties, but we'll leave it there.

I hope somebody finds this thoughts interesting. Feel free to share your ideas or correct me in case I'm wrong.

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    $\begingroup$ You might enjoy this answer I provided to explain Laplace transforms. You probably know that the Fourier transform of a function is just the values of its Laplace transform along the $s=j\omega$ imaginary axis, so understanding one greatly helps you understand the other. My discussion there is somewhat complementary to yours here. I discuss the matrix representation, orthogonality, an introduction to the functional analysis perspective, etc, so take a look $\endgroup$ – jnez71 May 3 '17 at 17:23
  • $\begingroup$ Thanks, I was about to move on to Laplace tranform, so that's a spot on. $\endgroup$ – piwox May 3 '17 at 17:53

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