I solved many problems on Fourier series,transforms and inverse fourier transforms as part of my academics. And i am aware that FT converts a time domain signal to frequency domain and IFT is vice versa.

How to visualise that FT really does convert a time domain signal to frequency domain?

My Approach:

Actually when i first thought about this i started with Fourier series. A function is expressed as sum of sine and cosine functions.Then i thought why only sine,cosine? which made me realise its related to right angle triangle (to get x and y co-ordinates of a point) and angle is related to the distance of point from origin. This is where omega*t creeps into theta of sine and cosine. And as x axis is time domain and t creeps in here.

Am i in the right path? Please guide me through this...

  • $\begingroup$ @SamL. So do you mean to say i can take fourier series of a function wrt to any two diff functions other than sine and cosine?? $\endgroup$
    – funtime
    Feb 18, 2013 at 15:39
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    $\begingroup$ You may like the betterexplained piece on FT $\endgroup$ Feb 18, 2013 at 15:44
  • $\begingroup$ @SamL. Ya i get it.. So i stepped onto a wrong path by assuming sine and cosine.. So there exist a relation between all the g_n 's you mentioned jus like sin2+cos2=1 ..am i right? so how do we go about fourier transforms domain conversion? $\endgroup$
    – funtime
    Feb 18, 2013 at 16:06
  • $\begingroup$ @alancalvitti That was really a very great article. Can you post it as answer $\endgroup$
    – funtime
    Feb 18, 2013 at 17:08
  • $\begingroup$ Thanks but it's just a link, not a proper answer. Although I will add if you really want to understand FT, especially visualizing it, it's instructive to compare it to transforms like Wavelet, which is a time-scale space rather than frequency-phase space technique. $\endgroup$ Feb 18, 2013 at 20:43

2 Answers 2


A Fourier transform represents the amount of oscillation of a particular frequency $\omega$ in a function. A function having one frequency is represented by a spike at that frequency. A periodic function is represented by spikes at an arithmetic sequence of frequencies. In general, higher frequencies represent faster variations in the original function. By performing a low-pass filter, we are cutting off the higher frequencies and "smoothing" out the function.

  • $\begingroup$ "particular frequency w in a function". How to determine this? Like in how do you visualise it... I understand this is basis of conversion of domains... so you start with a function in mind .and see at what frequency of that the sum matches to that of the value of original equation with some coefficient etc... Is this correct approach?? $\endgroup$
    – funtime
    Feb 18, 2013 at 16:19
  • $\begingroup$ I'm not really sure of what your are saying. I would just say that the FT measures the amount of "ringing" in the function. If the function is slowly varying, then its transform will be crowded in the low frequencies. (Thus, a wide gaussian becomes a narrow one.) If the function is quickly varying, then it will have significant high frequencies. (Thus, a narrow gaussian is transformed into a wide one.) $\endgroup$
    – Ron Gordon
    Feb 18, 2013 at 16:29
  • $\begingroup$ Not just the amount, but also the phase relation to each of the oscillations (basis functions). $\endgroup$ Feb 18, 2013 at 17:44

These are the links which made my visualisation complete..

For the ones with the same problem as i had u suggest them read the following in the same order for clarity

Start with this pdf for an intutution on why Fourier Transform works

Then this for a better clarity on imagination


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