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We know that Fourier transform $F(\omega)$ of function $f(t)$ is summation from $-\infty$ to $+\infty$ product of $f(t)$ and $e^{-j \omega t}$:

$$ F(\omega) = \int\limits_{-\infty}^{+\infty} f(t) \ e^{-j \omega t} \ dt $$

Here, what does the exponential term mean?

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It's a complex exponential that rotates forever on the complex plane unit circle:

$$e^{-j\omega t} = \cos(\omega t) + j \sin(\omega t).$$

You can think of Fourier transform as calculating correlation between $f(t)$ and a complex exponential of each frequency, comparing how similar they are. Complex exponentials like that have the nice quality that they can be time-shifted by multiplying them with a complex number of unit magnitude (a constant complex exponential). If the Fourier transform result at a particular frequency is a non-real complex number, then the complex exponential of that frequency can be multiplied by that complex number to get it shifted in time so that the correlation to $f(t)$ is maximized.

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If you don't like thinking about imaginary numbers, complex numbers and functions, you can alternatively think of the complex exponential in the FT as just shorthand for mashing together both a sinewave and a cosine wave (of the same frequency) into a single function that requires less chalk on the chalkboard to write.

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Whether it's the Fourier Transform or the Laplace Transform or the Z Transform, etc. the exponential is the eigenfunction of Linear and Time-invariant (LTI) operators. if an exponential function of "time" goes into an LTI, an exponential just like it (but scaled by the eigenvalue) comes out. what the F.T. does is break down a general function into a sum of these exponentials. that can be seen by looking at the inverse Fourier Transform.

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The Fourier Transform:

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(t)e^{i\omega t} dt\\ F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt$$

converts a function to an integral of harmonic functions. You can think of these as sins and cosines because $e^{i\theta} = cos(\theta) + i \sin(\theta)$. The Fourier Transform as a continuous form of the Fourier Series that transforms any periodic signal into a sum of other real periodic (harmonic) signals:

$$f(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega t) + b_n \sin(n\omega t)$$

In the Fourier Transform, you can think of the coefficients $a_n$ and $b_n$ going over the the values of a continuous function. To take the comparison further, there is a complex version of the series:

$$ f(t) = \sum_{n=-\infty}^{\infty} a_n e^{in\omega t} = \sum_{n=-\infty}^{\infty} a_n \cos(n \omega t) + b_n i \sin(n\omega t)$$

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  • $\begingroup$ Try to stick to one independent variable, either $t$ or $x$, but not both. Furthermore, please try to find a better word than 'hearken', which doesn't make any sense here. $\endgroup$ – Matt L. Apr 9 '15 at 15:31
  • $\begingroup$ You also miss $\omega$ in the arguments of the sinusoids and the exponential function: $\cos(n\omega t)$, etc. $\endgroup$ – Matt L. Apr 9 '15 at 16:43
  • $\begingroup$ @MattL. Do I need $\omega$? The Fourier Transform has $e^{i \omega t}$, but in the series, "$n$" takes the place of $\omega$. Isn't that right? $\endgroup$ – abalter Apr 9 '15 at 16:56
  • $\begingroup$ No, $\omega=2\pi/T$, where $T$ is the period of $f(t)$, i.e. unless $T=2\pi$ you need $\omega$. $\endgroup$ – Matt L. Apr 9 '15 at 17:32
  • $\begingroup$ Ok. I see what you mean. $\endgroup$ – abalter Apr 9 '15 at 18:12

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