2
$\begingroup$

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?

$\endgroup$
8
$\begingroup$

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.

| improve this answer | |
$\endgroup$
6
$\begingroup$

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.

| improve this answer | |
$\endgroup$
4
$\begingroup$

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.

| improve this answer | |
$\endgroup$
4
$\begingroup$

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:

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

Consider the case $\ f(t) = 2 \cos(\omega_0 t) = e^{+i \omega_0 t} + e^{-i \omega_0 t}.\ $ Then

$$ F(\omega) = \int\limits_{-\infty}^{+\infty} e^{i (-\omega + \omega_0) t} \ dt + \int\limits_{-\infty}^{+\infty} e^{i (-\omega - \omega_0) t} \ dt\\ $$

When $|\omega| \ne |\omega_0|$, both integrands oscillate around zero, and the integrals are effectively zero. The only non-zero results are

$$ F(\omega_0) = \int\limits_{-\infty}^{+\infty} e^{i (0) t} \ dt + \int\limits_{-\infty}^{+\infty} e^{i (-2 \omega_0) t} \ dt\ =\ \int\limits_{-\infty}^{+\infty} 1 \ dt\ +\ 0\\ F(-\omega_0) = \int\limits_{-\infty}^{+\infty} e^{i (2 \omega_0) t} \ dt + \int\limits_{-\infty}^{+\infty} e^{i (0) t} \ dt\ =\ 0\ +\ \int\limits_{-\infty}^{+\infty} 1 \ dt $$

which is often expressed as $F(\omega) = \delta(\omega - \omega_0) + \delta \big(\omega -(-\omega_0)\big) = \delta(\omega - \omega_0) + \delta(\omega + \omega_0).$

In words, for any given value of argument $\omega$, the $e^{-i\omega t}$ factor translates the component of $f(t)$ at that frequency to $0$, and all other components away from zero. Then the infinite integral produces a measure of the strength of the component at $0$.

Note that if $f(t) = e^{i\omega_0 t}$, then $F(\omega) = \delta(\omega - \omega_0)$. What this actually means is that the sign of $\omega_0$ can be unambiguously deduced from the function $e^{i\omega_0 t}$. It cannot be deduced from $\cos(\omega_0 t)$, because it is trigonometrically identical to $\cos(-\omega_0 t)$. The Fourier transform handles this ambiguity by giving non-zero responses at both $\omega=\omega_0$ and $\omega=-\omega_0$. That does not mean $\cos(\omega_0 t)$ contains both frequencies, because $\omega_0$ can have only one value. The correct interpretation is that $e^{i\omega_0 t}$ contains more information, not less, than $\cos(\omega_0 t)$. The formula $\ e^{+i \omega_0 t} + e^{-i \omega_0 t}\ $ looks like more information, but it is actually a cancellation of information.

| improve this answer | |
$\endgroup$
  • $\begingroup$ "That does not mean $cos(\omega_0 t)$ contains both frequencies, because $\omega_0$ can have only one value." No. The cosine is the sum of two complex pure tones of opposite frequencies (two distinct values). What you can't tell is the sign of $\omega_0$. Either is a valid interpretation, similar to picking a square root. So by convention, frequencies for real valued pure tones are considered positive. $\endgroup$ – Cedron Dawg Aug 25 at 22:31
  • $\begingroup$ @Cedron - Consider a function $f(x) = x^2 +ix$. $\ $And $\ \therefore\ f(-x) = x^2 -ix$ $\ x^2 = \tfrac{1}{2}(f(x) + f(-x))\ $ Should we conclude that $x^2$ is something more than just a function on the real number line? It is secretly made of two complex functions? If so, which two?... because I could just as easily have defined $f(x)$ as $x^2 +ix^3$. $\endgroup$ – Bob K Aug 26 at 0:00
  • $\begingroup$ This isn't about function decomposition. You could have just as readily said $f(x) = x^2 = x^{3/2} x^{1/2}$ for just as specious of an argument. The phrase "contains both frequencies" is in context of the FT (continuous in this case). If $cos$ only had one frequency there would only be one non-zero value in the spectrum. $\endgroup$ – Cedron Dawg Aug 26 at 0:10
  • 1
    $\begingroup$ I don't think it makes sense to argue how many frequencies a general signal contains, without agreeing about what "reasonable" decomposition into periodic functions is meant. A frequency is then just a shorthand expression for a periodic component of a frequency. A reasonable decomposition will not, for example, include components that completely cancel one another, or components that are identical. $\endgroup$ – Olli Niemitalo Aug 31 at 4:48
  • 1
    $\begingroup$ @Olli - Thanks for the editorial help with my deltas. I thought it didn't look quite right, but I didn't realize why. $\endgroup$ – Bob K Aug 31 at 11:18

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.