# Meaning of Real and Imaginary part of Fourier Transform of a signal

Say $f$ is a signal of time $t$, $F$ its Fourier transform of the variable $v$.

It is known that in polar coordinate, $|F(v)|$ tells us how much the frequency $v$ is present over the signal, and $Arg(F(v))$ tells us how much the contribution of this frequency is phase-shifted.

What information do its real and imaginary part tell us ?

Or if I reformulate my question : can we give an interpretation of the Fourier transform in Cartesian coordinate like we can do in polar coordinate ?

The real and imaginary parts of the Fourier transform of a signal $x(t)$ are the Fourier transforms of the signal's even and odd parts, respectively:

$$X_R(\omega)=\frac12[X(\omega)+X^*(\omega)]\Longleftrightarrow\frac12[x(t)+x^*(-t)]=x_e(t)\\ X_I(\omega)=\frac{1}{2j}[X(\omega)-X^*(\omega)]\Longleftrightarrow\frac{1}{2j}[x(t)-x^*(-t)]=-j\cdot x_o(t)$$

where $X_R(\omega)$ and $X_I(\omega)$ are the real and imaginary parts of $X(\omega)$, and $x_e(t)$ and $x_o(t)$ are the even and odd parts of $x(t)$, respectively.

• Sorry to be dense, but I still don't get it. What do you mean by "even and odd parts" of a signal? (I also am not sure what the double arrow means in your notation.) May 26 '16 at 19:16
• Update: perhaps this has something to do with even and odd functions, as outlined here: cs.unm.edu/~williams/cs530/symmetry.pdf? May 26 '16 at 20:14
• @natevw: The double arrow means that the functions to its left and right form a Fourier transform pair. Every signal can be decomposed in its even and odd parts: $x(t)=x_e(t)+x_o(t)$, where $x_e(t)$ is an even function, and $x_o(t)$ is an odd function. May 26 '16 at 20:21
• Thanks, that clarifies your answer combined with the intro slides of the "symmetry" presentation I linked above! May 26 '16 at 22:00
• And what is j in the imaginary/odd part? Jan 9 '20 at 11:14

If there are equal frequencies but one is the negative of the other they will cancel and there will be zero imaginary signal.

Fourier transform of a system is its transfer function and gives the multiplication factor when $e^{j\omega t}$ is the input. $\omega$ is frequency. If you consider the input as current, the transfer function or Fourier transform as impedance then the output is potential. If Fourier transform is impedance, then the real part of FT is resistive part of the impedance and imaginary part is the reactive part of the impedance.

• While your point about resistive part/reactive part in linear systems may be realy interesting, in the current form, your answer is messy and hardly understandable. I'm downvoting it Jun 9 '15 at 15:24