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<math> \Re ( \chi (- \omega )) - \jmath \Im ( \chi (- \omega )) = \Re ( \chi (- \omega )) + \jmath \Im ( \chi ( \omega ))</math> | <math> \Re ( \chi (- \omega )) - \jmath \Im ( \chi (- \omega )) = \Re ( \chi (- \omega )) + \jmath \Im ( \chi ( \omega ))</math> | ||
Therefore, this splits into the even, real equation: | Therefore, this splits into the even, real equation: | ||
| − | <math> \Re ( \chi (- \omega )) = \Re ( \chi ( | + | <math> \Re ( \chi (- \omega )) = \Re ( \chi ( \omega )) </math> |
And the odd, imaginary equation: | And the odd, imaginary equation: | ||
<math>- \Im ( \chi (- \omega )) = \Im ( \chi ( \omega )) </math> | <math>- \Im ( \chi (- \omega )) = \Im ( \chi ( \omega )) </math> | ||
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<math>\chi (- \omega ) = \int_{-\infty}^{\infty} x(t) e^{ \jmath (-\omega) t} dt </math> | <math>\chi (- \omega ) = \int_{-\infty}^{\infty} x(t) e^{ \jmath (-\omega) t} dt </math> | ||
| + | |||
<math>\chi (- \omega ) = \int_{-\infty}^{\infty} x(-\tau) e^{ \jmath \omega \tau } dt </math> | <math>\chi (- \omega ) = \int_{-\infty}^{\infty} x(-\tau) e^{ \jmath \omega \tau } dt </math> | ||
| + | |||
| + | <math>\chi (- \omega ) = \int_{-\infty}^{\infty} x(\tau) e^{ \jmath \omega \tau } dt </math> (This swap is possible due to x(t)'s evenness) | ||
| + | |||
| + | <math>\chi (- \omega ) = \chi ( \omega )</math> | ||
| + | |||
| + | Thus, if x(t) is even, its Fourier Transform is also even. However, since we proved earlier that <math>\chi ( \omega )</math> even component is real, and that its imaginary component is odd, the odd section must be zero, so that the equation: <math>\chi ( \omega ) | even = \Re ( \chi ( \omega )) | even + \Im ( \chi ( \omega )) | odd</math> | ||
Revision as of 16:21, 20 April 2018
Determining the Properties of a Signal Based on its Fourier Transform
As part of this course, it is important to be able to examine the Fourier Transform of a signal, and tell if the original signal is real, pure imaginary, even, or odd. This article contains proof of properties that can help with this determination, and a few short example problems.
Table of Properties
| Real X(w) | Imaginary X(w) | |
| Even X(w) | Real and Even x(t) | Imaginary and Even x(t) |
| Odd X(w) | Imaginary and Odd x(t) | Real and Odd x(t) |
Proofs
The first proof, that the Fourier Transform of a real and even signal is also real and even, was completed by Prof. Boutin, as part of the class notes. I have put it here for convenience and make no attempt to claim it as my own. The remaining proofs are my own.
Real and Even Signals have Real and Even Fourier Transforms
Since x(t) is real:
$ x*(t) = x(t) $
$ \mathfrak{F}(x*(t)) = \mathfrak{F}(x(t)) $
$ \chi * ( - \omega ) = \chi ( \omega ) $ , by conjugation property of CTFT
$ \Re ( \chi (- \omega )) - \jmath \Im ( \chi (- \omega )) = \Re ( \chi (- \omega )) + \jmath \Im ( \chi ( \omega )) $ Therefore, this splits into the even, real equation: $ \Re ( \chi (- \omega )) = \Re ( \chi ( \omega )) $ And the odd, imaginary equation: $ - \Im ( \chi (- \omega )) = \Im ( \chi ( \omega )) $
This means that when x(t) is real, the real part of $ \chi ( \omega ) $ will be even, and the imaginary part will be odd. If, also x(t) is even:
$ x(-t) = x(t) $
$ \chi (- \omega ) = \int_{-\infty}^{\infty} x(t) e^{ \jmath (-\omega) t} dt $
$ \chi (- \omega ) = \int_{-\infty}^{\infty} x(-\tau) e^{ \jmath \omega \tau } dt $
$ \chi (- \omega ) = \int_{-\infty}^{\infty} x(\tau) e^{ \jmath \omega \tau } dt $ (This swap is possible due to x(t)'s evenness)
$ \chi (- \omega ) = \chi ( \omega ) $
Thus, if x(t) is even, its Fourier Transform is also even. However, since we proved earlier that $ \chi ( \omega ) $ even component is real, and that its imaginary component is odd, the odd section must be zero, so that the equation: $ \chi ( \omega ) | even = \Re ( \chi ( \omega )) | even + \Im ( \chi ( \omega )) | odd $
