Difference between revisions of "ECE301Spring18-DeterminingSignalPropertiesThroughFourierExam" - Rhea

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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

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 Signals

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 + \jmath \Im ( \chi ( \omega )) | odd $ Thus, real and even signals must have real and even Fourier Transforms.

If x(t) is odd, -x(-t) = x(t). Since $ -\chi (- \omega ) = -\int_{-\infty}^{\infty} x(t) e^{ \jmath (-\omega) t} dt = \int_{\infty}^{-\infty} x(t) e^{ \jmath (-\omega) t} dt = \int_{-\infty}^{\infty} x(t) e^{ \jmath (-\omega) t} dt $, the right side of the equation of the proof doesn't change, and $ -\chi (- \omega ) = \chi ( \omega ) $ is the result. Similarly to the x(t) being even case, we now know that x(t) is odd when $ \chi ( \omega ) $ is odd, and in order to satisfy $ \chi ( \omega ) | odd = \Re ( \chi ( \omega )) | even + \jmath \Im ( \chi ( \omega )) | odd $, the real part of $ \chi ( \omega ) $ must be 0. Thus, real and odd signals must have purely imaginary and odd Fourier Transforms.

Pure Imaginary Signals

Since x(t) is purely imaginary: $ -x*(t) = x(t) $

$ -\mathfrak{F}(x*(t)) = \mathfrak{F}(x(t)) $ (Negative is carried outside the CTFT by Linearity Property of CTFT)

$ -\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 odd, real equation: $ \Re ( \chi (- \omega )) = \Re ( \chi ( \omega )) $ And the even, imaginary equation: $ \Im ( \chi (- \omega )) = \Im ( \chi ( \omega )) $