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 )) $

This means that when x(t) is purely imaginary, the real part of $ \chi ( \omega ) $ will be odd, and the imaginary part will be even.

We already know from the previous proof that an even x(t) will CTFT into an even $ \chi ( \omega ) $ , and an odd x(t) will CTFT into an odd $ \chi ( \omega ) $. So, following the formula of $ \chi ( \omega )= \Re ( \chi ( \omega )) | even + \jmath \Im ( \chi ( \omega )) | odd $, when $ \chi ( \omega ) $ is real and odd, x(t) is imaginary and odd, and when $ \chi ( \omega ) $ is imaginary and even, x(t) is imaginary and even. Thus, we have proved every square in the table at the start of the document.

Sample Problems

Problem 1: $ \chi ( \omega ) = tan(\omega) $

We know that tangent is the quotient of two real functions, $ tan(\omega) = sin(\omega)/cos(\omega) $, and is thus real. We can also use the $ tan(\omega) = sin(\omega)/cos(\omega) $ formula to determine that since $ tan(-\omega) = -sin(-\omega)/cos(-\omega) $, $ tan(\omega) = -tan(-\omega)</omega>, and tangent is odd. With the knowledge that the tangent function is real and odd, we can conclude the inverse CTFT of <math>tan(\omega) $ is imaginary and odd.

Problem 2: $ \chi (\omega) = |\omega| $

By definition, absolute value is real and even. Therefore, $ |\omega| $ has an inverse CTFT that is real and even.

Problem 3: $ \chi (\omega) = sqrt(\omega) $

The square root function signal only has real output values in the first quadrant of the coordinate grid, and only imaginary values in the second quadrant. The symmetry required for being odd or even is not present, and thus neither the real nor the imaginary part of $ sqrt(\omega) $ is odd or even. This means none of the conditions on the table can be satisfied, and the inverse CTFT of $ sqrt(\omega) $ is not real, pure imaginary, even, or odd.