Is the result of Fourier transform always complex?I mean, no matter the signal,$g(t)$, is real or image.In my opinion,i think the answer is "No",because fourier transform is $$ G(f)=\int^{\infty}_{-\infty} g(t)*e^{-j2\pi ft}dt $$ If $g(t)$ is $e^{j2\pi ft}$ ,complex and even ,then $G(f)$ is not complex.So if the $g(t)$ is complex ,$G(f)$ must not complex,and if the signal is real or image,$G(f)$ must be complex,is my opinion right?
Is the result of Fourier transform always complex?
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fourier-transform
1 Answers
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There are indeed functions for which th Fourier transform is only real valued. The most prominent example would be the Gaussian function \begin{align*} f(x) = \alpha \exp(-\beta x^2). \end{align*} The important fact here is that $f$ is symmetric around $0$. In the Wikipedia article on Characteristic Functions (another name for Fourier transforms in probability theory) you can find more information and examples: https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)