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Cheers, I am trying to prove that a real and odd function/signal has imaginary and odd Fourier Transform. Although it seems fairly easy, I can't find a way to achieve it, and searching online hasn't helped a lot.

Edit:

To prove that it is imaginary I tried:

$$ F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t}dt = \\ \int_{-\infty}^{\infty} f(t) [\cos(\omega t) - j\sin(\omega t)]dt = \\ 0 - j\int_{-\infty}^{\infty} f(t)\sin(\omega t)dt $$

and I think this proves that it's imaginary indeed, using the fact that the integral of a even and odd function zero.

How would I go about proving it? Should I use Fourier Transform or could I do something wit the Fourier Series? Thanks a lot for any help!

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2 Answers 2

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You've already shown that the real part of $F(\omega)$ is zero. And the imaginary part is odd because $\sin(\omega t)$ is an odd function, i.e., $\sin(\omega t)=-\sin(-\omega t)$. Done.

You should also know that the Fourier transform $F(\omega)$ of a real-valued time domain function $f(t)$ always satisfies

$$F(\omega)=F^*(-\omega)\tag{1}$$

So in the special case that $F(\omega)$ is real-valued, from $(1)$ it must be even. And, also from $(1)$, if $F(\omega)$ is imaginary, its imaginary part must be odd.

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A suggestion that is helpful for such parity (even, odd) properties is to remember that every function $f(t)$ can be decomposed into an even and an odd part:

$$ h(t) = \frac{1}{2}\left(h(t)+h(-t)\right) + \frac{1}{2}\left(h(t)-h(-t)\right)$$

Simply computing the Fourier transform of the odd function $f(t)-f(-t)$ and checking the nullity of the real part $F(\omega)+F(-\omega)$, along with the conjugation of $\overline{e^{-j\omega t}}=e^{j\omega t}$ may yield results without splitting the exponential into sine and cosine.

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