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Is the frequency spectrum of $x^*(t)$ the same as the one of $x(t)$.

I'm telling this because I found that $$X^\ast(f) = (X(-f))^\ast$$

So to find $X^\ast(f)$, intuitively i would just flip $X(f)$ because of the minus sign and reflip it because of the conjugate. Then flipping two times gives me the original spectrum.

Tell me if I'm doing something wrong.

Thank you very much.

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  • $\begingroup$ No, conjugation flips the sign of the imaginary part, not the the function itself, and so there is no "reflip because of the conjugate" as you seem to think. $\endgroup$ – Dilip Sarwate Jun 19 '15 at 2:46
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If $X(f)$ is the Fourier transform of $x(t)$, then the Fourier transform of $x^*(t)$ is given by $X^*(-f)$. That's why for real signals $X(f)=X^*(-f)$ holds.

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