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From the Fourier transform I know that if $x(t)$ is a real function then $$X(j\omega)=X^*(-j\omega)$$ and that $$|X(j\omega)|=|X(-j\omega)|$$ $$\angle{X(j\omega)}=-\angle{X(-j\omega)}$$ My question is whether the magnitude plot is always even, for any complex function $x(t)$ , since $$|X(j\omega)|=\sqrt{a(\omega)^2+b(\omega)^2}$$ for $X(j\omega)=a(\omega)+j b(\omega)$

So, regardless of whether $a(\omega), b(\omega)$ be even or odd, can the magnitude plot be always even function of $\omega$.

I also have similar dilemma for the phase function $$\angle{X(j\omega)}=tan^{-1}\Biggr(\frac{b(\omega)}{a(\omega)}\Biggr)$$

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  • $\begingroup$ One way to tackle this type of question is try some simple functions and see if it holds. In this case a simple counter example is a complex sinusoid $x(t) = e^{j\omega_0 t}$. The FT is $X(\omega) = \delta( \omega-\omega_0)$ which is neither odd or even for $\omega_0 \neq 0$ $\endgroup$ Commented yesterday
  • $\begingroup$ What is the difference between $X^*(-j\omega)$ and $X^*(j\omega)$ and $X(-j\omega)$? $\endgroup$ Commented 16 hours ago

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No, for both magnitude and phase.

You can, in fact, choose any arbitrary $X(\omega)$ and find a corresponding $x(t)$. It is only for $X(\omega)$ with specific properties that $x(t)$ is real.

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