MHB Property of real-valued Fourier transformation

[mathmari]

Hey!

When it is given that a signal $x(t)$ has a real-valued Fourier transformation $X(f)$ then is the signal necessarily real-valued?

I have thought the following:

$X_R(ω)=\frac{1}{2}[X(ω)+X^{\star}(ω)]⟺\frac{1}{2}[x(t)+x^{\star}(−t)]=x_e(t) \\ X_I(ω)=\frac{1}{2i} [X(ω)−X^{\star}(ω)]⟺ \frac{1}{2i}[x(t)−x^{\star}(−t)]=−i⋅x_o(t)$

where $X_R(ω)$ and $X_I(ω)$ are the real and imaginary parts of $X(ω)$, and $x_e(t)$ and $x_o(t)$ are the even and odd parts of $x(t)$, respectively.So the odd part of $x$ is $0$ and the even one is real-valued, and so the signal $x(t)$ is real-valued.Is everything correct? Are the above properties known or do we have to derive them? (Wondering)

 

[I like Serena]

Hey mathmari!

Wiki lists such a property.
If $X(\omega)$ is real, then $x(t)$ is Hermitian. That is, $x(-t)=x^*(t)$.
It still means that $x(t)$ can be imaginary, but the imaginary part must be odd. Additionally the real part must be even. (Nerd)

Consider for instance $X(\omega)=2\pi\delta(\omega-1)$. It's real isn't it?
Its inverse Fourier transform is $x(t)=\cos t+i\sin t$.
As you can see the real part is even and the imaginary part is odd.
Furthermore, the odd part $x_o(t)$ is indeed $0$, but the even part $x_e(t)$ has an imaginary component. (Worried)