I am relatively new to Fourier transforms, so I apologize for the rather basic nature of my question, but, despite much googling, I was not able to find a clear answer on-line, so I must be stuck with some misconception.

I do understand that the Fourier transform of a **real** valued function *f: R => R* is, in general, a **complex** valued function, which encodes both the amplitude and phase of the harmonics that make up the function itself. What I am having trouble with is understanding how the *inverse* transform of the Fourier transform of a real valued function *f* (which is complex valued) can return the original **real** valued function *f*. As far as I can tell, the inverse Fourier transform will, in general, return a complex valued function, not a real valued one. I am obviously missing something. Could anybody point me to the right direction and/or suggest some resources that would help me clarify this point?

Thank you so much for your help.

## Best Answer

If $f(t)$ is a real-valued function, then its Fourier transform $F(\omega)$ may be a complex function, but $F(\omega)$ has the property that $F(-\omega)$ is the complex conjugate of $F(+\omega)$. The inverse Fourier transform of any function having this property is real.

A related fact is that any real function that is symmetric about the origin, so that $f(t)=f(-t)$, has a Fourier transform $F(\omega)$ that is a real function.