Fourier Transforms, symmetry, real/imaginary

Question

I was hoping to clarify if the following was correct:

• A real function (neither even nor odd) in time exhibits conjugate symmetry in frequency, so the real part of the frequency response is even, and the imaginary portion is odd

• A real, even function has a frequency response that is strictly real and even

• An imaginary, odd function has a frequency response that is strictly imaginary and odd

-also: Can anything be said if the function were imaginary in time (neither odd nor even)? Would it be anti-conjugate symmetric, so the frequency response would be such that the real part is odd and the imaginary portion is even?

Answer 1

Your suspicion is correct. You can show this as follows:

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$

$$X^*(\omega)=\int_{-\infty}^{\infty}x^*(t)e^{j\omega t}dt\tag{2}$$

Now if $x(t)$ is purely imaginary we have $x^*(t)=-x(t)$, and, consequently,

$$X^*(\omega)=-\int_{-\infty}^{\infty}x(t)e^{j\omega t}dt=-X(-\omega)\tag{3}$$

Answer 2

Based on \begin{equation} x(-t) \rightarrow {X}(-\omega) \hspace{1cm} (time-reversal) \end{equation} \begin{equation} {x^*}(t) \rightarrow {X^*}(-\omega) \hspace{1cm} (conjugation) \end{equation}

if we were able to split a function $x(t)$ into real/imaginary parts and then further into even/odd functions,

\begin{equation} x(t) = {x_R^E}(t) + {x_I^E}(t) +{x_R^O}(t)+ {x_I^O}(t) \end{equation} \begin{equation} \hspace{1cm} \big\downarrow \hspace{1cm} \big\downarrow \hspace{2.0cm} \searrow \hspace{1.0cm} \swarrow \end{equation} \begin{equation} X(\omega) = {X_R^E}(\omega) + {X_I^E}(\omega) +{X_R^O}(\omega)+ {X_I^O}(\omega) \end{equation}

real functions exhibit conjugate symmetry and imaginary functions shows anti-conjugate symmetry