# Fourier Transforms, symmetry, real/imaginary

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?

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,

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

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

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

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

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