Fourier Transform Complex Conjugate Discussion

FT of function $f(t)$ is to take integration of the product of $f(t)$ and $e^{-jOmega t}$. By separating these two term into real and imaginary forms, the FT can be written as follow: 

$egin{align*}mathcal{F}Big( f(t) Big) &= int_{-infty}^{infty}f(t)e^{-jOmega t}dt\
&=int_{-infty}^{infty}ig[f_R(t)+if_I(t)ig]ig[cos(-Omega t)+isin(-Omega t)ig]dt\
&=int_{-infty}^{infty}Big{f_R(t)cos(-Omega t)-f_I(t)sin(-Omega t)+iBig[f_R(t)sin(-Omega t)+f_I(t)cos(-Omega t)Big]Big}dt\
&=int_{-infty}^{infty}f_R(t)cos(-Omega t)dt-int_{-infty}^{infty}f_I(t)sin(-Omega t)dt+iint_{-infty}^{infty}f_R(t)sin(-Omega t)dt+iint_{-infty}^{infty}f_I(t)cos(-Omega t)dt
end{align*}$

Now, consider a function $g(t)=f(-t)$, and take the FT on function $g(t)$:

$egin{align*}mathcal{F}Big( g(t) Big) &= int_{-infty}^{infty}g(t)e^{-jOmega t}dt\
&=int_{-infty}^{infty}f(-t)e^{-jOmega t}dt\
&=int_{infty}^{-infty}f(v)e^{-jOmega(-v)}d(-v) qquad letting v=-t\
&=int_{-infty}^{infty}f(v)e^{jOmega v}dv\
&=int_{-infty}^{infty}ig[f_R(v)+if_I(v)ig]ig[cos(Omega v)+isin(Omega v)ig]dv\
&=int_{-infty}^{infty}Big{f_R(v)cos(Omega v)-f_I(v)sin(Omega v)+iBig[f_R(v)sin(Omega v)+f_I(v)cos(Omega v)Big]Big}dv\
&=int_{-infty}^{infty}f_R(v)cos(Omega v)dv-int_{-infty}^{infty}f_I(v)sin(Omega v)dv+iint_{-infty}^{infty}f_R(v)sin(Omega v)dv+iint_{-infty}^{infty}f_I(v)cos(Omega v)dv\
&=int_{-infty}^{infty}f_R(v)cos(-Omega v)dv-int_{-infty}^{infty}f_I(v)sin(Omega v)dv-iint_{-infty}^{infty}f_R(v)sin(-Omega v)dv+iint_{-infty}^{infty}f_I(v)cos(Omega v)dvend{align*}$

Compare the derivations. Only if the function $f(t)$ is real ($f_I = 0$) can we receive the equations:

$egin{align*}
mathcal{F}Big(f(t)Big)
&=int_{-infty}^{infty}f_R(t)cos(-Omega t)dt+iint_{-infty}^{infty}f_R(t)sin(-Omega t)dt\
mathcal{F}Big( f(-t) Big)
&=int_{-infty}^{infty}f_R(t)cos(-Omega t)dt-iint_{-infty}^{infty}f_R(t)sin(-Omega t)dtend{align*}$

Which can be easily concluded that if $f(t)$ is real, the FT of $f(t)$ is complex conjugate to the FT of $f(-t)$

$color{red}{mathcal{F}Big(f(-t)Big) = F^{*}(jOmega) qquad for f(t) is real}$

Take FT on the complex conjugate function $f^{*}(t) = f_R(t) – if_I(t)$

$egin{align*}
mathcal{F}Big(f^*(t)Big)
&=int_{-infty}^{infty}f^*(t)e^{-jOmega t}dt\
&=int_{-infty}^{infty}Big[f_R(t)-if_I( t)ig]ig[cos(-Omega t)+isin(-Omega t)Big]dt\
&=int_{-infty}^{infty}Big{f_R(t)cos(-Omega t)+f_I(t)sin(-Omega t)+iBig[f_R(t)sin(-Omega t)-f_I(t)cos(-Omega t)Big]Big}dt\
&=int_{-infty}^{infty}Big{f_R(t)cos(Omega t)-f_I(t)sin(Omega t)+iBig[-f_R(t)sin(Omega t)-f_I(t)cos(Omega t)Big]Big}dt\
&=int_{-infty}^{infty}Big{f_R(t)cos(Omega t)-f_I(t)sin(Omega t)-iBig[f_R(t)sin(Omega t)+f_I(t)cos(Omega t)Big]Big}dt\
&=int_{-infty}^{infty}f_R(t)cos(Omega t)dt-int_{-infty}^{infty}f_I(t)sin(Omega t)dt-ileft{int_{-infty}^{infty}f_R(t)sin(Omega t)dt+int_{-infty}^{infty}f_I(t)cos(Omega t)dt ight}\
end{align*}$

Compare the equations. 

$egin{align*}
mathcal{F}Big(f(t)Big)
&=int_{-infty}^{infty}f_R(t)cos(-Omega t)dt-int_{-infty}^{infty}f_I(t)sin(-Omega t)dt+ileft{int_{-infty}^{infty}f_R(t)sin(-Omega t)dt+int_{-infty}^{infty}f_I(t)cos(-Omega t)dt ight}\
mathcal{F}Big(f^*(t)Big)
&=int_{-infty}^{infty}f_R(t)cos(Omega t)dt-int_{-infty}^{infty}f_I(t)sin(Omega t)dt-ileft{int_{-infty}^{infty}f_R(t)sin(Omega t)dt+int_{-infty}^{infty}f_I(t)cos(Omega t)dt ight}\
end{align*}$

The sign of $Omega$ and the sign of imaginary part have been changed. We can concluded that FT of the complex conjugate of function f is equal to the FT of the function f then do the complex conjugate and reverse on frequency domain.

$color{red}{mathcal{F}Big(f^*(t)Big) = F^*(-jOmega)}$