Fourier Transform Complex Conjugate Discussion

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

$\begin{align*}\mathcal{F}\Big( f(t) \Big) &= \int_{-\infty}^{\infty}f(t)e^{-j\Omega t}dt\\
&=\int_{-\infty}^{\infty}\big[f_R(t)+if_I(t)\big]\big[cos(-\Omega t)+isin(-\Omega t)\big]dt\\
&=\int_{-\infty}^{\infty}\Big\{f_R(t)cos(-\Omega t)-f_I(t)sin(-\Omega t)+i\Big[f_R(t)sin(-\Omega t)+f_I(t)cos(-\Omega t)\Big]\Big\}dt\\
&=\int_{-\infty}^{\infty}f_R(t)cos(-\Omega t)dt-\int_{-\infty}^{\infty}f_I(t)sin(-\Omega t)dt+i\int_{-\infty}^{\infty}f_R(t)sin(-\Omega t)dt+i\int_{-\infty}^{\infty}f_I(t)cos(-\Omega t)dt
\end{align*}$

 

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

$\begin{align*}\mathcal{F}\Big( g(t) \Big) &= \int_{-\infty}^{\infty}g(t)e^{-j\Omega t}dt\\
&=\int_{-\infty}^{\infty}f(-t)e^{-j\Omega t}dt\\
&=\int_{\infty}^{-\infty}f(v)e^{-j\Omega(-v)}d(-v) \qquad letting\ v=-t\\
&=\int_{-\infty}^{\infty}f(v)e^{j\Omega v}dv\\
&=\int_{-\infty}^{\infty}\big[f_R(v)+if_I(v)\big]\big[cos(\Omega v)+isin(\Omega v)\big]dv\\
&=\int_{-\infty}^{\infty}\Big\{f_R(v)cos(\Omega v)-f_I(v)sin(\Omega v)+i\Big[f_R(v)sin(\Omega v)+f_I(v)cos(\Omega v)\Big]\Big\}dv\\
&=\int_{-\infty}^{\infty}f_R(v)cos(\Omega v)dv-\int_{-\infty}^{\infty}f_I(v)sin(\Omega v)dv+i\int_{-\infty}^{\infty}f_R(v)sin(\Omega v)dv+i\int_{-\infty}^{\infty}f_I(v)cos(\Omega v)dv\\
&=\int_{-\infty}^{\infty}f_R(v)cos(-\Omega v)dv-\int_{-\infty}^{\infty}f_I(v)sin(\Omega v)dv-i\int_{-\infty}^{\infty}f_R(v)sin(-\Omega v)dv+i\int_{-\infty}^{\infty}f_I(v)cos(\Omega v)dv\end{align*}$

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

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

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

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

 

 

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

$\begin{align*}
\mathcal{F}\Big(f^*(t)\Big)
&=\int_{-\infty}^{\infty}f^*(t)e^{-j\Omega t}dt\\
&=\int_{-\infty}^{\infty}\Big[f_R(t)-if_I( t)\big]\big[cos(-\Omega t)+isin(-\Omega t)\Big]dt\\
&=\int_{-\infty}^{\infty}\Big\{f_R(t)cos(-\Omega t)+f_I(t)sin(-\Omega t)+i\Big[f_R(t)sin(-\Omega t)-f_I(t)cos(-\Omega t)\Big]\Big\}dt\\
&=\int_{-\infty}^{\infty}\Big\{f_R(t)cos(\Omega t)-f_I(t)sin(\Omega t)+i\Big[-f_R(t)sin(\Omega t)-f_I(t)cos(\Omega t)\Big]\Big\}dt\\
&=\int_{-\infty}^{\infty}\Big\{f_R(t)cos(\Omega t)-f_I(t)sin(\Omega t)-i\Big[f_R(t)sin(\Omega t)+f_I(t)cos(\Omega t)\Big]\Big\}dt\\
&=\int_{-\infty}^{\infty}f_R(t)cos(\Omega t)dt-\int_{-\infty}^{\infty}f_I(t)sin(\Omega t)dt-i\left\{\int_{-\infty}^{\infty}f_R(t)sin(\Omega t)dt+\int_{-\infty}^{\infty}f_I(t)cos(\Omega t)dt\right\}\\
\end{align*}$

Compare the equations. 

$\begin{align*}
\mathcal{F}\Big(f(t)\Big)
&=\int_{-\infty}^{\infty}f_R(t)cos(-\Omega t)dt-\int_{-\infty}^{\infty}f_I(t)sin(-\Omega t)dt+i\left\{\int_{-\infty}^{\infty}f_R(t)sin(-\Omega t)dt+\int_{-\infty}^{\infty}f_I(t)cos(-\Omega t)dt\right\}\\
\mathcal{F}\Big(f^*(t)\Big)
&=\int_{-\infty}^{\infty}f_R(t)cos(\Omega t)dt-\int_{-\infty}^{\infty}f_I(t)sin(\Omega t)dt-i\left\{\int_{-\infty}^{\infty}f_R(t)sin(\Omega t)dt+\int_{-\infty}^{\infty}f_I(t)cos(\Omega t)dt\right\}\\
\end{align*}$

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.

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