Fourier Transform 2 Real Functions with 1 DFT

First we recall the symmetry properties of the DFT. The DFT of $x[m]=x_r[m]+jx_i[m]$ is defined as

$$X[n] = \sum_{m=0}^{N-1} x[m]e^{-j2\pi mn/N}=\sum_{m=0}^{N-1} [x_r[m]+jx_i[m]] [cos(2\pi mn/N)-j\;sin(2\pi mn/N)]$$

$$= \sum_{m=0}^{N-1} [x_r[m]cos(\frac{2\pi mn}{N})+x_i[m]sin(\frac{2\... ... +j\sum_{m=0}^{N-1} [x_i[m]cos(\frac{2\pi mn}{N})-x_r[m]sin(\frac{2\pi mn}{N})]$$

$$= X_r[n]+jX_i[n]$$

where $X_r[n]$ and $X_i[n]$ are the real and imaginary part of the spectrum respectively. If $x[m]$ is real, i.e., $x_i[m] \equiv 0$, then we have

$$\left\{ \begin{array}{c} X_r[-n]=X_r[n] X_i[-n]=-X_i[n] \end{array} \right.$$

or

$$X[-n]=X_r[-n]+jX_i[-n]=X_r[n]-jX_i[n]=X^*[n]$$

If $x[m]$ is imaginary, i.e., $x_r[m] \equiv 0$, then we have

$$\left\{ \begin{array}{c} X_r[-n]=-X_r[n] X_i[-n]=X_i[n] \end{array} \right.$$

or

$$X[-n]=X_r[-n]+jX_i[-n]=-X_r[n]+jX_i[n]=-X^*[n]$$

Next we show how an arbitrary function $f(x)$ can be decomposed into the even and odd components $f_e(x)$ and $f_o(x)$:

$$\left\{ \begin{array}{c} f_e(x)=(f(x)+f(-x))/2=f_e(-x) f_o(x)=(f(x)-f(-x))/2=-f_o(-x) \end{array} \right.$$

and

$$f_e(x)+f_o(x)=f(x)$$

Now we are ready to show how to Fourier transform two real functions $x_1[m]$ and $x_2(m)$ to get their spectra $X_1[n]$ and $X_2[n]$ by one DFT.

1. Define a complex function $x[m]$ by the two real functions:
   
   $$x[m] \stackrel{\triangle}{=}x_1[m]+jx_2[m]$$
   
   
   Notice here that we impose $j$ on $x_2[m]$ to make it imaginary.
2. Find the DFT of $x[m]$
   
   $$DFT[x[m]]=X[n]=X_r[n]+jX_i[n]$$
   
   

3. Separate $X[n]$ into $X_1[n]$ and $X_2[n]$, the spectra of $x_1[m]$ and $x_2[m]$, using the symmetry properties discussed previously.
   • Since $x_1[m]$ is real, the real part of its spectrum $X_1[n]$ is the even component of $X_r[n]$ and the imaginary part of $X_1[n]$ is the odd component of $X_i[n]$, i.e.,
     
     $$X_1[n]=X_{1r}[n]+jX_{1i}[n]=\frac{X_r[n]+X_r[-n]}{2} +j\frac{X_i[n]-X_i[-n]}{2}$$
     
     

   • Since $jx_2[m]$ is imaginary, the real part of its spectrum $jX_2[n]$ is the odd component of $X_r[n]$ and the imaginary part of $jX_2[n]$ is the even component of $X_i[n]$, i.e.,
     
     $$jX_2[n]=j[X_{2r}[n]+jX_{2i}[n]]=\frac{X_r[n]-X_r[-n]}{2} +j\frac{X_i[n]+X_i[-n]}{2}$$
     
     
     Dividing both sides by $j$, we get
     
     $$X_2[n]=X_{2r}[n]+jX_{2i}[n]=\frac{X_i[n]+X_i[-n]}{2} -j\frac{X_r[n]-X_r[-n]}{2}$$
     
     

   Note that $X[-n]=X[N-n]$ because $X[n]$ is a periodic function.