Fourier Transform 2 Real Functions with 1 DFT

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

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

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

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

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

where $X_r[n]$ and $X_i[n]$ are two real functions for the real and imaginary parts of the spectrum, respectively. Consider the following two cases:

• If $x[m]=x_r[m]$ is real, i.e., $x_i[m]=0$, 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 $x[m]=j x_i[m]$ is imaginary, i.e., $x_r[m]=0$, then we have
  
  $$\left\{ \begin{array}{c} X_r[-n]=-X_r[n] X_i[-n]=X_i[n] \end{array} \right.$$
  
  
  or
  
  $$X[-n]=X_r[-n]+jX_i[-n]=-X_r[n]+jX_i[n]=-X^*[n]$$
  
  

We note that half of the computation for the DFT of a real-valued signal $x[m]$ is wasted, because

• in the time domain, the imaginary components are all zero carrying no information.
• in the frequency domain, due to the symmetry (even or odd), half of the spectrum is redundant also carrying no independent information.

This waste of computation can be avoided by the following algorithm by which the DFTs of two real signals can be obtained by only one DFT.

We first note that an arbitrary function $f(x)$ can be decomposed into the even and odd components $f_e(x)$ and $f_o(x)$:

$$f_e(x)=\frac{f(x)+f(-x))}{2}=f_e(-x),\;\;\;\;\;\;\;\; f_o(x)=\frac{f(x)-f(-x))}{2}=-f_o(-x)$$

and

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

Now we can show how to obtain the DFT spectra $X_1[n]={\cal F}[x_1[m]]$ and $X_2[n]={\cal F}[x_2[m]]$ of two real functions $x_1[m]$ and $x_2[m]$ by carrying out only 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. Obtain the DFT of $x[m]$
   
   $${\cal F}[x[m]]=X[n]=X_r[n]+jX_i[n]$$
   
   

3. Separate $X[n]={\cal F}[x[m]]$ into $X_1[n]={\cal F}[x_1[m]]$ and $X_2[n]={\cal F}[x_2[m]]$, based on 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.