An Example

Consider a discrete signal $x[n]$ of $N=8$ complex components:

$${\bf x}=\left[ \begin{array}{c} x[0] x[1] x[2] x[3] ... ...3,0) (4,0) (0,0) (0,0) (0,0) \end{tabular} \right]$$

Note that Fourier transform is a complex transform. In general, the product of two complex numbers $A=A_r+j A_i$ and $B=B_r+j B_i$ is also complex:

$$A \times B=(A_r+j A_i)(B_r+j B_i)=(A_rB_r-A_iB_i)+j(A_rB_i+A_iB_r)$$

The foreword discrete Fourier transform is:

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

or

$$\left\{ \begin{array}{l} X_r[n]=\sum_{m=0}^{N-1} x_r[m]cos(2... ..._i[m]cos(2\pi mn/N)-x_r[m]sin(2\pi m n/N) \end{array} \right.$$

The inverse transform is:

$$x[m]=x_r[m]+jx_i[m]=\frac{1}{N}\sum_{n=0}^{N-1} X[n] e^{j2\pi m n/N}$$

or

$$\left\{ \begin{array}{l} x_r[m]=\sum_{n=0}^{N-1} X_r[n]cos(2... ..._i[n]cos(2\pi mn/N)+X_r[n]sin(2\pi m n/N) \end{array} \right.$$

These transforms can be expressed in matrix forms. For example, the forward transform is:

$$\left[ \begin{array}{c} X[0] X[1] X[2] X[3] X[4] ... ...] x[3] x[4] x[5] x[6] x[7] \end{array} \right]$$

For this specific example of 8-point DFT, we have

$$w[m,n]=e^{-j2\pi mn/N}=cos(2\pi mn/N)-j\;sin(2\pi mn/N) =cos(\pi mn/4)-j\;sin(\pi mn/4)$$

and

$$\left[ \begin{tabular}{r} (9.0,0.0) (-6.1,-4.1) (2.0,... ...,0) (4,0) (0,0) (0,0) (0,0) \end{tabular} \right]$$

If the signal $x[m]$ is real, then its spectrum $X[n]$ has the following properties:

• The real part of $X[0]$ is the average or DC component of the signal, the imaginary part of $X[0]$ is zero;
  
  $$Re\{X[0]\}=\frac{1}{N} \sum_{m=0}^{N-1}x[m] cos(2\pi m 0/N) =\frac{1}{N}\sum_{m=0}^{N-1} x[m]$$
  
  
  
  
  $$Im\{X[0]\}=-\frac{1}{N} \sum_{m=0}^{N-1} x[m] sin(2\pi m 0/N)=0$$
  
  

• The real part of $x[N/2]$ is the difference between the even components and odd components, the imaginary part of $X[N/2]$ is zero:
  
  $$Re\{X[N/2]\}=\frac{1}{N} \sum_{m=0}^{N-1}x[m] cos(2\pi m N/2N)= \frac{1}{N} \sum_{m=0}^{N-1}x[m] (-1)^m$$
  
  
  
  
  $$Im\{X[N/2]\}=-\frac{1}{N} \sum_{m=0}^{N-1}x[m] sin(2\pi m N/2N)=0$$
  
  

• When $0 < n <N/2$, the real part of $X[n]$ is even symmetric and the imaginary part of $X[n]$ is odd symmetric:
  
  $$Re\{X[N-n]\}=\frac{1}{N} \sum_{m=0}^{N-1}x[m] cos(2\pi m (N-n... ... =\frac{1}{N} \sum_{m=0}^{N-1}x[m] cos(2\pi m n/N)=Re\{X[n]\}$$
  
  
  
  
  $$Im\{X[N-n]\} = -\frac{1}{N} \sum_{m=0}^{N-1}x[m] sin(2\pi (N... ...\frac{1}{N} \sum_{m=0}^{N-1}x[m] sin(2\pi n m /N)=-Im\{X[n]\}$$