The Length 2 DFT

The length $2$ DFT is particularly simple, since the basis sinusoids are real:

$$\begin{eqnarray*} \sv_0 &=& (1,1) \\ \sv_1 &=& (1,-1) \end{eqnarray*}$$

The DFT sinusoid $\sv_0$ is a sampled constant signal, while $\sv_1$ is a sampled sinusoid at half the sampling rate.

Figure 6.4 illustrates the graphical relationships for the length $2$ DFT of the signal $\underline{x}=[6,2]$.

Figure 6.4: Graphical interpretation of the length 2 DFT.

Analytically, we compute the DFT to be

$$\begin{eqnarray*} X(\omega_0) &=& \left<\underline{x},\sv_0\right> = 6\cdot 1 + ... ...=& \left<\underline{x},\sv_1\right> = 6\cdot 1 + 2\cdot (-1) = 4 \end{eqnarray*}$$

and the corresponding projections onto the DFT sinusoids are

$$\begin{eqnarray*} {\bf P}_{\sv_0}(\underline{x}) &\isdef & \frac{\left<\underlin... ...6\cdot 1 + 2 \cdot (-1)}{1^2 + (-1)^2} \sv_1 = 2 \sv_1 = (2,-2). \end{eqnarray*}$$

Note the lines of orthogonal projection illustrated in the figure. The ``time domain'' basis consists of the vectors $\{\underline{e}_0,\underline{e}_1\}$, and the orthogonal projections onto them are simply the coordinate axis projections $(6,0)$ and $(0,2)$. The ``frequency domain'' basis vectors are $\{\sv_0, \sv_1\}$, and they provide an orthogonal basis set that is rotated $45$ degrees relative to the time-domain basis vectors. Projecting orthogonally onto them gives ${\bf P}_{\sv_0}(\underline{x}) = (4,4)$ and ${\bf P}_{\sv_1}(\underline{x}) =(2,-2)$, respectively. The original signal $\underline{x}$ can be expressed either as the vector sum of its coordinate projections (0,...,x(i),...,0), (a time-domain representation), or as the vector sum of its projections onto the DFT sinusoids (a frequency-domain representation of the time-domain signal $\underline{x}$). Computing the coefficients of projection is essentially ``taking the DFT,'' and constructing $\underline{x}$ as the vector sum of its projections onto the DFT sinusoids amounts to ``taking the inverse DFT.''

In summary, the oblique coordinates in Fig.6.4 are interpreted as follows:

$$\begin{eqnarray*} \underline{x}\;=\; (6,2)&=& (4,4)+(2,-2)=4\cdot(1,1)+2\cdot(1,... ..._0 + \frac{X(\omega_1)}{\left\Vert\,\sv_1\,\right\Vert^2}\sv_1 \end{eqnarray*}$$