Orthogonality of the DFT Sinusoids
We now show mathematically that the DFT sinusoids are exactly orthogonal. Let
![$\displaystyle s_k(n) \isdef e^{j\omega_k nT} = e^{j2\pi k n /N} = \left[W_N^k\right]^n,
\quad n=0,1,2,\ldots,N-1,
$](http://www.dsprelated.com/josimages_new/mdft/img1025.png)
![$ k$](http://www.dsprelated.com/josimages_new/mdft/img20.png)
![$ k=0:N-1$](http://www.dsprelated.com/josimages_new/mdft/img1026.png)
![\begin{eqnarray*}
\left<s_k,s_l\right> &\isdef & \sum_{n=0}^{N-1}s_k(n) \overlin...
...i (k-l) n /N}
= \frac{1 - e^{j2\pi (k-l)}}{1-e^{j2\pi (k-l)/N}}
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/mdft/img1027.png)
where the last step made use of the closed-form expression for the sum
of a geometric series (§6.1). If , the
denominator is nonzero while the numerator is zero. This proves
![$\displaystyle \zbox {s_k \perp s_l, \quad k \neq l.}
$](http://www.dsprelated.com/josimages_new/mdft/img1029.png)
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Orthogonality of Sinusoids