Sign in

username:

password:



Not a member?

Search Online Books



Search tips

Free Online Books

Click Here!

Chapters

See Also

Embedded Systems

Mathematics of the DFT
    Derivation of the Discrete Fourier Transform (DFT)
       Orthogonality of the DFT Sinusoids

Chapter Contents:

Search Mathematics of the DFT

  

Book Index | Global Index

Would you like to be notified by email when Julius Orion Smith III publishes a new entry into his blog?

  

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, $

denote the $ k$th DFT complex-sinusoid, for $ k=0:N-1$. Then

\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*}

where the last step made use of the closed-form expression for the sum of a geometric series (§6.1). If $ k\neq l$, the denominator is nonzero while the numerator is zero. This proves

$\displaystyle \zbox {s_k \perp s_l, \quad k \neq l.} $

While we only looked at unit amplitude, zero-phase complex sinusoids, as used by the DFT, it is readily verified that the (nonzero) amplitude and phase have no effect on orthogonality.

Order a Hardcopy of Mathematics of the DFT

Previous: DFT Sinusoids
Next: Norm of the DFT Sinusoids
written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


Comments

No comments yet for this page


Add a Comment
You need to login before you can post a comment (best way to prevent spam). ( Not a member? )