Studying DSP on my own time on Coursera.

Was given a proof to why the Fourier basis is orthogonal, but I can't figure it out. Here is how it is proof goes.

Consider the Fourier basis $$ \left\lbrace \mathbf w^{(k)} \right\rbrace_{k=0,...,N−1} $$

defined as:

$$ \mathbf w^{(k)}_n = e^{−j\frac{2\pi}{N} nk} $$

Let us compute the inner product, that is

\begin{align} <\mathbf w^{(k)}, \mathbf w^{(h)} > &= \sum_{n=0}^{N-1} \mathbf w^{*(k)}[n] \mathbf w^{(h)}[n]\\ &= \sum_{n=0}^{N-1} e^{j\frac{2\pi}{N} nk} e^{−j\frac{2\pi}{N} nh}\\ & = \sum_{n=0}^{N-1} e^{-j\frac{2\pi}{N} n (h-k)}\\ &= \begin{cases} N & \text{if}\quad k=h\\[2ex] 0 & \text{otherwise} \end{cases} \end{align}

I do understand all of the derivation except the last step. How does one arrive at the conclusion that when $k\neq h$ the inner product is $0$. Would appreciate any help with that.


1 Answer 1


Just use the formula for the geometric series (I use $l=h-k\neq mN$):

$$\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nl}=\frac{1-e^{-j\frac{2\pi}{N}Nl}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-1}{1-e^{-j\frac{2\pi}{N}l}}=0,\quad l\neq mN$$

  • $\begingroup$ How does one solve this sum for $l=0$? It looks like it is undefined, $$\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nl}=\frac{1-e^{-j\frac{2\pi}{N}Nl}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-1}{1-1}=undefined$$ $\endgroup$
    – flashburn
    Jan 17, 2018 at 0:36

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