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

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

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.

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.

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

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

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.

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

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

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.