Discrete sine transformΒΆ

As we did to derive the discrete cosine transform, we consider a shifted discrete Fourier transform of \(2 N\) real numbers. Here, instead of the even symmetry, we consider a signal satisfying the odd symmetry:

\[x_{2 N - 1 - n} = - x_n\]

under \(\seq{n}{0}{1}{2 N - 1}\). Due to

\[\sum_{n = N}^{2 N - 1} x_n \twiddle{- 2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} = - \sum_{n = 0}^{N - 1} x_n \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N},\]
Derivation

With \(m = 2 N - 1 - n\), we have

\[ \begin{align}\begin{aligned}\sum_{m = N - 1}^{0} x_{2 N - 1 - m} \twiddle{- 2 \pi}{\left( 2 N - 1 - m + \frac{1}{2} \right) k}{2 N} & = - \sum_{n = 0}^{N - 1} x_n \exp \left( - 2 \pi k I \right) \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N}\\& = - \sum_{n = 0}^{N - 1} x_n \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N}.\end{aligned}\end{align} \]

we obtain

\[X_k = - 2 I \sum_{n = 0}^{N - 1} x_n \stwiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N}.\]

It is clear that \(X_k \in \mathbb{I}\). Note that

\[X_{2 N - k} = X_k,\]
Derivation
\[ \begin{align}\begin{aligned}X_{2 N - k} & = - 2 I \sum_{n = 0}^{N - 1} x_n \stwiddle{2 \pi}{\left( n + \frac{1}{2} \right) \left( 2 N - k \right)}{2 N}\\& = - 2 I \sum_{n = 0}^{N - 1} x_n \sin \left( \pi \left( 2 n + 1 \right) - 2 \pi \frac{\left( n + \frac{1}{2} \right) k}{2 N} \right)\\& = - 2 I \sum_{n = 0}^{N - 1} x_n \stwiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} \,\, \left( \because \cos \left\{ \pi \left( 2 n + 1 \right) \right\} = -1 \right)\\& = X_k.\end{aligned}\end{align} \]

and thus it is sufficient to consider \(\seq{k}{0}{1}{N}\). Additionally we easily see that \(X_0 \equiv 0\).

The inverse transform can be found as follows; due to

\[\sum_{k = N}^{2 N - 1} X_k \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} = - \sum_{k = 1}^{N - 1} X_k \twiddle{- 2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} + \left( -1 \right)^n I X_N,\]
Derivation
\[ \begin{align}\begin{aligned}\sum_{k = N}^{2 N - 1} X_k \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} & = \sum_{k = N}^{2 N - 1} X_{2 N - k} \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N}\\& = \sum_{l = N}^{1} X_l \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) \left( 2 N - l \right)}{2 N} \,\, \left( l \equiv 2 N - k \right)\\& = \sum_{k = 1}^{N} X_k \exp \left\{ \pi \left( 2 n + 1 \right) I \right\} \twiddle{- 2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N}\\& = - \sum_{k = 1}^{N} X_k \twiddle{- 2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} \,\, \left( \because \exp \left\{ \pi \left( 2 n + 1 \right) I \right\} = -1 \right)\\& = - \sum_{k = 1}^{N - 1} X_k \twiddle{- 2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} + \left( -1 \right)^n I X_N.\end{aligned}\end{align} \]

we obtain

\[ \begin{align}\begin{aligned}x_n & = \frac{1}{2 N} \left\{ \sum_{k = 0}^{N - 1} X_k \twiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} - \sum_{k = 1}^{N - 1} X_k \twiddle{- 2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} + \left( -1 \right)^n I X_N \right\}\\& = \frac{I}{N} \left\{ \sum_{k = 1}^{N - 1} X_k \stwiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} + \left( - 1 \right)^n \frac{X_N}{2} \right\}.\end{aligned}\end{align} \]

By substituting \(X_k\) with \(- I X_k\), we find the forward transform:

\[X_k = 2 \sum_{n = 0}^{N - 1} x_n \stwiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N}\]

and the backward transform:

\[x_n = \frac{1}{N} \left\{ \sum_{k = 1}^{N - 1} X_k \stwiddle{2 \pi}{\left( n + \frac{1}{2} \right) k}{2 N} + \left( - 1 \right)^n \frac{X_N}{2} \right\}.\]

To eliminate the case of \(k = 0\) (\(X_0 \equiv 0\)), we shift \(k\) and finally we obtain the discrete sine transform of type II and type III:

\[X_k = 2 \sum_{n = 0}^{N - 1} x_n \stwiddle{ 2 \pi }{ \left( n + \frac{1}{2} \right) \left( k + 1 \right) }{ 2 N }\]

for \(\seq{k}{0}{1}{N - 1}\), and

\[x_n = \frac{1}{N} \left\{ \sum_{k = 0}^{N - 2} X_k \stwiddle{2 \pi}{\left( n + \frac{1}{2} \right) \left( k + 1 \right)}{2 N} + \left( - 1 \right)^n \frac{X_{N - 1}}{2} \right\}\]

for \(\seq{n}{0}{1}{N - 1}\), respectively.

Note that both transforms are \(\mathbb{R}^N \rightarrow \mathbb{R}^N\).