Given the expressions for the DFT, the DCT, and the MDCT,
$$
\begin{aligned}
& X_{D F T}(k)=\frac{1}{\sqrt{2 M}} \sum_{n=0}^{2 M-1} x(n) e^{-j \pi n k / M}, 0 \leqslant k \leqslant 2 M-1 \\
& X_{D C T}(k)=c(k) \sqrt{\frac{2}{M}} \sum_{n=0}^{M-1} x(n) \cos \left[\frac{\pi}{M}\left(n+\frac{1}{2}\right) k\right], 0 \leqslant k \leqslant M-1
\end{aligned}
$$
where $c(0)=1 \sqrt{2}$, and $c(k)=1$ for $1 \leqslant k \leqslant M-1$
$$
\begin{aligned}
& X_{M D C T}(k)=\sqrt{\frac{2}{M}} \sum_{n=0}^{2 M-1} x(n) \underbrace{\sin \left[\left(n+\frac{1}{2}\right) \frac{\pi}{2 M}\right]}_{w(n)} \\
& \cos \left[\frac{(2 n+M+1)(2 k+1) \pi}{4 M}\right], \text { for } 0 \leqslant k \leqslant M-1
\end{aligned}
$$
FIGURE CANT COPY
Write the three transforms in matrix form as follows
$$
\mathbf{X}_{\mathbf{T}}=\mathbf{H x} \text {, }
$$
where $\mathbf{H}$ is the transform matrix, and $\mathbf{x}$ and $\mathbf{X}_{\mathrm{T}}$ denote the input and transformed vector, respectively. Note the structure in the transform matrices.