Question

Compute y = F₈c by the three steps of the Fast Fourier Transform if c = (1, 0, 1, 0, 1, 0, 1, 0). Repeat the computation with c = (0, 1, 0, 1, 0, 1, 0, 1).

Solution

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Answered 2 years ago
Answered 2 years ago
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The first step\textbf{The first step} is to separate c=(1,0,1,0,1,0,1,0)c=(1,0,1,0,1,0,1,0) into cc^{'} and cc^{''}.

c=(1,1,1,1)c=(0,0,0,0)c^{'}=(1,1,1,1)\quad\land\quad c^{''}=(0,0,0,0)

The second step\textbf{The second step} is to compute the values of yy^{'} and yy^{''}.

y=F4c=[11111ii2i31i2i4i61i3i6i9][1111]=[11111i1i11111i1i][1111]=[4000]y=F4c=[11111ii2i31i2i4i61i3i6i9][0000]=[0000]\begin{align*} y^{'}&=F_{4}c^{'}\\ &=\begin{bmatrix}1 & 1 & 1 & 1\\1 & i & i^2 & i^3\\1 & i^2 & i^4 & i^6\\1 & i^3 & i^6 & i^9\end{bmatrix}\begin{bmatrix}1\\1\\1\\1\end{bmatrix}\\ &=\begin{bmatrix}1 & 1 & 1 & 1\\1 & i & -1 & -i\\1 & -1 & 1 & -1\\1 & -i & -1 & i\end{bmatrix}\begin{bmatrix}1\\1\\1\\1\end{bmatrix}\\ &=\begin{bmatrix}4\\0\\0\\0\end{bmatrix}\\\\ y^{''}&=F_{4}c^{''}\\ &=\begin{bmatrix}1 & 1 & 1 & 1\\1 & i & i^2 & i^3\\1 & i^2 & i^4 & i^6\\1 & i^3 & i^6 & i^9\end{bmatrix}\begin{bmatrix}0\\0\\0\\0\end{bmatrix}\\ &=\begin{bmatrix}0\\0\\0\\0\end{bmatrix} \end{align*}

The third step\textbf{The third step} is to combine yy^{'} and yy^{''} into the vector yy.

y=(y,y)=(4,0,0,0,0,0,0,0)y=(y^{'},y^{''})=(4,0,0,0,0,0,0,0)

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