Paper 2, Section II, A

Numerical Analysis | Part II, 2010

The inverse discrete Fourier transform Fn1:RnRn\mathcal{F}_{n}^{-1}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} is given by the formula

x=Fn1y, where x=j=0n1ωnjyj,=0,,n1\mathbf{x}=\mathcal{F}_{n}^{-1} \mathbf{y}, \quad \text { where } \quad x_{\ell}=\sum_{j=0}^{n-1} \omega_{n}^{j \ell} y_{j}, \quad \ell=0, \ldots, n-1

Here, ωn=exp(2πi/n)\omega_{n}=\exp (2 \pi i / n) is the primitive root of unity of degree nn, and n=2p,p=1,2,n=2^{p}, p=1,2, \ldots

(1) Show how to assemble x=F2m1y\mathbf{x}=\mathcal{F}_{2 m}^{-1} \mathbf{y} in a small number of operations if we already know the Fourier transforms of the even and odd portions of y\mathbf{y} :

x(E)=Fm1y(E),x(O)=Fm1y(O)\mathbf{x}^{(\mathrm{E})}=\mathcal{F}_{m}^{-1} \mathbf{y}^{(\mathrm{E})}, \quad \mathbf{x}^{(\mathrm{O})}=\mathcal{F}_{m}^{-1} \mathbf{y}^{(\mathrm{O})}

(2) Describe the Fast Fourier Transform (FFT) method for evaluating x\mathbf{x} and draw a relevant diagram for n=8n=8.

(3) Find the costs of the FFT for n=2pn=2^{p} (only multiplications count).

(4) For n=4n=4, using the FFT technique, find x=F41y,\mathbf{x}=\mathcal{F}_{4}^{-1} \mathbf{y}, \quad for y=[1,1,1,1],\quad \mathbf{y}=[1,1,-1,-1], \quad and y=[1,1,1,1]\quad \mathbf{y}=[1,-1,1,-1]

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