**7.2 Appendix B**

10 end 11 end 12 end

06 *a* ← *convertToSH IA*(*i*)

08 *c* ← *convertFromSH IA*(*b*)

<sup>07</sup> *<sup>b</sup>* <sup>←</sup> <sup>10</sup>−*<sup>α</sup>* <sup>⊗</sup> *<sup>a</sup>*

09 *T*(*c*) ← *t*(*i*)

This section presents the mathematical proof of GOSH-FFT.

Notation: The symbol % will be employed to mean modular arithmetic. Let *B* = 7*m*, where *m* is a positive integer. *N* = *Bp*, where *p* is a positive integer.

Let *M* denote the compound transform *M*10*<sup>m</sup>* from SHIA.

Let 1,*e*,*e*2,*e*3, ...,*eN*−<sup>1</sup> represent *N* roots of unity, where multiplication of two arbitrary roots of unity denoted *e<sup>i</sup>* and *e<sup>j</sup>* is defined as: *e<sup>i</sup> e<sup>j</sup>* = *e*(*i*+*j*)%*N*.

$$\varepsilon(u) = e^{(u \% B) + (u/B)u} = 1, e^1, \dots, e^{B-1}, \dots, e^{(B-1)u}, \dots, e^{(B-1)u + (B-1)}, \text{ for } p > 1 \text{ and } = 1, e^1, \dots, e^{B-1}, \text{ for } \mathbf{p} = 1.$$

*<sup>F</sup>*(*u*) = *<sup>F</sup>u*%*<sup>B</sup> <sup>u</sup>*/*<sup>B</sup>* <sup>=</sup> *<sup>F</sup>*<sup>0</sup> <sup>0</sup> , ..., *<sup>F</sup>B*−<sup>1</sup> <sup>0</sup> , ..., *<sup>F</sup>*<sup>0</sup> *<sup>n</sup>*−1, ...*FB*−<sup>1</sup> *<sup>n</sup>*−<sup>1</sup> , denote a sequance of *<sup>N</sup>* Fourier components.

*<sup>f</sup>*(*x*) = *<sup>f</sup> <sup>x</sup>*%*<sup>B</sup> <sup>x</sup>*/*<sup>B</sup>* <sup>=</sup> *<sup>f</sup>* <sup>0</sup> <sup>0</sup> , ... *<sup>f</sup> <sup>B</sup>*−<sup>1</sup> <sup>0</sup> , ..., *<sup>f</sup>* <sup>0</sup> *<sup>n</sup>*−1, ..., *<sup>f</sup> <sup>B</sup>*−<sup>1</sup> *<sup>n</sup>*−<sup>1</sup> , denote a sequence of *<sup>N</sup>* points in the input signal.

The proof is by induction on p. When p=1, GOSH-FFT is simply a DFT. Assume the GOSH-FFT computes a Fourier transform for all levels less than *p*.

$$M^{-1}(f\_{x/B}^{\chi\emptyset B}) = f\_0^0 \dots .f\_{n-1}^0 \dots .f\_0^{B-1} \dots .f\_{n-1}^{B-1} .$$

We now have *B* sub-signals, each of which is composed of *n* points. Then, by the induction hypothesis, we can apply GOSH-FFT to obtain *B* individual transforms of the *B* sub-signals; which yields:

Let

18 Will-be-set-by-IN-TECH

01 function T = *m*10(*t*, *α*) 02 *f ori* = 0 : *n* − 1

<sup>04</sup> *<sup>b</sup>* <sup>←</sup> <sup>10</sup>*<sup>α</sup>* <sup>⊗</sup> *<sup>a</sup>*

06 *T*(*c*) ← *t*(*i*)

07 end 08 end

03 *a* ← *convertToSH IA*(*i*)

<sup>02</sup> *subSignalSize* <sup>←</sup> *baselevel*

 *wLB* ← *i* × *base lB* ← *i* × *base* + *tab uB* ← *lB* + *base* − 1 *f oru* = *lB* : *uB*

<sup>02</sup> *subSignalSize* <sup>←</sup> *baselevel*

04 for s = 0: numSubSignals-1 05 for i =0: subSignalSize-1 06 *a* ← *convertToSH IA*(*i*)

08 *c* ← *convertFromSH IA*(*b*)

<sup>07</sup> *<sup>b</sup>* <sup>←</sup> <sup>10</sup>−*<sup>α</sup>* <sup>⊗</sup> *<sup>a</sup>*

09 *T*(*c*) ← *t*(*i*)

10 end 11 end 12 end

**7.2 Appendix B**

12 end 13 end 14 end 15 end

05 *c* ← *convertFromSH IA*(*b*)

01 function F = localDFT( f, w, level, n, base )

<sup>11</sup> *<sup>F</sup>*(*u*) <sup>←</sup> <sup>∑</sup>*base*−<sup>1</sup> *<sup>x</sup>*=<sup>0</sup> *<sup>f</sup>*(*<sup>x</sup>* <sup>+</sup> *lB*) <sup>×</sup> *<sup>w</sup>*(*<sup>x</sup>* <sup>+</sup> *wLB*)

01 function T = localM10Inverse( t, level, n, base )

This section presents the mathematical proof of GOSH-FFT.

is a positive integer. *N* = *Bp*, where *p* is a positive integer. Let *M* denote the compound transform *M*10*<sup>m</sup>* from SHIA.

Notation: The symbol % will be employed to mean modular arithmetic. Let *B* = 7*m*, where *m*

03 *numSubSignals* ← *n*/*subSignalSize*

 *numSubSignals* ← *n*/*subSignalSize f ors* = 0 : *numSubSignals* − 1 *tab* ← *s* × *subSignalSize f ors* = 0 : *subSignalSize* − 1

$$\begin{split} \hat{f} &= \hat{\mathbf{f}}\_{0}^{0}, \dots, \hat{\mathbf{f}}\_{n-1}^{0}, \dots, \hat{\mathbf{f}}\_{0}^{B-1}, \dots, \hat{\mathbf{f}}\_{n-1}^{B-1}, \\ M(\hat{f}) &= \hat{\mathbf{f}}\_{0}^{0}, \dots, \hat{\mathbf{f}}\_{0}^{B-1}, \dots, \hat{\mathbf{f}}\_{n-1}^{0}, \dots, \hat{\mathbf{f}}\_{n-1}^{B-1} \\ M(e(u)) &= e((u/B) + (u\%oB)n) \\ &= 1, e^{n}, \dots, e^{(B-1)n}, e^{1}, e^{n+1}, \dots, e^{(B-1)n+1}, \dots, e^{n-1}, e^{2n-1}, \dots, e^{Bn-1}) \end{split}$$

Perform a local DFT on each of the *n* groups of *B* points. The general term is:

$$\sum\_{q=0}^{B-1} \hat{F}\_{u/B}^q e^{q\left(\left(\mu/B\right) + \left(\left(\mu \% B\right)n\right)\right) \% N} \tag{8}$$

$$=\sum\_{q=0}^{B-1} \left[ \sum\_{r=0}^{n-1} f\_r^q e^{Br\left(u/B\right)\% N} \right] e^{q\left(u/B\right) + \left(\left(u\% B\right)n\right)\left\|\% N} \tag{9}$$

$$=\sum\_{q=0}^{B-1} \left[ \sum\_{r=0}^{n-1} f\_r^q e^{Bn\left(\mathbf{u}/B\right)\% N} e^{\left(\mathbf{u}\% B\right)rBn\% N} \right] e^{q\left(\left(\mathbf{u}/B\right) + \left(\left(\mathbf{u}\% B\right)\right)n\right)\% N} \tag{10}$$

$$\mathbf{x} = \sum\_{q=0}^{B-1} \left[ \sum\_{r=0}^{n-1} f\_r^q e^{Br\left( (\mathbf{u}/\mathcal{B}) + ((\mathbf{u}\% \mathcal{B})\mathbf{u}) \right) \% \mathcal{N}} \right] e^{q\left( (\mathbf{u}/\mathcal{B}) + ((\mathbf{u}\% \mathcal{B})\mathbf{u}) \right) \% \mathcal{N}} \tag{11}$$

$$=\sum\_{q=0}^{B-1} \left[ \sum\_{r=0}^{n-1} f\_r^q e^{(Br+q)(\left(u/B\right)+\left(\left(u^q \% B\right)n\right))\% N} \right] \tag{12}$$

$$=\sum\_{s=0}^{N-1} f\_{s/B}^{s\%B} e^{s\left(\left(\mu/B + \left((\mu\%B)n\right)\right)\right)\%N} \tag{13}$$

$$=F((
u/B)+(
u\%B)n))\tag{14}$$

= *F*(*u*)

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