A multidimensional analog of the Cooley-Tukey FFT algorithm Текст научной статьи по специальности «Математика»

operator to translate the codes of WD values read from the IML.

Introduction results of the automation procedures. The developed software has completely solved the tasks in view of independent working IML, and thanks to the automation procedures, allowing:

- to automate the process of independent working IMLs to 90 %, leaving the operator only the performance and the analysis of specific checks;

- to reduce time spent for working out concrete IMLs, from several weeks to 1-2 working days;

- to spend simultaneous working to 8 IMLs as a part of a workplace, connected among each other on interblock sockets and connected to the CPM;

- to check the working IML capacity during irregular situations, by their modelling;

- to independently fulfill each complete the IML set (basic/reserve) connected to each complete the CPM set (basic/reserve);

- to fulfill the BM in gathering, with use regular cables as the IML connections;

- to use the PM at any stage of REE tests thanks to flexibility and universality.

Currently, the given software - the independent working IML is used in space vehicle management blocks -“Monsoon”, “Glonass-to”, “Amos-5”; and is used at the working SV “Luch-5”. During the tests, the correctness of the construction of the software and correctness of the

approach connected with design of the automated procedures have been confirmed.

Thus, the developed software has proved its reliability, universality, and simplicity in use, thanks to what it is applicable for the working of the subsequent IML management block of perspective SVs. The procedures of the automated software and their algorithms are applicable for designing the software of workings REEs.

References

1. Prudkov V. V. The working place of autonomous processing of interfacial modules in the conjugation bloc of perspective SV control // Modern instrumental systems, information technologies and innovations: Materials of the VII International (distance) scientific conf. / Ozerski technological institute MIFI. Kursk, 2010. P. 150-152.

2. Pichkalev A. V. Testing radio-electric equipment at a workout laboratory complex // Reshetnev Conference : materials of the XII International science conf. ; SibSAU. Krasnoyarsk, 2008. P. 158-159.

3. Prudkov V. V. Particularities of constructing software for the autonomous performance of subsystems in the control bloc of perspective SV // Reshetnev Conference : Materials of the XII International science conf. ; SibSAU. Krasnoyarsk, 2009. P. 531-532.

© Prudkov V. V., 2010

A. V. Starovoitov Siberian Federal University, Russia, Krasnoyarsk

A MULTIDIMENSIONAL ANALOG OF THE COOLEY-TUKEY FFT ALGORITHM

In this article a recurring sequence of orthogonal basis in the n-dimensional case has been applied to derive

formulas of n-dimensional fast Fourier transform algorithm, which uses

2n -1 2n

Nn log2N complex multiplication and

nNn log2N complex addition; where N = 2s - is a number of counts on one of the axes.

Keywords: space of signals, orthogonal basis sequence, multidimensional discrete Fourier transform.

Recurrent sequence of orthogonal bases in space of signals is well studied [1] and has numerous applications, including the derivation of Fourier’s formulas of fast transformation.

In this article the recurrent sequence of orthogonal bases to a n-dimensional case is applied in order to derive formulas of a fast n-dimensional Fourier

transformation variant, using

2n -1 2n

Nn log2N complex

multiplication and nNn log2N complex addition, where N = 2s - is a number of counts on one of the axes

(known in studies as in [2]). This variant n FFT contains a smaller number of complex multiplication operations than other algorithms, where the multidimensional Fourier transformation is carried out by repeated application of one-dimensional FFT (for example, see [3; 4]).

Furthermore, we give definitions and basic statements from the theory of multidimensional signals, which are used in the article.

To construct n-dimensional recurrent sequence of orthogonal bases we use the scheme of the statement, given in [1] for a one-dimensional case.

1. The space of periodic n-dimensional signals. Definition 1. With a fixed N, the n-dimensional periodic signal shall be a periodic complex function of integer argument, with the period N on each variable.

Define operations of adding the two signals x1, x2 and multiplying the signal x by a complex number c :

У( j) = Xi( j) + x2( j); y( j) = c • x( j ),

where x( j) - is the count of a signal x at point j є Zn. Then, a set of signals CnN becomes a linear complex space. A zero element in CnN is the signal O such, that O( j) = 0 for all is j є Zn. Scalar produce and norm of space CnN are:

<x,y) = X x(j)y(j),

jeBn(N)

||x||= < x, x)1/2,

where Bn (N) - is a set of integer vectors from [0, N - 1]n .

Definition 2. The unit n -dimensional periodic impulse, with the period N on each variable, is a signal 8N such, that s; (j) = 1, if each coordinate of a vector j divided by N and s; (j) = 0 otherwise.

The Following statements are true for a unit impulse.

1) s;(ji,..., jn) = sn(|ji |,...,| jn |);

2) s; (ji,..., jn ) = s; (ji) • ...-s; (jn);

3) For x e CN the equality is true:

x( j)= S x(t)S"n ( j - t),

teBn ( N )

for any j є В" (N).

,2га.

Let wN = exp(-N).

Lemma 1. Then

w

( j ,t )

(2)

teB" ( N )

where ( j, t) - is the scalar product of vectors j and t. Equation is checked (2) by direct calculation. Definition З. The n-dimensional discrete Fourier transform is called a depiction: FN :C"N ^ C"N take each signal x to a signal X , where:

X( j)= S x(t)wNj■'>, j є Bn (N).

tEB" (N)

Note that, for DFT the formula of inversion is true:

x(t)=Nn S x ( j ) •

N jєBn ( N )

w

( j,t )

and the Parseval identity:

if X = Fn(x), Y = Fn(y),

< x, y) = X, Y ).

2. The recurrent sequences of orthogonal bases.

Let N = 2s, Nv = 2s

Av = 2v . We shall construct

recurrent sequence of basesf0,f,...,fs, where f - /-th basis, consisting of Nn signals f (k), k e Bn (N). We will denote a value of a signal f (k) at

c°unt j = (j^.. jn^ j e Bn(N) by f(k; j).

Let B1(N) by a set of integer vectors from [0, Nv - l]n and Bn2(N) by a set of integer vectors from [0, Av - 1]n. We will define the sequence of orthogonal bases as:

fo (k; j) = SN(j - k) = SN(ji - kl) X xSN (j2-k2)... S^ (jn-kn), k, j eB(J.

fv(li +aiAv + ^1Av+1, l2 +°2Av+ P2Av+i,‘"", ln +CT„Av + PnAv+l) =

1 1 STi (li + aiAv}

= S... S

w;

fv-1:

Ti=0 Tn =0

X (li + 2AVPi +XiAv,...,ln + 2AVPn +TnAvx (3)

where p = Cp^..^ pn ) e Bn(N), l = (l1,..., ln ) e Bn2(N), CT,.

is equal to 0 or i for all i = i,...,n,v = i,...,s .

For studying the properties of recurrent sequence of bases, we can use reverse rearrangement [i].

Let j by an integer from set J = {0,i,...,2V - i} be equal to jV-i2V-' +... + ji2 + j0 in a binary system, where (i) ji =0,i for alli = 0,...,V-i. A vector (j^,...,ji,j0)2 is called a binary code of number j . We compare number ji e J with number j, which is set by a binary code(j0, ji,..., j-i)2. Rearrangement revV(j) = ji for set J is called reverse rearrangement. For reverse rearrangements the following equalities are true:

2revV-! (q) = revV (q);

2revV-! (q) +1 = revV (Av + q).

Using a reverse rearrangement we can prove that:

fV (l1 + PlAv+1,..., ln + pnAV+1) =

(4)

Av+1-1 Av+1-1

Sjirevv (Q; )

= S ... S w/A=+1 f0(q1 + PlЛv+l,...,q" +pn

Q1=0 q" =0

where p = (Pl,..., pn ) є Bn(;), l =(11,...Л )є B"2(N):

v = 1,...,s .

In particular, if v = s we have:

N-1 N-1 Sl‘reVv (q‘} S1™' (* S

fs(l; і) = S... S wN=1 (j1- 9l,..., j»- 9» ) = wi .

0[=0 qn =0

12S

Theorem 1. For all v = 0,..., s, k e Bn (N), a set of

signals f, = f, (k) is orthogonal and || f (k) ||2 = 2nv .

The solution. Let v = 0. Then:

<fo(k), fo(k')> = X 8N (j - k) -SN (j - k')

jeBn (N)

the last sum can be distinct from zero only when k = k', in other case it is equal to 1 and the theorem is proved.

Let’s now v = 1,...,s andk,k' e Bn(N), which are presented in the following way:

k =(k1,..kn ) = (l1 + PlAv + 1,■■.ln + Pn Av + lK

k' = (kl' ...^n ) = (l1 + piAv+1,--.ln + p'nAv+1) ,

where l = (lj,..,ln), l' = (l/,..., ln) belongs to B2(N), and

P = (P1.....Pn), P' = (P[,-,P'n) belongs to Bn(N). Then:

<fv (kX fv (k= <fv (l1 + PlAv+1,...,ln + PnAv+1). fv (l1 + P1Av+1,..., ln + P'nAv+1)> =

/q(?1 + PlAv+l,•••, + PnAv+lX

Av+l- l Av+i-l X‘ireVv

=<x ••• X <vo+i

qi=o qn = 0

N

'v+l-l A 'v+l-i X‘irevv(q}

x,~ X wA=Q+i

/0 (qi+pî Av+i >•••> q'n + Pn Av+i)) =

ql=Q qn =Q

Av+1 -1 Av+1-1Av+1 -1 Av+1-1 XlireVv (?i)-,ireVv (q‘)

= £... £ £... £ <+1 X

qi=0 qn =0 q1=0 q'n =0

X SN (q1 -q1 +(P1 - Pl' ^l — qn -q'n +(Pn - Pn )Av+1).

Arguments of a unit impulse SnN on the module do not exceed N -1. For Pt = p' and at the some t arguments

are distinct from zero for all qt, q' e 0 : Av+1 -1,

i = 1,..., N, as | qt - qi |< Av+1 -1. Therefore

<fv(k),fv' (k ')> = 0, if Pj * pj.

Let Pi = p' . For all j = 1,...,n then:

X^ -li )revv (qi)

</v (k), /v (k ')) = X ••• X <+i

Av+l-l Av+l-l

q1=0 qn =0

n

Av+1-1 Av+1-1 £qi

= £ ... £ <+1 = Av+1...Av+1sa v+1 (l1 - i;,..., ln - ln).

qi=0 qn=0

From the last formula it is concluded, that the scalar product <fv (k, fv (k')> is distinct from zero only, if pj = pj, lt = l/, where i, j = 1,...,n . In the

last case || fv(k;,...,kn) || = Av+;...Av+; = 2nv for all k;,..., kn = 0: N -1. The theorem is now proved.

3^ Sequence application of orthogonal bases to denote the fast discrete Fourier transform

Letx(j) = x(;!,•••, jn) e CN, j e Bn(N) • We compare

a signal x( j) a to signal Xq( j) = x(rev^ ( jl),•••, revs ( jn ))

and we will spread out x0 (j) on basis /v :

xo=:èv X xv(k)-fv(k) (5)

2 keBn (N)

(^V - is a normalizing multiplier) • Multiplying both parts (5) by /v (l) scalar l e Bn (N) • Then

<Xq, /v (l)) = Xv (l),

Xv (k )= X Xo( j ) /v (k ) =

jeBn( N )

= X x(revs(jl),•••,revs(jn)).fv(k;jX

jeBn( N )

and coefficients xv (k ) in (5) are determined^

In particular, for v = 0 we have from (l):

x0(k)= X x(revs (jl),•••, revs ( jn )8N (j - k) =

jeBn( N )

= x(revs(ki),•••,revs (kn))•

From (3) we get the following:

xv (ll +SlAv + PA+l —ln + SnAv + P„Av+l) =

= < x0 = L (ll +SlAv + PA^"1 +8„ Av + Pn Av+l)) =

n

l l XTi (li +n' Av}

=X^X

(6)

t+i < X0 , fv-ldl + 2Av Pi + TlAv ,-,ln +

Tl=0 =0

(7)

l l XTi (li +n' Av}

+2AvPn + T A )) = X ■■■ X Wv+l X

Tl=0 =0

xxv-i(li + 2A^,Pi + TA,-, ln + 2Av p„ + TAX

where p = (Pl,•••, Pn ) e Bii(N), l = (ll,-Jn )e Bn2(N),

v = 1,„.,s and (^•••^ are equal to 0 or L As

Xs (k ) = x(ki,:; kn ) =

n -Xkirevs ( ji)

= X x(revs (jl),•••, revs (jn)) • WNi=0 =

jeB„ ( N )

n

-XLk'ji

= X x(j)Wn'=q = x(k),

jeBn ( N )

where k e Bn (N) and coefficients xs (k) define components of a spectrum for a signal x on a basic period^

From (6) and (7) we have received the recurrent scheme for the calculation of a spectrum for a signal x e CN :

X) (k) = x(revs (k;),..., revs (kn));

Xv (l1 + CT1Av + PlAv+1,...,ln + CTnAv + PnAv+1) = (8)

n

; ; £Ti (li+niAv)

= £... £ <v+i • xv-i x

T1=0 Tn =0

X (li + 2Av Pi +TiAv ,..., ln + 2Av Pn +Tn Av ),

where p ^Pp..^ pn ) e Bn(N), l = (ll,■■■, ln ) e Bn2(N),

v = 1,...,s and CT;,...,CTn are equal to 0 or i.

Let’s find a number of complex addition and multiplication operations necessary for finding a spectrum of the signal for scheme (8).

Lemma 2. For some r vectors t = (t;,...,tr) and

ct =(a1,..,CTr), where tt, ui e 0,1. Then the calculation of all the values of some function:

s (ct) = £f (t)(-i)<CT,t >

t

requires r • 2r additions (subtractions).

The solution. To prove we apply an induction onr.

Let r = 2. Then:

S(ct) = S(CTi,CT2) = £ £f(ti, t2) • (-i)CTltl+CT2t2 = t; =0t2 =0

= f (0,0) + f(i,0)(-i)CTl + f (0,i)(-i)CT2 +

+f (i,i)(-i)CT1+CT2.

Let’s define:

S;(ct) = S (0,0) = f (0,0) + f (1,0) + f (0,1) + f (1,1) =

= (f (0,0) + f (1,0)) + (f (0,1) + f (1,1)) = Si* + S3*;

S2 (ct) = S (1,0) = f (0,0) - f (1,0) + f (0,1) - f (1,1) =

= (f (0,0) - f (1,0)) + (f (0,1) - f (1,1)) = S2* + S*; S3(ct) = S (0,0) = f (0,0) + f (1,0) - f (0,1) - f (1,1) =

= (f (0,0) + f (1,0)) - (f (0,1) + f (1,1)) = Si* - S3*;

Si (ct) = S(0,0) = f (0,0) - f (1,0) - f (0,1) + f (1,1) =

= (f (0,0) - f (1,0)) - (f (0,1) - f (1,1)) = S* - S*, where:

Si* = f (0,0) + f (1,0), S* = f (0,0) - f (1,0),

S3* = f (0,1) + f (1,1), S* = f (0,1) - f (1,1).

For calculating S*, i = 1,2,3,4 it is required to apply

4 additions (subtractions); to calculate all values S it requires 8 such operations and then the statement of the lemma is correct.

Let the statement of the lemma be correct, if r = k ; for any function g (t) i. e. all values of the function:

S (стl,..., CTk ) = £g(t )(-i)<CT,t >

t

are calculated by 2 • 2k additions (subtractions), where t = (tl,..., tk).

Let’s consider r = k +1.

s(CT;,...CTk+i)=£... £ f(t;,..., tk+i) • (-i)CTiti+..+CTk+itk+1 =

t1=0 tk+1=0

=£... £ f (ti,..., tk ,0) • (-i)CTiti+..+CTktk + t; =0 tk =0

+£... £ f (t;,..., tk ,i) • (-i)CTiti+..+CTktk+CTk+1. ti=0 tk=0 Let’s denote:

S (CT; , CT2,..., CTk ,0) = S; (CT; , CTj,..., CTk ) =

= £... £ f (t;,..., tk ,0) • (-i)CTiti+.+CTktk + t; =0 tk =0

+£... £ f (ti,..., tk ,i) • (-i)CTiti+.+CTktk = t; =0 tk =0

= £... £( f (t;,..., tk ,0) + f (t;,..., tk ,1)) • (-1)CTltl+..+CTktk ; t; =0 tk =0

S (CTP CT2 ,..., CTk ,;) = S2 (CT1, CT2,..., CTk ) =

=£... £ f (t;,..., tk ,0) • (-i)CTiti+.+CTktk -t; =0 tk =0

-£... £ f (ti,..., tk ,i) • (-i)CTiti+.+CTktk = t; =0 tk =0

= £... £(f (t;,..., tk ,0) - f (t;,..., tk ,i)) • (-i)CTiti+..+CTktk.

t;=0 tk =0

Let’s define how many operations of addition (subtraction) are required for calculating S; and S2.

Let’s denote f+ (t;,..., tk ) = f (t;,..., tk ,0) + ,

+ f(t;,...,tk,1) f-(t;,...,tk) = f(t;,...,tk,0)-f(t;,...,tk,1).

For calculating all values f + required are 2k additions

and to calculate S; = £...£f+ (t;,...,tk)• (-i)CTlt1+-+CTktk t; =0 tk =0

required are k 2k according to the induction assumption.

Then the total sum of addition (subtraction) operations for calculation S; requires k2k + 2k. Such a number of operations it is necessary for S2 .

Therefore, to calculate of all values S (k + 1)2k+1 additions (subtractions) are required; this needed to be proved.

Theorem 2. To calculate the recurrent scheme for a signal spectrum x e CN (N = 2s)

x0(k) = x(revs (k;),..., revs (kn);

Xv (l1 +CT1 Av + Pl Av+1,..., ln + CTnAv + PnAv+1 ) = (9)

n

; ; £Ti (li+CTiAv)

- £••.£wa:!i • Xv-iX

X;=0 Tn =0

X(l1 + 2Av Pl +T1Av ,..., ln + 2Av Pn +TnAv X

2n -1

demands ^ Nn log2N complex multiplications and nNn log2N complex additions.

The solution. First, we will find a number of complex Let’s find the amount of complex addition in the multiplications. Complex multiplication is multiplication algorithm For fixed V* and l * = (l* l *)

n

Yj‘(li+ctav) p* = (p*,...,pD we have:

only by wj=0 . We shall define the number of

Xv(l1 + CT1A* + AA * +,,..., l„ + CTn Av* + PA* +1) =

products required in (9) for all parameters

v\l* = (ll*,...,ln*) andp* = (pl*,...,pn). For this purpose ^ ^ £(,+0iA.)

we shall consider products: = £ £ wi=0 • x X (i2)

T1=0 T„ =0 v +1

5*(li+CiAv*) x X (l1* + 2p*Av +тlAv*,...,l„ + 2p„Av+x„Av*).

w*=0 • X *

v +1

v -1

(l1* + 2P*Av + T1 Av* ,..., l„ + 2P„Av +T„Av*),

From (10) follows, that for calculation (12) we need to calculate expressions:

where ct* e 0,1. Y'il*

1T7- u + + f(t) = wA=0 x * (l* + 2p*A +t,A l* + 2p*A +t A *)

We have to notice that: JK J Av*+ v*-r1 v 1 v*’ 9„ Fn v „ v*'

Yt(l*+CTi av*) Y

V/ +tctA *

kA * rvA *

v +1 v +1

which depend only ont = (t1,..., Tn), where t, e 0:1. Then (12) can be presented as:

(10) xv*(li* +CT1Av * + PlAv * + 1,..., l* + CTn Av* + p'n Av* +1) =

Yv* Yctav* Yx'CTi Yv*

= wA=0 • w;A=0 =(-1)“'=0 • w;A=0

v +1 v +1 v +1

That is product:

TiCTi

1 1 Y = y...Y (-1)'=° • f(t).

t1=0 t„=0

n This way, complex addition is required from Lemma 2

£(l*+CTiAv*) „ „ to calculate:

WT! • x*1(li* + 2pi*Av+TiA „,...,l* + 2p„*Av+T„A,).

v +1 - x*(li +CT;Av* + P;Av*+l,...,l„ + CTnAv* + PnAv*+1) n2 .

As parameterv e1: s , vectors l = (l1,...,l„)

Let parameters v ,l =(l;,...,ln) and p =(p;,...,pK) by fixed; then product (10) can possibly be replaced with complex product: andp = (pi,..,pn), where = 0,1,...,N -1

(Nv = N / 2 v), l,. =0,1,..., Av -1 (Av =2v-1), then the sum of complex additions are equal to:

Yv*

wA=! • X * 1(l1* + 2P1*Av+T1A .,... = ln* + 2Pn*Av+TnA *) (11)

v +1 v -1 v v

N 1

s(—• 2v )n • (n2n ) = nNnlog2N. The theorem is now

£ n

as (-1)i=0 can accept values only ±1. proved.

The number of complex products of a kind (11) is References

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___ 1 n 1. Malozemov V. N., Macharskii S. M. Basises of

where t, e 0,1, i. e. 2n -1, since we have a product of diskrete harmonic analysis. S. 2. St. Petersburg : NIIMM.

real numbers for r = 0. As parameter vel: s , vectors 2°°3.

l = (l l ) and p = ( p p ) (where 2. Dudgeon D. E., Mersereau R. M. Multidimensional ’ n n Digital Signal Processing : Transl. from English. M. :

pt =0,1,..., Nv-1 (N = N /2v)), f =0,1,..., Av-1 Mir, 1988.

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Processing : Transl. from English. M. : Mir, 1989.

equal to: s(— • 2v-1)n • (2n -1) = —-Nnlog2N. 4. Oppenheim A. V., Schafer R. W. Digital Signal

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© Starovoitov A. V., 2010