ABSTRACT
Errors in the computation of Fourier Transforms arise from the truncation of the integral and the oscillation of the trigonometric factor in the integral. Chapters I and II of this report describe attempts to find weighting functions which will give faster convergence of the "finite approximation"to the infinite integral. In Chapter Ill a quadrature formula is derived for the numerical evaluation of the finite approximation.
TABLE OF CONTENTS Abstract CHAPTER 1 CHAPTER 2 CHAPTER 3 BIBLIOGRAPHY ii i 1 5 15 37
CHAPTER I
1. 1. In recent years, engineering theory has begun to take ad-vantage of the concepts of Fourier Analysis, in particular those of Fourier Transformation. Generally, the function g( ) of which the transform is to be found is given either only empirically or may be of such a complex nature as to make analytical evaluation of the integral extremely difficult. In either of these cases the transform is often calculated by numerical integration using a digital computing device.
b 1. 2 In practice, a digital computer evaluates an integral I=-f( )d by dividing the interval (a, b) into n+l (usually equally spaced)
sub-interval ( j i+1), i=0, I, . . ., n and then considering the increments AI = f(g) dg. The desired integral I =bg(g) dg is then
fi a
obtained by using the recursion formula
i+1 i I i= 0,1,...,n
where 10 = 0. Hence it is necessary to calculate the quantities Ali Now restrict attention to the Fourier Transform in which
case f (g) = g(g) e iag and I =0 g(g) e iag dg( with I T g(g)e ing dg.
-oo'- -T.
1
converge to I. The question then arises: "would it be possible to determine a function w(t), which for obvious reasons will be called a weighting function, such that for a given T:
0 0 w111 where wn = w((u) and
n +
- 4ia)j~
l~n
4~(a)f n(a) = g(g) e ingdg. " Stated in mathematically
~00 more precise terms, the question is:
"Does thezve exist, for any given T and g(g) in a certain class K, a function w(t) such that: max T(t)
ol 10T g()ei dg-(a)
or equivalently, a W(t) such that
T . max I W() g(g) ea dg -G< T for every a. ~(a) ~ T g(C)eiagdt
-T
I ... of.Upon consideration of special forms of g( ), the only function found which had the desired properties, was the identity, i. e., W(t) = 1. An explanation is offered in the following section.
- 4 (al
T icz
gMe d, - O(a)
Definition: A set D of function is called symmetric if x i D=> -x E D.
As an example, let Dp r( c[ I r(g) IP dg ]1/p,
<cI~,p~
I. - 00 Obviously, In the following, D is symmetric. let F(f) = -00 f() eia d Theorem:*If fEL is the function to be transformed,
let g() = { 0 () T 1 ST 2 <T)T 2 0 T and r(g) = 1(x) < T1 2
Also let some norm be chosen (such as the least upper bound) which is both homogeneous and subadditive, and denote the norm
of h( 7)by II hil. Assume r kndi*n'to--be a member of aasyti-metric set D and to be such that
II
F(r) II is bqunded.Then if F(wg) is used to approximate F(f),, :be -largest possible rfo
MA.
f .9+F()(
is rinimized by w=1. Proof: f( ) = g( ) + r((). ThenF(f)= F(g) + F(r) since the operation of Fourier Trans-formation is linear. Adding -F(wg) to both sides:
F(f) - F(wg) = F(g) - F(wg) + F(r) = F(g-wg) + F(r).
Hence
I F(g-wg) + F(r) may be considered as an equivalent expression *The proof of this theorem was suggested by Professor Calderon.
for the error. The problem then is to take, for each choice of w((), the least upper bound of the norm of the error and then minimize this by a proper choice of w((), i. e., find:
min {max II F(g-wg) + F(r)
II
}
W(9) #Pik>
Now due to the linearity of Fourier Transformation and thefact that r is a member of a symmetric set, it follows that F(r) must be a member of some symmetric set A. Since II F(r)
II
was assumed to be bounded, there exist a seq{
rn I rn ED } such that11 F(rn)II - lub I1F(r) I1I Then r ED
[F(g-wg) + F(rn)j -
1F(g-wg)
- F(rn) E 2 F(r )and
I IF(g-wg) + F(r 11 + I I F(G-wg) + F(-rnII2IIFr)1
since the rnorm- was chosen homogeneous and subadditive, Then
max II F(g-wg)+F(r )114 i{ IIF(g-wg)+F(rn))IWWI(g-wg)+F(-rn)11} IP(rn)1II
rad mao QF(g-wg)F(r)1 l.1ubIF(r)J
and then twv (A)
since A is symmetric. The left member of t (A) t, is minimized when w() =1. Thus it is seen that the maximum possible error is
minimized by choosing w(g)=L. It does not necessarily follow that w(g)=l is the only possible weighting function, but we may conclude that there is none which gives better results.
CHAPTER II 2.1 Introduction
In the last chapter it was shown that in general it is not possible to find a function W(g) such that for any given T:
T I T
max W(9)g() eiagd( -
4
(a) < Tg() eiagdg -4
(a)A -T (-T
T.
and for which: lim W()g() eladg = (a).
T-oo T
However, in the following section it will be shown that if certain assumptions are made regarding the nature of the transform (a), that there exists a weighting function K&)such that
RK(a) =5 K( ) g(g) eiag d9 4(a) (A) 00
as T - oo and that by properly specializing the function K(1) the T infinite integral may be reduced to a finite integral. It should be noted that the use of such a weighting function effects a certain smoothing of the function g(g) which is to be transformed. In the case of an empirically determined g(g), this smoothing may in itself be desirable since the function which is to'be transformed. may contain a certain amount of static, which in turn would mean that the exact transform would be unrealistic.
2. 2 This section considers an approximation theorem which is quite similar to one given by Bochner.
As discussed in 2. 1, consider the expression: -00
RK- () ia T( d
and K(!) has the following properties: T 1)( L T 2) K(O)=1 3) K() is continuous 'T 4) K(l) is an even function. T 5) K( ) e diagd = HO(a). -00 i.e. , K(!)= K( -) T T
Then it follows that K( ) eia d(
7Ted = TH0(Ta). If also H0 (a) E L1, then the inversion:
1 r0
27 00
TH0(Ta) eia da is valid everywhere and in
particular:
1= K(O)
=-2 7 -00
TH0(Ta)da
6) K(() is such that there exists H(a) where i) IHO(a)
I
S H(a)ii) H(a) is monotomically decreasing in 0 5 a 5 00 00
iii) 5aH(a)da = M< o, i. e., the integral is finite..
where g() EL1I
00
Note also that if K( ) is even then also Ha) is even, for: H(a) = K(9) eia dg =2
O
K( ) cos agd(
r- 00
.oo
K(M) cos -agd( = Hi-a)Now consider the following: Lemma 1:
where 'a(M = are as above.
RK(a) - (a) -(d
)
~(a ~)+ 4i(a-~) - ~k(a) 2
Proof: Consider 0 .(a) =
TK(THOdg
d RK(a), 4 (a) and HO (a)
00 f (()eia dg,
j
= 1, 2. -00Then from a theorem of Fourier Integrals, it follows that
Then in (A) of 2. 1
let 0 (9) = -g(g)eiag and f2(K) = (
0 2() = TH(Tg) RK(a) 1 00 and RK(a) 1(00 - 00 Hence f(() 1 p(a-9) TH Tg) dg and - 4(a) = 0
where use has been made of the fact that 1 = - S
C
TH(T9)dg. 2 7 -0 0 00 -'00 Loo0 and fl1(9) 62 2(dg. {O (a-g)- 0(d) } TH TE) dgBut also H(T9) is an even function, thus it follows that
RK(a) - (a) _ 1 0 O(a + 9) + (a - 9)-(a) } THST()d(
which was to be proven.
Lemma 2: Let H(M) b6 as discussed before.
ThenGogH(()dg = M< co,
00
where q: 0, implies tq+lH(t) - 0
as t - o.
t
for 0 : t <o since H() is positive and monotonely decreasing.
[1 ()q+0 1 >
tq+1-(t)F
H(g)dg - 0 as t -o, hence-tq+1 H(t)- 0 as t -- o.
The essential characteristics of the approximations are contained in the following:
Theorem
Let K(4) and H(a) be as discussed before with00 H()d
00
bounded, and if in addition (a) is such that 1
pl+ q
0
-
( I d9!-*0
0 al
as p-* 0; then:
T qIRK(a) - 4 (a) 1 - 0 as T - co for q _ 0. Proof: 00o '2 t/2 But t / 2 gq H(g)dg -:I=H(t)
for 0/,cdntalinuous-, q A -lat least zere-cifferentiab1e, q ie-at least one.
Proof: By Lemma 1 IRK(a) =0 1 a THO(T9) d9 I IH0(T ) I
S
00 Td( H(Tg)TdIn order to show the desired convergence, split up the last integral in the following manner:
0 /(() I H(Tg)Tq+1dg=- I (g)
I
H(T1)Tq+d+()IH(T9)Tqdg.Now letP)O
' 0
(t) I dt. Then dP(g) =
V
a()I d( and substituting in the first integral on the right yields:- H(T)Tq+11 ;( ) d9 =I xOr " ,-% x
L.
H(Tg 0 )Tq+ldP( ) Integrating by parts 1 00 7T 0 71 2T-0 Then Tq I RK(a) -()2 K-Note that I (q)/ H(TE) Tq+1 dL x Jo 7 r 1 {(xT) q+1H(xT)} P(x)x q+ 1 I- T q+1 P(9)dH(TE) V0 )q+l
Lemma 2 shows that (xT)qH(xT) - 0 as T- co hence the inte-grated term tends to zero. Since lim =0, there exists for
g- q+
any E > 0 a gi such that This, together
with the monotonatzfof H(T9) permits P(() to be majorized under the integral, thus obtaining
q+lq+l =
{
T q+1lq+H(T)}
7r
- (q+1) E.. T q+1 q H(TE)d(
Again by Lemma 2, it follows that the integrated terms tend to 0 as T--oo. For the last integral:
0 Tx
= e 0
where ( = TE.
000
Since ( H()d, is finjte and H(,) is even, " 00
it follows that s H( )d( is it, Thus q+1 ThTxq H( )d
0r
is bounded as T - 00 and may be made as small as desired by a
I
P((1< 5+1 for all <i.proper choice of e. So far it has been shown that
ox
I ()H(T()Tq+1dg may be made arbitrarily small as T-oo
0 by a proper choice of x. Regarding-is true: 00 x I the following 1 it 00{ 14 (a+g) I + 10(a-) I 2 +
1p(a)
I
} H(TM)Tq+ld( Then: 2Ip(a+()
I H(T)T q+1dg-1 Tq+lH(Tx)j rx 2 7r x I q(a+~) I d~. 00 x \I p (a+() I d( < <o for any x and
(a00) d dl q+1x q H(Tx)<+l
x
Since (Tx)q+lH(Tx) -. 0 as T- oo, then also
as T - 00 for x# 0. where M ko Tq+1xq+1H(Tx)M-0 2 x q+1 The integral 1 0i
2- 4 (a-) I Tq+1 H(TI)d( is treated in the same manner.
q (a) L LTq+H(Tx) 2 q For a(0)1 H(T)T q+1 dt, c x 1I Oa() I H(T9) Tq+d( x 2x
Lastly:
Iq(a) I H(IM)T d ()I T q+lH(Tg)d( x
Sh)
Then let T(=t.
o() * T q+1qH(Tg)d =
j
p(a)Ixx )x x q
integral goes to 0 as T--oo for x 0. that
\T
t H(t)dt and the last T xTThus it has been shown
lim {T
IRK(a)
- (a)) =0.T-oo
2. 3 In this section we will determine a polynomial
which satisfies the conditions of K(() as given in the last section and which in particular vanishes outside a symmetric compact set B. As before, let F(f) = Of(() eiagd . Then:
00
Lemma I
Let f E L1 , of bounded variation, and vanish outside a
compact set B. M
Then I F(f) I - where M is some positive number. a
Proof:
F(f) = B f()einagd = l {[f(9)e ia B Beiagdf(g)
IT
I
x
but the terms in brackets are bounded, hence IF(f)<
-Corollary
k-1
Let f be of C with-the drivative of o-rder-kof baunded variatian. Then IF(fM)1 N /a+1, and f ,v=0,1,...,k EL1.
Pr'oof: If F(f) =t (a) then we have F(f') -iaikta) for f and hence
F(fk) = (-ia)k / (a). But from lemma 1 we know that I F(fk) I< N for some N sinceffk is of 'bounded 4ariation.
N
HenceI& (a)N
a
Consider now the class of functions of the form
S{ P(4) for
I
1 P() an even polynamialT 0 elsewhere
Since K( ) - 0 for I> T, then whatever the value of j, we must
T
have K(9 / T)=0 for I !T. Hence if we require K(ECk-lwe T
must have K (+1) = K (-l) 0 for
j
- 0, 1,...,k-1. But if is an mth order zero of K ( it then is an m +j
- 1 order zero ofK(). Hence if +T, -T are first order zeros of K he simplest form of K(-), Which also satisfies the requirements as stated in
T
the last section, is:
K(- ) = { %
0 elsewhere
Note that K(-) has a discontinuity in the kth derivative. Hence IF(K)6 M from previous lemma. Then it follows that q<k
insures that a q+l F(K)-O as a-oo and from the Theorem of section 2. 2 we have that
T IRK(a)- q (a) -0 0as T-oo
T [ 2 k !eiag if RK(a) T T ()k g( ) eiad(withk> g.
As stated before, it should be noted *that d
ag4oye Q 1. . Generally, more than this can beti besaid for the transform of empirically determined functions. Thus an exponent k i! 2 would effect convergence. However, the choice of k would also be affected by the degree of smoothing desired, with large k bringing about greater smoothigg of g(g) than a small k.
CHAiTER III 3. 1 Introduction
In this chapter a quadrature formula forithe finite approximation Ia T g( O)eia d
-T
00
to the infinite Fourier integral I= qi (a)= g(g)ea d( is ohbtdned.
-00
This approximation may contain a weighting function such as was discussed in the last chapter, or may result from a truncation of the infinite integral. It is calculated by considering a finite set of points: g0 l'. .. n-+1 where
%
-Tn+=T and such thati~l -=h, a constant; and the corresponding functional values
g0 gl ... Pgn+1, where gi=g(gi). If g(g) is approximated by a third
degree polynomial over some interval, an approximation of the integral of g(g) over this interval may be obtained by analytic methods. If then the integral la is approximated by a sum of
such integrals, the following expression will be obtained:
n+l n+l
'a * *>CjRg( cos a ( + iAl SiRg(( )sin a + E
j='0 j=0
n+l
Ajg(9) e'ig + E j=0
R R
where E denotes the error and the CR, S and A are functions
jj J
only of p = ah. 3. 2
b
In general, when an integral I f(()d( is to be
a
evaluated by numerical methods, a partition P of the interval is formed, where P is the finite set of points: a= (0 < 1 2 ...
< (ntl=b, and which have the corresponding functional values fo 1) ' n+1 where fi= i) . The function f-is then
approxi-mated in some manner over the interval
(,
+ by a function fi(0) where fi() is a member of some class generally chosen in such a way that the definite integral of its members is readily evaluated by analytic methods. The integral I= bf(9)d maya
then be approximated by the sum of the integrals of the function
fi(), calculated over their respective intervals of approximation, i. e.
n
I a f (x)dx = l i + E
a i=O0
where ~ x.+1
li reX +fi(g)d g and E is the error. x.i
1
Clearly, in the case of the Fourier Transform, where f() = g(g)eia, it would be very difficult to obtain a good approxima-tion for the funcapproxima-tion f((), for no matter how smooth the funcapproxima-tion g(g)
may be, the factor e ia introduces a degree of oscillation depending upon the parameter a. We may, however, be able to obtain a good approximation gi(97 to 'the function g(() over the interval ( in which case we can consider
ia.( T'ei a iag
fi(() = e gi(). The integral I -Tg()e d may then
be approximated as:
h
IIR + il M= (Ij + iI-) + E j=0
T T
where IR
5
g() cos agd M= TI g() sin agd(IR = () cos agd, IM g+lgj( ) sin agd
and the error is dependent only upon the way in which g.(9) approximates g(g) over the interval (gi* i+1) and not upon the parameter a.
In the following, only the determination of the approximation of IR, the real part of the integral, will be
dis-M
cussed in detail. The approximation of IM is arrived at in an analogous manner and the results are cited at the end of this chapter.
3.3
Consider a partition of the interval (-T, T) such that ( 1- i =h, where h is constant for i=0,1,...,nand let
g , gl' 0 1 ' g' ngdenote r~j the corresponding functional values,
i.e., g =gR ). The function g() which is to be transformed, will be approximated over each interval by a third degree
poly-nomial of the form () = ai0 + ailE+ ai2 2 + ai3 3. To obtain
the coefficients for the approximation over the first interval
(N0
1); use is made of Newton's Forward Difference Formula:6g h 6 63 g)0 +E( ) where 6 n+1g E() = 4 30) 4 i= = 6 ng -ngand 6 gi gg+1-gi V AV
Then the approximation used is g0(E) = g(E)
-expression takes on a simpler form if a dimensionless variable
S is introduced, where 32e
h .Then:
g0(S) = g0 + S6g0
S(S-) 62 gS(S-)(S-2) 3
2 g
2 3
a00 + a01S + a0 2 +a03s,OS
4'r
which for 0 s I- 1, approximates g() over the interval
(%O
. 6g0 =:g 1 g0 g() = g0 with The 2 g 2h2Hence it follows that: a0 0 0 11 3~ 1 a0 1 -- 190+ 3g1 2 -1 3 6 2 3 5 a0 2 g0 -g 1 +2g 2 2 a03 0 + 1 g2 + 3 6 2 2 6
Similarly, in order to obtain the coefficients corresponding to the last interval (n , n+1), Newton's Backward Difference Formula
is used in the form:
ag 2 gn1+ ;n+1 g+1 h n-l n 2h2 3 n+1nn where
a
gi+1x1 i ag, i+1i+1i 4 3with E(M) g II (x-xn+1-i). The approximation used is 4! i=0
gn () = g() - E(). If the aforementioned dimensionless variable S
nn
is introduced in the form S - ,the expression for gn(9) takes h
gn(S) = gn+1 + (Srl)3 8 gn+1 = anO +a n0n1S+ which for 0 - S 1 S(S-1) a2g _i S(S2-1) 3 2n+1l+ an2 s an3S3
approximates g(g) over the interval (gnP n+1) Expanding the backward differences shows that:
a gn+1 = gn+1
a2 gn+1
n+1
-from whence it follows that:
anO n 5 a =-g -nI - n+1 6 gn+1
an
2 -2 an+1 an3 6 2 gn+ n -1 gn 2 gn + 1 2 6 n-2 1 2 1 6 n + n-1 2In the intermediate intervals ( ) i=l, 2, n-1, the polynomial approximation is obtained by considering Gauss' formula
2
g() = gi i i+
h g 2h
3
6h
where again 6 n+gi = 6 ni+1- 6 ni, 6 gi gi+1-i and
E(h) hg ()(-+ i-9-2) The approximation
to be used gi(g) = g(g) - E(g). Here again, if we substitute
S ,the expression may be simplified to:
2
~+S(S-1) 2 S(S -1) 3
g (S) = gi + S6 gi + i-1 6 6i-1
2 6
= ai0 + ail S+ ai2S2 + ai3S3 which for O S-_1Igives the desired approximation to g(g) over 'the interval (gi, i+1) As
before, expansion of the forward differences gives:
6 gi= i+1 - i 2
6 2gi+1=+1 - 2gi+
aind Lhe poI-Li.k , al coefficients are determined to be
ai0 i
1 i+.
ail
i-1 -- 1+13 2
a i2 i-1- 2gi i+1
- i- 1 1 1 ai3 i 2i+ 6 2 2 n To determine li i=O iR li+1 variable: S
which estimates IR, the integrals
(() cos agdr must be evaluated. Making the change in
i , we obtain IR = g (S) cos a (Sh + )dS.
h 0
If gi(S) = ai0 + ailS + aiUS2 + ai3S, this integral may be evaluated2
with the result,
a. + 2a a--Pa 32 -p )1 cos a ( + h) p a. +a. io ii p 22 +ai2 P + - 6ai3 P pil T ]jco 2 + ai3(P- ) sin a P3 1 a 2ai2 -
1
i0 P P 6 i+2 6 (C. + h)where = ah .Upon substituting for the a.. the equivalent
ex-pressions in terms of the ordinates gi, a rearrangement of the terms gives:
-
g1() cos agd( = BO, 0g0 + B0, 1g A1+ B0, 22 + B0 3g3 0.+i() cos agd = Bi, -1gi-1+B i Oi+Bi, 1gi+1+Bi, 2i+2
i=L12,..., n-I
n+l g(() cos agd=B gg B +
gn csa g n, - 29n- 2 + n, -1 n-1 + n, Ogn n, 1 n+1 n
where the B m. n are functions of (m, p. M.9 Explicit expressions for the B nnare given-in Appendix.A.
By again regrouping the terms, the approximation may be expressed as: n
i=0
ilk g0o[Bo + Bl-1
+ g [B0 1 + B1 0 + B2, 1 ]
+ g2 [ Bo, 2 + BI, 1 + B2,O + B3y -10
+3 [ B + Bl,2 + B2,1,3 1 tB3.,0 +B 4,0-1 +g [ B i i- 2.,2 2+ Bi ;+ B. +B. ] i l' i 1, O 1+1,-(i=4p5,...,n-3) +gn- 2 [ B n-4. 2 + B n- 3, 1 + Bun-2, O + Bn-1, -1 + Bn., -2 n-1 [Bun-3, 2 + Bn-2, 1 + Bn-1, O + Bn 1 gn [ Bn-72, 2 + B n-1, 1I + B n, O +g [B n-1, 2 + Bn, I+
Upon simplifying the sums Bj, k' this expression reduces to the form:
h n+1 n+l
jR
=LhCj g((jcos a i+ S Rg(Q ) sin aj=0 j=O j=0
R R
where C , SR are dependent only on p = ah.
jSj
I M = CjMg( )cos a + SjMg( ) sin at
j=0 j=0 j=0
for which C = S R CM. MS Expressions for C R S as functions of p are given in Appendix. B.
R M R M
Since C. = S , S.= -C. , it should be noted
that the integral I= Tg(()edia d may be approximated as:
-T
n n+1
(IR + iIM)= A gQ) einj where A = C -SR
j=0 j=0
Error
We considered gi(g) = g() -E() where in all three cases the error was expressed as t(n) P() with '1S <i+l
and where P(x) is a polynomial of degree four which has an upper bound M. Then the error of the estimate to the integral over the interval is:
E = ei+leE(()d 5 9(9)e g+l g +lgi() e iadg
i i i
4 4
hence: g (I)Mh
Then if I g 4(H)I =< m for 1 (i+l it follows that IE< h4Mm
1Ei
4! a
and for n intervals, the total error in the integration is less than 4
h 4Mmn It should be pointed out that this is an extremely con-4! a
servative estimate and in practice much better results will generally be achieved.
The value of the above quadrature formula lies in the fact that the coefficients are functions only of p=ah, and hence once they are tabulated for various values of p, the finite trigono-metric integral may be evaluated by calculating a single finite sum. The expressions for C R, SiA are such that they may easily be programmed for a digital computer. Using a device such as the IBM 650, a complete set of constants may be calculated for a single choice of p in-about five minutes.
AFFENDIX
1. B 0,k P 0,n, 1, cos a(g 0 +h)+ P 0, n,2 sin a(g 0 +h)
+ pOn, 3 cos a
+ +0, n, 4 sin ag
0
for k = 1, 2, 3, 4
and where the P0, n.
j
0, 0,1I
6p
are defined as follows: 1 3p 2_6 T4 6 2 -5 2 P0,0, 2 6p[-6p p3 p 0 0, 3 0l 0, 4 0,1, 1 11 2 6 2 p 1 p 1 -p Ii I - 2 1 3 26 p 2 3p p0,1,2 =[ 3 p 5 2 P 1 6 2 p I (P .p 4 1 2
0 ,1, 3 0 , 1, 4 p 0, 2, 1 -3 2p 0,o 2.,2 2 p0, 2,32 2p2 4 P0, 2,4 =2 P P0 , 3P1 ... 2 3p 0., 3, 3 = 0, 3, 4 3 p4 ~1 2 +
]
(3 p _6 4-p 1 2(Zj~)I
p p2 -3 p p 4I
I
1 6 ~1 6 3p 2_-6 p42L~)
P4 3 1 =0Bi k p1cos a(f +ti ) n+ 2 sin a(9 +h)+Pn
3cosag +Pn 4 sina
for i =l,2,o...,,n-1 and k=-1,0, 1, 2, where
1-72 3p 1 P- 1, 2 1 3 + p 1I -1, 3 P-1, 4 -P 6 (P 26 6 p 4 22 (P ) 3 p [1 3 6 (p )] 6 p 1 P I 1 2 - [ -I p P + 9 1 P0, 2 2p 2 1 3p2-6 2 P4 2-2 p I p2-6 2 p PO, 3 Po, 4 1
2
1 4 - [ 1 + -p 3 pl I p 2 p 1 3p 2-6 2 p 1 2 3 p Ip p 2 p 1 p 2 p T 3 p 2p 112 2, 1 6p21 2,2 6 p =2+3[ 2, 3 2 p 2o 4 1 +1 6 1 6 3p 2-6 p 2 3 p 1 p
I
= 0 Bn, k =n. k,1 cos a( +h)+P n.k. 2sin a (n+hn)+Pnk3cos a n+Pn, k. 4sinan for k = -2, -1, 0, 1 and where11 2 n,1 1 6p 5 n,, 1.2 6 2 I 6 3p 2-6 4) p 2 p3 l 2 pp
3
1,4 P P P P P 1 6 2 P 6 p3p 6 P n nO,2p1 -
5
n, 0, 2 n, 0, 3 n, 0, 4 .. 1 2p 1 -pI p 2 2. 1 2 ) -1 ( 2 1 26 p 32w
p 1 2 --[1+ -P P p p 2p + n, -1,21
-2
n, -1, 3 n, -1, 4 1 2 p 2 1 26 p3 3 p=-0
n, 113
P n, 1., 4 I 2 p P B + 1 2 )P I6p n, -2, 2 6p P n, -2, 3 n, -2, 4 1 6 1 6 1 =-
[ 1
1 6 3p -_6 p 2 (p p3 p =0 APPENDIX B 0, 1 +P0, 3) cos -P + (P0, 1) cos 2p + P0 0, 30, 2 +P 0p 4) sin p + (P0, 2) sin 2pi}
0, 0, 1 P, 3) sin p - PO, 1 cos 2p + Pox 0, 4 sin 2p} = { (PO, 1, 3 +0, 1 P- 1,3) cospP 1,1 cos 2p p 1 4 0, ,l4) sin p + P-1, 2sin 2p + (P0 1 1 00+P3)
1,4 +0pO, 2 P- -+ 4) Cos p + P-l1 2 cos 2p + (PO, i, 3-o, 1-P1I, 3) sin p - P -1sin 2p+(PO,
1.,2 0,4) = f(p0, + (PO, coR soR 1 0. 0,2 + (P0 2 .. =, sIR
+.P0 1- P-11 3) cos p + (P0, 2,3+P 1 )cos 2p 0, 2, 1 + 1, 3
p 1, 4-p 0, 2-P- 1,
4)sin p - (Po 2, 4 -1, 2) sin
2p s2R = 0 2, + 0, 2, 1 2 1,4 0, 1,3 0, 1 4 P-1 4) sos p + (PO, 2, -P-1, 3) sin p + (POO 2, 3 4 +P-1, 2)cos 2p - 1 1) sin 2p 1, 3 p 0, 1 -1 3) cos p + (P0 3,1 +P 22,+P3, 4)cos 2p + (P0, 2 +-1 4 - , 4 2,2 ) sin p + (P- 1, 2--P 2,4 -p 0,3, 2) sin 2p
1,I10,3)0,p3, 3 cos 3p - P0,3,4 sin 3p}
2 1,p 4+0,2 + P-1,4) )Cos p + (PO,3p2 + 2, 1 - 1p, 3 - 2, 1 p2 4 + -1, 2)Cos 2p -1, 3) sin p + (P0 , 3,1 2, -1 ) sin 2p p0, 3,4 cos 3p + P op 3,.3 sin 3p + P1,2 +p0 ,4 ( 0.2, 2 + ( 1 +0po 3)
}
{ 2, 1 SR {P2. 2pIfP_3-0, 1 -1, 3)cos p+',(P2, 3 - ( 2 1,4 P -P- 1 4) sin p - (P 2 4 - P1 2 ) sin 2p +Pl 1+p0 3 } i = {(p2,2 +p 1, 4 p0, 2 i-(22, 1 4, 5, 6,. . . n-3 S Cos p + (P -, 2 cos 2p p 0, 1 - 1, 3) sin p +(P2, 3 p-1, I sin 2p + (P,2+P 0, 4) } i = 4,5,6, .. .,
n-3
c n 2R ={(P 3 +P0 1+P1 3 +P 2) cos p + (Pn +(P- 1, 4+p0, 2 -1, 4 2, 2) sin p + (Pn, 3, 3 +P 0 1 +P2 3)cos 2p 4 +P-1 2 p 2 , 4) sin 2p i P cos 3p + P sin 3p + (P0 3 +P1 1 )} n, 3, 1 n, 3,,20,3 11 -1, 4+P0, 2+P1 4+P 2) Cos p + (P+3,4-1,2 2, 4) cos 2p -(P- 1, 3 ,+PA1p1, 3 p2,l1) sin p - (P 3 3 -1,0 +-P3,2 cos 3p - P3, 1 sin 3p + (P0,4 1 2 ,3 ) sin 2p + 1., 2) } 34 Sn-2 CiR p2,1 1, 3 -1 )cos 2p SiR= ((P, 2,3 +P 01+P1 3+P ) cos p + (P, 11CosP 2P)+osn + (P 2, 4 +P0, 2-P 2 sin p (P 2-P ) sin 2p n,2,2 2, 4)ip + (P 0, 3 + P 1, 1)
}
= n, 2,4 0, 2, 1 2, 3)sin 2p (p0, 4+P1, 2)= n{(P 1,1+P 1, 3.3+P Cos p +P2o 3 cos 2p +(Pu,"
+ n, 1, 2 1 , 4-2,2) sin p -P 2, 4 sin 2p } SnR n, 1,2 +1,3+2,2) Cos p + P2,4 cos 2p + (Pn, 1,4+ l, 2 pn, 1, 1,3 20 1) sin p + P 20 3 sin 2p} R n)c P
c n+i = {(P n ,3+P 2.,1 Csp P2v,3 cos 2 p +P nlI
S n, 0, 4 2, 2,) sin p - P2, 4 sin 2p
}
35 CR 1 , 1) a-1 201+P 2, ) cos 2ps.-l'
2 +P1,)4+2.,2 ) Cos p +(P n, 2, 2+P2.,4 ) cos 2p 2.,3 +p0. ,I-p .3- 2, 1) sin p -(Pn,.R
S = {(P +P ) cos p +P cos 2p + P
n (1 n,0,4S2,12 2,34 n, 0, 2
BIBLIOGRAPHY
1. Bochner, S. and Chandrasekharan, K., Fourier Transforms, Annals of Mathematics Studies, No. 19, Princeton University Press, Princeton, N. J., 1949.
2. Hildebrand, F. B. , Introduction to Numerical Analysis, McGraw-Hill Book Co, Inc., New York, N.Y., 1956.
3. Ross, Douglas T. , Improved Computational Techniques for Fourier Transforms , Report No. 7138-R-5, Servomechanisms Laboratory, Massachusetts Institute of Technology, Cambridge, Massachusetts, 1954.