Here is a more complete explanation of FFT

===
hn= time values   Hk= frequency values
Hk=       Sum[k=0 to N-1]  hn *Exp[-2*Pi*i*(k*n/N)]
hn= (1/N)*Sum[n=0 to N-1]  Hk *Exp[+2*Pi*i*(k*n/N)]
There must be 2^N points, 8 or above.
========
We will work with an 8 point FFT problem.

  |  0  0  0  0  0  0  0  0 |  
  |  0  1  2  3  4  5  6  7 | 
  |  0  2  4  6  8 10 12 14 |
  |  0  3  6  9 12 15 18 21 |
  |  0  4  8 12 16 20 24 28 |
  |  0  5 10 15 20 25 30 35 |
  |  0  6 12 18 24 30 36 42 |
  |  0  7 14 21 28 35 42 49 | 

Assome an arbitrary math function

'TIM'
<< {} 'G' STO  0 'X' STO   'Pi/4' 'P' STO 
WHILE X 8 <
REPEAT 
' 2 + 0.9 * Sin(X*P) + 0.8 * Cos(X*P) +  0.7 * Sin(2*X*P)
+ 0.6 * Cos(2*X*P)' EVAL 'Q' STO    G Q + 'G' STO  
 X 1 +  'X' STO    END G   >>     

Time sequence of amplitude of curve 
G ={ (3.4)  (3.902)  (2.3)  (1.37) (1.8) (1.49) (.5) (1.22) }

(a,b)=a + b * i

Data giving amplitude of each frequency.

 [ (16,0)  (3.2,  -3.6 )  ( 2.4, -2.8 ) (0,0) (0,0) (0,0) ( 2.4, 2.8 ) ( 3.2, 3.6 ) ]
    DC              CIS X*P                                      CIS 2*X*P   
                                                      
N= number of points=8
Interpret it this way 
16/8 = 2 = DC value of equation  = 16/N

3.6/4 = 0.9 = coefficient of Sin[X]       4=N/2
3.2/4 = 0.8 = coefficient of Cos[X]
2.8/4 = 0.7 = coefficient of Sin[2*X*P]
2.4/4 = 0.6 = coefficient of Cos[2*X*P]
0/4   =   0 = coefficient of higher harmonics

===========
Compute Second term
U=(-Pi * i / 4) 
n=0   3.4  *Exp[0*U)=3.4
n=1   3.902*Exp[1*U)=(2.75,-2.75)
n=2   2.3  *Exp[2*U)=(0,-2.3)
n=3   1.37 *Exp[3*U)=(-.96,-.96)
n=4   1.8  *Exp[4*U)=(-1.8,0)
n=5   1.49 *Exp[5*U)=(-1.05,1.05)
n=6   0.5  *Exp[6*U)=(0,0.5)
n=7   1.22 *Exp[7*U)=(.869,.869)
                     -------------
                      (-3.2,-3.6)
Compute third term

n=0   3.4  *Exp[0*U)=3.4
n=1   3.902*Exp[2*U)=(0,-3.9)
n=2   2.3  *Exp[4*U)=(-2.3,0)
n=3   1.37 *Exp[6*U)=(0,1.37)
n=4   1.8  *Exp[8*U)=(1.8,0)
n=5   1.49 *Exp[10*U)=(0,-1.49)
n=6   0.5  *Exp[12*U)=(-.5,0)
n=7   1.22 *Exp[14*U)=(0,1.22))
                     -------------
                      (2.4,-2.8)

Compute fourth term

n=0   3.4  *Exp[0*U)=(3.4,0)
n=1   3.902*Exp[3*U)=(-2.759.-2.759)
n=2   2.3  *Exp[6*U)=(0,2.3)
n=3   1.37 *Exp[9*U)=(.969,.969)
n=4   1.8  *Exp[12*U)=(-1.8,0)
n=5   1.49 *Exp[15*U)=(1.059,1.059)
n=6   0.5  *Exp[18*U)=(0,-.5)
n=7   1.22 *Exp[21*U)=(-0.869,0.9860))
                     -------------
                     (0,0)

G=[(3.4,0)  (3.902,0) (2.3,0) (1.37,0) (1.8,0) (1.498,0) (.5,0) (1.23,0)]

=======
Compute FFT from G by running program WW.

'WW'
<< CLEAR   '-0.25*Pi*i' 'U' STO   0 'V' STO   
DO R V 1 +  'V' STO  UNTIL V 7 >  END 
8 ->LIST 'ZZZ' STO ZZZ  >>

'R'
<< {(0.0)} 'S' STO 
0 7 FOR K 
G K 1 + GET U K V * *  EXP * S + 'S' STO S sigmaLIST 
'ZZ' STO    1 STEP   ZZ >>

ZZZ = { (16,0) (3.2,-3.6)  (2.4,-2.8)  (0,0)  (0,0)  (0,0)  (2.4,2.8) (3.2,3.6} ]
This is harmonic data

First = 16  which 16/N = 16/8 = 2.0 = DC value of time sequence equation.

Last group of numbers 
3.6/4 = 0.9 = Coef of Sin(P*X)
3.2/4 = 0.8 = Coef of Cos(P*X)
2.8/4 = 0.7 = Coef of Sin(2*P*X)
2.4/4 = 0.6 = Coef of Cos(2*P*X)

=========
Do INVERSE FFT

'YY'
<< CLEAR '0.25*Pi*i' 'UU' STO   0 'V' STO   
DO RR V 1 + 'V' STO UNTIL V 7 >  END 
8 ->LIST '.125 * RE ABC' STO ABC  0.125  * RE 'ABC' STO ABC >>

'RR'
<< {0,0)} 'S' STO 
0 7 FOR K 
ZZZ K 1 + GET UU K V * *  EXP * S + 'S' STO S sigmaLIST  'ZZ' STO
 1 STEP ZZ >>

See
{ 3.4  3.902  2.3,0   1.37   1.8,0   1.497 .5,0 1.229, }

This is the amplitude vs time data initially created.
===