# Discrete Fourier Transform and the Fast Fourier Transform

### 5.26.3 Discrete Fourier Transform and the Fast Fourier Transform

For any integer N, the Discrete Fourier Transform (DFT) is a transformation FN defined on the set of periodic sequences of period N; it depends on a choice of a primitive N-th root of unity ωN. For sequences with complex coefficients, we take:

ωN=e

2 i π
N

If x is a periodic sequence of period N, defined by the vector x=[x0,x1,… xN−1] then FN(x)=y is a periodic sequence of period N, defined by:

(FN(x))k=yk=

N−1
∑
j=0

xjωN−k· j,

for k=0..N−1.

The Discrete Fourier Transform FN is bijective with inverse

FN−1 =

1
N

FN,ωN

on ℂ

i.e.:

(FN,ωN−1(x))k=

1
N

N−1
∑
j=0

xjωNk· j

The Fast Fourier Transform (FFT) is an efficient way to compute the discrete Fourier transform; faster than computing each term individually. Xcas implements the FFT algorithm to compute the discrete Fourier transform when the period of the sequence is a power of 2.

The fft command computes the discrete Fourier transform.

- fft takes one argument:
  x, a list or sequence regarded as one period of a periodic sequence.
- fft(x) returns FN(x), the discrete Fourier transform of x.
  If x has length which is a power of 2, then FN(x) is computed with the Fast Fourier Transform.

The ifft command computes the inverse discrete Fourier transform.

- ifft takes one argument:
  x, a list or sequence regarded as one period of a periodic sequence.
- ifft(x) returns FN−1(x), the inverse discrete Fourier transform of x.
  If x has length which is a power of 2, then FN−1(x) is computed with the Fast Fourier Transform.

#### Examples.

- Input:
  fft(0,1,1,0)
  Output:

  | ⎡ ⎣ | 2.0,−1.0−i,0.0,−1.0+i | ⎤ ⎦

- Input:
  ifft([2,-1-i,0,-1+i])
  Output:

  | ⎡ ⎣ | 0.0,1.0,1.0,0.0 | ⎤ ⎦

#### The properties of the Discrete Fourier Transform

Definitions. Let x and y be two periodic sequences of period N.

- The Hadamard product (notation ·) is defined by:

  (x · y)k = xk yk

- the convolution product (notation *) is defined by:

  (x * y)k=

  N−1
  ∑
  j=0

  xjyk−j

Properties.

FN(x · y)

=

⎛
⎜
⎜
⎝

1
N

⎞
⎟
⎟
⎠

FN(x) * FN(y)

FN(x * y) | = | FN(x) · FN(y)

#### Applications

1. Value of a polynomial
   Define a polynomial P(x)=∑j=0N−1cjxj by the vector of its coefficients c:=[c0,c1,..cN−1], where zeroes may be added so that N is a power of 2 (so the Fast Fourier Transform can be used).
   - Compute the values of P(x) at

     x=ak=ωN−k=exp(

     −2ikπ
     N

     ), k=0..N−1

     This is just the discrete Fourier transform of c since

     P(ak)=

     N−1
     ∑
     j=0

     cj(ωN−k)j=FN(c)k

     #### Example.

     Find the values of P(x+x2) at x=1,i,−1,−i.
     Input:

     P(x):=x+x^2

     Here the coefficients of P are [0,1,1,0], N=4 and ω=exp(2iπ/4)=i.
     Input:

     fft([0,1,1,0])

     Output:

     | ⎡ ⎣ | 2.0,−1.0−i,0.0,−1.0+i | ⎤ ⎦

     Hence:

     - P(1)=2,
     - P(−i)=P(ω−1)=−1−i,
     - P(−1)=P(ω−2)=0,
     - P(i)=P(ω−3)=−1+i.
   - Compute the values of P(x) at

     x=bk=ωNk=exp(

     2ikπ
     N

     ), k=0..N−1

     This is N times the inverse fourier transform of c since

     P(ak)=

     N−1
     ∑
     j=0

     cj(ωNk)j=NFN−1(c)k

     #### Example.

     Use this method to find the values of P(x+x2) at x=1,i,−1,−i. Input:

     P(x):=x+x^2

     Again, the coefficients of P are [0,1,1,0], N=4 and ω=exp(2iπ/4)=i.
     Input:

     4*ifft([0,1,1,0])

     Output:

     | ⎡ ⎣ | 2.0,−1.0+i,0.0,−1.0−i | ⎤ ⎦

     Hence:

     - P(1)=2,
     - P(i)=P(ω1)=−1+i,
     - P(−1)=P(ω2)=0,
     - P(−i)=P(ω3)=−1−i.

     You find of course the same values as above.

2. Trigonometric interpolation
   Let f be periodic function of period 2π and let fk=f(2kπ/N) for k=0..(N−1). Find a trigonometric polynomial p that interpolates f at xk=2kπ/N, that is find pj, j=0..N−1 such that

   p(x)=

   N−1
   ∑
   j=0

   pj xj, p(xk)=fk

   Replacing xk by its value in p(x) we get:

   N−1
   ∑
   j=0

   pj exp(i

   j2kπ
   N

   ) = fk

   In other words, (fk) is the inverse DFT of (pk), hence

   (pk)=

   1
   N

   FN( (fk) )

   If the function f is real, p−k=pk, hence depending whether N is even or odd:

   p(x)=p0+ 2 ℜ(

   N
   2

   −1

   ∑
   k=0

   pkexp(ikx)) +ℜ(p

   N
   2

   exp(i

   Nx
   2

   ))

   if N is even and

   p(x)=p0+ 2 ℜ(

   N−1
   2

   ∑
   k=0

   pkexp(ikx))

   if N is odd.
3. Fourier series
   Let f be a periodic function of period 2π and let yk= f(xk) where xk=2kπ/N for k=0..N−1. Suppose that the Fourier series of f converges to f (this will be the case if for example f is continuous). If N is large, a good approximation of f will be given by:

   ∑

   −

   N
   2

   ≤ n<

   N
   2

   cn exp(inx)

   Hence we want a numeric approximation of

   cn=

   1
   2π

   ∫

   2π
   0

   f(t)exp(−int)dt

   The numeric value of the integral ∫02πf(t)exp(−int)dt can be computed by the trapezoidal rule (note that the Romberg algorithm would not work here because the Euler Mac Laurin development has its coefficients equal to zero, since the integrated function is periodic, hence all its derivatives have the same value at 0 and at 2π). If c_n is the numeric value of cn obtained by the trapezoidal rule, then

   c_n=

   1
   2π

   2π
   N

   N−1
   ∑
   k=0

   ykexp(−2i

   nkπ
   N

   ), −

   N
   2

   ≤ n<

   N
   2

   Indeed, since xk=2kπ/N and f(xk)=yk:

   f(xk)exp(−inxk)

   =

   ykexp(−2i

   nkπ
   N

   ),

   f(0)exp(0)=f(2π)exp(−2i

   nNπ
   N

   )

   =

   y0=yN

   Hence:

   [c0,..c

   N
   2

   −1

   ,c

   N
   2

   +1

   ,..cN−1]=

   1
   N

   FN([y0,y1… y(N−1)])

   since
   - if n≥0, cn=yn
   - if n<0 cn=yn+N
   - ωN=exp(2iπ/N), so ωNn=ωNn+N

   Properties.

   - The coefficients of the trigonometric polynomial that interpolates f at x=2kπ/N are

     pn=cn, −

     N
     2

     ≤ n<

     N
     2

   - If f is a trigonometric polynomial of degree m≤ N/2, then

     f(t)=

     m−1
     ∑
     k=−m

     ckexp(2ikπ t)

     the trigonometric polynomial that interpolates f is f itself, the numeric approximation of the coefficients are in fact exact (cn=cn).
   - More generally, you can compute cn−cn.
     Suppose that f is equal to its Fourier series, i.e. that:

     f(t)=

     +∞
     ∑
     m=−∞

     cmexp(2iπ mt),

     +∞
     ∑
     m=−∞

     |cm|<∞

     Then:

     f(xk)=f(

     2kπ
     N

     )=yk=

     +∞
     ∑
     m=−∞

     cmωNkm, c_n=

     1
     N

     N−1
     ∑
     k=0

     ykωN−kn

     Replace yk by its value in c_n:

     c_n=

     1
     N

     N−1
     ∑
     k=0

     +∞
     ∑
     m=−∞

     cmωNkmωN−kn

     If m≠ n (mod N ), ωNm−n is an N-th root of unity different from 1, hence:

     ωN(m−n)N=1,

     N−1
     ∑
     k=0

     ωN(m−n)k=0

     Therefore, if m−n is a multiple of N (m=n+l· N) then ∑k=0N−1ωNk(m−n)=N, otherwise ∑k=0N−1ωNk(m−n)=0. By reversing the two sums, you get

     c_n

     =

     1
     N

     +∞
     ∑
     m=−∞

     cm

     N−1
     ∑
     k=0

     ωNk(m−n)

     =

     +∞
     ∑
     l=−∞

     c(n+l· N)

     | = | … cn−2· N+cn−N+cn+cn+N+cn+2· N+…

     Conclusion: if |n|<N/2, then c_n−cn is a sum of cj with large indices (at least N/2 in absolute value), hence is small (depending on the rate of convergence of the Fourier series).

   #### Example.

   Input:

   f(t):=cos(t)+cos(2*t)
   x:=f(2*k*pi/8)$(k=0..7)

   Output:

   2,

   √

   2

   2

   ,−1,−

   √

   2

   2

   ,0,−

   √

   2

   2

   ,−1,

   √

   2

   2

   Input:

   fft(x)

   Output:

   ⎡ ⎣ | 0.0,4.0,4.0,0.0,0.0,0.0,4.0,4.0 | ⎤ ⎦

   Dividing by N=8, you get

   c0=0,c1=0.5,c2=0.5,c3=0.0,
   c−4=0.0,c−3=0.0,c−2=0.5,=c−1=0.5

   Hence bk=0 and ak=c−k+ck equals 1 for k=1,2 and 0 otherwise.

4. Convolution Product
   If P(x)=∑j=0n−1ajxj and Q(x)=∑j=0m−1bjxj are given by the vectors of their coefficients a=[a0,a1,..an−1] and b=[b0,b1,..bm−1], you can compute the product of these two polynomials using the DFT. The product of polynomials is the convolution product of the periodic sequence of their coefficients if the period is greater or equal to (n+m). Therefore we complete a (resp. b) with m+p (resp. n+p) zeros, where p is chosen such that N=n+m+p is a power of 2. If a=[a0,a1,..an−1,0..0] and b=[b0,b1,..bm−1,0..0], then:

   P(x)Q(x)=

   n+m−1
   ∑
   j=0

   (a*b)jxj

   If you know FN(a) and FN(b), then a∗ b=FN−1(FN(a)· FN(b)), since

   FN(x * y)=FN(x) · FN(y)