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Matrix Equation for the DTFS

Module by: Roy Ha. E-mail the author

Summary: This module looks at writing our DTFS equations in matrix form to make calculations and displaying the basis easier.

Why another representation?

There are many ways to understand the Discrete Fourier Transform, another name for the Discrete Time Fourier Series. One method that proves extremely useful is the matrix representation, since this avoids much of the concern of convergence that plagues the continuous time version of the Fourier Series. We will now proceed to evaluate and understand this representation and its implications

Deriving the DFT

The DTFS is just a change of basis in CN N . To start, we have fn f n in terms of the standard basis.

fn=f0 e 0 +f1 e 1 ++fN1 e N - 1 =k=0n1fkδkn f n f 0 e 0 f 1 e 1 f N 1 e N - 1 k 0 n 1 f k δ k n
(1)
f0f1f2fN1=f0000+0f100+00f20++000fN1 f 0 f 1 f 2 f N 1 f 0 0 0 0 0 f 1 0 0 0 0 f 2 0 0 0 0 f N 1
(2)
Taking the DTFS, we can write fn f n in terms of the sinusoidal Fourier basis
fn=k=0N1 c k ei2πNkn f n k 0 N 1 c k 2 N k n
(3)
f0f1f2fN1= c 0 1111+ c 1 1ei2πNei4πNei2πN(N1)+ c 2 1ei4πNei8πNei4πN(N1)+ f 0 f 1 f 2 f N 1 c 0 1 1 1 1 c 1 1 2 N 4 N 2 N N 1 c 2 1 4 N 8 N 4 N N 1
(4)
We can form the basis matrix (we'll call it WW here instead of BB) by stacking the basis vectors in as columns
W=( b 0 n b 1 n b N - 1 n )=( 1111 1ei2πNei4πNei2πN(N1) 1ei4πNei8πNei2πN2(N1) 1ei2πN(N1)ei2πN2(N1)ei2πN(N1)(N1) ) W b 0 n b 1 n b N - 1 n 1 1 1 1 1 2 N 4 N 2 N N 1 1 4 N 8 N 2 N 2 N 1 1 2 N N 1 2 N 2 N 1 2 N N 1 N 1
(5)
with b k n=ei2πNkn b k n 2 N k n

Note:

the entry in the k-th row and n-th column is W j , k =ei2πNkn= W n , k W j , k 2 N k n W n , k
So, here we have an additional symmetry (W=WT)(WT¯=W¯=1NW-1) W W W W 1 N W (since b k n b k n are orthogonal)
  • WW has the following properties:
    • WW is Vandermonde: the nnth column of WW is a polynomial in W N n W N n
    • WW is symmetric: W=WT W W
    • 1NW 1 N W is unitary: (1NW)1NWH=1NWH(1NW)=I 1 N W 1 N W H 1 N W H 1 N W I
    • 1NW¯=W-1 1 N W W -1 , the IDFT matrix.
  • For NN a power of 2, the FFT can be used to compute the DFT using about N2log2N N 2 2 N rather than N2 N 2 operations.

We can now rewrite the DTFS equations in matrix form where we have:

  • f f = signal (vector in CN N )
  • c c = DTFS coeffs. (vector in CN N )

Table 1
"synthesis" f=Wc f W c fn=c, b n ¯ f n c b n
"analysis" c=WT¯f=W¯f c W f W f ck=f, b k c k f b k

We can therefore conclude that finding (and inverting) the DTFS is just matrix multiplication. We are able to avoid much of the concern of convergence that plagues the continuous time version of the Fourier Series.

Note:

Everything in CN N is clean: no limits, no convergence questions, just good ole matrix arithmetic.

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