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Matrix Equation for the DTFS

Module by: Roy Ha

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

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 +…+fN-1 e N - 1 =∑k=0n-1fkδk-n 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)
f0f1f2⋮fN-1=f000⋮0+0f10⋮0+00f2⋮0+…+000⋮fN-1 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=0N-1 c k ⅇⅈ2πNkn f n k 0 N 1 c k 2 N k n (3)
f0f1f2⋮fN-1= c 0 111⋮1+ c 1 1ⅇⅈ2πNⅇⅈ4πN⋮ⅇⅈ2πNN-1+ c 2 1ⅇⅈ4πNⅇⅈ8πN⋮ⅇⅈ4πNN-1+… 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=111…11ⅇⅈ2πNⅇⅈ4πN…ⅇⅈ2πNN-11ⅇⅈ4πNⅇⅈ8πN…ⅇⅈ2πN2N-1⋮⋮⋮⋮⋮1ⅇⅈ2πNN-1ⅇⅈ2πN2N-1…ⅇⅈ2πNN-1N-1 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)

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 )

"synthesis" f=Wc f W c fn=<c,bn¯> f n c b n
"analysis" c=WT¯f=W¯f c W f W f ck=<f,bk> c k f b k

Finding (and inverting) the DTFS is just matrix multiplication.

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

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