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Ecuación de Matriz para la DTFS

Module by: Roy Ha. E-mail the authorTranslated By: Fara Meza, Erika Jackson

Based on: Matrix Equation for the DTFS by Roy Ha

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Summary: Este modulo ve como se escribe la ecuación de matriz para la DTFS para hacer los calculos y mostrar las bases mas facilmente.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

La DTFS es nada mas un cambio de bases en ℂN N . Para comenzar, tenemos fn f n en términos de la base estándar.

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)
Tomando la DTFS, podemos escribir fn f n en términos de la base de Fourier senosoidal
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)
Podemos formar la matriz base (llamaremos esto WW envés deBB) al acomodar los vectores bases en las columnas obtenemos
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)
con b k n=ⅇⅈ2πNkn b k n 2 N k n

note:

la entrada k-th fila y n-th columna es W j , k =ⅇⅈ2πNkn= W n , k W j , k 2 N k n W n , k
Así, aquí tenemos una simetría adicional W=WT⇒WT¯=W¯=1NW-1 W W W W 1 N W (ya que b k n b k n son ortonormales)

Ahora podemos rescribir la ecuación DTFS en forma de matriz, donde tenemos:

  • f f = señal (vector en ℂN N )
  • c c = coeficientes DTFS (vector en ℂN N )

Tabla 1
"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

Encontrar (e invertir) la DFTS es nada mas una multiplicación de matrices.

Todo lo que se encuentra en ℂN N esta limpio: no se utilizan límites, no se usan preguntas de convergencia, nada mas se utilice aritmética de matrices.

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