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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

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

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 ++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)
Tomando la DTFS, podemos escribir fn f n en términos de la base de Fourier senosoidal
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)
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 )=( 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)
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, 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

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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