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Extensión Periódica de las DTFS

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

Based on: Periodic Extension to DTFS by Roy Ha

Summary: Este Modulo ve la extensión Ppriódica de los coeficientes de la DTFS y como los coefcientes juegan un papel critico para manipular las señales.

Introducción

Ya que contamos con entendimiento de lo que son las series discretas de Fourier (DTFS), podemos considerar la extensión periódica de c⁢k c k (coeficientes discretos de Fourier). Las figures básicas mostradas a continuación muestran una simple ilustración de como nosotros podríamos representar una secuencia en forma de señales periódicas graficadas sobre un numero infinito de intervalos.

Figura 1
(a) vectores
Figura 1(a) (fig1a.png)
(b) secuencias periodicas
Figura 1(b) (fig1b.png)

Exercise 1

¿Porqué una extensión periódica de los coeficientes de la DFTS c⁢k c k tiene sentido?

Solution

Aliasing: b k =ej⁢2⁢πN⁢k⁢n b k 2 N k n

b k + N =ej⁢2⁢πN⁢(k+N)⁢n=ej⁢2⁢πN⁢k⁢n⁢ej⁢2⁢π⁢n=ej⁢2⁢πN⁢n= b k b k + N 2 N k N n 2 N k n 2 n 2 N n b k
(1)
→ Coeficientes DTFS también son periódicos con periodo NN.

Ejemplos

Ejemplo 1: Función Cuadrada Discreta

Figura 2
Figura 2 (fig2.png)

Calcule la DTFS c⁢k c k usando:

c⁢k=1N⁢∑n=0N−1f⁢n⁢e−(j⁢2⁢πN⁢k⁢n) c k 1 N n 0 N 1 f n 2 N k n
(2)
Como en las series de Fourier continuas, podemos tomar la sumatoria en cualquier intervalo, para tener
c k =1N⁢∑n=− N 1 N 1 e−(j⁢2⁢πN⁢k⁢n) c k 1 N n N 1 N 1 2 N k n
(3)
Deje que m=n+ N 1 m n N 1 ,(así tenemos una serie geométrica que empieza en 0)
c k =1N⁢∑m=02⁢ N 1 e−(j⁢2⁢πN⁢(m− N 1 )⁢k)=1N⁢ej⁢2⁢πN⁢k⁢∑m=02⁢ N 1 e−(j⁢2⁢πN⁢m⁢k) c k 1 N m 0 2 N 1 2 N m N 1 k 1 N 2 N k m 0 2 N 1 2 N m k
(4)
Ahora, usando la “formula parcial de sumatoria”
∑n=0Man=1−aM+11−a n 0 M a n 1 a M 1 1 a
(5)
c k =1N⁢ej⁢2⁢πN⁢ N 1 ⁢k⁢∑m=02⁢ N 1 e−(j⁢2⁢πN⁢k)m=1N⁢ej⁢2⁢πN⁢ N 1 ⁢k⁢1−e−(j⁢2⁢πN⁢(2⁢ N 1 +1))1−e−(j⁢k⁢2⁢πN) c k 1 N 2 N N 1 k m 0 2 N 1 2 N k m 1 N 2 N N 1 k 1 2 N 2 N 1 1 1 k 2 N
(6)
Manipulándola para que se vea como el sinc (distribuya)
c k =1N⁢e−(j⁢k⁢2⁢π2⁢N)⁢(ej⁢k⁢2⁢πN⁢( N 1 +12)−e−(j⁢k⁢2⁢πN⁢( N 1 +12)))e−(j⁢k⁢2⁢π2⁢N)⁢(ej⁢k⁢2⁢πN⁢12−e−(j⁢k⁢2⁢πN⁢12))=1N⁢sin2⁢π⁢k⁢( N 1 +12)Nsinπ⁢kN= digital sinc c k 1 N k 2 2 N k 2 N N 1 1 2 k 2 N N 1 1 2 k 2 2 N k 2 N 1 2 k 2 N 1 2 1 N 2 k N 1 1 2 N k N digital sinc
(7)

note:

¡es periódica! figura 3, figura 4, y figura 5 muestran como nuestra función y coeficientes para distintos valores de N 1 N 1 .

Figura 3: N 1 =1 N 1 1
(a) Grafíca de f⁢n f n . (b) Grafíca de c⁢k c k .
Figura 3(a) (dtfs1.png)Figura 3(b) (dtfs1a.png)
Figura 4: N 1 =3 N 1 3
(a) Grafíca de f⁢n f n . (b) Grafíca de c⁢k c k .
Figura 4(a) (dtfs2.png)Figura 4(b) (dtfs2a.png)
Figura 5: N 1 =7 N 1 7
(a) Grafíca de f⁢n f n . (b) Grafíca de c⁢k c k .
Figura 5(a) (dtfs3.png)Figura 5(b) (dtfs3a.png)

Ejemplo 2: Sonido de un Pájaro

Figura 6
Figura 6 (nofig.png)

Ejemplo 3: Análisis Espectral de la DFTS

Figura 7
Figura 7 (nofig.png)

Ejemplo 4: Recuperando la Señal

Figura 8
Figura 8 (nofig.png)

Ejemplo 5: Compresión (1-D)

Figura 9
Figura 9 (nofig.png)

Ejemplo 6: Compresión de Imágenes

Figura 10: We've cut down on storage space by > 90% (while incurring some loss)
(a) Imagen de 256 por 256 (65,636 pixeles)
Figura 10(a) (nofig.png)
(b) Imagen reconstruida usando 6000 coefícientes
Figura 10(b) (nofig.png)

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