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Desplazamientos Circulares

Module by: Justin Romberg. E-mail the authorTranslated By: Fara Meza, Erika Jackson

Based on: Circular Shifts by Justin Romberg

Summary: Este modulo ve el desplazamiento circular y como se puede usar para representar el desplazamiento de secuencias periodicas.

Las muchas propiedades de la DFTS se convierten sencillas (muy similares a las de las Series de Fourier) cuando entendemos el concepto de: desplazamientos circulares.

Desplazamientos Circulares

Podemos describir las secuencias periódicas teniendo puntos discretos en un círculo como su dominio.

Figura 1
Figura 1 (fig01.png)

Desplazar mm, fn+m f n m , corresponde a rotar el cilindro mm puntos ACW (en contra del reloj). Para m=-2 m -2 , obtenemos un desplazamiento igual al que se ve en la siguiente ilustración:

Figura 2: para m=-2 m -2
Figura 2 (fig2.png)
Figura 3
Figura 3 (fig03.png)

Para ciclar los desplazamientos seguiremos los siguientes pasos:

1) Escriba fn f n en el cilindro, ACW

Figura 4: N=8 N 8
Figura 4 (fig4.png)

2) Para ciclar por mm, gire el cilindro m lugares ACW fnf (( n + m )) N f n f (( n + m )) N

Figura 5: m=-3 m -3
Figura 5 (fig5.png)

Ejemplo 1

Si fn=01234567 f n 0 1 2 3 4 5 6 7 , entonces f (( n - 3 )) N =34567012 f (( n - 3 )) N 3 4 5 6 7 0 1 2

Es llamado desplazamiento circular, ya que nos estamos moviendo alrededor del círculo. El desplazamiento común es conocido como “desplazamiento linear” (un movimiento en una línea).

Notas para el desplazamiento circular

f (( n + N )) N =fn f (( n + N )) N f n Girar por NN lugares es lo mismo que girar por una vuelta completa, o no moverse del mismo lugar.

f (( n + N )) N =f (( n - ( N - m ) )) N f (( n + N )) N f (( n - ( N - m ) )) N Desplazar ACW mm es equivalente a desplazar CW Nm N m

Figura 6
Figura 6 (fig6.png)

f (( - n )) N f (( - n )) N La expresión anterior, escribe los valores de fn f n para el lado del reloj.

Figura 7
(a) fn f n (b) f (( - n )) N f (( - n )) N
Figura 7(a) (fig7a.png)Figura 7(b) (fig7b.png)

Desplazamientos Circulares y el DFT

Theorem 1: Desplazamiento Circular y el DFT

Si fn DFT Fk f n DFT F k entonces f (( n - m )) N DFT e(i2πNkm)Fk f (( n - m )) N DFT 2 N k m F k (Por ejemplo. Desplazamiento circular en el dominio del tiempo= desplazamiento del ángulo en el DFT)

Proof

fn=1Nk=0N1Fkei2πNkn f n 1 N k 0 N 1 F k 2 N k n
(1)
así que el desplazar el ángulo en el DFT
fn=1Nk=0N1Fke(i2πNkn)ei2πNkn=1Nk=0N1Fkei2πNk(nm)=f (( n - m )) N f n 1 N k 0 N 1 F k 2 N k n 2 N k n 1 N k 0 N 1 F k 2 N k n m f (( n - m )) N
(2)

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