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Series de Fourier y los Sistemas LTI

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

Based on: Fourier Series and LTI Systems by Justin Romberg

Summary: (Blank Abstract)

Introduciendo las Series de Fourier a los Sistemas LTI

Antes de ver este modulo, usted debería familiarizarse con los conceptos de Eigenfunciones de los sistemas LTI. Recuerde, para ℋ ℋ sistema LTI tenemos la siguiente relación

Figura 1: Señales de entrada y salida para nuestro sitema LTI.
Figura 1 (simpleLTIsys.png)

donde es⁢t s t es una eigenfunción de ℋ ℋ. Su eigenvalor correspondiente H⁢s H s pueden ser calculado usando la respuesta de impulso h⁢t h t H⁢s=∫−∞∞h⁢τ⁢e−(s⁢τ)d τ H s τ h τ s τ

Así, usando la expansión de las series de Fourier para f⁢t f t periódica donde usamos la entrada f⁢t=∑n c n ⁢ej⁢ ω 0 ⁢n⁢t f t n c n ω 0 n t en el sistema,

Figura 2: Sistema LTI
Figura 2 (Transferfunc.png)

nuestra salida y⁢t y t será y⁢t=∑nH⁢j⁢ ω 0 ⁢n⁢ c n ⁢ej⁢ ω 0 ⁢n⁢t y t n H ω 0 n c n ω 0 n t Podemos ver que al aplicar las ecuaciones de expansión de series de fourier, podemos ir de f⁢t f t a c n c n y viceversa, y es lo mismo para la salida, y⁢t y t

Efectos de las Series de Fourier

Podemos pensar de un sistema LTI como el ir moldeando el contenido de la frecuencia de la entrada. Mantenga en mente el sistema básico LTI que presentamos en figura 2. El sistema LTI, ℋ ℋ, multiplica todos los coeficientes de Fourier y los escala.

Dado los coeficientes de Fourier de la entrada c n c n y los eigen valores del sistema H⁢j⁢ w 0 ⁢n H w 0 n , las series de Fourier de la salida, es H⁢j⁢ w 0 ⁢n⁢ c n H w 0 n c n (una simple multiplicación de termino por termino).

note:

los eigenvalores, H⁢j⁢ w 0 ⁢n H w 0 n describen completamente lo que un sistema LTI le hace a una señal periódica con periodo T=2⁢π⁢ w 0 T 2 w 0

Ejemplo 1

¿Qué hace este sistema?

Figura 3
Figura 3 (fslti_f1.png)

Ejemplo 2

Y, ¿esté sistema?

Figura 4
(a) (b)
Figura 4(a) (fslti_f2.png)Figura 4(b) (fslti_f3.png)

Examples

Ejemplo 3: El circuito RC

h⁢t=1R⁢C⁢e−tR⁢C⁢u⁢t h t 1 R C t R C u t

¿Qué es lo que este sistema hace a las series de fourier de la f⁢t f t ?

Calcula los eigenvalores de este sistema

H⁢s=∫−∞∞h⁢τ⁢e−(s⁢τ)d τ =∫0∞1R⁢C⁢e−τR⁢C⁢e−(s⁢τ)d τ =1R⁢C⁢∫0∞e(−τ)⁢(1R⁢C+s)d τ =1R⁢C⁢11R⁢C+s⁢e(−τ)⁢(1R⁢C+s)| τ =0∞=11+R⁢C⁢s H s τ h τ s τ τ 0 1 R C τ R C s τ 1 R C τ 0 τ 1 R C s 1 R C 1 1 R C s τ 0 τ 1 R C s 1 1 R C s
(1)

Ahora, decimos que a este circuito RC lo alimentamos con una entrada f⁢t f t periódica (con periodo T=2⁢π⁢ w 0 T 2 w 0 ).

Vea los eigen valores para s=j⁢ w 0 ⁢n s w 0 n |H⁢j⁢ w 0 ⁢n|=1|1+R⁢C⁢j⁢ w 0 ⁢n|=11+R2⁢C2⁢ w 0 2⁢n2 H w 0 n 1 1 R C w 0 n 1 1 R 2 C 2 w 0 2 n 2

El circuito RC es un sistema pasa bajas: pasa frecuencias bajas n n alrededor de 0 0) atenúa frecuencias altas ( n n grandes).

Ejemplo 4: Pulsó cuadrado a través del Circuito RC

  • Señal de entrada : tomando las series de Fourier f⁢t f t c n =12⁢sinπ2⁢nπ2⁢n c n 1 2 2 n 2 n 1t 1 t en n=0 n 0
  • Sistema : Eigenvalores H⁢j⁢ w 0 ⁢n=11+j⁢R⁢C⁢ w 0 ⁢n H w 0 n 1 1 R C w 0 n
  • Señal de salida: tomando las series de Fourier de y⁢t y t d n =H⁢j⁢ w 0 ⁢n⁢ c n =11+j⁢R⁢C⁢ w 0 ⁢n⁢12⁢sinπ2⁢nπ2⁢n d n H w 0 n c n 1 1 R C w 0 n 1 2 2 n 2 n

d n =11+j⁢R⁢C⁢ w 0 ⁢n⁢12⁢sinπ2⁢nπ2⁢n d n 1 1 R C w 0 n 1 2 2 n 2 n y⁢t=∑ n d n ⁢ej⁢ w 0 ⁢n⁢t y t n d n w 0 n t

¿Qué podemos decir sobre y⁢t y t de d n d n ?

  1. ¿Es y⁢t y t real?
  2. ¿ Es y⁢t y t simétrico par? ¿simétrico impar?
  3. ¿Comó se y⁢t y t ¿es mas “suave” que f⁢t f t ? (el radio de descomposición de d n d n vs. c n c n )

d n =11+j⁢R⁢C⁢ w 0 ⁢n⁢12⁢sinπ2⁢nπ2⁢n d n 1 1 R C w 0 n 1 2 2 n 2 n | d n |=11+R⁢C⁢ w 0 2⁢n2⁢12⁢sinπ2⁢nπ2⁢n d n 1 1 R C w 0 2 n 2 1 2 2 n 2 n

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