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Teorema de Nyquist

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

Based on: Nyquist Theorem by Justin Romberg

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Summary: Este modulo introduce el Teorema de Nyquist.

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.

Introducción

Anteriormente habia estado expuesto a los conceptos detras del muestreo y el teorema de muestreo. Mientras aprendía estas ideas, debio haber empezado a notar que si muestreamos a muy bajo valor, hay una oportunidad que nuestra señal original no sea únicamente definida por nuestra señal muestreada. Si esto sucede, entonces no es garantia de que recontruyamos correctamente la señal. Como resultado de esto, el Teorema de Nyquist ha sido creado. A continuación veremos exactamente lo que este torema nos dice.

Teorema de Nyquist

Sea TT igual a nuestro período de muestreo (distancia entre las muestras). Después sea Ωs=2πT Ωs 2 T (frecuencia de muestreo radianes/seg). Hemos visto que si ft f t es limitado en banda en -ΩBΩB ΩB ΩB y muestreamos con período T<πΩb(2πΩs<πΩBΩs>2ΩB) T Ωb 2 Ωs ΩB Ωs 2 ΩB entonces podemos reconstruir ft f t de sus muestras.

Theorem 1: Teorema de Nyquist ("Teorema Fundamental de Procesamiento Digital de Señales DSP")

Si ft f t es limitado en banda a -ΩBΩB ΩB ΩB , podemos reconstruirlo perfectamente de sus muestras fsn=fnT fs n f n T para Ωs=2πT>2ΩB Ωs 2 T 2 ΩB

ΩN=2ΩB ΩN 2 ΩB es llamada la "frecuencia Nyquist " para ft f t . Para la reconstrucción perfecta de ser posible Ωs2ΩB Ωs 2 ΩB donde Ωs Ωs es la frecuancia de muestreo y ΩB ΩB es la frecuencia más alta en la señal.

Figura 1: Illustración de la Frecuencia Nyquist
Figura 1 (nyq_f1a.png)

Ejemplo 1: Ejemplos:

  • El oído humano oye frecuencias hasta 20 kHz → CD el valor de la muestra es 44.1 kHz.
  • La linea telefónica pasa frecuencias de hasta 4 kHz → la muestra de la compañia de telefonos es de 8 kHz.

Reconstrucción

La formula de la reconstrucción en el dominio del tiempo se ve como ft=n=-fsnsinπTtnTπTtnT f t n fs n T t n T T t n T Podemos concluir, desde antes que n,n:sinπTtnTπTtnT n n T t n T T t n T es una base para el espacio de -ΩBΩB ΩB ΩB funciones limitadas en banda, ΩB=πT ΩB T . Los coeficientes de expansión para esta base son calculados muestreando ft f t en el valor 2πT=2ΩB 2 T 2 ΩB .

nota:

La base también es ortogonal. Para hacerla ortonormal, necesitamos un factor de normalización de T T .

La gran Pregunta

Exercise 1

¿Que pasa si Ωs<2ΩB Ωs 2 ΩB ? ¿Qué sucede cuando muestreamos abajo del valor de Nyquist?

Solution

Vayase a través de los pasos: (véase la figura 2)

Figura 2:
Figura 2 (nyq_f2a.png)

Finalmente, ¿Qué le pasara ahora a F s ω F s ω ? Para contestar esta última pregunta, necesitamos ver el concepto de aliasing.

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