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La Transformada Inversa de Z

Module by: Benjamin Fite. E-mail the authorTranslated By: Fara Meza, Erika Jackson

Based on: Inverse Z-Transform by Benjamin Fite

Summary: (Blank Abstract)

Cuando usamos la transformada-z

X⁢z=∑ n =−∞∞x⁢n⁢z−n X z n x n z n
(1)
Es de gran ayuda el poder encontrar x⁢n x n dado X⁢z X z . Existen al menos 4 métodos diferentes para hacer esto:

El Método de Inspección

Este "método" es el familiarizarse con la tabla para los pares de la transformada-z y el usar “ingeniería inversa”.

Ejemplo 1

Dado X⁢z=zz−α X z z z α Con un ROC de |z|>α z α Al “inspeccionar” podemos determinar que x⁢n=αn⁢u⁢n x n α n u n

Método de Expansión de Fracciones Parciales

Cuando tratamos con sistemas lineares de tiempo invariante la transformada-z tiene la forma

X⁢z=B⁢zA⁢z=∑ k =0M b k ⁢z−k∑ k =0N a k ⁢z−k X z B z A z k 0 M b k z k k 0 N a k z k
(2)
Que también se puede expresar como
X⁢z= a 0 b 0 ⁢∏ k =1M1− c k ⁢z-1∏ k =1N1− d k ⁢z-1 X z a 0 b 0 k 1 M 1 c k z k 1 N 1 d k z
(3)
donde c k c k representa los ceros con valor no ceros de X⁢z X z y d k d k representa los polos con valores no cero

Si M<N M N entonces X⁢z X z se puede representar por

X⁢z=∑ k =1N A k 1− d k ⁢z-1 X z k 1 N A k 1 d k z
(4)
Esta forma ayuda el invertir cada termino de la suma usando el método de inspección y la tabla de transformadas. Por lo tanto el numerador es un polinomio se vuelve necesario usar la expansión de fracciones parciales para poder poner X⁢z X z en la forma descrita arriba. Si M≥N M N entonces X⁢z X z se puede expresar como
X⁢z=∑ r =0M−N B r ⁢z−r+∑ k =0N−1 b k ' ⁢z−k∑ k =0N a k ⁢z−k X z r 0 M N B r z r k 0 N 1 b k ' z k k 0 N a k z k
(5)

Ejemplo 2

Encuentre la transformada inversa-z de X⁢z=1+2⁢z-1+z-21−3⁢⁢z-1+2⁢z-2 X z 1 2 z z -2 1 -3 z 2 z -2 Donde el ROC es |z|>2 z 2 . En este caso M=N=2 M N 2 , así que tenemos que la división larga para obtener X⁢z=12+12+72⁢z-11−3⁢⁢z-1+2⁢z-2 X z 1 2 1 2 7 2 z 1 -3 z 2 z -2 Tenemos que factorizar el denominador. X⁢z=2+-1+5⁢z-1(1−2⁢z-1)⁢(1−z-1) X z 2 -1 5 z 1 2 z 1 z Ahora utilizamos la expansión de fracciones parciales. X⁢z=12+ A 1 1−2⁢z-1+ A 2 1−z-1=12+921−2⁢z-1+-41−z-1 X z 1 2 A 1 1 2 z A 2 1 z 1 2 9 2 1 2 z -4 1 z Ahora cada término se puede invertir usando el método de inspección y la tabla de transformada-z. ya que la ROC es |z|>2 z 2 , x⁢n=12⁢δ⁢n+92⁢2n⁢u⁢n−4⁢⁢u⁢n x n 1 2 δ n 9 2 2 n u n -4 u n

Método de Expansión de Series de Potencia

Cuando la transformada-z es definida con una serie de potencia en la forma de

X⁢z=∑ n =−∞∞x⁢n⁢z−n Xz n x n z n
(6)
entonces cada termino de la secuencia x⁢n x n se puede terminar al ver los coeficientes del respectivo poder z−n z n .

Ejemplo 3

Ve la siguiente transformada -z de la secuencia de tamaño finito.

X⁢z=z2⁢(1+2⁢z-1)⁢(1−12⁢z-1)⁢(1+z-1)=z2+52⁢z+12+−z-1 X z z 2 1 2 z 1 1 2 z 1 z z 2 5 2 z 1 2 z
(7)
En este caso, ya que no hay polos, multiplicamos los factores de X⁢z X z . Al inspecciónala, se vuele claro que x⁢n=δ⁢n+2+52⁢δ⁢n+1+12⁢δ⁢n+−δ⁢n−1 x n δ n 2 5 2 δ n 1 1 2 δ n δ n 1 .

Una de las ventajas de las series de potencia es que muchas funciones usadas en funciones de ingeniería tienen tabuladas sus series de potencia. Por lo tanto funciones como el Log, seno, exponencial, seno h, etc., se pueden invertir fácilmente.

Ejemplo 4

Suponga X⁢z=log n (1+α⁢z-1) X z n 1 α z Al notar que log n (1+x)=∑ n =1∞-1n+1⁢xnn n 1 x n 1 -1 n 1 x n n Entonces X⁢z=∑ n =1∞-1n+1⁢αn⁢z−nn X z n 1 -1 n 1 α n z n n Por lo tanto X⁢z={-1n+1⁢αnn  if  n≥10  if  n≤0 X z -1 n 1 α n n n 1 0 n 0

Método de Integración del Contorno

Sin entrar en detalles

x⁢n=12⁢π⁢j⁢∮rX⁢z⁢zn−1d z x n 1 2 z r X z z n 1
(8)
donde r r es el contorno del lado contrario del reloj en la ROC de X⁢z X z circulando el origen del plano-z. Para expandir este método para encontrar la inversa se necesita conocimiento de teoría de variables complejas y esto no se vera en este modulo.

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