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# MODULE-5 - MODULE 5 Z-TRANSFORM One and Two-Sided...

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MODULE 5 Z-TRANSFORM One- and Two-Sided Z-Transforms Convergence Inversion Stability and Causality Page 5.1

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INDEX Z-Transform Properties of One-Sided Z-Transform Geometric Series Example - DE Solution with I.C.'s via Z-Transform Two-Sided Z-Transform Convergence of Two-Sided Z-Transform Example of Region of Convergence Importance of ROC Specification ROC of Different Types of Sequences Inversion by Residue Theorem Example - Inversion by Residue Theorem Inversion by Partial Fraction Expansion Two-Sided Z-Transform Properties Stability, Causality, and the Z-Transform General BIBO Stability Condition Exercise - Stability Triangle MAIN INDEX Page 5.2
5. Z-TRANSFORM index READ : Chapter 3 of Oppenheim & Schafer. Work as many problems as possible. The Z-Transform plays the same role as that played by the Laplace Transform in continuous system theory. It is a linear transformation that is useful for analyzing LTI systems and for solving LCCDE's. One-Sided Z-Transform Definition - The one-sided Z-transform of x ( n ) is given by Z 1 { x ( n ) } = X ( z ) = 0 n x ( n ) z -n - The variable z is complex - and hard to assign a physical meaning to, except when z = e j ϖ . - X ( z ) is a power series - it may or may not converge . The one-sided Z-transform is primarily useful for solving LCCDE's with specified or known initial conditions . Page 5.3

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Properties of One-Sided Z-Transform index Linearity : Z 1 { a·x ( n ) + b·y ( n ) } = a·X ( z ) + b·Y ( z ) Shifting : Z 1 { x ( n +1) } = = 0 n x ( n +1) z -n = = 1 m x ( m ) z -m +1 = = 0 m x ( m ) z -m +1 - z · x (0) = z [ X ( z ) - x (0) ] Generally : Z 1 { x ( n + M ) } = z M [ X ( z ) - - = 1 0 M m z -m x ( m ) ] Page 5.4
Geometric Series index Basic analysis tool used in Z-transform evaluation and inversion is the geometric series formula . If s ( N ) = - = 1 0 N n a ( n ) = a (0) + a (1) + · · · + a ( N -1) where a ( i +1) a ( i ) = r = common ratio then s ( N ) = a (0) - = 1 0 N n r n = a (0)· 1 - r N 1 - r and if | r | < 1 (convergence criteria) s ( ) = a (0) = 0 n r n = a (0) 1 - r . We will often use it in sums of the form: = 0 n a n z -n = 1 1 - az -1 (provided | a/z | < 1) hence Z 1 { a n u ( n ) } = 1 1 - az -1 . Page 5.5

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index Assume i.c. y (0) = 1. Consider the LCCDE: y ( n +1) + 2 y ( n ) = u ( n ). Apply the one-sided Z-transform to both sides: z [ Y ( z ) - y (0) ] + 2 Y ( z ) = U ( z ) where U ( z ) = = 0 n z -n = 1 1 - z -1 = z z - 1 for | z | > 1 Solve for Y ( z ): Y ( z ) ( z +2) = z z - 1 + z or Y ( z ) = z ( z - 1)( z + 2) + z ( z + 2) = z -1 (1 - z -1 )(1 + 2 z -1 ) + 1 1 + 2 z -1 = (1/3) 1 1 - z -1 + (2/3) 1 1 + 2 z -1 (by partial fraction expansion - more later) so y ( n ) = (1/3) u ( n ) + (2/3) (-2) n u ( n ). Page 5.6
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MODULE-5 - MODULE 5 Z-TRANSFORM One and Two-Sided...

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