lect11 - EL 625 Lecture 11 1 EL 625 Lecture 11...

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Unformatted text preview: EL 625 Lecture 11 1 EL 625 Lecture 11 Frequency-domain analysis of discrete-time systems Theorem : If S is a fixed linear discrete-time system, the zero-state response, S { z k } = H ( z ) z k (1) Proof : S { z k } = S { z k } z k | {z } H ( z,k ) .z k = H ( z, k ) z k (2) S { z k- k } = H ( z, k- k ) z ( k- k ) (3) = z- k S { z k } = z- k H ( z, k ) z k (4) = H ( z, k ) = H ( z, k- k ) for any k (5) Thus, H ( z, k ) is independent of k = H ( z, k ) = H ( z ). Thus, S { z k } = H ( z ) z k The response of a linear time-invariant discrete-time system to a complex exponential input, z k is the same complex exponen- tial with a change of complex magnitude (magnitude and phase). EL 625 Lecture 11 2 z-transform : F ( z ) = Z { f ( k ) } = X k =0 f ( k ) z- k Inverse z-transform : f ( k ) = 1 2 I C F ( z ) z k- 1 dz Output, y ( k ) = S { u ( k ) } = 1 2 I C U ( z ) S { z k- 1 } dz (6) = 1 2 I C U ( z ) H ( z ) z k- 1 dz (7) Y ( z ) = H ( z ) U ( z ) H ( z ) is the discrete transfer function matrix. The element at the ( i, j ) position is the discrete transfer function from the j th input to the i th output. Properties of z-transform : EL 625 Lecture 11 3 1. Linearity : Z { 1 f 1 ( k ) + 2 f 2 ( k ) } = 1 Z { f 1 ( k ) } + 2 Z { f 2 ( k ) } (8) 2. Convolution : f ( k ) = k X j =0 f 1 ( k- j ) f 2 ( j ) = Z { f ( k ) } = Z { f 1 ( k ) } Z { f 2 ( k ) } (9) 3. Time translation : Z { f ( k + m ) } = z m Z { f ( k ) } - m- 1 X i =0 f ( i ) z- i for m Z { f ( k- m ) } = z- m Z { f ( k ) } + m X i =1 f (- i ) z i for m (10) 4. Initial value : f (0) = lim z F ( z ) = lim z (1- z- 1 ) F ( z ) (11) 5. Final value : lim N f ( N ) = lim z 1 (1- z- 1 ) F ( z ) (12) provided (1- z- 1 ) F ( z ) has no poles for | z | 1....
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lect11 - EL 625 Lecture 11 1 EL 625 Lecture 11...

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