lecture2_small

lecture2_small - Discrete-time systems Discrete-time...

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Z -Transform Z -Transform Recall the Z -Transform deFnition: y ( z ) := ± k = -∞ y ( k ) z - k , where z ∈ C . This has an associated region of convergence: r 0 < | z | <R 0 . Example: y ( k )= ² 0 for k< 0 , e - akT for k 0 0 2 4 6 8 10 0.2 0.4 0.6 0.8 1.0 y(k) Index: k a =0 . 25 , T =1 Roy Smith: ECE 147b 2 :2 Discrete-time systems Discrete-time Systems: P ± ± u ( k ) y ( k ) Input and output signals are sequences: u = { u (0) ,u (1) (2) ,. .., u ( k ) ,... } and y = { y (0) ,y (1) (2) , . .., y ( k ) } . Causal LTI/LSI models can be described by di±erence equations: y ( k - a 1 y ( k - 1) - a 2 y ( k - 2) ... - a n y ( k - n )+ b 0 u ( k b 1 u ( k - 1) . . . b m u ( k - m ) . Note that the current output, y ( k ), depends only on current and past inputs, u ( k ), u ( k - 1), . . . , and past outputs, y ( k - 1), . . . . Roy Smith: ECE 147b 2 :1
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Transfer functions Transfer Functions Applying Z -transforms to the diFerence equations gives: y ( z )= ± k = -∞ y ( k ) z - k = ± k = -∞ ( - a 1 y ( k - 1) ... - a n y ( k - n )+ b 0 u ( k ) + b m u ( k - m )) z - k , = - a 1 ± k = -∞ y ( k - 1) z - k - a n ± k = -∞ y ( k - n ) z - k + b 0 ± k = -∞ u ( k ) z - k + b m ± k = -∞ u ( k - m ) z - k , = - a 1 z - 1 y ( z ) - a n z - n y ( z b 0 u ( z ) . . . b m z - m u ( z ) , Rearranging gives, ( 1+ a 1 z - 1 + ··· + a n z - n ) y ( z ( b 0 + b 1 z - 1 + + b m z - m ) u ( z ) . ±rom this we get the transfer function, P ( z y ( z ) u ( z ) = b 0 + b 1 z - 1 + + b m z - m a 1 z - 1 + + a n z - n = b 0 z n + b 1 z n - 1 + + b m z n - m z n + a 1 z n - 1 + + a n = b ( z ) a ( z ) . Roy Smith: ECE 147b 2 :4 Z -Transform Example (continued) y ( z ± k =0 e - akT z - k = ± k =0 ( e - aT z - 1 ) k = 1 1 - e - aT z - 1 , for e - aT < | z | < , = z z - e - aT , for e - aT < | z | < .
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lecture2_small - Discrete-time systems Discrete-time...

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