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# lecture2_small - Z-Transform Z-Transform Recall the...

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Unformatted text preview: Z-Transform Z-Transform Recall the Z-Transform definition: y ( z ) := ∞ summationdisplay k = −∞ y ( k ) z − k , where z ∈ C . This has an associated region of convergence: r < | z | < R . Example: y ( k ) = braceleftbigg for k < , e − akT for k ≥ 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 a27 a27 u ( k ) y ( k ) Input and output signals are sequences: u = { u (0) , u (1) , u (2) , . . . , u ( k ) , . . . } and y = { y (0) , y (1) , y (2) , . . . , y ( k ) , . . . } . Causal LTI/LSI models can be described by difference equations: y ( k ) = − a 1 y ( k − 1) − a 2 y ( k − 2) . . . − a n y ( k − n ) + b 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 Transfer functions Transfer Functions Applying Z-transforms to the difference equations gives: y ( z ) = ∞ summationdisplay k = −∞ y ( k ) z − k = ∞ summationdisplay k = −∞ ( − a 1 y ( k − 1) . . . − a n y ( k − n ) + b u ( k ) . . . + b m u ( k − m )) z − k , = − a 1 ∞ summationdisplay k = −∞ y ( k − 1) z − k . . . − a n ∞ summationdisplay k = −∞ y ( k − n ) z − k + b ∞ summationdisplay k = −∞ u ( k ) z − k . . . + b m ∞ summationdisplay k = −∞ u ( k − m ) z − k , = − a 1 z − 1 y ( z ) . . . − a n z − n y ( z ) + b u ( z ) . . . b m z − m u ( z ) , Rearranging gives, ( 1 + a 1 z − 1 + · · · + a n z − n ) y ( z ) = ( b + b 1 z − 1 + · · · + b m z − m ) u ( z ) ....
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lecture2_small - Z-Transform Z-Transform Recall the...

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