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lect2 - SOME USEFUL Z-TRANSFORMS We found earlier that a n...

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SOME USEFUL Z-TRANSFORMS We found earlier that a z z n U a n - ) ( . Equivalently, a z a z z z a n n n - = = - ; 0 . (2.1) Differentiating both sides with respect to a yields ( 29 a z a z z z na n n n - = = - - ; 2 0 1 . (2.2) Now, the sum on the left is the z-transform of the signal ) ( n U na n . Thus, we have the new transform pair ( 29 2 1 ) ( a z z n U na n - - with ROC a z (2.3) Differentiating (2.2) now yields ( 29 a z a z z z a n n n n n - = - = - - ; 2 ) 1 ( 3 0 2 , which gives us the transform pair ( 29 3 2 2 ) ( ) 1 ( a z z n U a n n n - - - with ROC a z (2.4) Differentiating again with respect to a yields the pair
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( 29 4 3 ) 3 )( 2 ( ) ( ) 2 )( 1 ( a z z n U a n n n n - - - - with ROC a z (2.4) Continuing this process, we find the general pair: ( 29 1 ! ) ( ) 1 ( ) 1 ( + - - + - - m m n a z z m n U a m n n n which can be re-written as ( 29 1 ) ( + - - m m n a z z n U a m n with ROC a z (2.5) If we repeat the above process starting with the anticausal pair a z z n U a n - - - - ) 1 ( , we arrive at the general transform pair ( 29 1 ) 1 ( + - - - - - m m n a z z n U a m n with ROC a z < (2.6) Example 2.1 a) With m = 1 (2.5) and (2.6) yield ( 29 2 1 ) ( a z z n U na n - -
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( 29 2 1 ) 1 ( a z z n U na n - - - - - . b) With m = 3, ( 29 4 3 ) ( 6 ) 2 )( 1 ( a z z n U a n n n n - - - - ( 29 4 3 ) 1 ( 6 ) 2 )( 1 ( a z z n U a n n n n - - - - - - - . SOME Z-TRANSFORM THEOREMS 1. The time shifting theorem ) ( ) ( z F z k n f k - - Proof: -∞ = - - -∞ = + - -∞ = - = = - - n n k n k n n n z n f z z n f z k n f k n f ) ( ) ( ) ( ) ( ) ( . Q.E.D. 2. The convolution theorem ) ( ) ( ) ( ) ( z G z F n g n f . Proof:
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-∞ = - = k k n g k f n g n f ) ( ) ( ) ( ) ( ∑ ∑ -∞ = - -∞ = -∞ = - -∞ = - = - k n n n n k z k n g k f z k n g k f ) ( ) ( ) ( ) ( -∞ = - = k k z z G k f ) ( ) ( (by the time shifting theorem) ) ( ) ( ) ( ) ( z F z G z k f z G k k = = -∞ = - . Q.E.D. 3. A conjugate sequence property - z F n f 1 ) ( . Proof: -∞ = - = n n z n f z F ) ( ) ( -∞ = = n n z n f z F ) ( ) ( 1 -∞ = - -∞ = - = = n n n n z n f z n f z F ) ( ) ( ) ( 1 . Q.E.D.
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4. Parseval’s formula - -∞ = = p p w w w p d e G e F n g n f j j n ) ( ) ( 2 1 ) ( ) ( . Proof: Let ) ( ) ( z F n f and ) ( ) ( z G n g . Then by the conjugate sequences property above, - z G n g 1 ) ( , and by the convolution theorem - z G z F n g n f 1 ) ( ) ( ) ( . Expanding the convolution sum yields - -∞ = z G z F n k g k f k 1 ) ( ) ( ) ( = - -∞ = - -∞ = z G z F z n k g k f n n k 1 ) ( ) ( ) ( = - -∞ = - -∞ = z G z F z n k g k f k n n 1 ) ( ) ( ) ( . Now substitute w j e z = to get ( 29 w w w j j k n j n e G e F e n k g k f -∞ = - -∞ = = - ) ( ) ( ) ( .
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Integrating both sides yields ( 29 - -∞ = - - -∞ = = - p p w w p p w w p w p d e G e F d e n k g k f j j k n j n ) ( 2 1 2 1 ) ( ) ( and, since ) ( 2 1 n d e n j d w p p p w = - - , we finally have ( 29 - -∞ = = p p w w w p d e G e
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