Challenge07_inverse-z

Challenge07_inverse-z - Digital Signal Processing Dr Fred J...

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Digital Signal Processing Dr. Fred J. Taylor, Professor Lesson Title: Inverse z-Transform Challenge Lesson Number: 07 Challenge Problem A signal having a z-transform: ( 29 ( 29 ( 29 2 2 3 1 1 1 3 11 - + + + = z z z z z X has a partial fraction expansion: ( 29 ( 29 ( 29 ( 29 2 3 2 1 0 1 1 1 - + - + + + = z z K z z K z z K K z X Q: What are the value of the 4-tuple { K 0 , K 1 , K 2 , K 3 }? Answer: First convert X ( z ) into X ( z )/z resulting in: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 = - + - + + + = - + + + = = 2 3 2 1 0 2 2 3 1 1 1 1 1 1 3 11 ' z K z K z K z K z z z z z z z X z X Solving for the individual terms yields: ( 29 ( 29 ( 29 1 1 1 1 3 11 0 2 2 3 0 0 = - + + + = = = = z z z z z z z z X z K ( 29 ( 29 ( 29 75 . 1 4 9 1 1 3 11 1 1 2 2 3 1 1 = = - + + = + = - = - = z z z z z z z z X z K ( 29 ( 29 ( 29 5 . 7 2 15 1 1 3 11 1 1 2 3 1 2 3 = = + + + = - = = = z z z z z z z z X z K ( 29 ( 29 ( 29 ( 29 25 . 8 4 33 1 1 2 3 22 11 1 1 3 11 1 1 2 2 2 3 4 1 2 3 1 2 2 = = + - - + + = + + + = - = = = = z z z z z z z z z dz z z z z d dz z z X z d K Therefore INVERSE_Z 1
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Digital Signal Processing
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Challenge07_inverse-z - Digital Signal Processing Dr Fred J...

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