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Lesson 31

# Lesson 31 - Lesson 31 Chapter 5.2 5.3 Challenge 30...

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Lesson 31 Lesson 31 Chapter 5.2 5.3 Challenge 30 z-transforms Challenge 31 Exam #3 – ave. ~ 77

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Lesson 31 Challenge 30 A signal has a z-transform: What is x[k] in the form: ( 29 ( 29 ( 29 2 2 3 1 1 1 3 11 - + + + = z z z z z X For academic purposes, consider performing a machine analysis using residuez , residue , and a manual Heaviside expansion. ] [ ] [ ] [ ) ( ] [ k ku K k u K k u K K k x k 3 2 1 0 1 + + - + =
Lesson 31 Challenge 30 To use MATLAB’s residuez function, you need to convert X(z) into a descending polynomial representation and perform the following operations: ( 29 ( 29 ( 29 3 2 1 3 1 2 2 3 1 1 1 1 1 3 11 1 1 1 3 11 z z z z z z z z z z X + - - + + = - + + + = - - -

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Lesson 31 Challenge 30 >> n=[11,3,0,1]; >> d=[1,-1,-1,1]; >> [R,P,K]= residuez (n,d) R = 1.7500 0.7500 + 0.0000i 7.5000 - 0.0000i P = -1.0000 1.0000 + 0.0000i 1.0000 - 0.0000i K = 1 ( 29 3 2 1 3 1 1 1 1 1 1 3 11 - - - - - + - - + + = z z z z z z X Using residuez
Lesson 31 Challenge 30 >> n=[11,3,0,1]; >> d=[1,-1,-1,1]; >> [R,P,K]= residuez (n,d) R = 1.7500 0.7500 + 0.0000i 7.5000 - 0.0000i P = -1.0000 1.0000 + 0.0000i 1.0000 - 0.0000i K = 1 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 1 1 1 1 5 . 7 1 75 . 0 1 75 . 1 1 1 5 . 7 1 75 . 0 1 75 . 1 1 - + - + + + = - + - + + + = - - - z z z z z z z z z z X 7.5z 2 /(z-1) 2  is not in a standard table. Using residuez

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Lesson 31 Challenge 30 >> n=[11,3,0,1]; >> d=[1,-1,-1,1, 0 ]; >> [R,P,K]= residue (n,d) R = 1.7500 8.2500 7.5000 1.0000 P = -1.0000 1.0000 1.0000 0 K = [] ( 29 1 2 3 4 2 3 1 1 3 11 z z z z z z z z X z X + - - + + = = ) ( ' ( 29 ( 29 ( 29 ( 29 2 1 5 . 7 1 25 . 8 1 75 . 1 1 - + - + + + = z z z z z z z X All terms appear in a standard table. Using residue: Form X’(z)=X(z)/z
Lesson 31 Challenge 30 Manual calculation ( 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 ( 29 2 3 2 1 0 1 1 1 - + - + + + = z z K z z K z z K K z X

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Lesson 31 Challenge 30 Manual calculations ( 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 ( 29 ( 29 ( 29 ( 29 2 3 2 1 0 1 1 1 - + - + + + = z z K z z K z z K K z X
Lesson 31 Challenge 30 Obtain (in residue form) Inverse x[k] = δ [k] + 1.75(-1) k u[k] + 8.75 u[k] + 7.5k u[k] ( 29 ( 29 ( 29 ( 29 2 1 5 . 7 1 25 . 8 1 75 . 1 1 - + - + + + = z z z z z z z X

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Lesson 31 Mini Exam Given X(z) = z 2 /(z-1) 2 , what is x[k]? X(z)=a 0 + a 11 z/(z-1) + a 12 z/(z-1) 2 a 0 = X(z)| z 0 = 0 a 12 = (z-1) 2 X(z)| z 1 = 1 a 11 = d((z-1) 2 X(z)/z)/dz| z 1 = dz/dz| z 1 = 1 Check: X(z)= 0 + z/(z-1) + z/(z-1) 2 = z 2 /(z-1) 2 . Solution x[k] = 0 δ [k] + 1 u[k] + 1 ku[k] = (1+k)u[k]
Lesson 31 Preview The Laplace transform was shown to be a viable modeling and analysis paradigm for linear continuous time signals and system.

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