{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ps2 - EE 429 Reading assignment Ch 3 PROBLEM SET 2 DUE 18...

This preview shows pages 1–2. Sign up to view the full content.

EE 429 PROBLEM SET 2 DUE: 18 Feb 2008 Reading assignment: Ch 3 Problem 6: (20 points) The one-sided Z transform of a sequence y ( k ) is defined as Z{ y ( k ) } = k =0 y ( k ) z - k , where y ( k ) = 0 for k < 0. Suppose that the sequences e ( k ), f ( k ) and g ( k ) have Z transforms E ( z ), F ( z ) and G ( z ) respectively. Show that 1. (5 points) Z{ e ( k + n ) } = z n E ( z ) - z n e (0) - z n - 1 e (1) · · · - ze ( n - 1) 2. (5 points) Z{ e ( k - n ) } = z - n E ( z ) 3. (5 points) lim k →∞ e ( k ) = lim z 1 ( z - 1) E ( z ) provided that the left-side limit exists. 4. (5 points) lim k 0 e ( k ) = lim z →∞ E ( z ). Problem 7: (20 points) Calculate the Z transform of the following sequences: 1. (5 points) δ ( k ) = 1 k = 0 0 k = 0 2. (5 points) u o ( k ) = 1 k 0 0 k < 0 3. (5 points) a k u o ( k ) 4. (5 points) ka k u o ( k ) The relationship k =0 a k = 1 1 - a , for | a | < 1, may be of use. Problem 8: (20 points) 1. (6 points) If applicable, use the initial and final value theorems to calculate e (0) and lim k →∞ e ( k ) for each E ( z ) below: (a) (2 points) E ( z ) = 0 . 5 z ( z 2 - 1 . 6 z + 0 . 6) (b) (2 points) E ( z ) = 12 z 2 - 18 z 2 z 2 - 5 z + 2 (c) (2 points) E ( z ) = 18 z 9 z 2 - 6 z + 1 2. (12 points) For each E ( z

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern