Quiz_1_sol

# Quiz_1_sol - z ) = z-2 + 2 z + 1 z ( z-1)( z-. 7) = z-2 + 1...

This preview shows page 1. Sign up to view the full content.

ECE 483 Quiz # 1 Solution Problem 1. (6 points) For each of the following z -transforms, determine the initial value f (0) and the ±nal value f ( ) of the corresponding signal f ( k ), if such values exist. (a) (2 points) F ( z ) = 0 . 5 ( z 2 +1)( z +0 . 7) . Solution: f (0) = 0 and f ( ) does not exist (due to the poles at ± j ). (b) (2 points) F ( z ) = z 2 +4 2( z +0 . 9)( z - 1) . Solution: f (0) = 0 . 5 and f ( ) = lim z 1 ( z - 1) F ( z ) = 5 / 2 / 1 . 9 = 1 . 3158. (c) (2 points) F ( z ) = ( z +0 . 1) 2 z ( z - 0 . 5)( z +1 . 2) + 2. Solution: f (0) = 2 and f ( ) does not exist (due to the pole at - 1 . 2). Problem 2. (14 points) Given the z -transform F ( z ) = z - 2 + 2 z + 1 z ( z - 1)( z - 0 . 7) , (a) (6 pts) Find the inverse z -transform f ( k ) for k = 0 , 1 , 2 only. Solution: Using the long division approach, we have f (0) = 0 , f (1) = 0 , f (2) = 3 . (b) (8 pts) Find the inverse z -transform f ( k ) for all k = 0 , 1 , 2 , . . . . Solution: Find the partial fraction expansion: F (
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: z ) = z-2 + 2 z + 1 z ( z-1)( z-. 7) = z-2 + 1 . 4286 z + 10 z-1-11 . 4286 z-. 7 . Therefore, f ( k ) = 1 . 4286 · δ ( k-1) + δ ( k-2) + 10 · 1( k-1) + 11 . 4286 · (0 . 7) k-1 · 1( k-1) , k = 0 , 1 , 2 , . . . . Alternatively, we ±nd the partial fraction expansion of F ( z ) /z as F ( z ) z = z-3 + 2 z + 1 z 2 ( z-1)( z-. 7) = z-3 + 6 . 3265 z + 1 . 4286 z 2 + 10 z-1-16 . 3265 z-. 7 . Therefore, F ( z ) = z-2 + 6 . 3265 + 1 . 4286 z + 10 z z-1-16 . 3265 z z-. 7 , and f ( k ) = 6 . 3265 · δ ( k ) + 1 . 4286 · δ ( k-1) + δ ( k-2) + 10 · 1( k )-16 . 3265 · (0 . 7) k , k = 0 , 1 , 2 , . . . ....
View Full Document

## This note was uploaded on 04/15/2011 for the course ECE 483 taught by Professor Evens during the Spring '08 term at Purdue.

Ask a homework question - tutors are online