This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECE 310, Spring 2005, HW 4 solutions Problem E3.5 a) Using Definition 3.4.1 , (1 1 n ) r n 1 ( A ) + 1 n I A ( ω n ) = n 1 n × 1 n 1 n 1 X i =1 I A ( ω i ) + 1 n I A ( ω n ) = 1 n n X i =1 I A ( ω i ) = r n ( A ) / b) From part a), r n ( A ) r n 1 ( A ) = 1 n ( I A ( ω n ) r n 1 ( A )). ⇒  r n ( A ) r n 1 ( A )  = 1 n  I A ( ω n ) r n 1 ( A )  6 1 n c) No. Consider the following construction of a binary sequence as a counter example: Let x 1 = 1 and r i = 1 i ∑ j = i j =1 x j For i > 2 if r i > 2 / 3, then x i +1 = 0 else if r i 6 1 / 3, then x i +1 = 1 else if 1 / 3 < r i < 2 / 3 and x i = 1, then x i +1 = 1 else if 1 / 3 < r i < 2 / 3 and x i = 0, then x i +1 = 0 The sequence of relative frequencies will oscillate between 1/3 and 2/3. Problem E3.8 The polynomial has the following roots: p 1 = 2, p 2 = 2 √ 1, p 3 = 2 √ 1, p 4 = . 5, and p 5 = . 5 . The fifth root, p 5 = 0 . 5, is the only that belongs to the interval [0 , 1], therefore p = 0 . 5 is the answer.5 is the answer....
View
Full
Document
This homework help was uploaded on 09/26/2007 for the course ECE 3100 taught by Professor Haas during the Spring '05 term at Cornell.
 Spring '05
 HAAS
 Following, Trigraph, Christopher Nolan

Click to edit the document details