This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECE 3250 HOMEWORK III SOLUTIONS Fall 2009 1. Note that h has finite duration. (a) By Criterion 1, D h = F Z because h has finite duration. (b) S ( x ) = Conv h ( x ), so for every n Z , S ( x )( n ) = X k = h ( k ) x ( n k ) = 6 X k =0 3(5 k ) = 3(1 5 7 ) 1 1 / 5 = 15 4 (5 7 1); . Note that S ( x ) is a constant signal. (c) S ( x ) = Conv h ( x ), so for every n Z , S ( x )( n ) = X k = h ( k ) x ( n k ) = 6 X k =0 5 k u ( n k ) = 8 < : if n < P n k =0 5 k if 0 n < 7 P 6 k =0 5 k if n 7 = 8 > < > : if n < 1 5 ( n +1) 1 1 / 5 if 0 n < 7 1 5 7 1 1 / 5 if n 7 = 8 < : if n < 5 4 (5 ( n +1) 1) if 0 n < 7 5 4 (5 7 1) if n 7 . (d) Since x has duration 13, we can find N 1 so that x ( n ) = 0 when n < N 1 and x ( n ) = 0 when n > N 1 + 12. Now S ( x ) = Conv h ( x ), so for each n Z S ( x )( n ) = X k = h ( k ) x ( n k ) = 6 X k =0 5 k x ( n k ) . The sum features terms x ( n ), x ( n 1), . . . , x ( n 6). All these are zero if n < N 1 . Similarly, all are zero if n 6 > N 1 + 12, which is the same as n > N 1 + 18. It follows that S ( x )( n ) = 0 when n < N 1 and S ( x )( n ) = 0 when n > N 1 + 18. The maximum duration interval for S ( x ) is therefore N 1 n N 1 + 18, so the maximum possible duration for S ( x ) is 19. (Its easy to find x so that S ( x ) has duration exactly 19. Just suppose x ( n ) = 1 on its duration interval.) 2. (a) I just wanted you to think about what the averager does. What you need to do is find a signal x so that the average value of x ( k ) over any kinterval of length 5 is equal to 11. A reasonable way to do this is to find 5 numbers that average to 11 and just let x be the signal that cycles periodically through those numbers....
View
Full
Document
This note was uploaded on 03/26/2010 for the course ECE 3250 at Cornell University (Engineering School).
 '07
 DELCHAMPS

Click to edit the document details