ECE 101 problem set solution2

ECE 101 problem set solution2 - ECE 101 Linear Systems...

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ECE 101 – Linear Systems Problem Set # 2 Solutions January 24, 2008 (send questions/comments to [email protected]) 1.27 (e) () ( 2 ) , () 0 0, ( ) 0 yt xt =+− < Memoryless – NO Time Invariant – YES Linear – NO Causal – YES Stable – YES (g) d dt = Memoryless – NO Time Invariant – YES Linear – YES Causal – YES or NO Stable – NO d dt x t , recall, is defined as ( ) 0 lim . xth xt h h + Whichever way you implement the system, you must take the current time value x(t) and some other time value x(t+h) in order to compute the derivative. For example, given just the sample x( 0 ) at time t = 0, you would need more information about the signal to compute the derivative at t = 0; you’d need at least one other time sample. Thus, this system does have memory. For causality, if you take h , in the limit, from the left side of 0, then you are taking x(t) and a past value to compute d dt x t , so the system is causal. On the other hand, if you take h from the right of 0, then you are taking x(t) and a future value to compute d dt x t , so the system is not causal. So, causality is determined based on how you design your derivative system. 1.28 (a) [] [ ] yn x n = Memoryless – NO Time Invariant – NO Linear – YES Causal – NO Stable – YES (d) { } [ 1 ] Ev xn = 1 Memoryless – NO Time Invariant – NO Linear – YES Causal – NO Stable – YES
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2.21(a) = = (for ) −∞ = = k k n h k x n y ] [ ] [ ] [ = n k k n k b a 0 = n k k n b a b 0 ) / ( 0 n From here, we can use the power series formula you proved in Problem 1.54(a) with a bit of algebraic manipulation to obtain ] [ ] [ 1 1 n u a b a b n y n n = + + (b) This is a special case of part (a) above, where now b = a .
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This note was uploaded on 03/16/2009 for the course ECE 101 taught by Professor Siegel during the Spring '08 term at UCSD.

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ECE 101 problem set solution2 - ECE 101 Linear Systems...

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