630F09_hw1 - /5 H j ϖ (e ) Figure : P-2(b) 1 Problem 3...

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ENEE 630 F’09 Homework #1 Due: Wednesday September 23, 2009 at 12:30 PM Problem 1 For the system in Fig. P-1, find an expression for y ( n ) in terms of x ( n ). Simplify the expression as best as you can. 3 2 2 3 y(n) x(n) Figure : P-1 Problem 2 Show that the two systems shown in Fig. P-2(a) (where k is some integer) are equivalent (that is, y 0 ( n ) = y 1 ( n ) ) when h k ( n ) = h 0 ( n ) cos (2 πkn/L ). L H k (z) y(n) x(n) y 0 (n) x(n) L y(n) H (z) o y (n) 1 u(n) X cos 2 π L kn Figure : P-2(a) This is a structure where filtering followed by cosine modulation has the same effect as filtering with the cosine modulated impulse response. (This is not true in all situations; see next problem). Now consider the example where L = 5, and k = 1. Let X ( e ) and H 0 ( e ) be as sketched in Fig. P-2(b). Give sketches of Y ( e ), Y 0 ( e ) and U ( e ). 0 2 π ϖ X(e j ϖ ) 0 2 π ϖ 1 π /5 π
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Unformatted text preview: /5 H j ϖ (e ) Figure : P-2(b) 1 Problem 3 Show that the two systems shown in Fig.P-3 are no equivalent, that is, y ( n ) and y 1 ( n ) are not necessarily the same, even if h k ( n ) = h ( n ) cos (2 πkn/L ). H k (z) y(n) y (n) y(n) H (z) o y (n) 1 X cos 2 π L kn Figure : P-3 Problem 4 Consider a sequence x ( n ) with X ( e jω ) as shown in Fig. P-4(a). Suppose we generate the sequences y ( n ) and s ( n ) from x ( n ) as in Fig. P-4(b), where H ( e jω ) = ( 1 for | ω | < π/ 2 for π/ 2 ≤ | ω | ≤ π Plot the quantities Y ( e jω ) and S ( e jω ). 1 π 2 π ϖ X (e j ϖ ) Figure: P-4(a) H(z) 2 2 4 H(z) s(n) y(n) x(n) Figure: P-4(b) 2...
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This note was uploaded on 10/31/2010 for the course EE 630 taught by Professor Wu during the Spring '10 term at Aarhus Universitet, Aarhus.

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630F09_hw1 - /5 H j ϖ (e ) Figure : P-2(b) 1 Problem 3...

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