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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.
- Spring '10