# soln9 - EECS 451 SOLUTIONS TO PROBLEM SET #9 1a. | H ( j...

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Unformatted text preview: EECS 451 SOLUTIONS TO PROBLEM SET #9 1a. | H ( j Ω) | = 1 / | Ω | where s = j Ω. H ( j 0) = ∞ and H ( j ∞ ) = 0. 1b. H ( z ) = H a ( s = 2 2 z- 1 z +1 ) = z +1 z- 1 = Y ( z ) X ( z ) → y [ n ] − y [ n − 1] = x [ n ]+ x [ n − 1]. 1c. H ( e jω ) = e jω +1 e jω- 1 = e jω/ 2 e jω/ 2 e jω/ 2 + e- jω/ 2 e jω/ 2- e- jω/ 2 = − j cot( ω/ 2) ≈ 2 jω . 1d. For small ω , both look like 1 / | ω | : a hyperbola. 2a. | H ( j Ω) | = a √ Ω 2 + a 2 where s = j Ω. H ( j 0) = 1 and H ( j ∞ ) = 0. 2b. H ( z ) = H a ( s = 2 2 z- 1 z +1 ) = a/ [ z- 1 z +1 + a ] = a ( z +1) ( a +1) z +( a- 1) . 2c. H ( e jω )= a ( e jω +1) ( a +1) e jω +( a- 1) . H ( e j )= a (1+1) ( a +1)+( a- 1) =1. H ( e jπ )= a (1- 1) ( a- 1)- ( a +1) =0. 2d. Both are lowpass filters with DC gain=1 and high-frequency gain=0. 2e. a = Ω o = 2 2 tan( ω o 2 ) → ω o = 2 tan- 1 ( a ). 3a. braceleftbigg h IDEAL [ n ] = 1 2 π integraltext π- π ( jω ) e jωn dω Then h [ n ] = h IDEAL [ n ] w [ n ]...
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## This note was uploaded on 12/10/2009 for the course EECS 451 taught by Professor Andrewyagle during the Spring '08 term at University of Michigan.

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soln9 - EECS 451 SOLUTIONS TO PROBLEM SET #9 1a. | H ( j...

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