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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 highfrequency 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.
 Spring '08
 ANDREWYAGLE
 Digital Signal Processing, Signal Processing

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