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Unformatted text preview: Problem 1: It is easily obtained that π ϖ ϖ ϖ 525 . 2 = + = s p c , π ϖ ϖ ϖ 15 . = = ∆ s p , 005 . } , min{ = = s p δ δ δ . Therefore we have dB A 02 . 46 005 . log 20 10 = = . For Hamming window, we can calculate that 44 3 . 44 15 . / 64 . 6 ≈ = = π π M . Therefore length of the filter will be 45. The window function and filter coefficients can be expressed as ≤ ≤ = otherwises 44 ) 44 2 cos( 46 . 54 . ] [ n n n π ϖ ≤ ≤ = = otherwises 44 )] 44 2 cos( 46 . 54 . [ ) 22 ( )] 22 ( 525 . sin[ ] [ ] 2 [ ] [ n n n n n M n h n h d π π π ϖ . The magnitude response is as follows. For Kaiser window, we can calculate that . 36 3 . 35 285 . 2 8 091 . 4 ) 21 ( 07886 . ) 21 ( 5842 . 4 . ≈ = ∆ ⋅ = = + = ϖ β A M A A Therefore the window function and filter coefficients can be expressed as ≤ ≤ = otherwises 36 ) 091 . 4 ( } ] 18 / ) 18 [( 1 091 . 4 { ] [ 2 n I n I n ϖ ≤...
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 Fall '10
 ShuHungLeung
 Digital Signal Processing, Signal Processing, 8 m, β, Finite impulse response, filter coefficients, 36 m, magnitude response

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