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By BEERNARD KLAR

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E. 707… . For a low pass filter signals with a frequency higher than f3dB are attenuated by more than 3 dB. com  frequency High Pass frequency Band-Pass frequency Band-Stop 3 General Digital Filters 35 SystemView will give the user the facility to design a filter for an arbitrary frequency response by “sketching” graphically the magnitude of its transfer function: Gain 0 frequency User Defined Frequency Response There are two types of linear digital filters, FIR (finite impulse response filter) and IIR (infinite impulse response filter).

Com  (104) 4 Finite Impulse Response (FIR) Filter 61 To find the phase response, we first calculate: cos ω – j sin ω + a = --------------------------------------------------e –j ω + a = --------------------------------------------------( a + cos ω ) – j sin ω H ( e jω ) = ----------------------– ω j + – + a cos ω – ja sin ω 1 cos sin 1 a ja ω ω 1 + ae (105) and therefore: – sin ω a sin ω ∠H ( e jω ) = tan–1  ------------------------ + tan–1  --------------------------- 1 + a cos ω a + cos ω (106) For small values of x the approximation tan–1 x ≈ x , cos x ≈ 1 and sin x ≈ x hold.

The bottom line for all FFT algorithms is, however, that they remove redundancy from the direct DFT computational algorithm of Eq. 53. We can highlight the existence of the redundant computation in the DFT by inspecting Eq. 53. First, for notational simplicity we can rewrite Eq. 53 as: K–1 X(n) = ∑ x ( k )WK–nk for n = 0 to K – 1 (54) k=0 where W K = e j2π ⁄ K = cos 2π ⁄ K + j sin 2π ⁄ K . Using the DFT algorithm to calculate the first four components of the DFT of a (trivial) signal with only 8 samples requires the following computations: X( 0) = x( 0) + x( 1) + x( 2) + x(3 ) + x( 4) + x( 5) + x( 6) + x(7) X ( 1 ) = x ( 0 ) + x ( 1 )W 8–1 + x ( 2 )W 8– 2 + x ( 3 )W 8– 3 + x ( 4 )W 8– 4 + x ( 5 )W 8– 5 + x ( 6 )W 8– 6 + x ( 7 )W 8– 7 X ( 2 ) = x ( 0 ) + x ( 1 )W 8–2 + x ( 2 )W 8– 4 + x ( 3 )W 8– 6 + x ( 4 )W 8– 8 + x ( 5 )W 8– 10 + x ( 6 )W 8– 12 + x ( 7 )W 8– 14 (55) X ( 3 ) = x ( 0 ) + x ( 1 )W 8–3 + x ( 2 )W 8– 6 + x ( 3 )W 8– 9 + x ( 4 )W 8– 12 + x ( 5 )W 8– 15 + x ( 6 )W 8– 18 + x ( 7 )W 8– 21 However note that there is redundant (or repeated) arithmetic computation in Eq.

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