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Challenge21_Window_FIR

# Challenge21_Window_FIR - Basis...

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Digital Signal Processing Dr. Fred J. Taylor, Professor Lesson Title: FIR Filter Challenge Challenge: [21] Response The frequency response is technically given by: ( 29 ( 29 2 / 1 ; 0 4 . 0 ; - = = = - N otherwise e e H g c j j d g τ π ω ω ωτ ω The expected form of the impulse response is: [ ] ( 29 ( 29 ( 29 g g c k k k h τ π τ ω - - = sin The computer coefficients are h[k]={0.093, 0.303, 0.4, 0.303, 0.093}. The magnitude frequency response, phase response, and group delay are shown below. Notice that as ω c become smaller, more coefficients will be required to maintain acceptable filer action (vice-versa).

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Unformatted text preview: Basis: sin(0.4*pi*(1*10^-8))/(pi*(1*10^-8))=0.4 sin(0.4*pi*(3-2))/(pi*(3-2))=0.302731 sin(0.4*pi*(4-2))/(pi*(4-2))=0.0935489 FIR - 1 0Hz 0Hz 0Hz 2-side mag. Freq. response 2-side phase response 2-side group delay response Digital Signal Processing Dr. Fred J. Taylor, Professor x=[0.0935, 0.3027,0.4,0.3027,0.0935] X=fft(x&zeros(1024)) graph(mag(X),phs(X),grp(X)) FIR - 2...
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