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Unformatted text preview: hz 1 = 1 Design a FIR bandpass filter using rectangular, blackman and chebychev windows to satisfy the following specifications  fs = 1000 Hz, center at 100 Hz, with bandwidth (pass band) of 20 Hz , stopband attenuation > 20 dB is acceptable Pass band ripple is a dont care specification. stop bands starts 20 Hz away from center frequency. which is the best design and why ? analog frequencies f center 100 = f stop1 80 = f s 1000 = f pass1 90 = f stop2 120 = f pass2 110 = Digital frequencies: center 2 f center f s = center 0.628 = Radians/Sample !!!! pass1 2 f pass1 f s = pass1 0.565 = pass2 2 f pass2 f s = pass2 0.691 = stop1 2 f stop1 f s = stop1 0.503 = stop2 2 f stop2 f s = stop2 0.754 = c1 pass1 stop1 + 2 = c1 0.534 = c2 pass2 stop2 + 2 = c2 0.723 = To determine the number of the filter coefficients the transition bandwidth,stopband attenuation is required stop 25 = pass1 stop1 = 0.063 = M 1 0.92 = considering the highest even number of coefficients 0.92 M transition band rectangular M 1 46 = M 1 57 = DEFINING THE BANDPASS FILTER IMPULSE RESPONSE h bpf n ( ) sin c2 n ( 29 n sin c1 n ( 29 n  n if c2 c1  n = if = n M 1 M 1 .. = w r n ( ) 1 = 100 50 50 100 0.999 0.9995 1 1.0005 1.001 Rectangular Window Samples Window coefficients w r n ( ) n n M 1 M 1 .. = hh bpfr n ( ) w r n ( ) h bpf n ( ) = 100 50 50 100 1 0.5 0.5 1 Multiplication of rect window and bpf Samples Coefficients hh bpfr n ( ) n Computing the DTFT of the windowed bandpass filter N P 3000 = i 2N P 1 .. = i i 2 N P = H bpfr ( ) M 1 M 1 n hh bpfr n ( ) ( 29 e i n = = Hbpdf i H bpfr i ( 29 = 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.120....
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This note was uploaded on 02/21/2011 for the course EE 456 taught by Professor Dr.yang during the Fall '09 term at SUNY Buffalo.
 Fall '09
 Dr.Yang

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