Challenge_21-1

# Challenge_21-1 - h = 1.0000 2.2470 2.8019 2.2470 1.0000 Therefore H(z = 1 2.24 z-1 2.80z-2 2.24 z-3 z-4 1 EEL 3135 Signals and Systems Dr Fred J

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EEL 3135: Signals and Systems Dr. Fred J. Taylor, Professor Lesson Title: Filters Lesson Number: 21 (Section 7-7 ti 7.10) Challenge Design a heuristic lowpass FIR having a passband frequencies of ω = { -2 π /7, 0, 2 π /7}. Q: What is H(z)? Is the filter linear or non-linear phase? If linear phase, what is the FIRs group delay? Response The design of the heuristic FIR requires that the pole-zero cancellation occurs at the passband frequencies of ω = { -2 π /7, 0, 2 π /7} and filter zeros exist at ω = { -6 π /7, -4 π /7, 4 π /7, 6 π /7}. This is accomplished as shown below (using MATLAB). » h2=[1,-(cos(4*pi/7)+i*sin(4*pi/7))]; { ω =4 π /7} » h3=[1,-(cos(6*pi/7)+i*sin(6*pi/7))]; { ω =6 π /7} » h4=[1,-(cos(8*pi/7)+i*sin(8*pi/7))]; { ω =-6 π /7 = 8 π /7 } » h5=[1,-(cos(10*pi/7)+i*sin(10*pi/7))]; { ω =-4 π /7 = 10 π /7 } » h=conv(conv(conv(h2,h3),h4),h5); » y=conv(h,zeros(1,100)); » plot(abs(fft(y))) » y=conv(h,[1,zeros(1,100)]); » plot(abs(fft(y))) » zplane(h,1)

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Unformatted text preview: h = 1.0000 2.2470 2.8019 2.2470 1.0000 Therefore H(z) = 1 + 2.24 z-1 + 2.80z-2 +2.24 z-3 + z-4 1 EEL 3135: Signals and Systems Dr. Fred J. Taylor, Professor The filter response and pole zero distribution are shown below. Figure 1: Magnitude frequency response (two-sided) and pole zero distribution. The heuristic 5 th order filter has a symmetric impulse response and is therefore linear. According to theory the group delay is τ =(N-1)/2 = 2 samples. The theoretical linear phase response is θ ( ω )=-τω + φ . The phase response is computed as shown below. >> [h,w]=freqz(h,1); >> h=[1, 2.24, 2.8, 2.24, 1]; >> [h,w]=freqz(h,1); >> plot(180*angle(h)/pi); >> plot(w,180*angle(h)/pi); and graphically interpreted in Figure 2. The estimate slope is m=-2 which agrees with the theoretical predication. Figure 2: Phase response 2 Passband Passband π/2 π Slope m=-2...
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## This note was uploaded on 04/15/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Challenge_21-1 - h = 1.0000 2.2470 2.8019 2.2470 1.0000 Therefore H(z = 1 2.24 z-1 2.80z-2 2.24 z-3 z-4 1 EEL 3135 Signals and Systems Dr Fred J

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