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1 function [SeneffPreemphasis, SeneffFilterBank, SeneffForward, SeneffBackward] ...
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2 = SeneffEarSetup(fs)
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3
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4 % This m-function is based on data from the following paper:
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5 % Benjamin D. Bryant and John D. Gowdy, "Simulation of Stages
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6 % I and II of Seneff's Auditory Model (SAM) Using Matlab", and
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7 % published in the Proceedings of the 1993 Matlab User's Group
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8 % Conference.
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9 % Thanks to Benjamin Bryant for supplying us with his filter
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10 % coefficients and the initial organization of this implementation.
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11
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12 % (c) 1998 Interval Research Corporation
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13
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14 % Set the following variable to a non-zero value to see a summary
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15 % of the filter bank's behaviour.
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16 plotTests = 0;
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17
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18 % The following values were taken from Figure 2 of Bryant's paper.
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19 PreemphasisRTheta = [0.86 3.1148863;0.99 0; 0.5 0; 0.95 3.14159];
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20
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21 % The following values were taken from Table 1 of Bryant's paper.
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22 % They represent the cascade zeros (R-z and Theta-z), and the
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23 % second order poles (radius and theta) and zeros (radius and theta/2).
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24 %
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25 % R-z Theta-z Radius Theta R-z2
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26 FilterBankRTheta = [
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27 0 3.14159 0.740055 2.633909 0.8
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28 0.86 2.997077 0.753637 2.178169 0.8
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29 0.86 2.879267 0.775569 1.856744 0.8
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30 0.86 2.761458 0.798336 1.617919 0.8
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31 0.86 2.643648 0.819169 1.433496 0.8
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32 0.86 2.525839 0.837158 1.286795 0.8
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33 0.8 2.964876 0.852598 1.167321 0.8
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34 0.86 2.408029 0.865429 1.068141 0.8
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35 0.86 2.29022 0.876208 0.984489 0.8
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36 0.86 2.17241 0.885329 0.912985 0.8
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37 0.86 2.054601 0.893116 0.851162 0.8
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38 0.86 1.936791 0.899823 0.797179 0.8
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39 0.8 2.788161 0.906118 0.749633 0.8
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40 0.86 1.818981 0.911236 0.70744 0.8
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41 0.86 1.701172 0.915747 0.669742 0.8
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42 0.86 1.583362 0.919753 0.635858 0.8
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43 0.86 1.465552 0.923335 0.605237 0.8
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44 0.86 1.347743 0.926565 0.57743 0.8
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45 0.8 2.611447 0.929914 0.552065 0.8
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46 0.86 1.229933 0.932576 0.528834 0.8
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47 0.86 1.112123 0.944589 0.487783 0.75
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48 0.86 0.994314 0.957206 0.452645 0.660714
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49 0.86 0.876504 0.956548 0.42223 0.672143
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50 0.86 0.758694 0.956653 0.395644 0.682143
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51 0.8 2.434732 0.956518 0.372208 0.690966
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52 0.86 0.640885 0.956676 0.351393 0.69881
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53 0.86 0.523075 0.956741 0.316044 0.712143
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54 0.8 2.258018 0.956481 0.287157 0.723052
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55 0.8 2.081304 0.956445 0.263108 0.732143
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56 0.8 1.904589 0.956481 0.242776 0.739835
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57 0.86 0.405265 0.958259 0.217558 0.749384
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58 0.8 1.727875 0.963083 0.197086 0.757143
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59 0.8 1.55116 0.969757 0.175115 0.769048
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60 0.8 1.374446 0.97003 0.153697 0.780662
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61 0.8 1.197732 0.970382 0.134026 0.791337
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62 0.8 1.021017 0.970721 0.118819 0.799596
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63 0.8 1.5 0.970985 0.106711 0.8
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64 0.8 1.2 0.971222 0.096843 0.8
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65 0.8 1 0.97144 0.088645 0.8
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66 0.8 0.9 0.971645 0.081727 0.8];
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67
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68 % Let's plot the cascade zero locations and the locations of the
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69 % pole and zeros in the resonator.
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70 if plotTests
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71 clf;
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72 subplot(3,3,1);
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73 plot(FilterBankRTheta(:,1).*exp(i*FilterBankRTheta(:,2)))
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74 axis([-1 1 0 1])
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75 title('Cascade Zero Locations')
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76
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77 subplot(3,3,2);
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78 plot([FilterBankRTheta(:,3).*exp(i*FilterBankRTheta(:,4)) ...
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79 FilterBankRTheta(:,5).*exp(i*FilterBankRTheta(:,4)/2)],'+')
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80 title('Resonator Pole/Zero')
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81 drawnow;
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82 end
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83
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84 % Convert r-theta form, first into a list of roots, then a polynomial
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85 roots=exp(i*PreemphasisRTheta(:,2)).*PreemphasisRTheta(:,1);
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86 SeneffPreemphasis=real(poly([roots;conj(roots)]));
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87
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88 % Plot the preemphasis filter response, if desired
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89 if plotTests
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90 subplot(3,3,3);
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91 freqScale=(0:255)/256*8000;
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92 freqresp = FreqResp(SeneffPreemphasis,[1], freqScale, 16000);
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93 semilogx(freqScale,freqresp)
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94 title('Preemphasis Response');
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95 axis([100 10000 -60 20])
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96 drawnow;
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97 end
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98
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99 % Now figure out the second order sections that make up the main
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100 % filter bank cascade. We put the zeros into the numerator (b's)
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101 % and there are no poles. Just to keep things simpler, we adjust
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102 % the gain of each filter to keep it unity gain at DC.
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103 [channels,width] = size(FilterBankRTheta);
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104 roots=exp(i*FilterBankRTheta(:,2)).*FilterBankRTheta(:,1);
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105 SeneffFilterBank = zeros(channels,5);
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106 for j=1:channels
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107 SeneffFilterBank(j,1:3) = poly([roots(j) conj(roots(j))]);
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108 SeneffFilterBank(j,1:3) = SeneffFilterBank(j,1:3)/sum(SeneffFilterBank(j,1:3));
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109 end
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110
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111 % Plot the cascade zero responses, if desired.
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112 if plotTests
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113 subplot(3,3,4);
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114 y=soscascade([1 zeros(1,511)],SeneffFilterBank);
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115 freqresp=20*log10(abs(fft(y(1:5:40,:)')));
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116 freqScale=(0:511)/512*16000;
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117 semilogx(freqScale(1:256),freqresp(1:256,:))
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118 axis([100 10000 -150 0]);
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119 title('Cascade Response');
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120 drawnow;
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121 end
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122
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123 % Now figure out the resonating filters. Each of these resonators
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124 % is a double pole-zero pair.
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125 zlocs = FilterBankRTheta(:,5).*exp(i*FilterBankRTheta(:,4)/2);
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126 plocs = FilterBankRTheta(:,3).*exp(i*FilterBankRTheta(:,4));
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127 SeneffForward = zeros(5,channels);
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128 SeneffBackward = zeros(5,channels);
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129
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130 for j=1:channels
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131 SeneffForward(:,j) = real(poly([zlocs(j) conj(zlocs(j)) ...
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132 zlocs(j) conj(zlocs(j))]))';
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133 SeneffBackward(:,j) = real(poly([plocs(j) conj(plocs(j)) ...
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134 plocs(j) conj(plocs(j))]))';
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135 end
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136
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137 % Now plot the frequency response of just the resonating filters.
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138 % These are all bandpass filters.
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139 if plotTests
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140 subplot(3,3,5);
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141 impulse = [1 zeros(1,255)];
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142 y=zeros(256,channels);
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143 for j=1:40
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144 y(:,j) = filter(SeneffForward(:,j),SeneffBackward(:,j),impulse)';
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145 end
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146 freqresp=20*log10(abs(fft(y(:,1:5:40))));
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147 freqScale=(0:255)/256*16000;
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148 semilogx(freqScale(1:128),freqresp(1:128,:))
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149 axis([100 10000 -30 40]);
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150 title('Resonators Response')
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151 drawnow;
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152 end
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153
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154 % The plot below shows the overall response of the preemphasis filters
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155 % along with the just-designed cascade of zeros.
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156 if plotTests
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157 subplot(3,3,6);
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158 impulse = [1 zeros(1,511)];
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159 y=soscascade(filter(SeneffPreemphasis, [1], impulse), ...
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160 SeneffFilterBank);
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161 freqresp=20*log10(abs(fft(y(1:5:40,:)')));
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162 freqScale=(0:511)/512*16000;
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163 semilogx(freqScale(1:256),freqresp(1:256,:))
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164 axis([100 10000 -100 25]);
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165 title('Preemphasis+Cascade');
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166 drawnow;
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167 end
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168
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169 % Now we need to normalize the gain of each channel. We run an impulse
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170 % through the preemphasis filter, and then through the cascade of zeros.
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171 % Finally, we run it through each of the resonator filters.
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172 impulse = [1 zeros(1,255)];
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173 y=soscascade(filter(SeneffPreemphasis, [1], impulse), ...
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174 SeneffFilterBank);
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175 for j=1:channels
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176 y(j,:) = filter(SeneffForward(:,j),SeneffBackward(:,j),y(j,:));
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177 end
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178
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179 % Now we have impulse responses from each filter. We can find the FFT
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180 % and then find the gain peak. We divide each forward polynomial by the
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181 % maximum gain (to normalize) and then multiply by the desired low
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182 % frequency roll-off. The Bryant paper says that the last 24 channels
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183 % should be cut at 6dB per octave and that this occurs at 1600 Hz, but
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184 % it looks to me like the gain change happens at 3200 Hz.
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185 freqresp=abs(fft(y'));
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186 gain = ones(1,channels)./max(freqresp);
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187 cfs = FilterBankRTheta(:,4)/pi*fs/2;
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188 rolloff = min(cfs/1600,1);
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189
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190 for j=1:channels
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191 SeneffForward(:,j)=SeneffForward(:,j)*gain(j)*rolloff(j);
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192 end
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193
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194 % All Done. The figure below should match Figure 3 of Bryant's paper.
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195 if plotTests
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196 subplot(3,3,8);
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197 impulse = [1 zeros(1,511)];
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198 y=soscascade(filter(SeneffPreemphasis, [1], impulse), ...
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199 SeneffFilterBank);
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200 for j=1:channels
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201 y(j,:) = filter(SeneffForward(:,j),SeneffBackward(:,j),y(j,:));
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202 end
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203
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204 freqresp=20*log10(abs(fft(y(1:5:40,:)')));
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205 freqScale=(0:511)/512*16000;
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206 plot(freqScale(1:256),freqresp(1:256,:))
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207 axis([100 10000 -120 0]);
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208 title('Magnitude Response vs. Linear Frequency');
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209 drawnow;
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210 end
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211
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212
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213 function mag=FreqResp(b,a,f,fs)
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214 cf = exp(i*2*pi*f/fs);
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215 num = 0;
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216 for i=1:length(b)
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217 num = num + b(end-i+1)*cf.^i;
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218 end
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219
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220 denom = 0;
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221 for i=1:length(a)
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222 denom = denom + a(end-i+1)*cf.^i;
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223 end
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224 mag = 20*log10(abs(num./denom));
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