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1 ### Copyright (C) 1996 John W. Eaton |
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2 ### |
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3 ### This file is part of Octave. |
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4 ### |
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5 ### Octave is free software; you can redistribute it and/or modify it |
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6 ### under the terms of the GNU General Public License as published by |
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7 ### the Free Software Foundation; either version 2, or (at your option) |
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8 ### any later version. |
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9 ### |
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10 ### Octave is distributed in the hope that it will be useful, but |
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11 ### WITHOUT ANY WARRANTY; without even the implied warranty of |
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12 ### MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU |
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13 ### General Public License for more details. |
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14 ### |
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15 ### You should have received a copy of the GNU General Public License |
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16 ### along with Octave; see the file COPYING. If not, write to the Free |
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17 ### Software Foundation, 59 Temple Place - Suite 330, Boston, MA |
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18 ### 02111-1307, USA. |
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19 |
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20 ## usage: [r, p, k, e] = residue (b, a) |
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21 ## |
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22 ## If b and a are vectors of polynomial coefficients, then residue |
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23 ## calculates the partial fraction expansion corresponding to the |
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24 ## ratio of the two polynomials. The vector r contains the residue |
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25 ## terms, p contains the pole values, k contains the coefficients of |
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26 ## a direct polynomial term (if it exists) and e is a vector containing |
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27 ## the powers of the denominators in the partial fraction terms. |
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28 ## Assuming b and a represent polynomials P(s) and Q(s) we have: |
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29 ## |
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30 ## P(s) M r(m) N |
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31 ## ---- = # ------------- + # k(n)*s^(N-n) |
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32 ## Q(s) m=1 (s-p(m))^e(m) n=1 |
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33 ## |
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34 ## (# represents summation) where M is the number of poles (the length of |
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35 ## the r, p, and e vectors) and N is the length of the k vector. |
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36 ## |
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37 ## [r p k e] = residue(b,a,tol) |
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38 ## |
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39 ## This form of the function call may be used to set a tolerance value. |
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40 ## The default value is 0.001. The tolerance value is used to determine |
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41 ## whether poles with small imaginary components are declared real. It is |
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42 ## also used to determine if two poles are distinct. If the ratio of the |
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43 ## imaginary part of a pole to the real part is less than tol, the |
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44 ## imaginary part is discarded. If two poles are farther apart than tol |
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45 ## they are distinct. |
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46 ## |
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47 ## Example: |
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48 ## b = [1, 1, 1]; |
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49 ## a = [1, -5, 8, -4]; |
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50 ## |
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51 ## [r, p, k, e] = residue (b, a) |
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52 ## |
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53 ## returns |
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54 ## |
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55 ## r = [-2, 7, 3]; p = [2, 2, 1]; k = []; e = [1, 2, 1]; |
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56 ## |
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57 ## which implies the following partial fraction expansion |
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58 ## |
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59 ## s^2 + s + 1 -2 7 3 |
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60 ## ------------------- = ----- + ------- + ----- |
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61 ## s^3 - 5s^2 + 8s - 4 (s-2) (s-2)^2 (s-1) |
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62 ## |
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63 ## SEE ALSO: poly, roots, conv, deconv, polyval, polyderiv, polyinteg |
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64 |
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65 function [r, p, k, e] = residue (b, a, toler) |
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66 |
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67 ## Written by Tony Richardson (amr@mpl.ucsd.edu) June 1994. |
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68 |
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69 ## Here's the method used to find the residues. |
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70 ## The partial fraction expansion can be written as: |
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71 ## |
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72 ## |
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73 ## P(s) D M(k) A(k,m) |
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74 ## ---- = # # ------------- |
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75 ## Q(s) k=1 m=1 (s - pr(k))^m |
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76 ## |
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77 ## (# is used to represent a summation) where D is the number of |
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78 ## distinct roots, pr(k) is the kth distinct root, M(k) is the |
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79 ## multiplicity of the root, and A(k,m) is the residue cooresponding |
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80 ## to the kth distinct root with multiplicity m. For example, |
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81 ## |
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82 ## s^2 A(1,1) A(2,1) A(2,2) |
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83 ## ------------------- = ------ + ------ + ------- |
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84 ## s^3 + 4s^2 + 5s + 2 (s+2) (s+1) (s+1)^2 |
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85 ## |
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86 ## In this case there are two distinct roots (D=2 and pr = [-2 -1]), |
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87 ## the first root has multiplicity one and the second multiplicity |
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88 ## two (M = [1 2]) The residues are actually stored in vector format as |
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89 ## r = [ A(1,1) A(2,1) A(2,2) ]. |
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90 ## |
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91 ## We then multiply both sides by Q(s). Continuing the example: |
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92 ## |
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93 ## s^2 = r(1)*(s+1)^2 + r(2)*(s+1)*(s+2) + r(3)*(s+2) |
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94 ## |
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95 ## or |
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96 ## |
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97 ## s^2 = r(1)*(s^2+2s+1) + r(2)*(s^2+3s+2) +r(3)*(s+2) |
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98 ## |
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99 ## The coefficients of the polynomials on the right are stored in a row |
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100 ## vector called rhs, while the coefficients of the polynomial on the |
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101 ## left is stored in a row vector called lhs. If the multiplicity of |
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102 ## any root is greater than one we'll also need derivatives of this |
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103 ## equation of order up to the maximum multiplicity minus one. The |
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104 ## derivative coefficients are stored in successive rows of lhs and |
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105 ## rhs. |
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106 ## |
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107 ## For our example lhs and rhs would be: |
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108 ## |
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109 ## | 1 0 0 | |
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110 ## lhs = | | |
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111 ## | 0 2 0 | |
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112 ## |
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113 ## | 1 2 1 1 3 2 0 1 2 | |
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114 ## rhs = | | |
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115 ## | 0 2 2 0 2 3 0 0 1 | |
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116 ## |
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117 ## We then form a vector B and a matrix A obtained by evaluating the |
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118 ## polynomials in lhs and rhs at the pole values. If a pole has a |
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119 ## multiplicity greater than one we also evaluate the derivative |
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120 ## polynomials (successive rows) at the pole value. |
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121 ## |
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122 ## For our example we would have |
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123 ## |
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124 ## | 4| | 1 0 0 | | r(1) | |
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125 ## | 1| = | 0 0 1 | * | r(2) | |
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126 ## |-2| | 0 1 1 | | r(3) | |
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127 ## |
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128 ## We then solve for the residues using matrix division. |
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129 |
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130 if (nargin < 2 || nargin > 3) |
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131 usage ("residue (b, a [, toler])"); |
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132 endif |
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133 |
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134 if (nargin == 2) |
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135 toler = .001; |
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136 endif |
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137 |
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138 ## Make sure both polynomials are in reduced form. |
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139 |
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140 a = polyreduce (a); |
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141 b = polyreduce (b); |
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142 |
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143 b = b / a(1); |
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144 a = a / a(1); |
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145 |
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146 la = length (a); |
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147 lb = length (b); |
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148 |
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149 ## Handle special cases here. |
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150 |
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151 if (la == 0 || lb == 0) |
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152 k = r = p = e = []; |
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153 return; |
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154 elseif (la == 1) |
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155 k = b / a; |
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156 r = p = e = []; |
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157 return; |
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158 endif |
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159 |
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160 ## Find the poles. |
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161 |
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162 p = roots (a); |
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163 lp = length (p); |
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164 |
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165 ## Determine if the poles are (effectively) real. |
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166 |
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167 index = find (abs (imag (p) ./ real (p)) < toler); |
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168 if (length (index) != 0) |
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169 p (index) = real (p (index)); |
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170 endif |
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171 |
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172 ## Find the direct term if there is one. |
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173 |
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174 if (lb >= la) |
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175 ## Also returns the reduced numerator. |
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176 [k, b] = deconv (b, a); |
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177 lb = length (b); |
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178 else |
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179 k = []; |
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180 endif |
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181 |
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182 if (lp == 1) |
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183 r = polyval (b, p); |
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184 e = 1; |
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185 return; |
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186 endif |
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187 |
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188 |
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189 ## We need to determine the number and multiplicity of the roots. |
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190 ## |
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191 ## D is the number of distinct roots. |
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192 ## M is a vector of length D containing the multiplicity of each root. |
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193 ## pr is a vector of length D containing only the distinct roots. |
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194 ## e is a vector of length lp which indicates the power in the partial |
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195 ## fraction expansion of each term in p. |
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196 |
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197 ## Set initial values. We'll remove elements from pr as we find |
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198 ## multiplicities. We'll shorten M afterwards. |
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199 |
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200 e = ones (lp, 1); |
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201 M = zeros (lp, 1); |
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202 pr = p; |
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203 D = 1; |
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204 M(1) = 1; |
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205 |
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206 old_p_index = 1; |
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207 new_p_index = 2; |
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208 M_index = 1; |
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209 pr_index = 2; |
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210 |
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211 while (new_p_index <= lp) |
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212 if (abs (p (new_p_index) - p (old_p_index)) < toler) |
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213 ## We've found a multiple pole. |
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214 M (M_index) = M (M_index) + 1; |
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215 e (new_p_index) = e (new_p_index-1) + 1; |
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216 ## Remove the pole from pr. |
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217 pr (pr_index) = []; |
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218 else |
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219 ## It's a different pole. |
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220 D++; |
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221 M_index++; |
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222 M (M_index) = 1; |
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223 old_p_index = new_p_index; |
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224 pr_index++; |
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225 endif |
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226 new_p_index++; |
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227 endwhile |
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228 |
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229 ## Shorten M to it's proper length |
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230 |
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231 M = M (1:D); |
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232 |
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233 ## Now set up the polynomial matrices. |
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234 |
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235 MM = max(M); |
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236 |
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237 ## Left hand side polynomial |
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238 |
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239 lhs = zeros (MM, lb); |
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240 rhs = zeros (MM, lp*lp); |
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241 lhs (1, :) = b; |
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242 rhi = 1; |
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243 dpi = 1; |
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244 mpi = 1; |
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245 while (dpi <= D) |
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246 for ind = 1:M(dpi) |
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247 if (mpi > 1 && (mpi+ind) <= lp) |
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248 cp = [p(1:mpi-1); p(mpi+ind:lp)]; |
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249 elseif (mpi == 1) |
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250 cp = p (mpi+ind:lp); |
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251 else |
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252 cp = p (1:mpi-1); |
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253 endif |
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254 rhs (1, rhi:rhi+lp-1) = prepad (poly (cp), lp); |
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255 rhi = rhi + lp; |
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256 endfor |
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257 mpi = mpi + M (dpi); |
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258 dpi++; |
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259 endwhile |
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260 if (MM > 1) |
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261 for index = 2:MM |
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262 lhs (index, :) = prepad (polyderiv (lhs (index-1, :)), lb); |
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263 ind = 1; |
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264 for rhi = 1:lp |
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265 cp = rhs (index-1, ind:ind+lp-1); |
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266 rhs (index, ind:ind+lp-1) = prepad (polyderiv (cp), lp); |
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267 ind = ind + lp; |
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268 endfor |
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269 endfor |
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270 endif |
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271 |
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272 ## Now lhs contains the numerator polynomial and as many derivatives as |
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273 ## are required. rhs is a matrix of polynomials, the first row |
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274 ## contains the corresponding polynomial for each residue and |
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275 ## successive rows are derivatives. |
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276 |
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277 ## Now we need to evaluate the first row of lhs and rhs at each |
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278 ## distinct pole value. If there are multiple poles we will also need |
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279 ## to evaluate the derivatives at the pole value also. |
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280 |
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281 B = zeros (lp, 1); |
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282 A = zeros (lp, lp); |
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283 |
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284 dpi = 1; |
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285 row = 1; |
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286 while (dpi <= D) |
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287 for mi = 1:M(dpi) |
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288 B (row) = polyval (lhs (mi, :), pr (dpi)); |
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289 ci = 1; |
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290 for col = 1:lp |
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291 cp = rhs (mi, ci:ci+lp-1); |
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292 A (row, col) = polyval (cp, pr(dpi)); |
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293 ci = ci + lp; |
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294 endfor |
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295 row++; |
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296 endfor |
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297 dpi++; |
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298 endwhile |
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299 |
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300 ## Solve for the residues. |
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301 |
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302 r = A \ B; |
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303 |
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304 endfunction |
