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1 # Copyright (C) 1995 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 the |
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7 # Free Software Foundation; either version 2, or (at your option) any |
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8 # later version. |
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9 # |
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10 # Octave is distributed in the hope that it will be useful, but WITHOUT |
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11 # ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or |
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12 # FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License |
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13 # 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, 675 Mass Ave, Cambridge, MA 02139, USA. |
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18 |
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19 function [r, p, k, e] = residue (b, a, toler) |
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20 |
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21 # usage: [r, p, k, e] = residue (b, a) |
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22 # |
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23 # If b and a are vectors of polynomial coefficients, then residue |
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24 # calculates the partial fraction expansion corresponding to the |
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25 # ratio of the two polynomials. The vector r contains the residue |
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26 # terms, p contains the pole values, k contains the coefficients of |
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27 # a direct polynomial term (if it exists) and e is a vector containing |
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28 # the powers of the denominators in the partial fraction terms. |
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29 # Assuming b and a represent polynomials P(s) and Q(s) we have: |
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30 # |
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31 # P(s) M r(m) N |
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32 # ---- = # ------------- + # k(n)*s^(N-n) |
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33 # Q(s) m=1 (s-p(m))^e(m) n=1 |
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34 # |
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35 # (# represents summation) where M is the number of poles (the length of |
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36 # the r, p, and e vectors) and N is the length of the k vector. |
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37 # |
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38 # [r p k e] = residue(b,a,tol) |
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39 # |
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40 # This form of the function call may be used to set a tolerance value. |
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41 # The default value is 0.001. The tolerance value is used to determine |
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42 # whether poles with small imaginary components are declared real. It is |
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43 # also used to determine if two poles are distinct. If the ratio of the |
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44 # imaginary part of a pole to the real part is less than tol, the |
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45 # imaginary part is discarded. If two poles are farther apart than tol |
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46 # they are distinct. |
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47 # |
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48 # Example: |
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49 # b = [1, 1, 1]; |
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50 # a = [1, -5, 8, -4]; |
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51 # |
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52 # [r, p, k, e] = residue (b, a) |
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53 # |
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54 # returns |
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55 # |
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56 # r = [-2, 7, 3]; p = [2, 2, 1]; k = []; e = [1, 2, 1]; |
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57 # |
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58 # which implies the following partial fraction expansion |
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59 # |
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60 # s^2 + s + 1 -2 7 3 |
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61 # ------------------- = ----- + ------- + ----- |
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62 # s^3 - 5s^2 + 8s - 4 (s-2) (s-2)^2 (s-1) |
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63 # |
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64 # SEE ALSO: poly, roots, conv, deconv, polyval, polyderiv, polyinteg |
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65 |
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66 # Written by Tony Richardson (amr@mpl.ucsd.edu) June 1994. |
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67 |
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68 # Here's the method used to find the residues. |
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69 # The partial fraction expansion can be written as: |
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70 # |
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71 # |
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72 # P(s) D M(k) A(k,m) |
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73 # ---- = # # ------------- |
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74 # Q(s) k=1 m=1 (s - pr(k))^m |
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75 # |
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76 # (# is used to represent a summation) where D is the number of |
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77 # distinct roots, pr(k) is the kth distinct root, M(k) is the |
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78 # multiplicity of the root, and A(k,m) is the residue cooresponding |
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79 # to the kth distinct root with multiplicity m. For example, |
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80 # |
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81 # s^2 A(1,1) A(2,1) A(2,2) |
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82 # ------------------- = ------ + ------ + ------- |
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83 # s^3 + 4s^2 + 5s + 2 (s+2) (s+1) (s+1)^2 |
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84 # |
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85 # In this case there are two distinct roots (D=2 and pr = [-2 -1]), |
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86 # the first root has multiplicity one and the second multiplicity |
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87 # two (M = [1 2]) The residues are actually stored in vector format as |
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88 # r = [ A(1,1) A(2,1) A(2,2) ]. |
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89 # |
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90 # We then multiply both sides by Q(s). Continuing the example: |
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91 # |
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92 # s^2 = r(1)*(s+1)^2 + r(2)*(s+1)*(s+2) + r(3)*(s+2) |
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93 # |
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94 # or |
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95 # |
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96 # s^2 = r(1)*(s^2+2s+1) + r(2)*(s^2+3s+2) +r(3)*(s+2) |
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97 # |
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98 # The coefficients of the polynomials on the right are stored in a row |
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99 # vector called rhs, while the coefficients of the polynomial on the |
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100 # left is stored in a row vector called lhs. If the multiplicity of |
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101 # any root is greater than one we'll also need derivatives of this |
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102 # equation of order up to the maximum multiplicity minus one. The |
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103 # derivative coefficients are stored in successive rows of lhs and |
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104 # rhs. |
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105 # |
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106 # For our example lhs and rhs would be: |
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107 # |
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108 # | 1 0 0 | |
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109 # lhs = | | |
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110 # | 0 2 0 | |
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111 # |
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112 # | 1 2 1 1 3 2 0 1 2 | |
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113 # rhs = | | |
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114 # | 0 2 2 0 2 3 0 0 1 | |
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115 # |
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116 # We then form a vector B and a matrix A obtained by evaluating the |
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117 # polynomials in lhs and rhs at the pole values. If a pole has a |
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118 # multiplicity greater than one we also evaluate the derivative |
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119 # polynomials (successive rows) at the pole value. |
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120 # |
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121 # For our example we would have |
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122 # |
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123 # | 4| | 1 0 0 | | r(1) | |
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124 # | 1| = | 0 0 1 | * | r(2) | |
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125 # |-2| | 0 1 1 | | r(3) | |
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126 # |
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127 # We then solve for the residues using matrix division. |
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128 |
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129 if (nargin < 2 || nargin > 3) |
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130 usage ("residue (b, a [, toler])"); |
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131 endif |
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132 |
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133 if (nargin == 2) |
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134 toler = .001; |
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135 endif |
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136 |
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137 # Make sure both polynomials are in reduced form. |
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138 |
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139 a = polyreduce (a); |
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140 b = polyreduce (b); |
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141 |
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142 b = b / a(1); |
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143 a = a / a(1); |
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144 |
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145 la = length (a); |
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146 lb = length (b); |
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147 |
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148 # Handle special cases here. |
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149 |
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150 if (la == 0 || lb == 0) |
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151 k = r = p = e = []; |
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152 return; |
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153 elseif (la == 1) |
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154 k = b / a; |
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155 r = p = e = []; |
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156 return; |
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157 endif |
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158 |
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159 # Find the poles. |
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160 |
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161 p = roots (a); |
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162 lp = length (p); |
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163 |
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164 # Determine if the poles are (effectively) real. |
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165 |
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166 index = find (abs (imag (p) ./ real (p)) < toler); |
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167 if (length (index) != 0) |
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168 p (index) = real (p (index)); |
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169 endif |
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170 |
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171 # Find the direct term if there is one. |
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172 |
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173 if (lb >= la) |
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174 # Also returns the reduced numerator. |
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175 [k, b] = deconv (b, a); |
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176 lb = length (b); |
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177 else |
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178 k = []; |
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179 endif |
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180 |
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181 if (lp == 1) |
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182 r = polyval (b, p); |
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183 e = 1; |
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184 return; |
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185 endif |
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186 |
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187 |
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188 # We need to determine the number and multiplicity of the roots. |
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189 # |
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190 # D is the number of distinct roots. |
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191 # M is a vector of length D containing the multiplicity of each root. |
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192 # pr is a vector of length D containing only the distinct roots. |
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193 # e is a vector of length lp which indicates the power in the partial |
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194 # fraction expansion of each term in p. |
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195 |
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196 # Set initial values. We'll remove elements from pr as we find |
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197 # multiplicities. We'll shorten M afterwards. |
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198 |
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199 e = ones (lp, 1); |
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200 M = zeros (lp, 1); |
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201 pr = p; |
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202 D = 1; |
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203 M(1) = 1; |
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204 |
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205 old_p_index = 1; |
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206 new_p_index = 2; |
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207 M_index = 1; |
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208 pr_index = 2; |
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209 |
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210 while (new_p_index <= lp) |
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211 if (abs (p (new_p_index) - p (old_p_index)) < toler) |
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212 # We've found a multiple pole. |
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213 M (M_index) = M (M_index) + 1; |
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214 e (new_p_index) = e (new_p_index-1) + 1; |
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215 # Remove the pole from pr. |
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216 pr (pr_index) = []; |
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217 else |
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218 # It's a different pole. |
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219 D++; |
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220 M_index++; |
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221 M (M_index) = 1; |
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222 old_p_index = new_p_index; |
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223 pr_index++; |
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224 endif |
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225 new_p_index++; |
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226 endwhile |
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227 |
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228 # Shorten M to it's proper length |
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229 |
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230 M = M (1:D); |
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231 |
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232 # Now set up the polynomial matrices. |
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233 |
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234 MM = max(M); |
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235 |
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236 # Left hand side polynomial |
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237 |
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238 lhs = zeros (MM, lb); |
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239 rhs = zeros (MM, lp*lp); |
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240 lhs (1, :) = b; |
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241 rhi = 1; |
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242 dpi = 1; |
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243 mpi = 1; |
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244 while (dpi <= D) |
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245 for ind = 1:M(dpi) |
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246 if (mpi > 1 && (mpi+ind) <= lp) |
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247 cp = [p(1:mpi-1); p(mpi+ind:lp)]; |
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248 elseif (mpi == 1) |
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249 cp = p (mpi+ind:lp); |
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250 else |
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251 cp = p (1:mpi-1); |
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252 endif |
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253 rhs (1, rhi:rhi+lp-1) = prepad (poly (cp), lp); |
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254 rhi = rhi + lp; |
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255 endfor |
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256 mpi = mpi + M (dpi); |
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257 dpi++; |
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258 endwhile |
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259 if (MM > 1) |
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260 for index = 2:MM |
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261 lhs (index, :) = prepad (polyderiv (lhs (index-1, :)), lb); |
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262 ind = 1; |
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263 for rhi = 1:lp |
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264 cp = rhs (index-1, ind:ind+lp-1); |
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265 rhs (index, ind:ind+lp-1) = prepad (polyderiv (cp), lp); |
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266 ind = ind + lp; |
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267 endfor |
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268 endfor |
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269 endif |
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270 |
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271 # Now lhs contains the numerator polynomial and as many derivatives as |
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272 # are required. rhs is a matrix of polynomials, the first row |
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273 # contains the corresponding polynomial for each residue and |
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274 # successive rows are derivatives. |
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275 |
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276 # Now we need to evaluate the first row of lhs and rhs at each |
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277 # distinct pole value. If there are multiple poles we will also need |
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278 # to evaluate the derivatives at the pole value also. |
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279 |
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280 B = zeros (lp, 1); |
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281 A = zeros (lp, lp); |
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282 |
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283 dpi = 1; |
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284 row = 1; |
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285 while (dpi <= D) |
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286 for mi = 1:M(dpi) |
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287 B (row) = polyval (lhs (mi, :), pr (dpi)); |
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288 ci = 1; |
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289 for col = 1:lp |
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290 cp = rhs (mi, ci:ci+lp-1); |
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291 A (row, col) = polyval (cp, pr(dpi)); |
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292 ci = ci + lp; |
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293 endfor |
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294 row++; |
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295 endfor |
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296 dpi++; |
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297 endwhile |
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298 |
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299 # Solve for the residues. |
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300 |
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301 r = A \ B; |
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302 |
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303 endfunction |
