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1 ## Copyright (C) 1996, 1997 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, Inc., 51 Franklin Street, Fifth Floor, Boston, MA |
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18 ## 02110-1301, USA. |
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19 |
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20 ## -*- texinfo -*- |
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21 ## @deftypefn {Function File} {} residue (@var{b}, @var{a}, @var{tol}) |
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22 ## If @var{b} and @var{a} are vectors of polynomial coefficients, then |
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23 ## residue calculates the partial fraction expansion corresponding to the |
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24 ## ratio of the two polynomials. |
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25 ## @cindex partial fraction expansion |
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26 ## |
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27 ## The function @code{residue} returns @var{r}, @var{p}, @var{k}, and |
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28 ## @var{e}, where the vector @var{r} contains the residue terms, @var{p} |
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29 ## contains the pole values, @var{k} contains the coefficients of a direct |
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30 ## polynomial term (if it exists) and @var{e} is a vector containing the |
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31 ## powers of the denominators in the partial fraction terms. |
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32 ## |
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33 ## Assuming @var{b} and @var{a} represent polynomials |
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34 ## @iftex |
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35 ## @tex |
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36 ## $P(s)$ and $Q(s)$ |
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37 ## @end tex |
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38 ## @end iftex |
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39 ## @ifinfo |
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40 ## P (s) and Q(s) |
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41 ## @end ifinfo |
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42 ## we have: |
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43 ## @iftex |
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44 ## @tex |
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45 ## $$ |
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46 ## {P(s)\over Q(s)} = \sum_{m=1}^M {r_m\over (s-p_m)^e_m} |
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47 ## + \sum_{i=1}^N k_i s^{N-i}. |
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48 ## $$ |
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49 ## @end tex |
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50 ## @end iftex |
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51 ## @ifinfo |
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52 ## |
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53 ## @example |
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54 ## P(s) M r(m) N |
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55 ## ---- = SUM ------------- + SUM k(i)*s^(N-i) |
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56 ## Q(s) m=1 (s-p(m))^e(m) i=1 |
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57 ## @end example |
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58 ## @end ifinfo |
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59 ## |
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60 ## @noindent |
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61 ## where @math{M} is the number of poles (the length of the @var{r}, |
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62 ## @var{p}, and @var{e} vectors) and @math{N} is the length of the |
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63 ## @var{k} vector. |
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64 ## |
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65 ## The argument @var{tol} is optional, and if not specified, a default |
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66 ## value of 0.001 is assumed. The tolerance value is used to determine |
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67 ## whether poles with small imaginary components are declared real. It is |
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68 ## also used to determine if two poles are distinct. If the ratio of the |
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69 ## imaginary part of a pole to the real part is less than @var{tol}, the |
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70 ## imaginary part is discarded. If two poles are farther apart than |
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71 ## @var{tol} they are distinct. For example, |
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72 ## |
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73 ## @example |
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74 ## @group |
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75 ## b = [1, 1, 1]; |
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76 ## a = [1, -5, 8, -4]; |
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77 ## [r, p, k, e] = residue (b, a); |
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78 ## @result{} r = [-2, 7, 3] |
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79 ## @result{} p = [2, 2, 1] |
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80 ## @result{} k = [](0x0) |
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81 ## @result{} e = [1, 2, 1] |
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82 ## @end group |
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83 ## @end example |
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84 ## |
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85 ## @noindent |
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86 ## which implies the following partial fraction expansion |
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87 ## @iftex |
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88 ## @tex |
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89 ## $$ |
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90 ## {s^2+s+1\over s^3-5s^2+8s-4} = {-2\over s-2} + {7\over (s-2)^2} + {3\over s-1} |
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91 ## $$ |
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92 ## @end tex |
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93 ## @end iftex |
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94 ## @ifinfo |
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95 ## |
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96 ## @example |
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97 ## s^2 + s + 1 -2 7 3 |
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98 ## ------------------- = ----- + ------- + ----- |
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99 ## s^3 - 5s^2 + 8s - 4 (s-2) (s-2)^2 (s-1) |
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100 ## @end example |
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101 ## @end ifinfo |
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102 ## @seealso{poly, roots, conv, deconv, polyval, polyderiv, and polyinteg} |
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103 ## @end deftypefn |
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104 |
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105 ## Author: Tony Richardson <arichard@stark.cc.oh.us> |
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106 ## Created: June 1994 |
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107 ## Adapted-By: jwe |
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108 |
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109 function [r, p, k, e] = residue (b, a, toler) |
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110 |
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111 ## Here's the method used to find the residues. |
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112 ## The partial fraction expansion can be written as: |
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113 ## |
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114 ## |
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115 ## P(s) D M(k) A(k,m) |
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116 ## ---- = # # ------------- |
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117 ## Q(s) k=1 m=1 (s - pr(k))^m |
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118 ## |
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119 ## (# is used to represent a summation) where D is the number of |
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120 ## distinct roots, pr(k) is the kth distinct root, M(k) is the |
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121 ## multiplicity of the root, and A(k,m) is the residue cooresponding |
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122 ## to the kth distinct root with multiplicity m. For example, |
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123 ## |
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124 ## s^2 A(1,1) A(2,1) A(2,2) |
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125 ## ------------------- = ------ + ------ + ------- |
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126 ## s^3 + 4s^2 + 5s + 2 (s+2) (s+1) (s+1)^2 |
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127 ## |
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128 ## In this case there are two distinct roots (D=2 and pr = [-2 -1]), |
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129 ## the first root has multiplicity one and the second multiplicity |
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130 ## two (M = [1 2]) The residues are actually stored in vector format as |
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131 ## r = [ A(1,1) A(2,1) A(2,2) ]. |
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132 ## |
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133 ## We then multiply both sides by Q(s). Continuing the example: |
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134 ## |
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135 ## s^2 = r(1)*(s+1)^2 + r(2)*(s+1)*(s+2) + r(3)*(s+2) |
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136 ## |
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137 ## or |
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138 ## |
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139 ## s^2 = r(1)*(s^2+2s+1) + r(2)*(s^2+3s+2) +r(3)*(s+2) |
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140 ## |
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141 ## The coefficients of the polynomials on the right are stored in a row |
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142 ## vector called rhs, while the coefficients of the polynomial on the |
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143 ## left is stored in a row vector called lhs. If the multiplicity of |
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144 ## any root is greater than one we'll also need derivatives of this |
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145 ## equation of order up to the maximum multiplicity minus one. The |
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146 ## derivative coefficients are stored in successive rows of lhs and |
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147 ## rhs. |
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148 ## |
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149 ## For our example lhs and rhs would be: |
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150 ## |
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151 ## | 1 0 0 | |
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152 ## lhs = | | |
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153 ## | 0 2 0 | |
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154 ## |
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155 ## | 1 2 1 1 3 2 0 1 2 | |
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156 ## rhs = | | |
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157 ## | 0 2 2 0 2 3 0 0 1 | |
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158 ## |
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159 ## We then form a vector B and a matrix A obtained by evaluating the |
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160 ## polynomials in lhs and rhs at the pole values. If a pole has a |
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161 ## multiplicity greater than one we also evaluate the derivative |
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162 ## polynomials (successive rows) at the pole value. |
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163 ## |
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164 ## For our example we would have |
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165 ## |
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166 ## | 4| | 1 0 0 | | r(1) | |
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167 ## | 1| = | 0 0 1 | * | r(2) | |
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168 ## |-2| | 0 1 1 | | r(3) | |
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169 ## |
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170 ## We then solve for the residues using matrix division. |
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171 |
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172 if (nargin < 2 || nargin > 3) |
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173 usage ("residue (b, a, toler)"); |
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174 endif |
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175 |
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176 if (nargin == 2) |
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177 toler = .001; |
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178 endif |
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179 |
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180 ## Make sure both polynomials are in reduced form. |
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181 |
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182 a = polyreduce (a); |
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183 b = polyreduce (b); |
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184 |
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185 b = b / a(1); |
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186 a = a / a(1); |
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187 |
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188 la = length (a); |
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189 lb = length (b); |
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190 |
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191 ## Handle special cases here. |
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192 |
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193 if (la == 0 || lb == 0) |
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194 k = r = p = e = []; |
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195 return; |
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196 elseif (la == 1) |
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197 k = b / a; |
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198 r = p = e = []; |
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199 return; |
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200 endif |
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201 |
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202 ## Find the poles. |
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203 |
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204 p = roots (a); |
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205 lp = length (p); |
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206 |
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207 ## Determine if the poles are (effectively) zero. |
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208 |
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209 p(abs (p) < toler) = 0; |
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210 |
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211 ## Determine if the poles are (effectively) real. |
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212 |
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213 index = (abs (p) >= toler && (abs (imag (p)) ./ abs (p) < toler)); |
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214 p(index) = real (p(index)); |
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215 |
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216 ## Find the direct term if there is one. |
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217 |
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218 if (lb >= la) |
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219 ## Also returns the reduced numerator. |
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220 [k, b] = deconv (b, a); |
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221 lb = length (b); |
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222 else |
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223 k = []; |
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224 endif |
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225 |
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226 if (lp == 1) |
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227 r = polyval (b, p); |
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228 e = 1; |
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229 return; |
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230 endif |
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231 |
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232 |
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233 ## We need to determine the number and multiplicity of the roots. |
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234 ## |
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235 ## D is the number of distinct roots. |
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236 ## M is a vector of length D containing the multiplicity of each root. |
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237 ## pr is a vector of length D containing only the distinct roots. |
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238 ## e is a vector of length lp which indicates the power in the partial |
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239 ## fraction expansion of each term in p. |
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240 |
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241 ## Set initial values. We'll remove elements from pr as we find |
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242 ## multiplicities. We'll shorten M afterwards. |
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243 |
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244 e = ones (lp, 1); |
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245 M = zeros (lp, 1); |
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246 pr = p; |
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247 D = 1; |
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248 M(1) = 1; |
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249 |
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250 old_p_index = 1; |
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251 new_p_index = 2; |
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252 M_index = 1; |
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253 pr_index = 2; |
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254 |
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255 while (new_p_index <= lp) |
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256 if (abs (p (new_p_index) - p (old_p_index)) < toler) |
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257 ## We've found a multiple pole. |
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258 M (M_index) = M (M_index) + 1; |
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259 e (new_p_index) = e (new_p_index-1) + 1; |
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260 ## Remove the pole from pr. |
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261 pr (pr_index) = []; |
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262 else |
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263 ## It's a different pole. |
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264 D++; |
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265 M_index++; |
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266 M (M_index) = 1; |
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267 old_p_index = new_p_index; |
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268 pr_index++; |
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269 endif |
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270 new_p_index++; |
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271 endwhile |
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272 |
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273 ## Shorten M to it's proper length |
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274 |
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275 M = M (1:D); |
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276 |
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277 ## Now set up the polynomial matrices. |
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278 |
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279 MM = max(M); |
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280 |
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281 ## Left hand side polynomial |
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282 |
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283 lhs = zeros (MM, lb); |
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284 rhs = zeros (MM, lp*lp); |
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285 lhs (1, :) = b; |
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286 rhi = 1; |
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287 dpi = 1; |
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288 mpi = 1; |
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289 while (dpi <= D) |
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290 for ind = 1:M(dpi) |
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291 if (mpi > 1 && (mpi+ind) <= lp) |
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292 cp = [p(1:mpi-1); p(mpi+ind:lp)]; |
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293 elseif (mpi == 1) |
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294 cp = p (mpi+ind:lp); |
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295 else |
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296 cp = p (1:mpi-1); |
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297 endif |
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298 rhs (1, rhi:rhi+lp-1) = prepad (poly (cp), lp, 0, 2); |
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299 rhi = rhi + lp; |
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300 endfor |
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301 mpi = mpi + M (dpi); |
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302 dpi++; |
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303 endwhile |
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304 if (MM > 1) |
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305 for index = 2:MM |
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306 lhs (index, :) = prepad (polyderiv (lhs (index-1, :)), lb, 0, 2); |
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307 ind = 1; |
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308 for rhi = 1:lp |
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309 cp = rhs (index-1, ind:ind+lp-1); |
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310 rhs (index, ind:ind+lp-1) = prepad (polyderiv (cp), lp, 0, 2); |
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311 ind = ind + lp; |
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312 endfor |
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313 endfor |
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314 endif |
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315 |
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316 ## Now lhs contains the numerator polynomial and as many derivatives as |
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317 ## are required. rhs is a matrix of polynomials, the first row |
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318 ## contains the corresponding polynomial for each residue and |
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319 ## successive rows are derivatives. |
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320 |
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321 ## Now we need to evaluate the first row of lhs and rhs at each |
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322 ## distinct pole value. If there are multiple poles we will also need |
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323 ## to evaluate the derivatives at the pole value also. |
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324 |
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325 B = zeros (lp, 1); |
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326 A = zeros (lp, lp); |
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327 |
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328 dpi = 1; |
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329 row = 1; |
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330 while (dpi <= D) |
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331 for mi = 1:M(dpi) |
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332 B (row) = polyval (lhs (mi, :), pr (dpi)); |
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333 ci = 1; |
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334 for col = 1:lp |
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335 cp = rhs (mi, ci:ci+lp-1); |
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336 A (row, col) = polyval (cp, pr(dpi)); |
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337 ci = ci + lp; |
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338 endfor |
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339 row++; |
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340 endfor |
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341 dpi++; |
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342 endwhile |
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343 |
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344 ## Solve for the residues. |
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345 |
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346 r = A \ B; |
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347 |
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348 endfunction |
