Page 149 - Modern Control of DC-Based Power Systems
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Control Approaches for Parallel Source Converter Systems     113


                 The focus here lies on the input output Linearization. For the envis-
              aged controller design, the form of the above system description in (5.1)
              is not favorable. A suitable transformation is therefore needed. For this
              purpose the Lie derivative is employed, which is defined as the gradient
              of a scalar function h(x) multiplied with a vector field f(x), i.e.:

                                                     T
                              L f hxðÞ 5  @hxðÞ  fxðÞ 5 grad hxðÞf ðxÞ    (5.4)
                                        @x
                 Considering the system presented in (5.1), and applying the Lie
              derivative yields:

                                      L a cxðÞ 5  @cxðÞ                   (5.5)
                                               @x  axðÞ

                 Now forming the time derivative of the output y one obtains:

                       _ ytðÞ 5  dc x ðÞ  5  @cx ðÞ  _ x 1 1 .. . 1  @cx ðÞ  _ x n x 5  @cx ðÞ  _ x  (5.6)
                              dt      @x             @x n       @x
                 Replacing now:

                                                                          (5.7)
                                      _ xtðÞ 5 axðÞ 1 bx uðÞ
                 Results in:

                                 _ ytðÞ 5  @cxðÞ axðÞ 1  @cxðÞ  bxðÞu     (5.8)
                                        @x         @x
                 Which can also be described as the Lie derivative:
                                    _ ytðÞ 5 L a cxðÞ 1 L b cxðÞu         (5.9)

                 In most technical systems according to [6] is L b 5 0, such that:
                                                                         (5.10)
                                         _ ytðÞ 5 L a cxðÞ
                                                  ̈
                 The next step would be to calculate y while continuing from (5.10):
                    ÿ 5  dL a cx ðÞ  5  @L a cx ðÞ  _ x 5  @L a cx ðÞ  axðÞ 1  @L a cx ðÞ  bxðÞ
                          dt       @x         @x          @x             (5.11)
                                           2
                                           a
                    5 L a L a cxðÞ 1 L b L a cxðÞ 5 L cðxÞ
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