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Biosystems Analysis and Optimization      49

               where z , z ,…, z  are defined as the zeros of G(s), values of s that
                      1  2    m
               annihilate the rational polynomial, and p , p ,…, p  are the poles of
                                                   1  2   n
               G(s), values, making the rational polynomial infinite. Poles as well as
               zeros can be a real or a complex number. In the later case, each pole or
               zero has a complex conjugate.
                   If the input to the system is u(t) = A sin(ωt), a sine wave of amplitude
               A and of frequency ω rad/s, then the Laplace transform of the input is
                                            Aω
                                     Us () =                        (2.38)
                                          s + ω  2
                                           2
               The Laplace transform Y(s) of the output y(t) of the system is then
               calculated as
                                        K ( −ω  s z )( −  s z )
                                                s z ) ... ( −
                      Ys () =  G s U s () =  A  1   2      m        (2.39)
                            ()
                                                           −
                                                   −
                                           2 2
                                      s (  2  + ω )(sp−  )(sp  ) ...  (sp  )
                                                1     2       n
               This expression can be rewritten as the partial fraction expansion
               (Schwarzenbach and Gill 1978):
                      Ys ()  Gs ()ω  A      A      B        B
                          =       =   1  +   2  +   1  +  ... +  n n  (2.40)
                                                  −
                                                            −
                            2
                       A    s + ω 2  s −  jω  s +  jω  sp  sp
                                                     1         n
               where A , A , B , B ,…, B  are constants. Taking the inverse Laplace
                       1  2  1  2    n
               transform of Eq. (2.40) leads to the following time response:
                          yt()
                              =  Ae  jt  +  A e −  jt  +  B e 1  +  ... +  B e n  (2.41)
                                          ω
                                   ω
                                                pt
                                                          pt
                           A     1     2       1        n
               All individual terms in the right-hand side of Eq. (2.41) are called
               modes of the solution, which contain a system pole or an input signal
               pole in their exponent. Provided the system is stable and linear, poles
               p , p ,…, p  all have negative real parts and all of the modes in Eq. (2.41)
                1  2   n
               decay to zero with increasing time t except the first two. The terms that
               decay to zero represent the transient response (or complementary
               function). The first two terms will not die out and thus represent the
               particular integral component of the solution (also called the steady-
               state response, particular solution, or stationary solution):
                                ⎡ yt()⎤     ω      −  ω
                                ⎢   ⎥  =  Ae  jt  +  A e  jt        (2.42)
                                                 2
                                          1
                                ⎣  A  t ⎦ →∞
               To determine coefficient A , both sides of Eq. (2.40) are multiplied by
                                     1
               (s – jω) and s is replaced by jω:
                            ⎡( s −  j ) (ω  G s)ω ⎤  ⎡ Gs ()ω ⎤  Gjω)
                                                          (
                        A =  ⎢          ⎥   =  ⎢   ⎥   =            (2.43)
                         1     s + ω  2       s +  jω
                            ⎣           ⎦ sjω  ⎣   ⎦ ⎦ =sjω  j 2
                                2
                                         =
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