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SIMILARITY TRANSFORMATION AND DIAGONALIZATION  375
            we use the modal matrix obtained as (E8.2.2) in Example 8.2 to make a substi-
            tution of variable
                                                     √
                                     x 1 (t)  1    1/ 2    w 1 (t)
                   x(t) = V w(t),          =         √                  (E8.3.2)
                                     x 2 (t)  0 −1/ 2      w 2 (t)
            which converts Eq. (E8.3.1) into

                                 V w (t) = AV w(t) + Bu s (t)           (E8.3.3)


              We premultiply (E8.3.3) by V −1  to write it in a decoupled form as


             w (t)=V  −1 AV w(t) + V  −1 Bu s (t)= w(t) + V  −1 Bu s (t) with w(0) = V  −1 x(0);


             w 1 (t)    0   0   w 1 (t)   1   1     0               u s (t)
                    =                 +       √       u s (t) =        √

             w 2 (t)    0  −1   w 2 (t)   0  − 2    1         −w 2 (t) −  2u s (t)
                                                                        (E8.3.4)

                                w 1 (0)   1    1      1      0
                          with         =       √         = √
                                w 2 (0)   0  − 2    −1        2
            where there is no correlation between the variables w 1 (t) and w 2 (t).Thenwe
            can solve these two equations separately to have

                     w 1 (t) = u s (t) with w 1 (0) = 0;
                                          1          1
                          sW 1 (s) − w 1 (0) =  ; W 1 (s) =  ; w 1 (t) = tu s (t)  (E8.3.5a)
                                          s          s 2
                                      √                  √

                     w 2 (t) =−w 2 (t) −  2u s (t) with w 2 (0) =  2;
                                                   √
                                                     2
                          sW 2 (s) − w 2 (0) =−W 2 (s) −  ;
                                                    s
                                       √         √      √
                             w 2 (0)     2         2   2 2
                     W 2 (s) =     −         =−     +      ,
                              s + 1  s(s + 1)     s    s + 1
                                 √
                                           −t
                          w 2 (t) =  2(−1 + 2e )u s (t)                (E8.3.5b)
            and substitute this into Eq. (E8.3.2) to get

                              √                      √
              x 1 (t)   1   1/ 2    w 1 (t)   1    1/ 2          t
                    =         √            =         √     √              u s (t)
                                                                      −t
              x 2 (t)   0 −1/ 2     w 2 (t)   0 −1/ 2        2(−1 + 2e )
                                −t
                        t − 1 + 2e
                    =          −t   u s (t)                             (E8.3.6)
                         1 − 2e
              This is the same result as Eq. (6.5.10) obtained in Section 6.5.1.
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