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Chapter 3  Elementary Signals


                                   ft  G' t –  a =  fa  G' t –  a –  f ' a  G t –  a





        Then, for this example,
                                  2            2              d 2
                                  t G' t –   3 =  t  G' t –  3 –  -----t  G t –  3




                                                t =  3       dt   t =  3
                                                   =
                                              9G' t – 3    6G t –  3
                                                            –
        Example 3.9
        a. Express the voltage waveform vt    shown in Figure 3.21 as a sum of unit step functions for the
           time interval  1   –  t  7  . s
        b. Using the result of part (a), compute the derivative of vt    and sketch its waveform.

                                    V
                              vt

                               3

                               2


                                1
                            1   0       1    2    3      4     5   6     7
                                                                             ts


                                   1
                                   2

                                     Figure 3.21. Waveform for Example 3.9

        Solution:
        a. We first derive the equations for the linear segments of the given waveform. These are shown in
           Figure 3.22.

           Next, we express vt     in terms of the unit step function u t   , and we get
                                                                 0
                       vt   =  2t u t +  >  0  1 –  u t –  1 @     2u t –  1 – u t –  2 @
                                                            +



                                                       >

                                                         0
                                            0
                                                                   0
                                   t
                               +  – +  5 u t –  >     0  2 –  u t –  4 @     +  u t –  4 –  u t –  5 @     (3.52)
                                                                   >





                                                                          0
                                                    0
                                                                0
                                   t
                               +  – +  6 u t –  >     0  5 –  u t –  7 @



                                                    0
           Multiplying and collecting like terms in (3.52), we get
        3-16                                                 Circuit Analysis II with MATLAB Applications
                                                                                   Orchard Publications
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