Page 42 - Circuit Analysis II with MATLAB Applications
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Chapter 1  Second Order Circuits

        Example 1.7

        The circuit of Figure 1.19 a known as a Multiple Feed Back (MFB) active low-pass filter. For this cir-
        cuit, the initial conditions are v C1  =  v C2  =  0 . Compute and sketch v out  t     for t !  . 0



                                                     C 2  10 nF
                                          R 2  40 K:
                                      R 1        50 K:

                                  +           v 1       v 2
                                    200 K:        R 3                    +
                                 v in                                     v out
                                         C 1  25 nF

                                          v    =   6.25cos 6280tu t
                                              t
                                           in
                                                                 0
                                       Figure 1.19. Circuit for Example 1.7
        Solution:

        At node  :
                 v
                  1
                               v –  v     dv   v –  v    v –  v
                                    in
                                                    out
                                                1
                                1
                                                          1
                                            1
                                                              2
                               ----------------- +  C -------- +  -------------------- +  --------------- =  0t ! 0  (1.83)
                                         1
                                 R 1       dt     R 2      R 3
        At node  :
                 v
                  2
                                            v –  v     dv out
                                            --------------- =  C -------------                  (1.84)
                                             2
                                                 1
                                              R 3      2  dt
        We observe that v =  0  (virtual ground).
                         2
        Collecting like terms and rearranging (1.83) and (1.84) we get
                                                      dv
                                                           1
                                                                    1
                                  1
                                       1
                                           1 ·
                                §  ----- + ----- +  ----- v +  C -------- –  -----v  =  -----v  (1.85)
                                                        1
                                ©  R 1  R 2  R 3  ¹  1  1  dt  R 2  out  R 1  in
        and
                                                       dv
                                                         out
                                            v =  – R C -----------                              (1.86)
                                             1
                                                      2
                                                   3
                                                        dt
        Differentiation of (1.86) yields
                                                         2
                                          dv           dv out
                                             1
                                          -------- =  – R C -------------                       (1.87)
                                                    3
                                           dt         2  dt 2
        1-30                                                Circuit Analysis II with MATLAB Applications
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