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                              Then we can calculate the current over the resistor R .
                                                                        1
                                                                   V i
                                                              i =                               (5.305)
                                                               1
                                                                   R 1
                              Also note that no current would flow into the op-amp, then the same current must pass
                              through the R and C combination (noting Kirchoff’s current law),
                                         2
                                                             V (t)
                                                              i
                                                       i (t) =   = i (t) + i (t)                (5.306)
                                                                           c
                                                       1
                                                                    R2
                                                              R 1
                                                             V (t)
                                                              o
                                                           =      + i (t)                       (5.307)
                                                                    c
                                                              R
                                                               2
                                                                 t
                                                             1
                                                      V (t) =  C ∫ 0  i (  )d                   (5.308)
                                                       o
                                                                   c
                              where we use the fact that the current flow over the R and C is determined by the output
                                                                         2
                              voltage and the ground voltage at the negative input (inverting) terminal. We will add the
                              negative sign to the input–output voltage relation at the end of the derivation. If we take the
                              Laplace transform of the above equations, we can easily find the transfer function betwen
                              input voltage and output voltage,
                                                           V (s)  V (s)
                                                            i
                                                                   o
                                                     i (s) =   =      + Cs ⋅ V (s)              (5.309)
                                                     1                       o
                                                            R      R
                                                             1      2
                                                    V (s)   R 2    1
                                                     o
                                                         =−    ⋅                                (5.310)
                                                    V (s)   R 1  1 + R Cs
                                                                     2
                                                     i
                              where we added the negative sign to indicate that the sign relationship between input voltage
                              and output voltage is opposite.
                                                        V (s)   R 2    1
                                                         o
                                                             =−    ⋅                            (5.311)
                                                        V (s)   R   R Cs + 1
                                                         i       1   2
                              For the case of R = R = R we have
                                           1    2
                                                          V (s)  =−  1                          (5.312)
                                                           o
                                                          V (s)    RCs + 1
                                                            i
                                   The same reasoning applies for the derivation of the transfer function between input
                              voltage and output voltage for the high pass filter. The only difference is the definition of
                              the output voltage. In the above derivation, we would use
                                                            V (t) = R ⋅ i(t)                    (5.313)
                                                             o
                              and it can be shown, by following the same procedure for the low pass filter, that the transfer
                              function for the high pass filter is
                                                          V (s)     RCs
                                                           o
                                                               =−                               (5.314)
                                                          V (s)    RCs + 1
                                                            i
                       5.7 DIGITAL ELECTRONIC DEVICES
                              The logic ON state is represented by 1 and the logic OFF state is represented by 0. Two of the
                              most popular digital device types are transistor–transistor logic (TTL) and complementary
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