Page 318 - Mechatronics with Experiments
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JWST499-Cetinkunt
            JWST499-c05
                       304   MECHATRONICS  Printer: Yet to Come                     October 28, 2014 11:15 254mm×178mm
                              a high pass filter is shown in Figure 5.36. The frequency domain input–output voltage
                              relationship of a high pass op-amp filter is
                                                         V (jw)    j   ⋅ w
                                                           o         1
                                                                =                               (5.287)
                                                         V (jw)   1 +    ⋅ jw
                                                                      1
                                                          in
                              where the time constant of the first-order filter    = RC. Notice that the only difference
                                                                     1
                              between the low pass and high pass filter is the placement of the resistor and capacitor in
                              the (+) input terminal.
                                   Similarly, the input–output voltage relationship can be derived for the high pass filter.
                              Notice that the location of the resistor and capacitor on the circuit is swapped compared to
                              the low pass filter. Following a similar derivation process, it is straightforward to derive the
                              input–output voltage relationship.
                                                            V (t) = R ⋅ i(t)                    (5.288)
                                                             o
                                                            V (s) = R ⋅ i(s)                    (5.289)
                                                             o
                              The voltage across the capacitor,
                                                                       t
                                                                   1
                                                       V (t) − V (t) =  i(  )d                  (5.290)
                                                        i     o    C ∫ o
                                                                    1
                                                      V (s) − V (s) =  i(s)                     (5.291)
                                                       i
                                                             o
                                                                   Cs
                                                                1
                              By substituting the relationship for i(s) = V (s), it can be shown that
                                                               R  o
                                                           V (s)  =  RCs                        (5.292)
                                                            o
                                                           V (s)  1 + RCs
                                                            i
                                                          V (jw)    jRCw
                                                           o
                                                                =                               (5.293)
                                                          V (jw)  1 + jRCw
                                                           i
                              which represents the high pass filter transfer function. Again, notice that at w = w where
                                                                                               c
                                   1              1
                              w =  RC  rad∕sor f =  2  RC  Hz, the magnitude ratio is 0.707. Except that in the high
                                             c
                               c
                              pass filter, the filter passes the frequency content above this frequency and attenuates the
                              frequency content below that frequency. The low pass filter does the opposite.
                              Band Pass Filter Op-Amp      A band pass filter passes a selected narrow band of
                              frequencies, and suppresses the rest (Figure 5.36). The design parameters of the filter are
                              to select the frequency band to pass: center frequency and the width around that frequency,
                              w and Δw . It can be realized by a low pass and high pass filter in series.
                                      B
                               r
                              Band Reject Filter Op-Amp      A band reject filter passes all frequencies except a
                              selected narrow band of frequencies (Figure 5.36). This is also called a notch filter, and
                              basically does the opposite of the band pass filter. The design parameters of a notch filter are
                              to select the center frequency and the width of the frequency around the center frequency
                              that will be suppressed, w and Δw . It can be realized by a low pass and high pass filter in
                                                         B
                                                  r
                              parallel.
                              Instrumentation Op-Amp       An instrumentation op-amp is used to amplify small
                              sensor signals in noisy environments (Figure 5.37). It is a modified version of the differential
                              op-amp with improved performance characteristics (i.e., higher input impedance, easy
                              adjustment of gain). The instrumentation amplifier has a higher CMRR (common mode
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