Passive, active, and digital filters (3ed., CRC, 2009) - tiera.ru

17-6 Passive, Active, and Digital Filtersamount of charge is extracted but the integratingcapacitor is reduced due to theswitches connected to C 1 , thus relativelyhigh-slew-rate op-amps could be required.A serious drawback could be the increased φ 1 (φ 2 )offset in comparison with the standard SCv o inintegrators. However, in typical two integratorloop filters, the other integrator can be φ 2 (φ 1 )chosen to be offset and low dc gain compensatedas shown in Figure 17.6.The SC integrator performs the integration FIGURE 17.6during f 1 by means of C S and C F and thehold capacitor C H stores the offset voltage. The voltage across C Hcompensates the offset voltage and the dc gain error of theop-amp. Note that the SC integrator of Figure 17.6 can operateas a noninverting integrator if the clocking in parenthesis isemployed. C M provides a time-continuous feedback around theop-amp. The transfer function for infinite op-amp gain isC Mφ 1 φ C CS 2 FC Hφ 2–+φ 1Offset and gain compensated integrator.Vi e1Vi o2C 1φ 2φ 1 C 2C Fv o oH oo (z) ¼ Vo o (z)V o in (z) ¼C S 1C F 1 z 1(17:9)Vi e3C 3φ 2 φ 1v oφ 1 φ 2+–Next we discuss a general form of a first-order building block(see Figure 17.7). Observe that some switches can be shared. Theoutput voltage during f 1 can be expressed asFIGURE 17.7 General form of a firstorderbuilding block.Vo o ¼ C 1Vi e C 2 1C 1 F C F 1 z 1 Vi o 2þ C 3 z 1=2 C F 1 z 1 Vi e 3(17:10)Observe that the capacitor C 3 and switches can be considered as the implementation of a negative resistorleading to a noninverting amplifier. Also note that Vi o 2could be Vo e , this connection would make theintegrator a lossy one. In that case Equation 17.10 can be written asV o o 1 þ C2C Fz 1¼ C 1Vi e z 1 C 1þ C 3 z 1=2F C F z 1 Ve i 3, for Vi o 2¼ Vo o (17:11)The building block of Figure 17.7 is the basis of higher order filters.17.2.3 Switched-Capacitor Biquadratic SectionsThe circuit shown in Figure 17.8 can implement any pair of poles and zeros in the z-domain. ForC A ¼ C B ¼ 1 we can writeH ee (z) ¼ Ve o (z)Vin e (z) ¼ (C 5 þ C 6 )z 2 þ (C 1 C 2 C 5 2C 6 )z þ C 6z 2 þ (C 2 C 3 þ C 2 C 4 2)z þ (1 C 2 C 4 )(17:12)