Filters and Equalizers 79123.2.3.5 BandpassThe impedance characteristics of the series and parallel
resonant elements can be used to form a bandpass filter
as in Fig. 23-11. The frequencies f 1 and f 2 are the cutoff
frequencies of the pass band. The design equations for
the component values are
(23-29)(23-30)(23-31)(23-32)where,
f 1 is the lower cutoff frequency,
f 2 is the upper cutoff frequency,
Z 0 is the line impedance.
23.2.3.6 Band RejectThe configuration for a band reject filter using series
and parallel resonant elements is shown in Fig. 23-12.
The configuration is the reverse of the bandpass
T-network filter. In this case the frequencies f 1 and f 2 are
at the edge of the reject band. The design equation for
the component values are
(23-33)(23-34)(23-35)(23-36)23.2.3.7 Ladder NetworksPassive filters of arbitrary length may be constructed by
adding RC, RL, or LC L-type sections into a network of
arbitrary length called a Cauer network. The interaction
between the various stages in this topography starts to
become important as the impedance of one section loads
the next.23.2.4 Filter DesignAs the number of components in a filter increases, the
number of possible transfer functions also increases.
Increasing the order of a filter by adding more of the
same sections will not necessarily produce the optimum
results. Consider chaining two low-pass filters with a
cutoff frequency of fc. The attenuation at the cutoff is
3 dB, so with two sections in series, the attenuation at fcFigure 23-11. T network bandpass filter.L 1Z 0
2 S f 2 – f 1= ---------------------------L 2f 2 – f 1 Z 0
2 Sf 1 f 2= ---------------------------C 1f 2 – f 1
2 Sf 1 f 2 Z 0= ----------------------C 2 1
2 S f 2 – f 1 Z 0= ----------------------------------A. Configuration.B. Transmission characteristics.L 1C 2 L 2C 1PassRejectf 1 f 2f 1 d fm f 2 dC 1RejectL 1L 1f 2 – f 1 Z 0
2 Sf 2 f 1= -------------------------Figure 23-12. T network band-reject filter.C 1 1
2 S f 2 – f 1 Z 0= --------------------------------L 2Z 0
2 S f 2 – f 1=--------------------------C 2f 2 – f 1
2 Sf 1 f 2 Z 0= ---------------------A. ConfigurationB. Transmission characteristics..Zl ZlL 1 L 12L 22C 2C 1 C 1± ±Pass PassRejectf 1 f 2f 1 d f 2 d
fm