Page 286 - Complete Wireless Design

Filter Design

Filter Design 285

If the low-pass filter is required to block DC, then a large-value capacitor
with an X of 1 ohm at its lowest frequency of interest can be inserted at the
C
filter’s input. This will have no effect to the low-pass filter’s response. The abil-
ity to block DC is especially valuable at the lower RF frequencies, where fer-
romagnetic cores are used in the filter’s inductors (the DC can affect the
permeability of the inductors’ cores by saturating them, and thus changing the
inductance of the coils, destroying the response of the filter).
The attenuation slope of the low-pass filter as presented can be approxi-
mated as an 18-dB increase in attenuation per octave for a 3-pole filter. In
other words, if the filter were designed to have an f of 1 MHz, then the low-
C
pass filter would have attenuated a 2-MHz signal by 18 dB. By 4 MHz, the sig-
nal would be down by 2 18 dB, or 36 dB. And if the low-pass filter were
increased to 6 poles, it would have an attenuation slope that dropped by 36 dB
per octave. With an f at 1 MHz, the signal is attenuated (has an insertion loss)
C
of approximately 3 dB at 1 MHz, while at 2 MHz the signal is down to about
36 dB, and at 4 MHz it has dropped to 72 dB.
When multiple half-sections are combined with low-pass filter designs, the
filter’s attenuation response becomes that of the amount of the L and C com-
ponents (poles) that result from the combination and joining of these half-sec-
tions. For instance, if a filter is created from three half-sections—which
contain six reactive components (poles)—the filter would now contain only
four reactive components (poles) after the appropriate components had been
combined. The components left after the combination of the half-sections will
be the indicator of the filter’s attenuation response.

High-pass filters. Shown in Fig. 6.24 are a 2-pole high-pass filter that incor-
porates a series capacitor and a shunt inductor, and another that has a shunt
inductor and a series capacitor. Both are half-sections. This is exactly the
opposite component layout of that of the low-pass filter style above.
For the high-pass filter, as it was with the low-pass filter design, cascading
these half-sections will produce a high-pass filter with as many poles as required.
To design a Butterworth high-pass filter, first calculate the L and C values
of the single half-section of Fig. 6.24 by:

R 0 1
L 2 and C 2
4 f
C
C 4 f R 0
where R impedance at the filter’s input and output and f filter’s 3-dB
O C
cutoff frequency.
Take the half-section as calculated, and combine by dividing by two the com-
bined inductors in Fig. 6.25, or by dividing the combined capacitors by two in
Fig. 6.26. Unlike low-pass filter design, in which the combined components are
added, a high-pass filter’s combined components must be divided by 2. As for
the similar situation of the low-pass filter above, the high-pass sections must
not be combined as in Fig. 6.27.

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