ACTIVE_FILTERS_Theory_and_Design

Sallen–Key Filters 27
ω
For = 10 ∴ slope = 20 dB/ dec.
ω 2
ω
For = 2 ∴ slope = 6 dB/ oct.
∴
ω 2
A= 20 log K+
20 dB
s
2. For >> 1 ∴ H( jω)
= K \
ω
2
The slope is 0 dB/
dec.
s
3. For = 1 ∴
ω 2
A= 20 log H( jω) = 20 log K dB
K
K
H( jω) = ∴ H( jω)
= ∴
1 + j
2
K
A = 20 log = 20 log K− 20 log 2 = 20 log K−3
dB
2
Figure 2.5 shows the frequency response of the filter.
EXAMPLE 2.2
A first-order HP Butterworth filter must be designed with gain of 5 at a cutoff
frequency of 100 Hz.
Solution
From the Butterworth coefficients of Appendix C, we have (n = 1):
b = 1, hence:
A
dB
20 log K
20 log K − 3
0
ideal
20 dB
0.1 f 2 f 2
10 f 2
20 dB/dec or 6 dB/oct
a decade
log f (Hz)
FIGURE 2.5 Frequency response of first-order HPF.