ACTIVE_FILTERS_Theory_and_Design

Sallen–Key Filters 33
K
Hs ()=
2
⎛ s s
a
⎝ ⎜ ⎞
⎟ + ⎛
⎠ ⎝ ⎜ ⎞
⎟ + 1
ω ω ⎠
1
1
(2.30)
s
1. For << 1,
we have:
ω 1
H( jω)
≅ K
The slope is 0 dB/dec and
A= 20 log H( jω) = 20 log K dB
s
2. For >> 1,
we have:
ω 1
H( jω)
K
≅
⎛ ω
⎝ ⎜
⎞
⎟
ω ⎠
1
2
⎛ ω ⎞
= K ⎜ ⎟
⎝ ω ⎠
1
−2
∴
ω
For = 10
ω 1
∴ slope =−40
ω
For = 2
ω 1
∴ slope =−12
3. For
⎛ ω ⎞
⎛ ω ⎞
A= 20 log H( jω) = 20 log K⎜
⎟ = 20 log K−40
log⎜
⎟ dB
⎝ ω ⎠
⎝ ⎠
s
= 1 ∴
ω 1
1
−2
dB/
dec
dB/
oct
A= 20 log K−40
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
ω 1
Figure 2.10 shows the frequency response of the filter.