Implementing IIR/FIR Filters

V
0
X
L
||R
----- = ------------------------
V
i X
C + X
L
||R
j ΩLR/ ( jΩL + R)
= -------------------------------------------------------
1/jΩC + jΩLR/jΩL
+ R
Ω
=
2
–
Ω 2
------------------------------------------------
– + j Ω /RC + 1/LC
Ω
=
2
–
Ω 2
------------------------------------------
– + jdΩ
c
Ω+ Ω2 c
Let s = jΩ and define H(s) = V o /V i ; then,
V i
MOTOROLA 1-11
C
L R
V 0
X c = 1/jΩC
X L = jΩL
1
Ωc = -----------
LC
L
d =
R 2 -----------
C
( s/Ω
c)
H( s)
2
( s/Ω ) c
2
= ---------------------------------------------------
+ d( s/Ω) + 1
c
which is the s-domain transfer function. The gain, G(Ω), of the filter is:
G( Ω)
≡ H( s)H∗(
s)
s = jω
( Ω/Ω
c)
2
1 Ω 2 2
( – /Ω ) c
2
( dΩ/Ω
c)
2
= ---------------------------------------------------------------
+
where ∗ denotes complex conjugate.
The phase angle, φΩ, is the angle between the imaginary and
φ( Ω)
tan 1 –
≡ [ l{ H( s)
}/R{ H( s)
} ]
– 1 d( Ω/Ωc )
= π– tan
1 ( Ω/Ωc) 2
real components of H(s).
----------------------------for
Ω ≤ Ωc –
– 1 d( Ω/Ωc )
= – tan
1 ( Ω/Ωc) 2
------------------------------
–
Figure 1-3 s-Domain Analysis of Second-Order Highpass Analog Filter
for Ω > Ω c