Analog Circuits and Systems
Prof. K Radhakrishna Rao
Lecture 23: Passive Filters
1
Review
—
—
—
—
—
—
First order and second order low-pass filters
Butterworth (MFM) Filters
Chebyshev and inverse Chebyshev Filters
Elliptic Filters
Bessel’s Filters (Maximally Flat Delay)
Passive Filters (RC, RL and RLC)
2
Review (contd.,)
3
First Order High Pass Filters
4
First Order High Pass Filters (contd.,)
Vo
sCR
=
Vi 1 + sCR
Vo
Vi
2
Vo
=
Vi
*
sL
R
L
1+s
R
( ωCR )
2
⎛ Vo ⎞ ⎛ Vo ⎞
X
= ⎜ ⎟×⎜ ⎟ =
=
2
2
1 + ( ωCR ) 1 + X
⎝ Vi ⎠ ⎝ Vi ⎠
2
ω
1
where X =
and ω0 =
is the Bandwidth(BW)
ω0
CR
5
Magnitude and delay response
of the high pass filter (maximally
flat functions of first order)
Rate of attenuation of magnitude
at the edge of pass band
(X = 1) is 0.25
6
Second order RC low pass filter
For maximizing Q
R 2 ? R1
Qmax occurs when
C1R1 = C2R 2 and
Vo
1
=
Vi ⎡1 + s (C1R1 + C2 (R1 + R 2 ) ) + s2C1R1C2R 2 ⎤
⎣
⎦
C1R1
C2R 2
1
ω0 @
; =
+
C2R 2
C1R1
R1C1R 2C2 Q
1
⎡
R1 ⎤
⎢1 +
⎥
R2 ⎦
⎣
1
Qmax = . As Q is a
2
parameter that determines
the characteristics of the
filter a maximum value
of 0.5 leaves very little
scope for general design
of low-pass filters
7
Second order low pass RLC filter
Vo
Vi
and the quality factor Q
1
Q=
=
ω0CR
2
=
1
⎛ 1
⎞ 2
4
1 + ⎜ 2 - 2⎟ X + X
⎝Q
⎠
ω
1
where X =
where ω0 =
ω0
LC
L 1
CR
ω
X=
= ω LC where ω0 = 2πf0
ω0
8
Example: Second Order Low-pass filter
Bandwidth = 40 Hz; ω0 = 80π
If C = 1µF; L = 15.8H (!)
For maximally flat response Q=
1
2
1 L
This determines R =
= 1000 31.2 = 5.62kΩ
Q C
While the C and R values are reasonable the inductor with 15.6 H will
become too big to be accommodated
9
Frequency Transformation
—
—
Design of high pass, band pass and band stop filters
can be done starting from the corresponding low
pass prototype.
Low-pass filter function is expressed as a function of
X2
10
Low-pass to High-pass transformation
Low-pass
High-pass
2
Vo
1
ω 1
=
where
X
=
=
ω
0
2
Vi
1+ X
ω0 CR
11
Low-pass to High-pass transformation (contd.,)
X = 1 when ω0 = BW Bandwidth of the filter
Magnitude function of the high-pass filter
1
Replace X by
X
2
Vo
X
=
2
Vi
1+ X
2
R 1
L = CR and
=
= BW
L CR
2
12
LP-HP Transformation
Low-pass
High-pass
2
Vo
1
ω R
=
where X =
; = ω0
2
Vi
1+ X
ω0 L
13
LP-HP Transformation (contd.,)
X = 1 when ω0 = BW Bandwidth of the filter
1
Replace X by
X
2
Vo
X
=
2
Vi
1+X
2
L
R
1
C = 2 and
=
= BW
R
L CR
14
Low-pass to High-pass transformation
If X2 is replaced by 1/X2 then lowpass filter gets transformed to a
high-pass filter of the same band
width. The range of X2 from 0 to
∞ gets transformed to the range
∞ to 0 for 1/X2.
Square of magnitude vs frequency (First-order filter)
15
Example
—
—
—
Design a first-order high-pass filter with f0 = 40Hz; w0 = 80p
Let us design first-order a low-pass filter prototype with w0 = 80p
wRC = w/80p, If C = 1 mF, R = 5.7 kW, L=CR2= 15.7 H
16
Low-pass to Band-pass Filter
Vo
Vi
2
1
ω 1
=
where X =
;
= BW
2
ω0 CR
1+ X
2
K
Replace X by (X −
)
X
where the center frequency of
band-pass filter is K(BW)
2
X
2
4
Vo
K
=
2
4
Vi
⎛
X ⎞
⎧1
⎫X
⎜1 + ⎨ 2 − 2⎬ 2 + 4 ⎟
K ⎠
⎩K
⎭K
⎝
17
Low-pass to Band-pass Filter (contd.,)
—
If C replaced by a parallel
resonant circuit with
resonance frequency K(BW)
Band-width of the band-pass filter
remains the same as that of
low-pass filter
1
BW =
RC
K(BW) =
1
LC
and hence
1
L= 2
2
K (BW) C
18
Low-pass to Band-pass Transformation
Square of magnitude vs frequency
19
Example
Design a second-order band-pass filter with center
frequency = 5 kHz and a band-width of 1 kHz.
—  Start with the design of first-order a low-pass filter
prototype for a bandwidth = bandwidth of the bandpass filter (1kHz)
—  For a C of 1 mF R = 159 W
—  L for a resonance of 5 kHz = 0.987 mH
20
Example: Low-pass to Band-pass Transformation
21
Low-pass to Band-stop Filter
Vo
Vi
2
1
ω
1
=
where X =
;
= BW
2
ω0 CR
1+X
−1
⎛
K ⎞
Replace X by ⎜ X −
⎟
X ⎠
⎝
where the center frequency of
2
band-stop filter is K(BW)
2
Vo
Vi
2
4
X
X
1−2 2 + 4
K
K
=
2
4
⎛
X ⎞
⎧1
⎫X
⎜1 + ⎨ 2 − 2⎬ 2 + 4 ⎟
K ⎠
⎩K
⎭K
⎝
22
Low-pass to Band-stop Filter (contd.,)
—
Band-width of the band-stop filter remains
If C replaced by a series
the same as that of low-pass filter
resonant circuit with
resonance frequency K(BW)
1
BW =
RC
1
2
L ′ = CR =
2
(BW) C
1
As K(BW) =
L ′C′
1
C
C′ = 2
= 2
2
K (BW) L ′ K
23
Example
—
—
—
—
Design a 4th-order Chebyschev band-pass filter centred around
10 k Radians/sec with a bandwidth of 1000 Radians/sec.
The proposed BP filter can be designed from second-order
Chebyschev LP filter prototype with a bandwidth of 1000 Radians/
sec. It will have a Q of 1(> 1/ 2).
The second-order Chebyschev LP filter will be a RLC filter. If we
assume a values of 1 mF for C, from the relation: bandwidth = 1/RC,
the value of resistance is 1 kW
L = CR2=1 H
24
Frequency response of the LPF
25
Frequency response of the LPF (contd.,)
—
—
—
The corresponding BPF
L is replaced by L forming a series resonant circuit at 10 k Radians/
sec with a capacitor C/= 1/w02L = 10 nF for L = 1 H.
C is replaced by C forming a parallel resonance at the same
frequency 10 k Radians/sec. C of 1 mF results in an inductance L/
=1/w02C = 10 mH.
26
The frequency response
27
Example
—
—
—
—
Design a 4th-order Chebyschev band-stop filter centred around 10 k
Radians/sec with a bandwidth of 1000 Radians/sec.
The proposed BS filter can be designed from second-order
Chebyschev LP filter prototype with a bandwidth of 1000 Radians/
sec. It will have a Q of 1(> 1/ 2).
The second-order Chebyschev LP filter will be a RLC filter. If we
assume a values of 1 mF for C, from the relation: bandwidth = 1/RC,
the value of resistance is 1 kW
L = CR2=1 H
28
Frequency response of the LPF
29
Frequency response of the LPF (contd.,)
—
—
L is replaced by L forming a parallel resonant circuit at 10 k radians/
sec with a capacitor C/= 1mF for L/ = 10 mH.
C is replaced by C forming a series resonance at the same
frequency 10 k Radians/sec. C// of 10 nF results in an inductance L//
=1/w02C = 1H.
30
Frequency response of the LPF (contd.,)
31
Example
—
Let us consider using inverse Chebyshev filter as given in the figure
Vo
=
Vi
1 − N1Ω
2
⎛
1 ⎞
4
1 + Ω ⎜ −2 + 2 ⎟ + Ω
Q ⎠
⎝
ω
1
where Ω =
where ωp =
ωp
(L1 + L2 ) C
2
and the quality factor
1
Q=
=
ω0CR
(L1 + L2 ) 1
C
R
32
Example (contd.,)
ω
Ω=
= ω (L1 + L 2 ) C
ωp
If C = 1µF
where ωp = 2πf with f = 40Hz
For maximally flat response
2
⎛ ωp ⎞
N1 = ⎜ ⎟ where ωz is chosen as
⎝ ωz ⎠
2πfz where fz = 50Hz (interferance)
1
2N1 = 2 − 2
Q
(L1 + L2 ) = 15.6H
with a zero
Q = 1.18
which determines
1 L1 + L 2
R=
= 3.34kΩ
Q
C
33
Responses of the four low pass filters
—
While the behavior of second order RLC filters is better, in
view of the large inductance values they are impractical at
these lower frequencies. Active RC filters, having active
elements like transistors, Op Amps and transconductors and
RC elements, can overcome the limitations of passive filters
34
Conclusion
35