Post on 09-Mar-2021
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Chapter 14 Analog Filters
14.1 General Considerations14.2 First-Order Filters14.3 Second-Order Filters14.4 Active Filters14.5 Approximation of Filter Response
1
Outline of the Chapter
2CH 14 Analog Filters
Why We Need Filters
In order to eliminate the unwanted interference that accompanies a signal, a filter is needed.
3CH 14 Analog Filters
Filter Characteristics
Ideally, a filter needs to have a flat pass band and a sharp roll-off in its transition band.Realistically, it has a rippling pass/stop band and a transitionband.
4CH 14 Analog Filters
attenuated amplified
sufficientlynarrow
bad stopbandattenuation
Example 14.1: Filter I
5CH 14 Analog Filters
Solution: A filter with stopband attenuation of 40 dB
Problem : Adjacent channel interference is 25 dB above the signal. Determine the required stopband attenuation if Signal to Interference ratio must exceed 15 dB.
Example 14.2: Filter II
6CH 14 Analog Filters
Problem: Adjacent 60-Hz channel interference is 40 dB above the signal. Determine the required stopband attenuation to ensure that the signal level remains 20dB above the interferer level.
Solution: A high-pass filter with stopband attenuation of 60 dB at 60Hz.
Example 14.3: Filter III
A bandpass filter around 1.5 GHz is required to reject the adjacent Cellular and PCS signals.
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Classification of Filters I
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Classification of Filters II
Continuous-time Discrete-time
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1 2 1Q C V=
2 2 2Q C V=
1 2 2 1
2
if , absorbs charge from and delivers it to a resistor
V V C VV
>⇒ ≈
Classification of Filters III
Passive Active
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Summary of Filter Classifications
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Filter Transfer Function
Filter (a) has a transfer function with -20dB/dec roll-off.Filter (b) has a transfer function with -40dB/dec roll-off and provides a higher selectivity.
(a) (b)
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General Transfer Function
pk = pole frequencies
zk = zero frequencies
( )( ) ( )( )( ) ( )
1 2
1 2( ) m
n
s z s z s zH s
s p s p s pα
− − −=
− − −
13CH 14 Analog Filters
jσ ω= +
Example 14.4 : Pole-Zero Diagram
1 1
1( )1aH s
R C s=
+1 2
1 1 2
1( )( ) 1bR C sH s
R C C s+
=+ +
12
1 1 1 1 1
( )cC sH s
R L C s L s R=
+ +
14CH 14 Analog Filters
Impulse response contains
Example 14.5: Position of the poles
Poles on the RHPUnstable (no good)
Poles on the jω axisOscillatory(no good)
Poles on the LHPDecaying
(good)
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exp( ) exp( )exp( )k k kp t t j tσ ω=
Transfer Function
The order of the numerator m ≤ The order of the denominator nOtherwise, H(s)→∞ as s→∞.For a physically-realizable transfer function, complex zeros or poles occur in conjugate pairs.If a zero is located on the jω axis, z1,2=± jω1 , H(s) drops to zero at ω1.
16CH 14 Analog Filters
( )( ) ( )( )( ) ( )
1 2
1 2( ) m
n
s z s z s zH s
s p s p s pα
− − −=
− − −
1 1 1z jσ ω= + 2 1 1z jσ ω= −
( )( )The numerator contains a product such as which vanishes at
2 21 1 1
1
s j s j s ,s j
ω ω ωω
− + = +=
Imaginary Zeros
Imaginary zero is used to create a null at certain frequency.For this reason, imaginary zeros are placed only in the stop band.
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Sensitivity
PC
dP dCSP C
=
Sensitivity indicates the variation of a filter parameter due tovariation of a component value.
P=Filter Parameter
C=Component Value
18CH 14 Analog Filters
In simple RC filter, the -3dB corner frequency is given by 1/(R1C1)
Example:
Errors in the cut-off frequency: (a) the value of components varies with process and temperature
in ICs(b) The available values of components deviate from those
required by the design
Example 14.6: Sensitivity
( )
1
1/1
0
1
1
1
0
0
1211
0
110
−=
−=
−=
=
ω
ωω
ωω
RS
RdRdCRdR
dCR
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Problem: Determine the sensitivity of ω0 with respect to R1.
For example, a +5% change in R1 translates to a -5% error in ω0.
First-Order Filters
1
1( ) s zH s
s pα +
=+
First-order filters are represented by the transfer function shown above. Low/high pass filters can be realized by changing the relative positions of poles and zeros.
20CH 14 Analog Filters
Example 14.8: First-Order Filter I
R2C2 < R1C1 R2C2 > R1C1
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2 1 1
1 2 1 2 1 2
( 1)( )( )
out
in
V R R C ssV R R C C s R R
+=
+ + +
1 1 11 ( ),z R C= − / 11 1 2 1 2[( ) ]p C C R R −= − + ||
.
2
1 2
R0R R
ω → ⇒+
1
1 2
CC C
ω →∞⇒+
( )( )1 1 1 2 1 2
1 1R C C C R R
<+
2 1
1 2
C R1 1C R
+ < +
( )( )1 1 1 2 1 2
1 1R C C C R R
>+
2 1
1 2
C R1 1C R
+ > +
Example 14.9: First-Order Filter II
R2C2 < R1C1 R2C2 > R1C1
22
11
2 1 1
1 2 2
1( )( ) 1
11
out
in
RV C ssV R
C sR R C sR R C s
− ||=
||
+= − ⋅
+
22CH 14 Analog Filters
inI
virtual short
inI
2
1
R0R
ω → ⇒ − 1
2
CC
ω →∞⇒ −
2
2 2( )
nn
s sH ss s
Q
α β γω ω
+ +=
+ +
1,2 211
2 4n
np jQ Qω
ω= − ± −
Second-Order Filters
Second-order filters are characterized by the “biquadratic” equation with two complex poles shown above. When Q increases, the real part decreases while the imaginary part approaches ±ωn.=> the poles look very imaginary thereby bringing the circuit closer to instability.
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General transfer function
Second-Order Low-Pass Filter
( )
22
222 2
( )n
n
H j
Q
γωωω ω ω
=⎛ ⎞
− + ⎜ ⎟⎝ ⎠
α = β = 0
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Peak magnitude normalized to the passband magnitude: 2 11 (4 )Q Q −/ −
.
for 2Q2
>
2p Q2
H( )ω>
2p Q2
2n
H( )ω
γω
>
2( )0
H jωω
∂=
∂
Example 14.10: Second-Order LPF
2
2
3
/ 1 1/(4 ) 3
1 1/(2 )n n
Q
Q Q
Qω ω
=
− ≈
− ≈
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Problem: Q of a second-order LPF = 3.Estimate the magnitude and frequency of the peak in the frequency
response.
( )
22
2222 2
( )
3n
n n
H j γωωω ω
=⎛ ⎞
− + ⎜ ⎟⎜ ⎟⎝ ⎠
Second-Order High-Pass Filter
2
2 2( )
nn
sH ss s
Q
αω ω
=+ +
21 1 (2 )n Qω / − /Frequency of the peak:
Peak magnitude normalized to the passband magnitude: 2 11 (4 )Q Q −/ −
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The zero(s) must fall below the poles
for 2Q2
>
0β γ= =
Second-Order Band-Pass Filter
2 2( )
nn
sH ss s
Q
βω ω
=+ +
27CH 14 Analog Filters
0α γ= =
The magnitude approaches zero for both s 0 and s ∞, reaching a maximum in between
Example 14.2: -3-dB Bandwidth
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Problem: Determine the -3dB bandwidth of a band-pass response.
2 2( )
( )nn
sH ss s
Q
βω ω
=+ +
1,2 0 21 11
24 QQω ω
⎡ ⎤= + ±⎢ ⎥
⎢ ⎥⎣ ⎦
2 2 2 2
22 2 2 2 2( ) ( )n nn
Q
Q
β ω βω ωω ω ω
=− +
0BWQω
=
The response reaches times its peak value at -3dB frequency
1 / 2
LC Realization of Second-Order Filters
11 1 2
1 1 1
1( ) ||1
L sZ L sC s L C s
= =+
An LC tank realizes two imaginary poles at ±j/(L1C1)1/2 , which implies infinite impedance at ω=1/(L1C1)1/2.
29CH 14 Analog Filters
at and at
1
1 0
Z 0 0Z
ωω ω
= = ∞= ∞ =
0ω
Example 14.13: LC Tank
At ω=0, the inductor acts as a short.At ω=∞, the capacitor acts as a short.
30CH 14 Analog Filters
11 1 2
1 1 1
1( ) ||1
L sZ L sC s L C s
= =+
at and 1Z 0 0ω= = ∞
RLC Realization of Second-Order Filters
1 1 12 1 2 2
1 1 1 1 1 1 1
1 1 1 1
2 2 21 1 1 1 1 1
1 1 1 1
||1
1 1( ) ( )nn
L s R L sZ RLC s R LC s L s R
R L s R L s
R LC s s R LC s sRC LC Q
ω ω
= =+ + +
= =+ + + +
1,2 2
12
1 1 1 11 1
112 4
1 1 12 4
nnp j
Q Q
LjR C R CL C
ω ω= − ± −
= − ± −
With a resistor, the poles are no longer pure imaginary which implies there will be no infinite impedance at any ω.
31CH 14 Analog Filters
11
11 1
1 ,nCQ RLLC
ω = =
Voltage Divider Using General Impedances
( )out P
in S P
V ZsV Z Z
=+
Low-pass High-pass Band-pass32CH 14 Analog Filters
Low-pass Filter Implementation with Voltage Divider
( ) 12
1 1 1 1 1
out
in
V RsV R C L s L s R
=+ +
33CH 14 Analog Filters
1 as SZ L s s= →∞ →∞
11
1 0 as PZ R sC s
= || → →∞
Example 14.14: Frequency Peaking
( ) 12
1 1 1 1 1
ou t
in
V RsV R C L s L s R
=+ +
( )22 2 2 21 1 1 1 1Let D R R C L Lω ω= − +
34CH 14 Analog Filters
Voltage gain greater than unity (peaking) occurs when a solution exists for
22 2
1 1 1 1 1 1 1 12 2( )( )( )
0
d DR C L R R C L L
dω
ω= − − +
=
11
1
1Thus, when ,2
peaking occurs.
CQ RL
= ⋅ >
2 21 1 1 1
1 1 0L C 2R C
ω⇒ = − >
2 11
1
C2R 1L
> 11
1
CQ RL
=∵
Example 14.15: Low-pass Circuit Comparison
The circuit (a) has a -40dB/dec roll-off at high frequency.However, the circuit (b) exhibits only a -20dB/dec roll-off since the parallel combination of L1 and R1 is dominated by R1
because L1ω→∞, thereby reduces the circuit to R1 and C1.
Good Bad
35CH 14 Analog Filters
(a) (b)
High-pass Filter Implementation with Voltage Divider
( )2
1 1 1 1 12
1 1 1 1 11 11
( ) ||1( ) ||
o u t
in
V L s R L C R ssV R C L s L s RL s R
C s
= =+ ++
36CH 14 Analog Filters
2
2 2( ) for high pass filter
nn
sH ss s
Q
αω ω
=+ +
∵
Band-pass Filter Implementation with Voltage Divider
( )1
1 12
1 1 1 1 11 11
1( ) ||
1( ) ||
out
in
L sV C s L ssV R C L s L s RL s R
C s
= =+ ++
37CH 14 Analog Filters
Zp must contain both a capacitor and an inductor so that it approaches zero as s 0 or s ∞
pZ
2 2( )
nn
sH ss s
Q
βω ω
=+ +
∵
Summary
Analog filters prove essential in removing unwanted frequency components that may accompany a desired signal.The frequency response of a filter consists of a passband, stopband, and a transition band between the two. The passband and stopband may exhibit some ripple.Filters can be classified as LP, HP, BP, or BR topologies. They can be realized as continuous-time or discrete-time configurations, and as passive or active circuits.The frequency response of filters has dependences on various component values and, therefore, suffers from sensitivity to component variations.First-order passive or active filters can readily provide a LP or HP response, but their transition band is quite wide and stopband attenuation only moderate.Second-order filters have a greater stopband attenuation and are widely used. For a well-behaved frequency and time response, the Q of these filters is typically maintained below 2 / 2
Why Active Filter?
Passive filters constrain the type of transfer function.
They may require bulky inductors.
39CH 14 Analog Filters
Sallen and Key (SK) Filter: Low-Pass
( )( )2
1 2 1 2 1 2 2
11
out
in
V sV R R C C s R R C s
=+ + +
11 2
1 2 2
1 CQ R RR R C
=+ 1 2 1 2
1n R R C C
ω =
Sallen and Key filters are examples of active filters. This particular filter implements a low-pass, second-order transfer function.
40CH 14 Analog Filters
out XV V≈
out 2V C sY out 2 2 outV V C sR V= +
By applying KCL
2
2 2( )
nn
s sH ss s
Q
α β γω ω
+ +=
+ +
Example 14.16: SK Filter with Voltage Gain
( )3
4
2 31 2 1 2 1 2 2 2 1 1
4
1
1
out
in
RV RsV RR R C C s R C R C R C s
R
+=
⎛ ⎞+ + − +⎜ ⎟⎝ ⎠
41CH 14 Analog Filters
( )out 3 4 XV 1 R R V= +
Example 14.17: SK Filter Poles
The poles begin with real, equal values for andbecome complex for .
3
4
2 31 2 1 2 1 2 2 2 1 1
4
1( )
( ) 1
out
in
RV Rs RV R R C C s R C R C R C s
R
+=
+ + − +
31 2 2 2 1 1
2 1 1 1 2 2 4
1 RR C R C R CQ R C R C R C R= + −
3
4
1
2Q R
R
=−
3 4 0R R/ =3 4 0R R/ >
Problem: Assuming R1=R2, C1=C2, Does such a filter contain complex poles?
42CH 14 Analog Filters
Q ↑
Sensitivity in Low-Pass SK Filter
1 2 1 2
12
n n n nR R C CS S S Sω ω ω ω= = = = −
1 2
2 2
1 1
12
Q QR R
R CS S QR C
= − = − +
1 2
2 2 1 2
1 1 2 1
12
Q QC C
R C R CS S QR C R C
⎛ ⎞= − = − + +⎜ ⎟⎜ ⎟
⎝ ⎠
1 1
2 2
QK
R CS QKR C
=
43CH 14 Analog Filters
3 41K R R= +
the gain of filterout
X
VV
=
1 2 1 2
1n R R C C
ω =∵ n
1
ddRω
=31 2 2 2 1 1
2 1 1 1 2 2 4
1 RR C R C R CQ R C R C R C R= + −∵
Example 14.18: SK Filter Sensitivity I
1 2
1 2
1 12 31 22 3
3
Q QR R
Q QC C
QK
S SK
S SK
KSK
= − = − +−
= − = − +−
=−
44CH 14 Analog Filters
Problem: Determine the Q sensitivities of the SK filter for the common choice R1=R2=R, C1=C2=C.
1 2
1 2
With =1,
0
12
Q QR R
Q Q QKC C
K
S S
S S S
= =
= = =
Integrator-Based Biquads
( )2
2 2
out
ninn
V ssV s s
Q
αω ω
=+ + ( ) ( ) ( ) ( )
2
21.n n
out in out outV s V s V s V sQ s sω ωα= − −
It is possible to use integrators to implement biquadratictransfer functions.
45CH 14 Analog Filters
( )2
2n
outV ssω
− ( )1.noutV s
Q sω
−
KHN (Kerwin, Huelsman, and Newcomb) Biquads
( ) ( ) ( ) ( )2
21Comparing with .n n
out in out outV s V s V s V sQ s sω ωα= − −
5 6
4 5 31R R
R R Rα
⎛ ⎞= +⎜ ⎟+ ⎝ ⎠
64
4 5 1 1 3
1. . 1n RRQ R R R C Rω ⎛ ⎞
= +⎜ ⎟+ ⎝ ⎠
2 6
3 1 2 1 2
1.nRR R R C C
ω =
46CH 14 Analog Filters
21 1 2 2 1 2 1 2
1 1 1, X out Y X outV V V V VR C s R C s R R C C s
= − = − = 5 4 6 6
4 5 3 3
1in Xout Y
V R V R R RV VR R R R
⎛ ⎞+= + −⎜ ⎟+ ⎝ ⎠
Simplifieddiagram
Calculation of Vout with Simplified Circuit
AV
To obtain AV ,
in
4A XV 0
4 5
RV VR R=
=+ X
5A inV 0
4 5
RV VR R=
=+
in X
4 X 5 inA A AV 0 V 0
4 5
R V R VV V VR R= =
+= + =
+
To obtain for given we ground out A YV V , V
( ) 6Aout 3 6 A
3 3
RVV R R V 1R R
⎛ ⎞= + = +⎜ ⎟
⎝ ⎠
To obtain for given we ground out Y AV V , V
6Yout 6 Y
3 3
RVV R VR R
⎛ ⎞= − = ⎜ ⎟
⎝ ⎠
Versatility of KHN Biquads
( )2
2 2High-pass: out
ninn
V ssV s s
Q
αω ω
=+ +
( )2
2 2 1 1
1Band-pass: .X
ninn
V ssV R C ss s
Q
αω ω
−=
+ +
( )2
22 2 1 2 1 2
1Low-pass: .Y
ninn
V ssV R R C C ss s
Q
αω ω
=+ +
48CH 14 Analog Filters
outX
1 1
V 1VR sC
= −∵
XY
2 2
V 1VR sC
= −∵
Sensitivity in KHN Biquads
1 2 1 2 4 5 3 6, , , , , , , 0.5nR R C C R R R RSω =
1 2 1 2 4 5
3 6
5, , , ,
4 5
3 6 2 2,
5 3 6 1 1
4
0.5, 1,
2 1
Q QR R C C R R
QR R
RS SR R
R R R CQS R R R R CR
= = <+
−=
+
49CH 14 Analog Filters
3 6
3 6
,
if ,
then vanishesQR R
R R
S
=
Tow-Thomas Biquad
32 2 4 1 1
1 1out inout
V V R VR C s R R sC
⎛ ⎞⎛ ⎞⋅ + = −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
50CH 14 Analog Filters
X YV V= −
2 2( )Y outV V R C s= −1
inVR
2 2 4
1outVR C s R
⋅
Tow-Thomas Biquad
2 3 4 22
1 2 3 4 1 2 2 4 2 3Band-pass: .out
in
V R R R C sV R R R R C C s R R C s R
= −+ +
51CH 14 Analog Filters
3 42
1 2 3 4 1 2 2 4 2 3
1Low-pass: .Y
in
R RVV R R R R C C s R R C s R
=+ +
2 2( )
nn
sH ss s
Q
βω ω
=+ +
∵
2 2( )
nn
H ss s
Q
γω ω
=+ +
∵
BFP
LFP
Differential Tow-Thomas Biquads
An important advantage of this topology over the KHN biquadis accrued in integrated circuit design, where differential integrators obviate the need for the inverting stage in the loop.
52CH 14 Analog Filters
Example 14.20: Tow-Thomas Biquad
2 4 1 2
1n R R C C
ω =
Adjusted by R2 or R4
1 2 4 2
3 1
1 R R CQR C
− =
Adjusted by R3
Note that ωn and Q of the Tow-Thomas filter can be adjusted (tuned) independently.
53CH 14 Analog Filters
3 42
1 2 3 4 1 2 2 4 2 3
1Low-pass: .Y
in
R RVV R R R R C C s R R C s R
=+ +
( )H sα
=2s β+
2 2nn
s
s sQ
γω ω
+
+ +
Antoniou General Impedance Converter
1 35
2 4in
Z ZZ ZZ Z
=
It is possible to simulate the behavior of an inductor by using active circuits in feedback with properly chosen passive elements.
54CH 14 Analog Filters
1 3 5 XV V V V= = =
4 45
XX
VV Z VZ
= +
4 33
3Z
V VIZ−
= 4
5 3
XV ZZ Z
= ⋅
2 3 2 3ZV V Z I= −
42
5 3
XX
V ZV ZZ Z
= − ⋅ ⋅
2
1
XX
V VIZ−
=
2 4
1 3 5X
Z ZVZ Z Z
=
5
XVZ
3ZI
3ZIXI
Simulated Inductor
By proper choices of Z1-Z4, Zin has become an impedance that increases with frequency, simulating inductive effect.
Thus, in X Y
eq X Y
Z R R CsL R R C
=
=
55CH 14 Analog Filters
1 35
2 4in
Z ZZ ZZ Z
=∵
1Z 2Z
3Z 4Z 5Z
High-Pass Filter with SI
With the inductor simulated at the output, the transfer functionresembles a second-order high-pass filter.
( )2
12
1 1 1 1 1
out
in
V L ssV R C L s L s R
=+ +
56CH 14 Analog Filters
Example 14.22: High-Pass Filter with SI
4 1 Yout
X
RV VR
⎛ ⎞= +⎜ ⎟
⎝ ⎠
Node 4 can also serve as an output.
57CH 14 Analog Filters
V4 is better than Vout since the output impedance is lower.
1 3 5outV V V V= = =
5 4 4 51X Y
Y X X
R RV V V VR R R
⎛ ⎞= ⇒ = + ⇒⎜ ⎟+ ⎝ ⎠
Low-Pass Filter with Super Capacitor
( )1
1inX
ZCs R Cs
=+
1
2 21 1
11
out in
in in
X
V ZV Z R
R R C s R Cs
=+
=+ +
58CH 14 Analog Filters
How to build a floating inductor to derive a low-pass filter?Not possible. So use a super capacitor.
2 2( )
nn
H ss s
Q
γω ω
=+ +
1Z 2Z
3Z4Z 5Z
Example 14.23: Poor Low-Pass Filter
( )41 2out X out out XX
V V Cs R V V R CsR
⎡ ⎤⎛ ⎞= + + = +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
Node 4 is no longer a scaled version of the Vout. Therefore the output can only be sensed at node 1, suffering from a high impedance.
59CH 14 Analog Filters
1out
XV Cs
R⎛ ⎞
+⎜ ⎟⎝ ⎠
1 3 5outV V V V= = =
Summary
Continuous-time passive second-order filters employ RSC sections, but they become impractical at very low frequencies (because of large physical size of inductors and capacitors).Active filters employ op amps, resistors, and capacitors to create the desired frequency response. The Sallen and Key topology is an example.Second-order active (biquad) sections can be based on integrators. Examples include the KHN biquad and the Tow-Thomas biquad.Biquads can also incorporate simulated inductors, which are derived from the “general impedance converter” (GIC). The GIC can yield large inductor or capacitor values through the use of two op amps.
Frequency Response Template
With all the specifications on pass/stop band ripples and transition band slope, one can create a filter template that will lend itself to transfer function approximation.
61CH 14 Analog Filters
Butterworth Response
2
0
1( )
1n
H jωωω
=⎛ ⎞
+ ⎜ ⎟⎝ ⎠
ω-3dB=ω0, for all n.
The Butterworth response completely avoids ripples in the pass/stop bands at the expense of the transition band slope.
62CH 14 Analog Filters
To obtain the poles, we make a reverse substitution, / ,and set the denominarot to zero:
s jω =
Butterworth Response
2
01 0
ns
jω⎛ ⎞
+ =⎜ ⎟⎝ ⎠
( ) ( ) ( )( ) ( )
2 222 20 0
220
0 0
0
n nnn n
nnn
s j s j
s j
ω ω
ω
+ = ⇒ + =
+ − =
This polynomial has 2n roots given by
02 1exp exp , 1,2,..., 2
2 2kj kp j k n
nπω −⎛ ⎞= =⎜ ⎟
⎝ ⎠
For 2nd order filter,
1,2,3,4 0 0 0 03 5 7 9exp , exp , exp , exp4 4 4 4k
j j j jp π π π πω ω ω ω= =
exp4jπ− exp
4jπ
But only the roots having a negative real part are acceptable
negative real positive real
63CH 14 Analog Filters
02 1exp exp , 1, 2, ,
2 2kj kp j k n
nπω π−⎛ ⎞= =⎜ ⎟
⎝ ⎠
Poles of the Butterworth Response
64CH 14 Analog Filters
2nd‐Order nth‐Order
1 2
1 2
( )( ) ( )( )( )( ) ( )
n
n
p p pH ss p s p s p− − ⋅⋅⋅ −
=− − ⋅⋅⋅ −
where the factor in the numerator is included to yield H(s=0)=1
Example 14.24: Order of Butterworth Filter
The minimum order of the Butterworth filter is three.
22
164.2
nff
⎛ ⎞=⎜ ⎟
⎝ ⎠2 12f f=
65CH 14 Analog Filters
Specification: passband flatness of 0.45 dB for f < f1=1 MHz, stopbandattenuation of 9 dB at f2=2 MHz.
1( 1MHz) 0 95H f| = |= .
22
1
0
1 0 9521
nfπ
ω
= .⎛ ⎞
+ ⎜ ⎟⎝ ⎠
03, 2 (1 45MHz)n ω π= = × .
2( 2MHz) 0 355H f| = |= .
22
2
0
1 0 35521
nfπ
ω
= .⎛ ⎞
+ ⎜ ⎟⎝ ⎠
0.45 dB≈ − 9 dB≈ −
Example 14.25: Butterworth Response
12 22 *(1.45 )* cos sin3 3
p MHz jπ ππ ⎛ ⎞= +⎜ ⎟⎝ ⎠
32 22 *(1.45 )* cos sin3 3
p MHz jπ ππ ⎛ ⎞= −⎜ ⎟⎝ ⎠
2 2 *(1.45 )p MHzπ= RC section
2nd-order SK
Using a Sallen and Key topology, design a Butterworth filter for the response derived in Example 14.24.
66CH 14 Analog Filters
23π
43π
1,2,3 0 0 02 3 4exp , exp , exp3 3 3k
j j jp π π πω ω ω= =
0ω= −
Example 14.25: Butterworth Response (cont’d)
[ ]2
1 32 2
1 3
( )( ) [2 (1 45MHz)]( )( )( ) 4 (1 45MHz)cos(2 3) [2 (1 45MHz)]SK
p pH ss p s p s s
ππ π π
− − × .= =
− − − × . / + × .
22 (1 45MHz) and 1/ 2cos 13n Q πω π ⎛ ⎞= × . = = →⎜ ⎟
⎝ ⎠
1 2 2 1 21k , 54.9pF, and 4R R C C C= = Ω = =
3 33 3
1 2 (1 45MHz) 1k and 109.8pFR CR C
π= × . → = Ω =
67CH 14 Analog Filters
2
2 2( )
nn
s sH ss s
Q
α β γω ω
+ +=
+ +2C
1C
2C3R1R 2R
11 2
1 2 2
1 CQ R RR R C
=+
∵1 2 1 2
1n R R C C
ω =∵
2 21 1
1
4n
R Cω =
Chebyshev Response
( )2 2
0
1
1 n
H j
C
ωωεω
=⎛ ⎞
+ ⎜ ⎟⎝ ⎠
The Chebyshev response provides an “equiripple” pass/stop band response.
68CH 14 Analog Filters
: the amount of rippleε2
0( / ) : the "Chebyshev polynomial" of th ordernC nω ω
Chebyshev Polynomial
Chebyshev Polynomial
Chebyshev polynomial for n=1,2,3
Resulting transfer function forn=2,3
10
0 0
10
0
cos cos ,
cosh cosh ,
nC n
n
ω ω ω ωω ω
ω ω ωω
−
−
⎛ ⎞ ⎛ ⎞= <⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎛ ⎞
= >⎜ ⎟⎝ ⎠ 69CH 14 Analog Filters
2
11 ε+
1
0
H ωω⎛ ⎞⎜ ⎟⎝ ⎠
( )2 2
0
1
1 n
H j
C
ωωεω
=⎛ ⎞
+ ⎜ ⎟⎝ ⎠
Chebyshev Response
2 2 1
0
1( ) |
1 cos cosPBH j
n
ωωεω
−
| =⎛ ⎞
+ ⎜ ⎟⎝ ⎠
2 2 1
0
1( ) |
1 cosh coshSBH j
n
ωωεω
−
| =⎛ ⎞
+ ⎜ ⎟⎝ ⎠
70CH 14 Analog Filters
2dBPeak-to-peak Ripple 20 log 1 ε| = +
Example 14.26: Chebyshev Response
( )2 (2MHz) 0.116 18.7dBH j π = = −
ω0=2π (1MHz)
A third-order Chebyshev response provides an attenuation of -18.7 dB a 2MHz.
Suppose the filter required in Example 14.24 is realized with third-order Chebyshev response. Determine the attenuation at 2MHz.
2
1 0 95 0 3291
εε
= . → = .+
( )23
2
0 0
1
1 4 3
H jω
ω ωεω ω
=⎡ ⎤⎛ ⎞⎢ ⎥+ −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
71CH 14 Analog Filters
Example 14.27: Order of Chebyshev Filter
Specification: Passband ripple: 1 dBBandwidth: 5 MHzAttenuation at 10 MHz: 30 dBWhat’s the order?
72CH 14 Analog Filters
21 dB 20log 1 0 509ε ε= + → = .
0Attenuation at 2 10 MHz: 30 dBω ω= =
2 2 1
1 0 03161 0 509 cosh ( cosh 2)n −
= .+ .
2cosh (1 317 ) 3862 3 66 4n n n. = → > . → =
Example 14.28: Chebyshev Filter Design
( ) ( )1 10 0
2 1 2 11 1 1 1sin sinh sinh cos cosh sinh2 2k
k kp j
n n n nπ π
ω ωε ε
− −− −⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2,3 0 00.337 0.407p jω ω= − ±
SK21,4 0 00.140 0.983p jω ω= − ±
SK1
Using two SK stages, design a filter that satisfies the requirements in Example 14.27.
73CH 14 Analog Filters
Example 14.28: Chebyshev Filter Design (cont’d)
1 41
1 420
2 20 0
( )( )( )( )( )
0 9860 28 0 986
SKp pH s
s p s p
s sω
ω ω
− −=
− −
.=
+ . + .
1 00 993 2 (4 965MHz)nω ω π= . = × .
1 3 55Q = .
1 2 1 21 k , 50 4R R C C= = Ω = .
1 2
2 1
1 2 (4 965MHz)50 4
4.52 pF, 227.8 pFR C
C C
π= × ..
→ = =
2 32
2 3
20
2 20 0
( )( )( )( )( )
0 2790 674 0 279
SKp pH s
s p s p
s sω
ω ω
− −=
− −
.=
+ . + .
2 00 528 2 (2 64MHz)nω ω π= . = × .
2 0 783Q = . .
1 2
2 1
1 2 (2 64 MHz)2 45
38 5 pF, C 94.3 pFR C
C
π= × ..
→ = . =
1 2 1 21 k , 2 45R R C C= = Ω = .
74CH 14 Analog Filters
Summary
The desired filter response must in practice be approximated by a realizable transfer function. Possible transfer functions include Butterworth and Chebyshev responses.The Butterworth response contains n complex poles on a circle and exhibits a maximally-flat behavior. It is suited to applications that are intolerant of any ripple in the passbandThe Chebyshev response provides a sharper transition than Butterworth at the cost of some ripple in the passband and stopbands. It contains n complex poles on an ellipse.