chapter 14 analog filters - seoul national university · 2021. 2. 17. · 1 / 70 chapter 15 analog...
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Chapter 15 Analog Filters
15.1 General Considerations
15.2 First-Order Filters
15.3 Second-Order Filters
15.4 Active Filters
15.5 Approximation of Filter Response
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Outline of the Chapter
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Why We Need Filters
In order to eliminate the unwanted interference that
accompanies a signal, a filter is needed.
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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 transition
band.
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Example 15.1: Filter I
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.
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Example 15.2: Filter II
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.
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Example 15.3: 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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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 p
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Example 15.4 : Pole-Zero Diagram
1 1
1( )
1aH s
R C s
1 2
1 1 2
1( )
( ) 1b
R C sH s
R C C s
2
1 1 1
2
1 1 1 1 1
( )c
R L C sH s
R L C s L s R
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Impulse response contains
Example 15.5: Position of the poles
Poles on the RHP
Unstable
(no good)
Poles on the jω axis
Oscillatory
(no good)
Poles on the LHP
Decaying
(good)
exp( ) exp( )exp( )k k kp t t j t
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Transfer Function
The order of the numerator m ≤ The order of the denominator n
Otherwise, 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.
1 2
1 2
( )m
n
s z s z s zH s
s p s p s p
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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 dCS
P C
Sensitivity indicates the variation of a filter parameter due to
variation of a component value.
P=Filter Parameter
C=Component Value
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Example 15.6: Sensitivity
1
1
/1
0
1
1
1
0
0
1
2
11
0
110
RS
R
dRd
CRdR
d
CR
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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.
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First-Order Filters
1
1
( )s z
H ss p
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.
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Example 15.8: First-Order Filter I
R2C2 < R1C1R2C2 > R1C1
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2 1 1
1 2 1 2 1 2
( 1)( )
( )
out
in
V R R C ss
V R R C C s R R
1 1 11 ( ),z R C 1
1 1 2 1 2[( ) ]p C C R R
.
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Example 15.9: First-Order Filter II
R2C2 < R1C1 R2C2 > R1C1
2
2
1
1
2 1 1
1 2 2
1( )
( )1
1
1
out
in
RV C s
sV
RC s
R R C s
R R C s
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2
2 2
( )n
n
s sH s
s sQ
1,2 2
11
2 4
nnp j
Q Q
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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Second-Order Low-Pass Filter
22
22
2 2
( )
nn
H j
Q
α = β = 0
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Peak magnitude normalized to the passband magnitude: 2 11 (4 )Q Q
.
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Example 15.10: Second-Order LPF
2
2
3
/ 1 1/(4 ) 3
1 1/(2 )n n
Q
Q Q
Q
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Problem: Q of a second-order LPF = 3.
Estimate the magnitude and frequency of the peak in the frequency
response.
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Second-Order High-Pass Filter
2
2 2
( )n
n
sH s
s sQ
β = γ = 0
21 1 (2 )n Q Frequency of the peak:
Peak magnitude normalized to the passband magnitude: 2 11 (4 )Q Q
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Second-Order Band-Pass Filter
2 2
( )n
n
sH s
s sQ
α = γ = 0
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Example 15.2: -3-dB Bandwidth
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Problem: Determine the -3dB bandwidth of a band-pass response.
2 2
( )
( )nn
sH s
s sQ
1,2 0 2
1 11
24 QQ
2 2 2 2
22 2 2 2 2( ) ( )n n
n
Q
Q
0BWQ
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LC Realization of Second-Order Filters
11 1 2
1 1 1
1( ) ||
1
L sZ L s
C s L C s
An LC tank realizes a second-order band-pass filter with two imaginary poles at ±j/(L1C1)
1/2 , which implies infinite impedance at ω=1/(L1C1)
1/2.
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Example 15.13: LC Tank
At ω=0, the inductor acts as a short.
At ω=, the capacitor acts as a short.
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11 1 2
1 1 1
1( ) ||
1
L sZ L s
C s L C s
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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( ) ( )n
n
L s R L sZ R
L C s R L C s L s R
R L s R L s
R L C s s R L C s sR C L C Q
1,2 2
12
1 1 1 11 1
11
2 4
1 11
2 4
nnp j
Q Q
Lj
R C R CL C
With a resistor, the poles are no longer pure imaginary which implies there will be no infinite impedance at any ω.
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11
11 1
1,n
CQ R
LL C
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Voltage Divider Using General Impedances
( )out P
in S P
V Zs
V Z Z
Low-pass High-pass Band-pass
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Low-pass Filter Implementation with Voltage Divider
12
1 1 1 1 1
out
in
V Rs
V R C L s L s R
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1 as SZ L s s
1
1
10 as PZ R s
C s
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Example 15.14: Frequency Peaking
12
1 1 1 1 1
out
in
V Rs
V R C L s L s R
22 2 2 2
1 1 1 1 1Let D R R C L L
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Voltage gain greater than unity (peaking)
occurs when a solution exists for
2
2 21 1 1 1 1 1 1 12
2( )( )( )
0
d DR C L R R C L L
d
11
1
1Thus, when ,
2
peaking occurs.
CQ R
L
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Example 15.15: 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
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(a) (b)
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High-pass Filter Implementation with Voltage Divider
2
1 1 1 1 12
1 1 1 1 11 1
1
( ) ||
1( ) ||
out
in
V L s R L C R ss
V R C L s L s RL s RC s
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Band-pass Filter Implementation with Voltage Divider
1
1 12
1 1 1 1 11 1
1
1( ) ||
1( ) ||
out
in
L sV C s L s
sV R C L s L s RL s R
C s
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Why Active Filter?
Passive filters constrain the type of transfer function.
They may require bulky inductors.
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Sallen and Key (SK) Filter: Low-Pass
2
1 2 1 2 1 2 2
1
1
out
in
Vs
V R R C C s R R C s
11 2
1 2 2
1 CQ R R
R R C
1 2 1 2
1n
R R C C
Sallen and Key filters are examples of active filters. This
particular filter implements a low-pass, second-order transfer
function.
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Example 15.16: SK Filter with Voltage Gain
3
4
2 31 2 1 2 1 2 2 2 1 1
4
1
1
out
in
R
V Rs
V RR R C C s R C R C R C s
R
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Example 15.17: SK Filter Poles
The poles begin with real, equal values for and
become complex for .
3
4
2 31 2 1 2 1 2 2 2 1 1
4
1
( )
( ) 1
out
in
R
V Rs
RVR 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 C
Q R C R C R C R
3
4
1
2
QR
R
3 4 0R R
3 4 0R R
Problem: Assuming R1=R2, C1=C2, Does such a filter contain complex poles?
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Sensitivity in Low-Pass SK Filter
1 2 1 2
1
2n n n n
R R C CS S S S
1 2
2 2
1 1
1
2
Q QR R
R CS S Q
R C
1 2
2 2 1 2
1 1 2 1
1
2
Q QC C
R C R CS S Q
R C R C
1 1
2 2
QK
R CS QK
R C
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3 41K R R
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Example 15.18: SK Filter Sensitivity I
1 2
1 2
1 1
2 3
1 2
2 3
3
Q QR R
Q QC C
QK
S SK
S SK
KS
K
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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
1
2
Q QR R
Q Q QKC C
K
S S
S S S
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Integrator-Based Biquads
2
2 2
out
ninn
V ss
Vs s
Q
2
2
1.n n
out in out outV s V s V s V sQ s s
It is possible to use integrators to implement biquadratictransfer functions.
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KHN (Kerwin, Huelsman, and Newcomb) Biquads
2
2
1Comparing with .n n
out in out outV s V s V s V sQ s s
5 6
4 5 3
1R R
R R R
64
4 5 1 1 3
1. . 1n RR
Q R R R C R
2 6
3 1 2 1 2
1.n
R
R R R C C
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2
1 1 2 2 1 2 1 2
1 1 1, X out Y X outV V V V V
R 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 V
R R R R
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Versatility of KHN Biquads
2
2 2
High-pass: out
ninn
V ss
Vs s
Q
2
2 2 1 1
1Band-pass: .X
ninn
V ss
V R C ss s
Q
2
22 2 1 2 1 2
1Low-pass: .Y
ninn
V ss
V R R C C ss sQ
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Sensitivity in KHN Biquads
1 2 1 2 3 6, , , , ,0.5n
R R C C R RS
1 2 1 2 4 5
3 6
5, , , ,
4 5
3 6 2 2,
5 3 6 1 1
4
0.5, 1,
21
Q QR R C C R R
QR R
RS S
R R
R R R CQS
R R R R C
R
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Tow-Thomas Biquad
32 2 4 1 1
1 1out inout
V VR V
R C s R R sC
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Tow-Thomas Biquad
2 3 4 22
1 2 3 4 1 2 2 4 2 3
Band-pass: .out
in
V R R R C s
V R R R R C C s R R C s R
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3 4
21 2 3 4 1 2 2 4 2 3
1Low-pass: .Y
in
R RV
V R R R R C C s R R C s R
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Differential Tow-Thomas Biquads
An important advantage of this topology over the KHN biquad
is accrued in integrated circuit design, where differential
integrators obviate the need for the inverting stage in the loop.
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Example 15.20: Tow-Thomas Biquad
2 4 1 2
1n
R R C C
Adjusted by R2 or R4
1 2 4 2
3 1
1 R R CQ
R C
Adjusted by R3
Note that n and Q of the Tow-Thomas filter can be adjusted (tuned)
independently.
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Antoniou General Impedance Converter
1 35
2 4in
Z ZZ Z
Z Z
It is possible to simulate the behavior of an inductor by using active circuits in feedback with properly chosen passive elements.
CH 15 Analog Filters
1 3 5 XV V V V
4 4
5
XX
VV Z V
Z
4 33
3
Z
V VI
Z
4
5 3
XV Z
Z Z
2 3 2 3ZV V Z I
42
5 3
XX
V ZV Z
Z Z
2
1
XX
V VI
Z
2 4
1 3 5
X
Z ZV
Z Z Z
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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 Cs
L R R C
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High-Pass Filter with SI
With the inductor simulated at the output, the transfer function
resembles a second-order high-pass filter.
2
12
1 1 1 1 1
out
in
V L ss
V R C L s L s R
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Example 15.22: High-Pass Filter with SI
4 1 Yout
X
RV V
R
Node 4 can also serve as an output.
CH 15 Analog Filters
V4 is better than Vout since the output impedance is lower.
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Low-Pass Filter with Super Capacitor
1
1in
X
ZCs R Cs
1
2 21 1
1
1
out in
in in
X
V Z
V Z R
R R C s R Cs
CH 15 Analog Filters
How to build a floating inductor to derive a low-pass filter?
Not possible. So use a super capacitor.
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Example 15.24: Poor Low-Pass Filter
4
12out X out out X
X
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.
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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.
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Butterworth Response
2
0
1( )
1
nH j
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.
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Poles of the Butterworth Response
0
2 1exp exp , 1,2, ,
2 2k
j kp j k n
n
2nd-Order nth-OrderCH 15 Analog Filters
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Example 15.24: Order of Butterworth Filter
The minimum order of the Butterworth filter is three.
2
2
1
64.2
nf
f
2 12f f
CH 15 Analog Filters
Specification: passband flatness of
0.45 dB for f < f1=1 MHz, stopband
attenuation of 9 dB at f2=2 MHz.
1( 1MHz) 0 95H f .
2
2
1
0
10 95
21
n
f
.
03, 2 (1 45MHz)n .
2( 2MHz) 0 355H f .
2
2
2
0
10 355
21
n
f
.
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Example 15.25: Butterworth Response
1
2 22 *(1.45 )* cos sin
3 3p MHz j
3
2 22 *(1.45 )* cos sin
3 3p MHz j
2 2 *(1.45 )p MHz RC section
2nd-order SK
Using a Sallen and Key topology, design a Butterworth filter for the
response derived in Example 14.24.
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Example 15.25: Butterworth Response (cont’d)
2
1 3
2 2
1 3
( )( ) [2 (1 45MHz)]( )
( )( ) 4 (1 45MHz)cos(2 3) [2 (1 45MHz)]SK
p pH s
s p s p s s
.
. .
22 (1 45MHz) and 1/ 2cos 1
3n Q
.
1 2 2 1 21k , 54.9pF, and 4R R C C C
3 3
3 3
12 (1 45MHz) 1k and 109.8pFR C
R C .
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Chebyshev Response
2 2
0
1
1 n
H j
C
The Chebyshev response provides an “equiripple” pass/stop
band response.
Chebyshev Polynomial
CH 15 Analog Filters
Textbook error:
No ripple
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Chebyshev Polynomial
Chebyshev polynomial for
n=1,2,3
Resulting transfer function for
n=2,3
10
0 0
10
0
cos cos ,
cosh cosh ,
nC n
n
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Chebyshev Response
2 2 1
0
1( ) |
1 cos cos
PBH j
n
2 2 1
0
1( ) |
1 cosh cosh
SBH j
n
CH 15 Analog Filters
2
dBPeak-to-peak Ripple 20log 1
Textbook error:
No ripple
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Example 15.26: Chebyshev Response
2 (2MHz) 0.116 18.7dBH j
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
10 95 0 329
1
. .
2
3
2
0 0
1
1 4 3
H j
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Example 15.27: Order of Chebyshev Filter
Specification:
Passband ripple: 1 dB
Bandwidth: 5 MHz
Attenuation at 10 MHz: 30 dB
What’s the order?
CH 15 Analog Filters
21 dB 20log 1 0 509 .
0Attenuation at 2 10 MHz: 30 dB
2 2 1
10 0316
1 0 509 cosh ( cosh 2)n .
.
2cosh (1 317 ) 3862 3 66 4n n n. .
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Example 15.28: Chebyshev Filter Design
1 10 0
2 1 2 11 1 1 1sin sinh sinh cos cosh sinh
2 2k
k kp j
n n n n
2,3 0 00.337 0.407p j
SK2
1,4 0 00.140 0.983p j
SK1
Using two SK stages, design a filter that
satisfies the requirements in Example 14.27.
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Example 15.28: Chebyshev Filter Design (cont’d)
1 41
1 4
2
0
2 2
0 0
( )( )( )
( )( )
0 986
0 28 0 986
SK
p pH s
s p s p
s s
.
. .
1 00 993 2 (4 965MHz)n . .
1 3 55Q .
1 2 1 21 k , 50 4R R C C .
1 2
2 1
12 (4 965MHz)
50 4
4.52 pF, 227.8 pF
R C
C C
..
2 32
2 3
2
0
2 2
0 0
( )( )( )
( )( )
0 279
0 674 0 279
SK
p pH s
s p s p
s s
.
. .
2 00 528 2 (2 64MHz)n . .
2 0 783Q . .
1 2
2 1
12 (2 64 MHz)
2 45
38 5 pF, C 94.3 pF
R C
C
..
.
1 2 1 21 k , 2 45R R C C .
CH 15 Analog Filters