Chapter 14 Analog Filters - Seoul National University · 2018. 1. 30. · Continuous-time Discrete-time CH 14 Analog Filters 9 QCV 121= QCV 222= 122 1 2 if , absorbs charge from and

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.


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.


attenuated amplified

sufficientlynarrow

bad stopbandattenuation

Example 14.1: 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.

Example 14.2: 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.

Example 14.3: Filter III

A bandpass filter around 1.5 GHz is required to reject the adjacent Cellular and PCS signals.


Classification of Filters I


Classification of Filters II

Continuous-time Discrete-time


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


Summary of Filter Classifications


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)


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

− − −=

− − −


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=

+ +


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)


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.


( )( ) ( )( )( ) ( )

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.


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


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


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.


Example 14.8: First-Order Filter I

R2C2 < R1C1 R2C2 > R1C1


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

− ||=

||

+= − ⋅

+


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.


General transfer function

Second-Order Low-Pass Filter

( )

22

222 2

( )n

n

H j

Q

γωωω ω ω

=⎛ ⎞

− + ⎜ ⎟⎝ ⎠

α = β = 0


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

=

− ≈

− ≈


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 −/ −


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

=+ +


0α γ= =

The magnitude approaches zero for both s 0 and s ∞, reaching a maximum in between

Example 14.2: -3-dB Bandwidth


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.


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.


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.


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

=+ +


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= − +


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


(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

= =+ ++


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

= =+ ++


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.


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.


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

+=

⎛ ⎞+ + − +⎜ ⎟⎝ ⎠


( )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?


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

=


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

= − = − +−

= − = − +−

=−


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.


( )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 =


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

=+ +


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

= = <+

−=

+


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

⎛ ⎞⎛ ⎞⋅ + = −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠


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

= −+ +


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.


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.


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.


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

=

=


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

=+ +


Example 14.22: High-Pass Filter with SI

4 1 Yout

X

RV VR

⎛ ⎞= +⎜ ⎟

⎝ ⎠

Node 4 can also serve as an output.


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

=+

=+ +


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.


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.


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.


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


02 1exp exp , 1, 2, ,

2 2kj kp j k n

nπω π−⎛ ⎞= =⎜ ⎟

⎝ ⎠

Poles of the Butterworth Response


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=


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.


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 =


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.


: 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

−

| =⎛ ⎞

+ ⎜ ⎟⎝ ⎠


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

=⎡ ⎤⎛ ⎞⎢ ⎥+ −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


Example 14.27: Order of Chebyshev Filter

Specification: Passband ripple: 1 dBBandwidth: 5 MHzAttenuation at 10 MHz: 30 dBWhat’s the order?


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.


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 = .


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.

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