Analog Filters (Ch. 14) - chungbuk.ac.krbandi.chungbuk.ac.kr/~ysk/ckt14.pdf · 2011-03-13 · Contents 14.1 General Considerations 14.2 First-Order Filters 14.3 Second-Order Filters

Analog Filters (Ch. 14)g

김 영 석김 영 석

충북대학교 전자정보대학

2012.9.1.9.

Email: [email protected]

전자정보대학 김영석 14-1

Contents14.1 General Considerations

14.2 First-Order Filters

14.3 Second-Order Filters

14 4 A ti Filt14.4 Active Filters

14.5 Approximation of Filter Response


14.1 General ConsiderationsWhy We Need Filters?

In order to eliminate the unwanted interference that accompanies a signal, a filter is needed.


14.1.1 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.


Ex14.1/2: Filter I

Ex14.1

Given: Adjacent channel Interference is 25 dB above the signal

Design goal: Signal to Interference ratio of 15 dB

S l ti A filt ith t b d Att ti f 4 dBSolution: A filter with stop band Attenuation of 40 dB

Ex14.2

Given: Adjacent channel Interference is 40 dB above the signal

Design goal: Signal to Interference ratio of 20 dB

Solution: A filter with stop band of 60 dB at 60 Hz


Ex14.3: Filter III

A bandpass filter around 1.5 GHz is needed to reject the dj t C ll l d PCS i ladjacent Cellular and PCS signals.


14.1.2 Classification of Filters


14.1.3 Filter Transfer Function

Filter a) has a transfer function with -20dB/dec roll-off

A B

) s s

Filter b) has a transfer function with -40dB/dec roll-off, better selectivity.

Zm=m’th zero( )( ) ( )1 2( ) ms Z s Z s ZH s α

− − −=

L

Pn =n’th pole( )( ) ( )1 2( )

mH s

s P s P s Pα=

− − −L


Ex14.4 Pole-Zero Diagram


Ex14.5 Position of the Poles

Poles on the RHPUnstable

Poles on the jω axisOscillatory

Poles on the LHPDecaying

(no good)y

(no good)y g

(good)

Imaginary zero is used to create a null at certain frequency.g y s s q y


14.1.4 SensitivitySensitivity measures the variation of a filter parameter due to

dP dCvariation of a filter component.

P=ParameterPC

dP dCSP C

=P Parameter

C=ComponentP C C Component

Ex14 6: Sensitivity ( )/1=ω CREx14.6: Sensitivity ( )1

/1

20

110

−=

=ω

ω

CRdRd

CR

10

111

−=ω

RdRdCRdR

10

1

10

−=ω

ω

RS

R


14.2 First-Order Filters

1( ) s zH s α +=

1( )H s

s pα=

+

L /hi h filt b li d b h i th l tiLow/high pass filters can be realized by changing the relative positions of poles and zeros.


KnowhowPole 구하는 법: Vin=0, Vout=∞, Output Node에서의 시정수

))(||(1

2121 CCRRwp +

=

Zero 구하는 법: Vout= 0, Input Node에서의 시정수

))(||( 2121

1

11

1CR

wz =

|H(w=0)|=? C open, L short

2RVout =

|H(w=∞)|=? C short, L open

21 RRVin +=

|H(w ∞)| ? C short, L open

21

1

CCC

VV

i

out

+=


21 CCVin +

Ex14.8: First-Order Filter I

R1*C1: zeroR1 C1: zero

R2C2 < R1C1 R2C2 > R1C1


Ex14.9: First-Order Filter II

R1*C1: zeroR1 C1: zero

R2C2 < R1C1 R2C2 > R1C1


14.3 Second-Order Filters

22

2

)(wsws

sssHn ++

++=

γβα

nws

Qs ++

1,2 211

2 4n

np jQ Qω ω= − ± − 22 4Q Q

Second-order filters are characterized by the “biquadratic” equation with two complex poles shown above.


Second-Order Low-Pass Filter

α=β=022

)(ws

Qws

sHn ++

=γ

nQ0)( ,)0( 2 =∞= H

wH

n

γ


Ex14.10: Second-Order LPF

3Q2)(n

wQwH n

γ=

2

3

/ 1 1/(4 ) 3

Q

Q Q

=

− ≈

2

/ 1 1/(4 ) 3

1 1/(2 )n n

Q Q

Qω ω

≈

− ≈


Second-Order High-Pass Filter

2

2 2( )

n

sH s αω= β=γ=0

2 2nns s

Qω+ +

α=∞= )( ,0)0( HH


Second-Order Band-Pass Filter

2 2( )

n

sH s βω= α=γ=0

2 2nns s

Qω ω+ +

0)(,)( ,0)0( =∞== HwQwHHn

nβ


14.3.2 LC Realization of Second-Order Filters

L 1) ( 0)0( 11 shortLZ =

11 2

1 1 1L sZ

L C s=

+ ) ( 0)(

)2

1(

11

1

shortCZLC

fZ

=∞

∞==π

An LC tank realizes a second-order band-pass filter with two imaginary poles at ±j/(L1C1)1/2 , which implies infinite i d t 1/(L C )1/2impedance at ω=1/(L1C1)1/2.


RLC Realization of Second-Order Filters

1 12 2

R L sZ =2 21 1 1 1 1R L C s L s R+ +

11 1 1 Lp j= − ± −) ( 0)0( 12 shortLZ =1,2 2

1 1 1 11 11

2 4p j

R C R CL C= − ± −

)2

1( 111

2

12

RCL

fZ ==π

) ( 0)( 12 shortCZ =∞

With a resistor, the poles are no longer pure imaginary which implies there will be no infinite impedance at any ω.


Voltage Divider Using General Impedances

( )out PV Zs =( )in S P

sV Z Z+

Low-pass High-pass Band-pass


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 1 1 1 1in

0),(pen )0( :),0( :0

11

1

=∞==∞====

outsp

inouts

VZoLorZshortCfVVZshortLf


Ex14.14: Frequency Peaking

( ) 1outV Rs( ) 12

1 1 1 1 1

out

ins

V R C L s L s R=

+ +

1Q > Peaking exists2

Q > Peaking existsVoltage gain larger than unity


Ex14.15 Low Pass Circuit Comparison

Good BadGood Bad

pen, :0 11= shortCoLf pen,:0 11= shortCoLf0

p , 11

=outVf

방해유한 0,,p ,

1

11

== outVRButf

The circuit on the left has a sharper roll-off at high frequency than the circuit on the right.than the circuit on the right.


High-pass Filter Implementation with Voltage DividerDivider

( )2

1 1 12

1 1 1 1 1

out

in

V L C R ssV R C L s L s R

=+ +

out

VVoLshortCfVshortLoCf

=∞===

pen:0, pen, :0 11

inout VVoLshortCf =∞= ,pen, : 11


Band-pass Filter Implementation with Voltage Divider

2V L s( ) 12

1 1 1 1 1

out

in

V L ssV R C L s L s R

=+ +

, /:2

10,:0

11

1

==

==

inout

out

VVopenLCCL

f

VshortLf

π0, :

2

1

11

=∞= outVshortCfCLπ


14.4 Active Filters14.4.1 Sallen and Key (SK) Filter: Low-Pass

11 CQ R R 11 2

1 2 2Q R R

R R C=

+

1nω =

( )( )2

1 2 1 2 1 2 2

11

out

in

V sV R R C C s R R C s

=+ + +

1 2 1 2n R R C C

Sallen and Key filters are examples of active filters. This ti l filt i l t l d d t fparticular filter implements a low-pass, second-order transfer

function.


Ex14.16 Sallen and Key (SK) Filter: Band-pass

31 R+

Pass-band gain>1

( )3

4

2 3

1

1

out

in

V RsV RR R C C s R C R C R C s

+=

⎛ ⎞+ + +⎜ ⎟1 2 1 2 1 2 2 2 1 1

41R R C C s R C R C R C s

R+ + − +⎜ ⎟⎝ ⎠


Ex14.17: SK Filter Poles

C1=C2

R1=R21 2


Sensitivity in Band-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 1 212

Q QC C

R C R CS S QR C R C

⎛ ⎞= − = − + +⎜ ⎟⎜ ⎟

⎝ ⎠1 2 1 2 2

2 21Q Q R C

1 21 1 2 12C C R C R C⎜ ⎟

⎝ ⎠

Q R C1 2

2 2

1 1

12

Q QR R

R CS S QR C

= − = − + 1 1

2 2

QK

R CS QKR C

= K=1+R3/R4


Ex14.18: SK Filter Sensitivity I

1 2R R R= =

1 2C C C= =

1 2

1 12 3

Q QR RS S

K= − = − +

−

1 2

1 22 3

Q QC CS S

K= − = − +

−

3QK

KSK

=−


Ex14.19: SK Filter Sensitivity II

2Q =2K =

1

2 2

1 1

3 14

QR

R C SR C

= ⇒ =

1

1 1

2 2

1 58 4

QC

R C SR C

= ⇒ =2 2 8 4

81 5

QK

R C

S =1.5


14.4.2 Integrator-Based Biquads

( )2

2 2

out

nin

V ssV s s

αω ω

=+ +

( ) ( ) ( ) ( )2

21.n n

out in out outV s V s V s V sQ s sω ωα= − −

ns sQ

ω+ +

It is possible to use integrators to implement biquadratic transfer s p ss s g s p q s functions.

The block-diagram above illustrates how.


KHN Biquads

( ) ( ) ( ) ( )21ω ω( ) ( ) ( ) ( )2

1.n nout in out outV s V s V s V s

Q s sω ωα= − −

5 6

4 5 31R R

R R Rα

⎛ ⎞= +⎜ ⎟+ ⎝ ⎠

4

4 5 1 1

1.n RQ R R R Cω

=+

2 6

3 1 2 1 2

1.nRR R R C C

ω =4 5 3⎝ ⎠ 4 5 1 1Q


Versatility of KHN Biquads

2V sα

High-Pass2 1V sα

Band-Pass

( )2 2

out

ninn

V ssV s s

Q

αω ω

=+ +

( )2 2 1 1

1.X

ninn

V ssV R C ss s

Q

αω ω

−=

+ +

2 1V sα

Low-Pass

( ) 22 2 1 2 1 2

1.Y

ninn

V ssV R R C C ss s

Q

αω ω

=+ +


Sensitivity in KHN Biquads

0 5nR R C C R R R RSω = 0 5QS1 2 1 2 4 5 3 6, , , , , , , 0.5R R C C R R R RS

1 2 1 2, , , 0.5QR R C CS =

3 6

3 6 2 2,

5 3 6 1 12 1

QR R

R R R CQS R R R R CR

−=

+ 4 5

5,

4 51Q

R RRS

R R= <

+4R

Low sensitivities


Tow-Thomas Biquad

Merge the adder and the first integratorMerge the adder and the first integrator

3 42

1 2 3 4 1 2 2 4 2 3

1.Y

in

R RVV R R R R C C s R R C s R

=+ +

2 3 4 22

1 2 3 4 1 2 2 4 2 3.out

in

V R R R C sV R R R R C C s R R C s R

= −+ +

Band-Pass Low-Pass


Differential Tow-Thomas Biquads

By using differential integrators, the inverting stage is eliminated.


14.4.3 Simulated Inductor (SI)

1 35in

Z ZZ ZZ Z

= 52 4

in Z Z

It is possible to simulate the behavior of an inductor by usingIt is possible to simulate the behavior of an inductor by using active circuits in feedback with properly chosen passive elements.


Simulated Inductor I

2in X YZ R R Cs=

By proper choices of Z1-Z4, Zin has become an impedance that increases with frequency, simulating inductive effect.

Z5=RX, Z2=1/(Cs), Z1=Z3=Z4=RY


Ex14.21: Simulated Inductor II

1 2 3

1YZ Z Z R

Z

= = =

4

2

ZCs

Z R R Cs

=

=in X YZ R R Cs=


High-Pass Filter with SI

( )2

12

1 1 1 1 1

out

in

V L ssV R C L s L s R

=+ +

With the inductor simulated at the output, the transfer function resembles a second-order high-pass filter.

⎛ ⎞Node 4 is also an output node

4 1 Yout

X

RV VR

⎛ ⎞= +⎜ ⎟

⎝ ⎠

Node 4 is also an output node


⎝ ⎠

Low-Pass Filter with Super Capacitor

( )1

1inX

ZCs R Cs

=+( )X

Product of 2 C

=> Super Capacitor

1out inV Z= =

=> Super Capacitor

2 21 1 1 1in in XV Z R R R C s R Cs+ + +

Low-PassLow Pass


Ex14.23: Poor Low Pass Filter

( )4 2out XV V R Cs= +

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.p


14.5 Approximation of Filter ResponseFrequency Response Template

With all the specifications on pass/stop band ripples and t iti b d l t filt t l t th ttransition band slope, one can create a filter template that will lend itself to transfer function approximation.


14.5.1 Butterworth Response

12

1( )

1n

H jωω

=⎛ ⎞

+ ⎜ ⎟⎝ ⎠0ω⎜ ⎟⎝ ⎠

ω-3dB=ω0 for all n3dB 0,

The Butterworth response completely avoids ripples in the pass/stop bands at the expense of the transition band slope.pass/stop bands at the expense of the transition band slope.


Poles of the Butterworth Response

2 1j kπ ⎛ ⎞0

2 1exp exp , 1, 2, ,2 2kj kp j k n

nπω π−⎛ ⎞= =⎜ ⎟

⎝ ⎠L

2nd‐Order nth‐Order


Ex14.24: Butterworth Order

22 64 2

nf⎛ ⎞

=⎜ ⎟1

64.2f

=⎜ ⎟⎝ ⎠

2 12f f=2 1

n=3

The Butterworth order of three is needed to satisfy the filter response on the leftresponse on the left.


Ex14.25: Butterworth Response

12 22 *(1.45 )* cos sin3 3

p MHz jπ ππ ⎛ ⎞= +⎜ ⎟⎝ ⎠

RC section

32 22 *(1.45 )* cos sin3 3

p MHz jπ ππ ⎛ ⎞= −⎜ ⎟⎝ ⎠ 2 2 *(1.45 )p MHzπ=

2nd-Order SK


14.5.2 Chebyshev Response

( )2 2

1

1

H j

C

ωωε

=⎛ ⎞

+ ⎜ ⎟0

1 nCεω

+ ⎜ ⎟⎝ ⎠

Chebyshev Polynomial

The Chebyshev response provides an “equiripple” pass/stop band response.


Chebyshev Polynomial

Chebyshev Polynomial for n=1,2,3

Resulting Transfer function forn=2,3, ,

10

0 0cos cos ,nC nω ω ω ω

ω ω−⎛ ⎞ ⎛ ⎞

= <⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10

0cosh cosh ,n ω ω ω

ω−⎛ ⎞

= >⎜ ⎟⎝ ⎠


Ex14.26: Chebyshev Attenuation

( )23

2

1

1 0 329 4 3

H jω

ω ω

=⎡ ⎤⎛ ⎞⎢ ⎥+ ⎜ ⎟

2

0 01 0.329 4 3

ω ω⎢ ⎥+ −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

ω0=2π X (2MHz)

A third-order Chebyshev response provides an attenuation of -18.7 dB a 2MHz.


Ex14.27: Chebyshev Order

Passband Ripple: 1 dB

Bandwidth: 5 MHz

Attenuation at 10 MHz: 30 dBAttenuation at 10 MHz: 30 dB

What’s the order?

( )2 2 1

1 0.03161 0 509 cosh cosh 2n −

=+

n>3.66( )1 0.509 cosh cosh 2n+


Ex14.28: Chebyshev Response

( ) ( )1 12 1 2 11 1 1 1k kπ π− −⎛ ⎞ ⎛ ⎞( ) ( )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

− −⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

K=1 2 3 4K=1,2,3,42,3 0 00.337 0.407p jω ω= − ±

SK21,4 0 00.140 0.983p jω ω= − ±

SK1


Documents

Analog Filters (Ch. 14) - chungbuk.ac.krbandi.chungbuk.ac.kr/~ysk/ckt14.pdf · 2011-03-13 · Contents 14.1 General Considerations 14.2 First-Order Filters 14.3 Second-Order Filters