Chapter 14 Analog Filters

Chapter 14 Analog Filters

 14.1 General Considerations

 14.2 First-Order Filters

 14.3 Second-Order Filters

 14.4 Active Filters

 14.5 Approximation of Filter Response

Outline of the Chapter

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 transition band.

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

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

p_k = pole frequencies z_k = zero frequencies

    

    

1 2

1 2

( )

n

s z s z s z

H s   s ^ p s ^ p s ^ p

  





Example 14.4 : Pole-Zero Diagram

1 1

( ) 1

a 1

H s  R C s



1 2

1 1 2

( ) 1

( ) 1

b

R C s H s R C C s

 

 

1 2

1 1 1 1 1

c( ) H s C s

R L C s L s R

  

Impulse response contains

Example 14.5: 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( p t

)  exp( 

t ) exp( j 

t )

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, z^1,2=± jω¹ , H(s) drops to zero at ω¹.

    

     

1 2

1 2

( )

n

s z s z s z

H s   s ^ p s ^ p s ^ p

  





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

P C

dP dC S  P C

 Sensitivity indicates the variation of a filter parameter due to variation of a component value.

P=Filter Parameter

C=Component Value

Example 14.6: Sensitivity

 

1 1 /

1

0 1

1 1 0

0

1 2 1 1

0

1 1 0









 













S

R dR d

C R dR

d

C R

Problem: Determine the sensitivity of ω

with respect to R

.

 For example, a +5% change in R₁translates to a -5% error in

ω

First-Order Filters

1 1

( ) s z

H s   s ^ 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.

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 s

V s R R C C s R R

 

  

1

1 (

),

z    R C p

  [( C

 C R

)

 R

]

.

Example 14.9: First-Order Filter II

R₂C₂ < R₁C₁ R₂C₂ > R₁C₁

2

2

1

1

2 1 1

1 2 2

( 1 )

( ) 1

1 1

out in

V R C s

V s R

C s R R C s R R C s

 





   



2

2 2

( )

n

n

s s

H s

s s

Q

  

 

 



 

1,2 2

1 1

2 4

n

p j n

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 ± ω_n.

=> the poles look very imaginary thereby bringing the circuit closer to instability.

Second-Order Low-Pass Filter

 

2 2

2 2

2 2

( )

n n

H j

Q

 

   



 

   

 

α = β = 0

Peak magnitude normalized to the passband magnitude: Q 1 (4 Q²)^1

.

Example 14.10: Second-Order LPF

2

2

3

/ 1 1/(4 ) 3 1 1/(2 )

n n

Q

Q Q

 Q 



 

 

Problem: Q of a second-order LPF = 3.

Estimate the magnitude and frequency of the peak in the frequency response.

Second-Order High-Pass Filter

2

2 2

( )

n

n

H s s

s s

Q



 



 

1 1 (2 2)

n Q

    Frequency of the peak:

Peak magnitude normalized to the passband magnitude: Q 1 (4 Q²)^1

Second-Order Band-Pass Filter

2 2

( )

n

n

H s s

s s

Q



 



 

Example 14.2: -3-dB Bandwidth

Problem: Determine the -3dB bandwidth of a band-pass response.

2 2

( )

(

)

H s s

s s

Q



 



 

1,2 0 2

1 1

1 4 Q 2 Q

   ^    ^ 

 

 

2 2 2 2

2 2 2 2

2

(

) (

)

Q Q

  

 

  



 

BW

Q

 

LC Realization of Second-Order Filters

1

1 1 2

1 1 1

( ) || 1

1 Z L s 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/(L C )^1/2.

Example 14.13: LC Tank

 At ω=0, the inductor acts as a short.

 At ω=, the capacitor acts as a short.

1

1 1 2

1 1 1

( ) || 1

1 Z L s L s

C s L C s

 



RLC Realization of Second-Order Filters

1 1 1

2 1 2 2

1 1 1 1 1 1 1

1 1 1 1

2 2 2

1 1 1 1 1 1

1 1 1 1

|| 1

1 1

( ) (

)

L s R L s

Z 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 s

R C L C Q

 

 

  

 

   

1,2 2

1 2

1 1 1 1 1 1

1 1

2 4

1 1

2 1 4

n n

p j

Q Q

j L

R C L C R C

 

   

   

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

1 1 1 1 1

1 ,

n

Q R C L C L

  

Voltage Divider Using General Impedances

out ( ) P

in S P

V Z

V s  Z Z



Low-pass High-pass Band-pass

Low-pass Filter Implementation with Voltage Divider

  ₂ ¹

1 1 1 1 1

out in

V R

V s  R C L s L s R

 

1

as

Z

 L s   s  

1 1

1 0 as

Z

R s

 C s    

Example 14.14: Frequency Peaking

 

1 1 1 1 1

o u t in

V R

V s  R C L s L s R

 

 

2 2 2 2

1 1 1 1 1

Let D  R  R C L   L 

Voltage gain greater than unity (peaking) occurs when a solution exists for

2

2 2

1 1 1 1 1 1 1 1

2 2( )( )

( )

0

d D R C L R R C L L

d







1 1

1

Thus, when 1 ,

2 peaking occurs.

Q R C

  L 

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 1

2

1 1 1 1 1

1 1

1

( ) ||

( ) || 1

o u t i n

V L s R L C R s

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

C s

 

 



Band-pass Filter Implementation with Voltage Divider

 

1

1 1

2

1 1 1 1 1

1 1

1

( ) || 1

( ) || 1

out in

V L s C s L s

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

 

 



Why Active Filter?

 Passive filters constrain the type of transfer function.

 They may require bulky inductors.

Sallen and Key (SK) Filter: Low-Pass

 





1

1

out in

V s

V  R R C C s R R C s

  

1 2 1

1 2 2

1 C

Q R R

R R C

 

1

n

R R C C

 

 Sallen and Key filters are examples of active filters. This

particular filter implements a low-pass, second-order transfer

Example 14.16: SK Filter with Voltage Gain

 

3 4

2 3

1 2 1 2 1 2 2 2 1 1

4

1

1

out in

R

V R

V s R

R R C C s R C R C R C s R



  

     

 

Example 14.17: SK Filter Poles

 The poles begin with real, equal values for and become complex for .

3 4

2 3

1 2 1 2 1 2 2 2 1 1

4

1 ( )

( ) 1

out in

R

V R

s R

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





   

1 2 2 2 1 1 3

2 1 1 1 2 2 4

1 R C R C R C R

Q  R C  R C  R C R

3 4

1 2

Q R

R





3 4

0

R R   0

R R  

Problem: Assuming R¹=R², C¹=C^2, Does such a filter contain complex poles?

Sensitivity in Low-Pass SK Filter

1 2 1 2

1 2

n n n n

R R C C

S

 S

 S

 S

 

1 2

2 2 1 1

1 2

Q Q

R R

S S Q R C

     R C

1 2

2 2 1 2

1 1 2 1

1 2

Q Q

C C

R C R C

S S Q

R C R C

 

         

 

1 1 2 2 Q

K

S QK R C

 R C

3 4

1

K   R R

Example 14.18: SK Filter Sensitivity I

1 2

1 2

1 1

2 3

1 2

2 3 3

Q Q

R R

Q Q

C C

Q K

S S

K

S S

K S K

K

    



    



 

Problem: Determine the Q sensitivities of the SK filter for the common choice R¹=R²=R, C¹=C²=C.

1 2

With =1, 0

1

Q Q

R R

Q Q Q

C C K

K

S S

S S S

 

  

Integrator-Based Biquads

 

2

2 2

out in n

n

V s

V s s s

Q



 



         

2 2

.1

n n

out in out out

V s V s V s V s

Q s s

 



  

 It is possible to use integrators to implement biquadratic transfer functions.

KHN (Kerwin, Huelsman, and Newcomb) Biquads

       

2 2

Comparing with V_out s V_in s ⁿ .1V_out s ⁿ V_out s

Q s s

 



  

5 6

4 5 3

R 1 R

R R R



  

현재 이 이미지를 표시할 수 없습니다 .

4 6

4 5 1 1 3

. 1 . 1

n R R

Q R R R C R



   

  

2 6

3 1 2 1 2

. 1

n

R

R R R C C



2

1 1 2 2 1 2 1 2

1 1 1

X out, Y X out

V V V V V

R C s R C s R R C C s

     ^{5 4 6 6}

4 5 3 3

in X 1

out Y

V R V R R R

V V

R R R R

 

    

  

Versatility of KHN Biquads

 

2

2 2

High-pass: ^out

in n

n

V s

V s s s

Q



 



 

 

2

2 2 1 1

Band-pass: ^X . 1

in n

n

V s

V s s s R C s

Q



 

 

 

 

2

2 2 1 2 1 2 2

Low-pass: ^Y . 1

in n

n

V s

V s s s R R C C s

Q



 



 

Sensitivity in KHN Biquads

1ⁿ, 2, 1, 2, 4, 5, 3, 6

0.5

R R C C R R R R

S



1 2 1 2 4 5

3 6

5

, , , ,

4 5

3 6 2 2

, 5 3 6 1 1

0.5, 1,

2 1

Q Q

R R C C R R

Q R R

S S R

R R

R R R C

S Q

R R R R C

  



 



Tow-Thomas Biquad

3

2 2 4 1 1

1 1

out in

out

V V

R V

R C s R R sC

   

   

   

   

Tow-Thomas Biquad

2 3 4 2

2

1 2 3 4 1 2 2 4 2 3

Band-pass:

.

in

V R R R C s

V   R R R R C C s R R C s R

 

3 4

2

1 2 3 4 1 2 2 4 2 3

Low-pass:

. 1

in

V R R

V  R R R R C C s R R C s R

 

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.

Example 14.20: Tow-Thomas Biquad

2 4 1 2

1

n R R C C



Adjusted by R₂ or R₄

1 2 4 2

3 1

1 R R C

Q R C





Adjusted by R₃

Note that _n and Q of the Tow-Thomas filter can be adjusted (tuned) independently.

Antoniou General Impedance Converter

1 3

5 2 4 in

Z Z Z Z

 Z 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 X

V V V V

4 4

5 X

X

V V Z V

 Z 

4 3

3

3 Z

V V

I Z

 

4

5 3

VX Z Z Z

 

2 3 2 Z3

V V Z I

4 2

5 3

X X

V Z

V Z

Z Z

   

2 1 X X

V V

I Z

 

2 4 1 3 5 X

V Z Z

Z Z Z



Simulated Inductor

 By proper choices of Z₁-Z₄, Z_in 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





High-Pass Filter with SI

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

 

2 1 2

1 1 1 1 1

out in

V L s

V s  R C L s L s R

 

Example 14.22: High-Pass Filter with SI

4 _out 1 ^Y

X

V V R

R

 

   

 

Node 4 can also serve as an output.

 V₄ is better than V_out since the output impedance is lower.

Low-Pass Filter with Super Capacitor

 

1

in

1

X

Z  Cs R Cs



1

2 2

1 1

1

1

out in

in in

X

V Z

V 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.

Example 14.24: Poor Low-Pass Filter

 

4

1 2

out X out out X

X

V V Cs R V V R Cs

R

  

       

 

 

 Node 4 is no longer a scaled version of the V_out. Therefore the output can only be sensed at node 1, suffering from a high impedance.

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

1 H j









 

  

 

ω_-3dB=ω_0, for all n.

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

Poles of the Butterworth Response

0

2 1

exp exp , 1, 2, ,

2 2

k

j k

p j k n

n

  ^{ }  ^

   

  

2

-Order nth-Order

Example 14.24: Order of Butterworth Filter

 The minimum order of the Butterworth filter is three.

2 2 1

64.2 f n

f

 

  

 

2

2

f  f

Specification: passband flatness of 0.45 dB for f < f₁=1 MHz, stopband attenuation of 9 dB at f₂=2 MHz.

( 1 1MHz) 0 95 H f

   

2 2

1 0

1 0 95

1 2

f n





 

 

  

 

3, 0 2 (1 45MHz) n    

( 2 2MHz) 0 355 H f

   

2 2

2 0

1 0 355

1 2

f n





 

 

  

 

Example 14.25: Butterworth Response

1

2 2

2 *(1.45 ) * cos sin

3 3

p MHz  j 

 ^{ }

   

 

3

2 2

2 * (1.45 ) * cos sin

3 3

p MHz  j 

 ^{ }

   

 

2 2 *(1.45 )

p   MHz RC section

2^nd-order SK Using a Sallen and Key topology, design a Butterworth filter for the

response derived in Example 14.24.

Example 14.25: 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 p H s

s p s p s s



  

   

 

        

2 (1 45MHz) and 1 / 2 cos 2 1

n Q 3

       

1 2 1k , 2 54.9pF, and 1 4 2

R  R   C  C  C

3 3

3 3

1 2 (1 45MHz) R 1k and C 109.8pF

R C        

Chebyshev Response

 

2 2

0

1

1 _n

H j

C



 





 

  

 

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

Chebyshev Polynomial

Chebyshev Polynomial

Chebyshev polynomial for n=1,2,3

Resulting transfer function for n=2,3

1

0

0 0

1

cos cos ,

Cn  n 

 

 







   

 

   

   

 

Chebyshev Response

2 2 1

0

( ) | 1

1 cos cos

HPB j

n



 





 

 

  

 

2 2 1

0

( ) | 1

1 cosh cosh

HSB j

n



 





 

 

  

 

2

Peak-to-peak Ripple dB 20 log 1

Example 14.26: Chebyshev Response





H 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

1 0 95 0 329

1





    



 

3 2 2

0 0

1

1 4 3

H j

 

  



   

 

   

   

 

Example 14.27: Order of Chebyshev Filter

Specification:

Passband ripple: 1 dB Bandwidth: 5 MHz

Attenuation at 10 MHz: 30 dB What’s the order?

1 dB  20 log 1  

    0 509

Attenuation at   2 

 10 MHz: 30 dB

2 2 1

1 0 0316

1 0 509 cosh ( cosh 2) n

 

 

cosh (1 317 )

 n  3862  n   3 66  n  4

Example 14.28: Chebyshev Filter Design

 

 

0 0

2 1 1 1 2 1 1 1

sin sinh sinh cos cosh sinh

2 2

k

k k

p j

n n n n

 

 

 

 

     

     

   

2,3

0.337

0.407

p     j 

SK2

1,4 0.140 0 0.983 0

p  





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 4

1

1 4

2 0

2 2

0 0

( )( )

( ) ( )( )

0 986

0 28 0 986

SK

p p

H s

s p s p

s s



 

 

  

 

   

1 0 993 0 2 (4 965MHz)

n       

1 3 55

Q  

1 2 1 k , 1 50 4 2

R  R   C   C

1 2

1 2 (4 965MHz) 50 4

4.52 pF, 227.8 pF R C

C C



  



  

2 3

2

2 3

2 0

2 2

0 0

( )( )

( ) ( )( )

0 279

0 674 0 279

SK

p p

H s

s p s p

s s



 

 

  

 

   

2 0 528 0 2 (2 64MHz)

n       

2 0 783 Q   

1 2

2 1

1 2 (2 64 MHz) 2 45

38 5 pF, C 94.3 pF R C

C



  



   

1 2 1 k , 1 2 45 2

R  R   C   C