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

CH 15 Analog Filters

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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


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


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


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


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


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.


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.


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


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


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


(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


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


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


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


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.


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.


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


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.


60 / 70

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

61 / 70

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


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

.

62 / 70

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


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


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


68 / 70

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?


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. .

69 / 70

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 .


chapter 14 analog filters - seoul national university · 2021. 2. 17. · 1 / 70 chapter 15 analog...

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