Filters and Tuned Amplifierscc.ee.nchu.edu.tw/~aiclab/teaching/Electronics3/lect12.pdf · Filters...

Ching-Yuan Yang

National Chung-Hsing UniversityDepartment of Electrical Engineering

Filters and Tuned Amplifiers

Microelectronic Circuits

12-1 Ching-Yuan Yang / EE, NCHUMicroelectrics (III)

Outline

Filter Transmission, Types, and Specification

The Filter Transfer Function

Butterworth Filters and Chebyshev Filters

First-Order and Second-Order Filter Functions

The Second-Order LCR Resonator

Second-Order Active Filters Based on Inductor Replacement

Second-Order Active Filters Based on the two-Integrator-Loop Topology

Single-Amplifier Biquadratic Active Filters

Sensitivity

Switch-Capacitor Filters

Tuned Amplifiers


Filters

Electronic filters are an important building block of communication and instrumentation systems.

Filter design is one of the very few areas of engineering for which a complete design theory exists, starting from specification and ending with a circuit realization.


Filter Types (Implementation)

Passive LC filters

Oldest technology

Work well at high frequency

For low frequency application (DC ~ 100kHz), L’s are large and impossible to fabricate in ICs.

Inductorless filter

Active-RC filters

Switched-capacitor filters

Transconductance-C filters

…


Filter Transmission, Types, and Specification


Two-Port Network

The filters studied in this chapter are linear circuits represented by the general two-port network.

)()()()()(

)( ωφωω j

i

o ejTjTsVsV

sT =⇒≡Transfer function:magnitude phase

Gain function: )(log20)( ωω jTG = dB

Attenuation function: )(log20)( ωω jTA −= dB

Output spectrum: )()()( ωωω jVjTjV io =


Filter Specification

Ideal transmission characteristics of the four major filter types:


Specification of the Transmission Characteristics of a Low-Pass Filter

ωp : the pass band edge

Amax : the max. allowed variation in passband transmission

ωs : the stop band edge

Amin : the min. required stopband attenuation


Ideal LP:

Lower Amax

Higher Amin

Selectively ratio 1ωω

→s

p


Transmission Specification for a Bandpass Filter

Note that this bandpass filter has a monotonically decreasing transmission in the passband on both sides of the peak frequency.


The Filter Transfer Function


Filter Transfer Function

Filter transfer function

Poles and zeros must be real or complex conjugate pairs.

To obtain zero or small stopband transmission, zeros are usually placed on

the jω-axis at stopband frequencies.

01

1

01

1)(bsbs

asasasT N

NN

MM

MM

++++++

= −−

−−

N : the filter order (the degree of the denominator)

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

)(21

21

N

MM

pspspszszszsa

sT−−−−−−

= z1, z2, ⋅⋅⋅, zM : transfer function zeros

p1, p2, ⋅⋅⋅, pN : transfer function poles


Pole-zero pattern

N = 5, fifth-order LPF01

22

33

44

5

22

221

24 ))((

)(bsbsbsbsbs

ssasT ll

+++++++

=ωω


Pole-zero pattern for the bandpass filter (N = 6)

05

56

22

221

25 ))((

)(bsbs

sssasT ll

+++++

=ωω


Pole-zero pattern for the low-pass filter (N = 5)

012

23

34

45

0)(bsbsbsbsbs

asT

+++++=


Butterworth Filters


Butterworth Filter

N-th order Butterworth filter:

N

p

jT2

21

1)(

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

ωωε

ω

Butterworth filter exhibits a monotonically decreasing transmission with all the transmission zeros at ω = ∞.

At ω = ω p,

ε determines the maximum variation in passband

transmission,

21

1)(

εω

+=pjT

ωp: passband edge

1101log20 102max

max −=⇒+= AA εε

At the edge of the stopband, ω = ωs, the attenuation of the Butterworth filter is

[ ][ ] min

22

22

)/(1log10

)/(11log20)(

A

AN

ps

Npss

≥+=

+−=

ωωε

ωωεω

Magnitude response:


Maximally flat response

The first 2N−1 derivative of |T | relative to ω are zero at ω = 0.

Response is very flat near ω = 0.

Order N passband flatness

Attenuation at stopband edge ω =ωs

Filter order can usually be obtained from A(ωs ).

2 2

2 2

120

1

10 1

= −+

⎡ ⎤= + ≥⎣ ⎦

s Ns p

Ns p

A

A

ωε ω ω

ε ω ω min

( ) log( / )

log ( / )


Magnitude response of Butterworth filters of various order with ε = 1


Normalized Butterworth Polynomials

Find poles:

22 2

2 2

22

2

( ) ( )

1 1( ) ( ) ( )

1 1 ( 1)

1( )

1 ( 1)

=⎯⎯⎯→

− = = =⎛ ⎞ ⎛ ⎞

+ + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=⎛ ⎞

+ − ⎜ ⎟⎜ ⎟⎝ ⎠

s j

N N

N

p p

N

N

p

T s T j

T j T j T j

T ss

ω ω

ω ω ωω ωε εω ω

εω

2 2

22

11

⎛ ⎞ ⎛ ⎞ −= − − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

N NN

N

p p

s sεω ω ε

( )( )

; is even.poles where

; is odd.

2

2 2

2

2 2

1

0 1 2 2 11

+ ⋅

⋅

⎧− =⎪⎪= = −⎨⎪ =⎪⎩

j l

j l

eN

l Ne

N

π π

πε ε

ε ε

( )

( ), , , ,( )


E.g. N = 2, l = 1

;

poles

4

2

24 4

41

4 22

43

34 2

4

1

0 1 2 3

1 12 2

1 12 2

1 12 2

1 12 2

+ ⋅

⎛ ⎞+ ⋅⎜ ⎟⎝ ⎠

−

⎛ ⎞− +⎜ ⎟⎝ ⎠

⎛ ⎞− +⎜ ⎟⎝ ⎠

⎛ ⎞− +⎜ ⎟⎝ ⎠

⎛ ⎞= − =⎜ ⎟⎜ ⎟

⎝ ⎠

= =

⎧ ⎛ ⎞= = +⎪ ⎜ ⎟⎝ ⎠⎪

⎪ ⎛ ⎞= = − +⎜ ⎟⎝ ⎠

= ⎨⎛ ⎞= = − −⎜ ⎟⎝ ⎠

⎛ ⎞= = −⎜ ⎟⎝ ⎠

j l

p

lj

p

j

p p

j

p p

j

p p

j

p p

se

s e l

s e j

s e j

s e j

s e j

π π

π π

π

π π

π π

π π

ω

ω

ω ω

ω ω

ω ω

ω ω

( )

, , ,

⎪⎪

⎪⎪⎪⎪⎪⎩


Normalized polynomials for N = 2

Take left plane poles

Let ωp = 1 frequency normalization

Normalized Butterworth polynomials

for 2

2 3

22

1 12

1 21 1

1 2

= = =⎛ ⎞⎛ ⎞ + ++ +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= + +

T s Ns ss s

s s

B s s s

( ) .

( )

N

1

2

3

4

5

6

7

8

(s + 1)

(s2 + 1.414s + 1)

(s + 1)(s2 + s + 1)

(s2 + 0.765s + 1)(s2 + 1.848s + 1)

(s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)

(s2 + 0.518s + 1)(s2 + 1.414s + 1)(s2 + 1.932s + 1)

(s + 1)(s2 + 0.445s + 1)(s2 + 1.247s + 1)(s2 + 1.802s + 1)

(s2 + 0.390s + 1)(s2 + 1.111s + 1)(s2 + 1.663s + 1)(s2 + 1.962s + 1)

Factors of polynomial Bn(s)


Graphical Construction of Butterworth Filters

)())(()(

21

0

N

N

pspsps

KsT

−−−=

ω

Transfer function:

where K is a constant equal to the required dc gain of the filter.


Graphical Construction of Butterworth Filters

General case: Graphical construction for determining Butterworth filter of order N.

All the poles lie in the left half of the s-plane on a circle of radius ω0 = ωp(1/ε )1/N ,

where ε is the passband deviation parameter. ( )

Transfer function:)())((

)(21

0

N

N

pspsps

KsT

−−−=

ω


110 10max −= Aε



How to design a Butterworth filter

Determine ε.

Determine the required filter order N.

Determine the N natural modes.

Determine T(s).

1101log20 102max

max −=⇒+= AA εε

[ ] min22 )/(1log10)( AA N

pss ≥+= ωωεω

)())(()(

21

0

N

N

pspspsK

sT−−−

=ω


Example: Find the Butterworth Transfer Function

Low-pass filter specifications: fp = 10 kHz, Amax = 1 dB, fs = 15 kHz,

Amin = 25 dB, dc gain = 1.

Determine ε :

Determine N :

Determine the N natural modes:

Combining p1 with its complex conjugate p9 yields the factor (s2 + s0.3472ω0 + ω02 )

in the denominator of the transfer function. The same can be done for the other

complex poles.

[ ] min22 )/(1log10)( AA N

pss ≥+= ωωεω

5088.0110110 10110max =−=−= Aε

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

≥⇒

p

s

A

N

ωω

ε

log

110log

21 2

10min

76.8≥∴ N Select N = 9

The poles all have the same frequency:

rad/s 10773.6)5088.0/1(10102)/1( 49/1310 ×=××== πεωω N

p

The first pole: )9848.01736.0()80sin80cos( 001 jjp +−=+−= ωω


Determine T(s):

Poles of the ninth-order Butterworth filter

)3472.0)()(5321.1)(8794.1)(()( 2

0022

0022

0022

002

0

90

ωωωωωωωωωω

+++++++++=

ssssssssssT


Chebyshev Filters

Equiripple in the passband

Monotonically decreasing in the stopband

Odd-order filter exhibits |T(0)| = 1

Even-order filter exhibits |T(0)| = |T(ωp )| = Amax

The total no. of passband maxima and minima equals the order of the filter, N. All the transmission zeros at ω = ∞, making it an all-pole filter.


N-th order Chebyshev filter ωp: passband edge

At ω = ω p,

ε determines the passband ripple,

p

p

p

p

T jN

T jN

ω ω ωε ω ω

ω ω ωε ω ω

−

−

= ≤+

= ≥+

for

for

( )cos [ cos ( / )]

( )cosh [ cosh ( / )]

2 2 1

2 2 1

1

1

1

1

( )pT jωε

=+ 2

1

1

maxmax log AA ε ε= + ⇒ = −10220 1 10 1


The attenuation achieved by the Chebyshev filter at the stopband edge (ω = ωs ) is

The poles of the Chebyshev filter:

The transfer function of the Chebyshev filter:


min( ) log[ cosh ( cosh ( / ))]s s pA N Aω ε ω ω−= + ≥2 2 110 1

sin sinh sinh cos cosh sinh

, , ,

k p pk k

p jN N N N

k N

π πω ωε ε

− −− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

1 12 1 1 1 2 1 1 12 2

1 2

( )( )( ) ( )

Np

NN

KT s

s p s p s p

ωε −=

− − −11 22


How to design a Chebyshev filter

Determine ε from Amax.

Determine N, the number of the order, from given A(ωs ).

Determine the poles, pk, using Chebyshev equation.

Determine the transfer function, T(s).

where K is the dc gain.

sin sinh sinh cos cosh sinh

, , ,

k p pk k

p jN N N N

k N

π πω ωε ε

− −− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

1 12 1 1 1 2 1 1 12 2

1 2

( )( )( ) ( )

Np

NN

KT s

s p s p s p

ωε −=

− − −11 22


Example: Find the Chebyshev Transfer Function

Low-pass filter specifications: fp = 10 kHz, Amax = 1 dB, fs = 15 kHz,

Amin = 25 dB, dc gain = 1.

Determine ε :

Determine N :

To meet identical specifications, one requires a lower order for the Chebyshev than

the Butterworth filter. Alternatively, for the same order and the same Amax, the

Chebyshev filter provides greater stopband attenuation than the Butterworth filter.

Determine the N natural modes:

Determine T(s):

max .Aε = − = − =10 11010 1 10 1 0 5088

min( ) log[ cosh ( cosh ( / ))]s s pA N Aω ε ω ω−= + ≥2 2 110 1

Select N = 5

dB

dB

( ) .

( ) .s

s

N A

N A

ωω

= ⇒ == ⇒ =

4 21 6

5 29 9

, ( . . )

, ( . . ) ( . )p

p p

p p j

p p j p

ω

ω ω

= − ±

= − ± = −1 5

2 4 3

0 0895 0 9901

0 2342 0 6119 0 2895

( ). ( . )( . )( . . )

p

p p p p p

T ss s s s s

ωω ω ω ω ω

=+ + + + +

5

2 2 2 28 1408 0 2895 0 4684 0 1789 0 9883where ωp = 2π×104 rad/s


First-Order and Second-Order Filter Functions


First-Order filters

Transfer function:

The bilinear transfer function characterizes:

a pole (natural mode) at s = −ω0,

a transmission zero at s = −a0 /a1,

a high-frequency gain that approaches a1, i.e., T( j∞) = a1.

The numerator coefficients, a0 and a1, determine the type of filter (e.g., low-

pass, high-pass, etc.)

0

01)(ω++

=s

asasT


Low-pass (LP) filters

( )a

T ss ω

=+

0

0


High-pass (HP) filters

( )a s

T ss ω

=+1

0


General filters

0

01)(ω++

=s

asasT


First-Order All-Pass Filters

Special case of first-order filters

Transmission zero and pole are symmetrically located relative to jω-axis

Transmission is constant at all frequencies

Phase is not constant at all frequencies

Most often used in the design of delay equations


Second-Order Filters (Biquadratic Filters)

Transfer function:

where ω0 and Q determine the natural modes (poles) according to

We are usually interested in the case of complex-conjugate natural modes,

obtained for Q > 0.5.

The numerator coefficients (a0, a1, a2) determine the type of filter function

(i.e., LP, HP, etc.).

20

02

012

2)(ωω +⎟

⎠⎞

⎜⎝⎛+

++=

sQ

s

asasasT

200

21 41

12

,Q

jQ

pp −±−= ωω


Definition of the parameter ω0 and Q of a pair of complex conjugate poles:

Pole frequency ω0 : distance of pole (from origin).

Pole quality factor Q : distance of the poles from the jω-axis

Higher Q higher selectivity (pole is closer to jω-axis)

Q = ∞ poles are on the jω-axis

can yield sustained oscillation

Q is negative poles are in the right half of s-plane

unstable


2nd-order Low-pass Filters

Two transmission zeros are at s = ∞.

Magnitude response

The peak occurs only for .

The response obtained for is the Butterworth, or maximally flat, response.

2/1>Q

2/1=Q


2nd-order High-pass Filters

Two transmission zeros are at s = 0 (dc).

Magnitude response shows a peak for .

Duality between the LP and HP response.

2/1>Q


2nd-order Bandpass Filters

One transmission zero is at s = 0 (dc), and the other is at s = ∞.

Magnitude response peaks at ω = ω0. The center frequency is equal to the pole frequency ω0.

The selectivity of 2nd-order bandpass filter is measured by 3-dB bandwidth.

As Q increases, BW decreases and the bandpass filters become more selective.

QQ 24

11, 0

2021

ωωωω ±+=Q

BW 012

ωωω =−≡


Notch Filters

Transmission zeros are located on the jω-axis. (ωz = ±jωn )A notch in the magnitude response occurs at ω = ωn. (ωn : notch frequency)

3 cases of 2nd-order notch filters:Regular notch (ωn = ω0 )Low-pass notch (ωn > ω0 )High-pass notch (ωn < ω0 )

The transmission at dc and at s = ∞ is finite. (Because there are no transmission zeros at either s = 0 or s = ∞.)



2nd-order All-pass Filters

Two transmission zeros are in the right half of the s-plane, at the mirror-image locations of the poles.

Magnitude response is constant over all frequencies; the flat gain is equal to |a2|.

The frequency selectivity of the all-pass function is its phase response.


The Second-Order LCR Resonator


Second-Order LCR Resonator

Use LCR resonator to derive various second-order filter functions.

Replacing inductor L by a simulated inductance obtained using an OPAMP-

RC circuit results in an OPAMP-RC resonator.

The second-order parallel LCR resonator:

(without changing poles)


Two ways for exciting the resonator without change its natural structure:

The resonator is excited with a current source I connected in parallel.

The resonator is excited with a voltage source Vi.

LCCRss

Cs

RsC

sLYI

Vo

111111

2 ++=

++==

CRQLCo11 02 ==

ωω

CRQLC 001 ωω ==

The resonator poles are poles of Vo /I.

The resonator poles are poles of Vo /Vi.

LCCRss

LC

sCRsL

sCR

VV

i

o

11

1

1

1

2 ++=

⎟⎠⎞

⎜⎝⎛+

=

CRQLCo11 02 ==

ωω

CRQLC 001 ωω ==


Various Second-Order Function Using Resonator

Transmission zeros:

The values of s are at which Z2(s) is zero, provided that Z1(s) is not simultaneously zero.

The values of s are at which Z1(s) is infinite, provided that Z2(s) is not simultaneously infinite.

)()()(

)()(

)(21

2

sZsZsZ

sVsV

sTi

o

+==


Realization of the Low-Pass Function

LCCRss

LC

RsC

sL

sLYY

YZZ

ZVV

sTi

o

11

1

11

1

)(

2

21

1

21

2

++=

++=

+=

+=≡


Realization of the High-Pass Function

20

02

22)(

ωω ++=≡

Qss

saVV

sTi

o


Realization of the Bandpass Function

LCCRss

CRs

sCsLR

RYYY

YsT

CLR

R

11

1

11

1

)(2 ++

=++

=++

=


Realization of the Notch Functions

20

02

20

2

2)(ωω

ω

++

+=≡

Qss

sa

VV

sTi

o

General notch function:

2111

n

CLω

= The L1C1 tank circuit introduces a pair of transmission zeros at ±jωn.

LLLCCC ==+ 2121


Realization of the low-pass notch (LPN) function 0ωω >n

))(( 212111 CCLLCL +<

Satisfied with L2 = ∝, L1 = L.

2 2

22 20

0

( ) o n

i

V sT s a

V s sQ

ωω ω

+≡ =

+ +

where ωn2 =1/LC1, ω0

2 = 1/L(C1 + C2), ω0 /Q = 1/CR,

and the high frequency a2= C1 /(C1 + C2).


Realization of the High-Pass Notch (HPN) Function 0ωω <n

Satisfied with C2 = 0, C1 = C.

CLLCRss

CLs

VV

sTi

o

)(11

1

)(

21

2

1

2

++

+=≡

))(( 212111 CCLLCL +>


Realization of the All-Pass Function

20

02

20

02

)(ωω

ωω

++

+−=≡

Qss

Qss

VV

sTi

o

20

02

02

1)(ωω

ω

++−=⇒

Qss

Qs

sT

Bandpass filter with a center-frequency

gain of 2.

Allpass filter with a flat gain of 0.5:

20

02

0

5.0)(ωω

ω

++−=

Qss

Qs

sT


Realization of Various Second-Order Filter Functions Using LCR Resonator


Second-Order Active Filters Based on Inductor Replacement


The Antoniou Inductance-Simulation Circuit (1969)

Many inductor replacement circuit exists – Antoniou inductance simulation circuit is the best, i.e., it is very tolerant to the nonideal properties of the OPAMP, gain and bandwidth.


The design is usually based on selecting R1 = R2 = R3 = R5 = R and C4 = C, which leads to L = CR2 .

.2

5314

2

5314

1

1

RRRRC

LL

RRRRC

sIV

Zin

=

=≡

by given inductance an of that is h whic

is impedance input The


The Op Amp-RC Resonator

An LCR resonator

An op amp-RC resonator obtained by replacing the inductor L in the LCR resonator with a simulated inductance realized by the Antoiou circuit.

531

2

4

66660

25316460 /

11

RRRR

CC

RRCQ

RRRRCCLC

==

==

ω

ω

Selecting R1 = R2 = R3 = R5 = R and C4 = C6 = C, then

RR

Q

CR

6

01

=

=ω


Implementation of the buffer amplifier K

Note that not only does the amplifier K buffer the output of the filter, but it also allows the designer to set the filter gain to any desired value by appropriately selecting the value of K.


Realizations for the Various Second-Order Filter Functions Using the Op-Amp-RC Resonator



Homework

HW1

Problems 8, 12, 19, 35, 41


Second-Order Active Filters Based on the two-Integrator-Loop Topology


Two-Integrator-Loop Biquad

Derivation

High-pass:

Bandpass:

Low-pass:

Block diagram:

hp

i

V KsV s s

Qω ω

=+ +

2

2 200

hp hp hp

hp hp hp

i

i

V V V KVQ s s

V KV V VQ s s

ω ω

ω ω

⎛ ⎞⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

20 0

2

20 0

2

1

1

bp hp( / )

i i

V s VK sV Vs s

Q

ωωω ω

−= =

+ +

00

2 200

lp hp( / )

i i

V s VKV Vs s

Q

ωωω ω

= =+ +

2 2200

2 200


Universal active filter realizes LP, BP, and HP, simultaneously.

Versatile

Easy to design

Very popular

Block-diagram realization:

Performance limited by the finite BW of OPAMP.

200

20

bp )/()(

ωωω

++=

Qss

sKsT

200

2

20

lp )/()(

ωωω

++=

Qss

KsT


Integrator

Miller integrator

Ideal OPAMP

Actual OPAMP (one-pole example)

o

i

V sT s

V s sRC= = −

( )( )

( )1

OPAMP transfer function

, , ,

(i.e., unity-gain BW )

( )/

/

/

( )( )

i o o

u

AA s

s s

R R A A RC s

s RC

AT s

ssRCA

A s

ω

=+

→∞ = >> >>

= >>−

=⎛ ⎞+ +⎜ ⎟

⎝ ⎠

0

1

0 1

1

0

00 1

1

0 1 1

1

1 1


log f

dB

Open-loop gain A0

Real integrator

Ideal integrator

f11

RCA0

20log A0

20dB/dec

40dB/dec

( )

( )

uu

T s sRCsRC

A s

RC ss s s

RC RCω

ω

= − ++

= −⎡ ⎤⎛ ⎞+ + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

2 11

11

11

Since

( )( )

u

u u u

uu

A s s

RC

A RC

sRC A RC

T sRC

s sA RC

ω

ω ω ω

ωω

= >>

>>

>>

+ + ≈ ≈ +

= −⎛ ⎞

+ +⎜ ⎟⎝ ⎠

0 1 1

0

10

0

1

1

1 1

11


Circuit Implementation of Universal Filter

Kerwin-Huelsman-Newcomb (KHN) biquad

Arbitrarily and practically choose R1, Rf , R2, and R3 to meet the above relationship.

fR

R

RQ

R

⎧ =⎪⎪⎨⎪ = −⎪⎩

1

2

3

1

2 1


To obtain notch and all-pass functions, the three outputs are assumed with appropriate weight.

E.g. Notch filter: choose

hp bp lp hp bp lpF F F F F F

o iH B L H B L

R R R R R RV V V V V T T T

R R R R R R⎛ ⎞ ⎛ ⎞

= − + + = − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( / ) ( / ) ( / )( / )

o F H F B F L

i

V R R s s R R R RK

V s s Qω ω

ω ω− +

= −+ +

2 20 0

2 20 0

, H nB

L

RR

Rωω⎛ ⎞

= ∞ = ⎜ ⎟⎝ ⎠

2

0


Alternative Two-Integrator-Loop Biquad

Two-Thomas biquad (Single-ended fashion)

Universal filter


Two-Thomas biquad with input feedward paths

Realize all special second-order function

Design table

222

22

31

12

11

111

RCQCRss

RRCRRr

RCs

CC

s

V

V

i

o

++

+⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟

⎠⎞

⎜⎝⎛

−=


Single-Amplifier Biquadratic Active Filters


Single-OPAMP Biquad (SAB) – compared with two-integrator biquad

Economic

require 1 OPAMP instead of 3 or 4.

More sensitive to R and C variations

Greater dependence on limited gain and bandwidth

Single-amplifier biquads (SABs) are therefore limited to the less strigent filter

spec. E.g., pole Q < 10.

Biquads can be cascaded to construct high-order filters.


SAB Synthesis

Two-step procedure

Synthesis of a feedback loop that realizes a pair of complex conjugate

poles characterized by ω0 and Q.

Injecting the input signal in a way that realizes the desired transmission

zeros.

Synthesis of the feedback loop


The produce of the op amp gain A and t(s):

The characteristic equation: 1 + L(s) = 0

The poles sP of the close-loop circuit obtain as solutions to the equation

In the ideal case, A = ∞ and the pole are obtained from

The circuit poles are identical to the zeros of the RC network.

N(s): zeros of the RC network

D(s): poles of the RC network

( )( ) ( )

( )AN s

L s At sD s

= =

( )Pt sA

= −1

( )PN s = 0


Bridged-T RC Networks

open-circuited

open-circuited

b

a

V

Vst =)(


The pole polynomial of the active-filter circuit will equal to the numerator polynomial of the Bridged-T network.

s s s sQ C C R C C R Rω ω

⎛ ⎞+ + = + + +⎜ ⎟

⎝ ⎠2 2 20

01 2 3 1 2 3 4

1 1 1 1

C C R R

C C R RQ

R C C

ω

−

=

⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

01 2 3 4

1

1 2 3 4

3 1 2

1

1 1

and

Let

QC C C m Q RC

R R

RR

m

ω⎧ = = ⇒ = =⎪⎪

=⎨⎪⎪ =⎩

21 2

0

3

4

24

Q and ω0 can be used to determine the component values.


Injecting the input signal

To obtain transmission zero

Example: A bandpass filter

o

i

sV C RV

s sC C R C C R R

α−=

⎛ ⎞+ + +⎜ ⎟

⎝ ⎠

1 4

2

1 2 3 1 2 3 4

1 1 1 1The denominator polynomial is identical to the numerator polynomial of t(s).


Complementary Transformation

Complement of transfer function – interchanging input and ground

Note ac a c

bc b c

V V Vt

V V V−

= =−


Two-step procedure

Nodes of the feedback network and any of the op amp inputs that are connected to ground should be disconnected from ground and connected to the op amp output. Conversely, those nodes that were connected to the op amp output should be now connected to ground. That is, we simply interchange the op amp output terminal with ground.

The two input terminals of the op amp should be interchanged.

Example

01 =+ At0)1(

11 =−

+− t

A

ACharacteristic equation:

01 =+ Atsame poles


Ex. Appling the complementary transformation to the feedback loop

Sallen-and-Key SAB circuit: (HP function)

C1 = C2 = C, R3 = R, R4 = R/4Q2, CR = 2Q/ω0, and the value of C is arbitrarily chosen to be practically convenient.


Another example: LP filter obtained by injecting

1

214

2143

21430

11

1

−

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=

RRC

RRCCQ

RRCCω

Selecting R1 = R2 = R, C4 = C, C3 = C/m,

02 /24 ωQCRQm ==

Complementary Trans.

Injecting Vi through R1:


Sensitivity


Sensitivity

Real components deviate from their designed values

initially inaccurate due to fabrication tolerances.

drift due to environmental effects such as temperature and

humidity.

chemical changes which occurs as the circuit ages.

inaccuracies in modeling the passive and active devices, e.g.,

nonideal OPAMP and parasitics.

All coefficients, and therefore poles and zeros of H(s), depend on

circuit element.

The size of H(s) error depends on how large the component

tolerances are and how sensitive the circuit’s performance is to

these tolerances.


Sensitivity calculation, allow the designer

to select the better circuit from those in the literature.

to determine whether a chosen filter circuit satisfies and will keep satisfying the given specifications.

Component χ

Performance criterion p(χ), such as

Quality factor

Pole frequency

Zero frequency

or p(s, χ), if p is also a function of frequency and stands for

H(s),

magnitude of H(s),

phase of H(s).


Sensitivity

Taylor series

pSχ( , ) ( , )

( , ) ( , ) ( )P s P s

P s P s d dχ χ

χ χχ χ χ χχ χ

∂ ∂= + + +

∂ ∂0 0

22

0 212

P

P

d dPd

P sP s P s d P s d

P s P s dP s P s

P s d PP PS

P s d

P PS

χ

χ

χχ χ χ

χ

χχ χχ χ

χχ χ χ χ χχ

χ χχ χχ χ χ χ

χχχ χ χ χ χ

χχ

<<=

⎧ ∂Δ = + − ≈⎪ ∂⎪⎨

Δ ∂⎪ ≈⎪ ∂⎩

∂ ∂= = =

∂ ∂

Δ ∂<< ≈

∂

if and is small

if , then

( , )( , ) ( , ) ( , )

( , ) ( , )( , ) ( , )

( , ) (ln )/( , ) / (ln )

/

0

0

0 0 0

00

0 0 0

0 0

0 0 0

0

0

0

1

1χ χ/


Sensitivity is to predict the deviations from

the tolerances in component values

the finite op-amp gain

Definition

x denotes the value of a component (a resistor, a capacitor, or an amplifier gain) and y denotes a circuit parameter of interest (ω0 or Q).

For small changes

/lim

/yx

x

y y y xS

x x x yΔ →

Δ ∂≡ =

Δ ∂0

//

yx

y yS

x xΔ

≈Δ


Example: find the sensitivities of ω0 and Q relative to all passive components

Note that: A 10% increase in R3 results in a 5% decrease in the value of ω0 and a 5% increase in the value of Q.

C C R RS S S Sω ω ω ω= = = = −0 0 0 01 2 3 4

12

, ,Q Q Q QC C R RS S S S= = = = −

1 2 3 4

1 10

2 2

Assume C1 = C2,

s s s sQ C C R C C R Rω ω

⎛ ⎞+ + = + + +⎜ ⎟

⎝ ⎠2 2 20

01 2 3 1 2 3 4

1 1 1 1

C C R R

C C R RQ

R C C

ω

−

=

⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

01 2 3 4

1

1 2 3 4

3 1 2

1

1 1

1

2

1

1

2

2

1

1

2

2

11

−

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

C

C

C

C

C

C

C

CS Q

C

/lim

/yx

x

y y y xS

x x x yΔ →

Δ ∂≡ =

Δ ∂0


- The sensitivities relative to the amplifier gain

Assume the op amp have a finite gain A, the characteristic equation for the loop

1 + At(s) = 0

Using the obtained design, C1 = C2 = C, R3 = R, R4 = R/4Q2, and CR = 2Q/ω0,

1 + At(s) = 0

20

20

2

200

2

)12)(/(

)/()(

ωωωω

+++++

=QQss

Qssst

( ) 0)/()12)(/( 200

220

20

2 =++++++ ωωωω QssAQQss

01

21 2

0

202 =+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

++∴ ωωA

Q

Qss

⎪⎩

⎪⎨⎧

++=

=

)1/(21 2

00

AQ

QQa

a ωω

)1/(21

)1/(2

1 0 2

20

+++

⋅+

==AQ

AQ

A

ASS aa Q

AAω .12

2 22

>>>>≈ AQAA

QS aQ

A and for :Note


Switch-Capacitor Filters


Switched-Capacitor Filters

Switched capacitor performing as a simulated resistor

large area resistor smaller area capacitor

VLSI technology requirements

Good switch

Well-defined capacitor

OPAMP

First discussion on the equivalent resistance of a periodically switched capacitor methods in 1946.

Rapid evolution of practical SC is due to the implementation in MOS technology.

Followed by a rapid development and implementation of analog signal processing techniques.

SC circuits are sampled-data analog MOS integrated circuits.


RC Time Constant

Active-RC integrator

If τ = R1C2, then

The absolute accuracies of R and C is very poor.

dCd dRR C

ττ

= + 21

1 2


Basic Principle of the Switched-Capacitor Filter Technique

Switched-capacitor integrator

Operation

Nonoverlapping two-phase clock

iC vCq 11=

c

Cav T

qi 1=

1CT

iv

R c

av

ieq =≡

An equivalent time-constant for the integrator:

1

22 C

CTRC ceq ==τ determined by Tc C2 /C1 accuracy


Accurate RC time constant

SC as a resistor

If R1 is replaced by a switched capacitor resistance realization with a C1, then

If clock frequency is assumed to be constant, then

Relative accuracy is good for two capacitors fabricated in the same integrated circuit.

Equivalent C

C C C

Q CV V TI I R

T T R C f C= = = = = =1 2

1

CC

C C

C C dCd dT dCT

f C C T C Cτττ

= = = + −2 2 2 1

1 1 2 1

1

dCd dCC C

ττ

= − →2 1

2 1

0


Z-Transform


At cycle n, i.e., t = nTs, we have Q1(n) = C1vi (n) and Q2(n) = C2vo(n)At cycle n + 1/2, i.e., t = (n + 1/2)Ts,

Q1(n + 1/2) = 0 Q2(n + 1/2) = Q2(n) − Q1(n) = C2vo(n) − C1vi(n)At cycle n + 1, i.e., t = (n + 1)Ts,

Q1(n + 1) = C1vi (n + 1) Q2(n + 1) = C2vo(n + 1) = Q2(n + 1/2) = C2vo(n) − C1vi(n)

Thus, the time-domain difference equation is

C2vo(n + 1) = C2vo(n) − C1vi(n)In the z-domain

-1

1 12 2 1 -1

2 2

( ) 1( ) ( )- ( ) - -

( ) -1 1-o

o o ii

v z C C zzC v z C v z C v z

v z C z C z= ⇒ = × = ×


Stray Capacitance in SC Integrators

11

12

( )( ) 1

o S

i

v z C C zv z C z

−

−

⎛ ⎞+= − ×⎜ ⎟ −⎝ ⎠


Practical Circuits of the Switched-Capacitor integrators

Noninverting switched-capacitor integrator

Inverting switched-capacitor integrator

( )( )

( )o

i

V z C zH z

V z C z

−

−= =−

11

12 1

( )( )

( )o

i

V z CH z

V z C z −= = −−

11

2

11


Practical Circuits of the Switched-Capacitor Filters

1

43210 RRCC=ω


Substituting for R2 and R4 by their SC equivalent values, that is

Give ω0 of the SC biquad as

If and C1 = C2 = C , then C3 = C4 = KC, where K = ω0Tc

(from Eq. 12.93).

Determine C5 :

The center-frequency gain of the bandpass function:

12

430

43210

1

1CCCC

TRRCC c

== ωω

44

33

CT

RCT

R cc == and

14

23

CCT

CCT cc =

(12.93)

QC

TQKC

QC

CCTCT

Q cc

c0

45

4

5 // ω====

CTC

QCC

c0

6

5

6 ω

==gainfrequency -Center


Summary of Filters

Continuous-time filter

RLC passive

RC active

Sampled-Data filter

Switched-Capacitor Filter

Digital Filter

OperationsMultiplyDelayAdd


Tuned Amplifiers


Tuned Amplifiers

A special kind of frequency-selective network

RF and IF application

Response is similar to that of bandpass filters

Center frequency ranges from a few hundred kHz to a few hundred MHz

Skirt selectivity SB⎛ ⎞⎜ ⎟⎝ ⎠

Center frequency

dB bandwidthdB bandwidth

SB

−=

−303


Single-Tuned amplifier

(Transistor operates in class A mode)

Equivalent circuit

( )

( )

L o FET

L FET

R R r

C C C

=

= +


( / ) /o m

i

V g sV C s s CR LC

= −+ +2 1 1

2nd-order bandpass function

Center frequency gain om

i

BCRLC

Q CRB

Vg R

V ω ω

ω

ω ω

=

= =

≡ =

= −0

0

00

1 1

m i m io

L

g V g VV

Y sCR sL

− −= =

+ +1 1


Inductor Losses

Inductor equivalent circuits

The inductor Q factor: (typically 50 ~ 200)

Relationship between Q and Rp:

For Q >> 1,

s

LQ

rω

≡ 00

( / )( )

( / )s

j QY j

r j L j L Qω

ω ω+

= =+ +

00 2

0 0 0

1 1 1 11 1

( )p

Y j jj L Q j L LQ j L R

ωω ω ω ω

⎛ ⎞≈ + = + = +⎜ ⎟

⎝ ⎠0

0 0 0 0 0 0

1 1 1 1 1 11

p pL L L R LQQ

ω⎛ ⎞

= + ≈ =⎜ ⎟⎝ ⎠

0 020

11 Fig.(a) Fig.(b)


Use of Transformers

Impedance transformer

Examples

Allow using a high inductance and a smaller capacitance.

Equivalent circuit:


Increase the effective input impedance of the second stage.

Equivalent circuit:


Amplifiers with Multiple Tuned Circuits

Greater selectivity can be obtained.

Double-tuned amplifier

Miller capacitance will cause

1. detuning of the input circuit.

2. skewing of the response of the input circuit.


Cascode and CC-CB Cascade

Amplifier configurations do not suffer from the Miller effect.

Cascode

Common-collector common-base cascade


Synchronous Tuning

N identical tuned circuits (do not interact)

Bandwidth shrinkage

where is known as the bandwidth-shrinkage factor.

-NBQω

=1

0 2 1

/ -N12 1


Stagger-Tuning

Maximal flatness around ω0 (Center frequency)


The flat response can be obtained by transforming the response of a maximally flat (Butterworth) low-pass filter up the frequency axis to ω0.

⎟⎟⎠

⎞⎜⎜⎝

⎛−++⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+

=

200

200

1

41

124

11

2

)(

Qj

Qs

Qj

Qs

sasT

ωωωω


For a narrow-band filter, Q >> 1, and s in the neighborhood of +jω0, the transfer function of 2nd-order BP filter can be expressed in terms of its poles as

The magnitude response has a peak value of a1Q /ω0 at ω = ω0.

LP to BP transformation for narrow-band filter:

Qjs

a

jQs

asT

2/)(

2/

2/

2/)(

00

1

00

1

ωωωω +−=

−+≈ (Narrow-band approximation)

LP BP

02/

)(0 ωω

jspQp

KsT

−=+

=


Transform a maximally flat 2nd-order LP filter (Q = 1/ ) to obtain a maximally flat BP filter:

2

BQ

BB

B

BQ

BB

B

011002

011001

2

222

2

222

ωωω

ωωω

≈=−=

≈=+=

(4th-order)


Magnitude response:


Homework

HW2

Problems 49, 56, 65, 67,74

Filters and Tuned Amplifierscc.ee.nchu.edu.tw/~aiclab/teaching/Electronics3/lect12.pdf · Filters...

Documents

Transcript of Filters and Tuned Amplifierscc.ee.nchu.edu.tw/~aiclab/teaching/Electronics3/lect12.pdf · Filters...