3. Filtros Ativos

8/18/2019 3. Filtros Ativos

1/63


2/63

Prof. Roberto M. Finzi Neto 12/12/14 2

SUMÁRIOa. Introduçãob. Fundamentos de Filtros Passa Baixa

I.

Resposta de Butterworth.

II. Resposta de Tschebyscheff.

III.

Resposta de BesselIV.

Fator de Qualidade Q

c. Projeto de Filtros Passa BaixaI.

Primeira OrdemII. Segunda OrdemIII.

Ordens Superiores

d.

Projeto de Filtros Passa AltaI.

Primeira OrdemII. Segunda OrdemIII. Ordens Superiores


3/63


SUMÁRIO

e. Projeto de Filtros Passa BandaI.

Segunda Ordem

II.

Quarta Ordem

f. Projeto de Filtros Rejeita BandaI.

Filtro Active Twin-T


4/63


BIBLIOGRAFIA

! MANCINI, R. “Op Amps for Everyone: design reference”. 2nd Edition.Elsevier, 2003. ISBN-13:978-0-7506-7701-1 ISBN-10:0-7506-7701-5

! MALVINO, Albert Paul, Princípios de Eletrônica, v ol. 2, 4ª Ed.. Brasil:McGraw-Hill, 2000.


5/63


a. Introdução! Def.: circuito ou dispositivo que permite a passagem de sinais

elétricos AC em determinadas faixas de frequências, enquanto quebloqueia o sinal em outras faixas.

! Principais aplicações:

!

Telecomunicações: filtram sinais para que reste apenas a faixa audível dosom (0 – 20kHz).

!

Comunicação de dados em RF. Filtram faixas estreitas de sinais em altafrequência (centenas de MHz) para a seleção de canais de comunicaçãode dados digitais.

!

Fontes DC de Sistemas de Energia. Usam filtros rejeita-banda parasuprimir a componente de 50Hz/60Hz vinda da rede AC.

!

Sistemas de Monitoramento e Aquisição de dados. Remoção de ruído dosinal fora da faixa de frequência de interesse.


6/63


a. Introdução! A altas frequências (> 1MHz) esses filtros são constituídos de

dispositivos passivos (RLC). A frequências menores o indutor se tornamuito grande e pesado.

! Os filtro ativos são compostos de um ou mais amplificadoresoperacionais, resistências e capacitores. Podem gerar a mesmaquantidade de polos de um filtro passivo sem a necessidade deindutores.

!

Abordaremos filtros ativos com os três principais tipos de resposta(Butterworth, Tschebyscheff e Bessel) seguido por seções quedescrevem as principais aplicações (passa baixa, passa alta, passafaixa, rejeita faixa e passa tudo).

just add a linear phase shift to each frequency component, thus contributing to a constant

time delay. These are called all-pass filters.

At high frequencies (> 1 MHz), all of these filters usually consist of passive components

such as inductors (L), resistors (R), and capacitors (C). They are then called LRC filters.

In the lower frequency range (1 Hz to 1 MHz), however, the inductor value becomes very

large and the inductor itself gets quite bulky, making economical production difficult.

In these cases, active filters become important. Active filters are circuits that use an op-

erational amplifier (op amp) as the active device in combination with some resistors and

capacitors to provide an LRC-like filter performance at low frequencies (Figure 16-1 ).

L R

R1

V N ~ T

V O U T

V N

c

C2

R 1

T Vouz

F igu re 16 1. Second O rde r Pass ive Low Pass an d Second O rde r Act ive Low Pass

261

nal conditioning stages. System power supplies often use band-rejection filters to sup-

press the 60-Hz line frequency and high frequency transients.

In addition, there are filters that do not filter any frequencies of a complex input signal, but

just add a linear phase shift to each frequency component, thus contributing to a constant

time delay. These are called all-pass filters.

At high frequencies (> 1 MHz), all of these filters usually consist of passive c omponents

such as inductors (L), resistors (R), and capacitors (C). They are then called LRC filters.

In the lower frequency range (1 Hz to 1 MHz), however, the inductor value becomes very

large and the inductor itself gets quite bulky, making economical production difficult.

In these cases, active filters become important. Active filters are circuits that use an op-

erational amplifier (op amp) as the active device in combination w ith some resistors and

capacitors to provide an LRC-like filter performance at low frequencies (Figure 16-1).

L R R1

V N ~ T

V O U T

V N

c

C2

R 1

T Vouz

F igu re 16 1 . Second Orde r Pass i ve Low Pass and Second Orde r Ac ti ve Low Pass

261

!


7/63


b. Fundamentos de Filtros Passa Baixas! O filtro passa baixa mais simples é composto por uma única rede RC

e sua função transferência demonstra a existência de um único polo.

onde s = j ω=j2!f para qualquer sinal senoidal.! A frequência de corte do filtro ocorre quando o sinal sofre uma

atenuação de 3dB ( Vo(t) = Vi(t)* 0,708) e é definida como:

!

Normalizada, a função transferência do filtro se torna:

.

ebyscheff, and Bessel), followed by five sections describing the most common

applications: low-pass, high-pass, band-pass, band-rejection, and all-pass fil-

r than resembling just another filter book, the individual filter sections are writ-

okbook style, thus avoiding tedious mathematical derivations. Each section

the general transfer function of a filter, followed by the design equations to cal-

individual circuit components. The chapter closes with a section on practical

s for single-supply fi lter designs.

of Low Pas s F i l ters

imple low pass filter is the pas sive RC low pass network shown in Figure 16 2.

R

V IN ~ ~ V OUT

c

d e r P a s s i ve R C L o w - P a s s

function is:

A(S) =

1

a ~ 1

s + ~ 1 + sR C

RC

complex frequency variable, s =~+ (~ , allows for any time variable signals. For

aves, the damping constant, (~, becomes zero and s = j(o.

alized p resentation of the transfer function, s is referred to the fi lter's corner

o r - 3 dB frequency, (oc, and has these relationships:

s _ J~ f = j ~

s ~ Jrcc

16 .2 F u n d a men t a l s o f L o w P ass F i lt e rs

The most simple low pass filter is the p assive RC low pass network show

R

V I N ~ ~ V O U T

c

F i gu r e 1 6- 2. F i r s t - O r d e r P a s s i v e R C L o w - P a s s

Its transfer function is:

A(S) =

1

a ~ 1

s + ~ 1 + sR C

RC

where the complex frequency variable, s =~ +( ~, allows for any time var

pure sine waves, the dam ping constant, (~, becom es zero and s = j(o.

For a normalized presentation of the transfer function, s is referred to

frequency, o r - 3 dB frequency, (oc, and has these relationships:

s _ J~ f = j ~

s ~ Jrcc

With the corner frequency of the low-pass in Figure 16-2 being fc = 1/

s = sRC and the transfer function A(s) results in:

1

A(s ) = 1 + s

(01)

f c =1

2!" !

R!

C

#$ c =1

R!

C

(04)

(06)

The most simple low pass filter is the pa ssiv e RC low pass

R

V I N ~ ~ V O U T

c

F i gu r e 1 6 -2 . F i r s t - O r d e r P a s s i v e R C L o w - P a s s

Its transfer function is:

A(S) =

1

a ~ 1

s + ~ 1 + sR C

RC

where the complex frequency variable, s =~ +( ~, al lows fo

pure sine waves, the damping constant, (~, becomes zer

For a normalized presentation of the transfer function, s i

frequency, o r - 3 dB frequency, (oc, and has these relatio

s _ J~ f = j ~

s ~ Jrcc

With the corner frequency of the low-pass in Figure 16-2


1

A(s ) = 1 + s

The magnitude of the gain response is:

IAI = 1

r s w e genera rans er unc on o a er, o owe y e es gn equa ons o ca -

late the individual circuit components. The chapter closes with a section on practical

sign hints for single-supp ly fi lter designs.

enta ls o f Low Pass F i l te rs

e most simple low pass filter is the pa ssive RC low pass network shown in Figure 16 2.

R

V I N ~ ~ V O U T

c

i r s t - O r d e r P a s s i v e R C L o w - P a s s

transfer function is:

A(S) =

1

a ~ 1

s + ~ 1 + sR C

RC

ere the complex frequency variable, s =~ + (~ , allows for any time variable signals. For

re sine waves, the damp ing constant, (~, becom es zero and s = j(o.

r a normalized presentation of the transfer function, s is referred to the filter's corner

quency, o r -3 dB frequency, (oc, and has these relat ionships:

s _ J~ f = j ~

s ~ Jrcc

ith the corner frequency of the low-pass in Figure 16-2 being

fc = 1/2~RC s

becomes

= sRC and the transfer function A(s) results in:

(05)

dB = 20 ! log10V Out

V In

"

# $

%

& ' (03)


8/63


b. Fundamentos de Filtros Passa Baixas!

A magnitude da resposta da função de ganho do filtro torna-se:

! Quando Ω >> 1 (f >> fc) a função de ganho apresenta uma taxa de atenuação

de aproximadamente 20dB do década.! As equações (01) e (07) só são válidas se Iout = 0. Podemos alcançar essa

condição ao se incluir um Amplificador Seguidor de Tensão na saída docircuito. Podemos ainda aumentar a ordem do filtro serializando N estágios.

!

Cada estágio aumenta a taxa de atenuação em mais 20dB/década de freq.

s _ J~ f = j ~

s ~ Jrcc

With the corner frequency of the low-pass in Figure 16-2 being


1

A(s ) = 1 + s

The magnitude of the gain response is:

IAI = 1

, /1 +Q2

For frequencie s O. >> 1, the rolloff is 20 dB/d ecade . For a ste

can be connected in series as shown in Figure 16-3. To avoid

operating as impedance converters, separate the individual fil

262

(07)

Fundamentals o f Low Pass Filters

VIN

R

c_

VOUT

Figure 16 3. Fourth Order Passive RC Low Pass with Decoupling Amplifiers

The resulting transfer function is:

A s) = 1


9/63


b. Fundamentos de Filtros Passa Baixas! A resposta de ganho e fase, do filtro passa baixa para diferentes ordens, são

apresentadas abaixo. Um filtro de ordem N ideal apresenta f c exatamente em -3dB comdecaimento instantâneo a partir de f c. No real, a atenuação ocorre bem antes, comdecaimento mais suave.

-10

-20

m -30

o

I

._c -40

o

I 5o

_<

-60

-70

100

80

0.01

1st Order Lowpass

II

I[

4th Order L~ i ' ~ ~ ~1

~ 4th Or ~

0.1 1 10

F r e q u e n c y

k

Pass Filters

-

-20

m -30

o

I

._c -40

o

I 5o

_<

-60

-70

100

80

0.01

1st Order Lowpass

II

I[

4th Order L~ i ' ~ ~ ~1

~ 4th Or ~

0.1 1 10

F r e q u e n c y

k

- 9 0

P

180

m

1-

e

2 7 0

- 3 6 0

0.01

b

Order

Low'asth

Order

0,1 1 10

100

Frequency

- -

Note:

Curve1 1St-orderpartial ow-pass ilter, Curve 2: 4th-order overall low-pass ilter, Curve 3: Ideal 4th-order ow-pass ilte


10/63


b. Fundamentos de Filtros Passa Baixas! Conforme gráficos anteriores, o filtro real apresenta problemas tanto

na amplitude do ganho de saída quanto na fase do sinal.! Podemos otimizar a saída do filtro para algum dos critérios abaixo:

I. Saída com ganho maximamente plano para f < f c.

II.

Transição imediata e aguda entre a banda de passagem e de atenuação.III. Resposta linear de variação na fase do sinal.

! Para permitir essas otimizações, a função transferência deve permitira existência de polos complexos. A função transferência se torna:

onde A0 é o ganho DC do filtro e ai e bi são os coeficientes dos filtros emcada estágio.! As seções a seguir descrevem as três principais otimizações de filtros

que garantem os tipos de respostas (I, II e III) citadas.

A(s) =

tion (Curve 3).

In comparison to the ideal low-pass, the RC low-pass lacks in the following characteris-

tics:

The passband gain varies long before the corner frequency, fc, thus amplifying the

upper passband frequencies less than the lower passband.

The transition from the passband into the stopband is not sharp, but happens

gradually, moving the actual 80-dB roll off by 1.5 octaves above fc.

The phase response is not linear, thus increasing the amount of signal distortion

significantly.

The gain and phase response of a low-pass filter can be optimized to satisfy one of the

following three criteria:

1) A maximum passband flatness,

2) An immediate passba nd-to-stop band transition,

3) A linear phase response.

For that purpose, the transfer function must allow for complex poles and needs to be of

the following type:

A0 A 0

(1 + a ,s + b ,s2)(1 + a2s + b2 s2) ...(1 + anS

+ bn s2

i[[(1 + ais + bi s2)

where Ao is the passband gain at dc, and ai and b are the filter coefficients.

Since the denominator is a product of quadratic terms, the transfer function represents

a series of cascaded second-order low-pass stages, with ai and bi being positive real coef-

ficients. These coefficients define the complex pole locations for each second-order filter

stage, thus determining the behavior of its transfer function.

The follow ing three types of prede termined filter coefficients are available listed in table

(08)


11/63


12/63


b. Fundamentos de Filtros Passa BaixasII. Resposta de Tschebyscheff! Apresentam alta taxa de atenuação acima de f c, com ordem inicial

igual a 2. Entretanto, o ganho não é constante na faixa de frequênciade passagem.

Fundamentals of Low Pass

9 , - . . ~ , . . ~ . . . . . ~ , ~ :- . . . . . . , 9 . . - ~ , ~ , ~ ~

16.2.2 Tschebyscheff Low Pass Filters

The Tsc hebyscheff low-pass filters provide an even higher gain rolloff above fc. Ho

as Figure 16-6 shows, the passband gain is not monotone, but contains ripp

constant magnitude instead. F or a given filter order, the higher the passban d rippl

higher the filter's roiloff.

10

- 1 0

- 2 0

- 3 0

- 4 0

- 5 0

--60

0.01

m

d

Order

0.1 1 10 100

Frequency

Figure 16 6. Gain Responses of Tschebyscheff Low Pass Filters

!

Um ripple (oscilação) é evidenciado eacrescido a medida que a ordem do filtroaumenta.!

A taxa de atenuação também aumentacom o aumento da ordem do filtro.!

Filtros de ordem impar apresentam oripple com valores menores que 0dB

(ganho < 1) na faixa de passagem.!

R e q u e r i d o s o n d e a p r i n c i p a lpreocupação é a remoção de faixas defrequências indesejáveis. Ex: equalizadoresde som.


13/63


b. Fundamentos de Filtros Passa BaixasIII. Resposta de Bessel ! Apresentam resposta de variação linear na fase do sinal. Abaixo de f c

atraso de grupo das harmônicas (group delay ) é sempre constante. Pass Filters

360

0.01

L

01

0

I ~80

W

(=

a .

~ 270 ~

90

0.1 1 10 100

F r e q u e n c y

3 6 0

0.01

0

I ~80

W

(=

a .

~ 270 ~

1.4

.~ 1.2

0.8

2

0.6

0.4

~ 0.2

0

0.1 1 10 100

F r e q u e n c y

Figure 16-7. Comparison of Phase Responses of Fourth-Order Low-Pass Fi lters

0.01 0.1 1 10 100

Frequency - -


14/63


b. Fundamentos de Filtros Passa BaixasIII. Resposta de Bessel! Baixa taxa de atenuação acima de f c e não possui resposta plana

abaixo de f c.! A resposta em atraso constante o torna a melhor escolha para filtrar

sinais de comunicação digitais (onda quadrada).

!

O defasamento linear fazcom que as distorções, deum sinal não senoidal,sejam minimizadas. Ex:

N=6 e fc = 20kHz.


15/63


b. Fundamentos de Filtros Passa BaixasIV. Fator de qualidade – Q! Representa um parâmetro de projeto filtro filtros de ordem N .! Ao invés de se projetar um filtro de ordem N , pode reduzir o problema

obter um filtro com determinado fator de qualidade Q.

!

Para filtros passa baixa e passa alta, Q é definido através da chamada“qualidade do polo”, ou seja:

! Já para filtros passa/rejeita faixa, Q é definido como a razão entre afrequência média, f m, e a largura de banda entre os dois pontos de

frequência em -3dB, ou seja:

! Valores muito altos de Q tendem a instabilidade do filtro.

Q =bi

ai(09)

Q = f m

f 2 ! f

1

(10)


16/63


b. Fundamentos de Filtros Passa BaixasIV. Fator de qualidade – Q! Altos valores de Q podem ser apresentados, nos gráficos de ganho x

frequência de um filtro, como a distância do maior pico de ganho atéa linha de 0dB.

Fundamentals of Low Pass Filters

. . . . . . . . . . . . . . .

m

-o

I

c

m

(:1

I

i


17/63


c. Projeto de Filtros Passa Baixas (PB)! Através da serialização de filtros de 2ª, Eq. (11), e 1ª, Eq. (12),

ordens, podemos construir filtros PB de qualquer ordem.

In addition, the ratio v ~ ~ _- Q is defined as the pole quality. T he higher the Q value, the

ai

more a filter inclines to instability.

-Pass Filter Design

Equation 16-1 represents a cascad e of secon d-orde r low-pass filters. The transfer func-

tion of a single stage is:

Ai(s ) = A~ (16-2)

(1 + ais + bis2 )

For a first-order filter, the coefficient b is always zero (bl=0), thus yielding:

A~

16-3)

A(s) = 1 +a ls

The first-order and s econd-order filter stages are the building blocks for higher-order fil-

ters.

Often the filters operate at unity gain (Ao=l) to lessen the stringent demands on the op

amp s open-loop gain.

Figure 16-11 shows the cascading of filter stages up to the sixth order. A filter with an even

order number consists of second-order stages only, while filters with an odd order number

include an additional first-order stage at the beginning.

16.3 Low-Pass Filter Design

Equation 16-1 represents a cascade of second-order low-pass filters

tion of a single stage is:

Ai(s ) = A~

(1 + ais + bis2 )

For a first-order filter, the coefficient b is always zero (bl=0), thus yi

A~

A(s ) = 1 +a ls

The first-order and second-order filter stages are the building blocks

ters.

Often the filters operate at unity gain (Ao=l) to lessen the stringent

amp s open-loop gain.

Figure 16-11 shows the cascading of filter stages up to the sixth order.

order number consists of second-order stages only, while filters with a

include an additional first-order stage at the beginning.

(11) (12)ow Pass

Filter Design

1st order o ]

1st order

a=l

2nd orde r~ J 2nd order

al , bl

3rd order o ;

1st order

al

2nd order

a2, b2

4th order o

5th order o ;

6th order o

2nd order

al

b

1st order

al

2nd order

al , bl

2nd order

a2 b 2

2nd order

a2, b2

2nd order

a2, b2

2nd order I

a3 b3

2nd order

a3, b3

Figure 16 11. Cascading Filter Stages for Higher Order Filters

!

Na maioria dos casos, A0 = 1 paraque Av(OL) do AO não tenha de ser muitoalto.!

C a d a e s t á g i o p r e c i s a s e rdimensionado de maneira que o Q dosestágios anteriores sejam menores do

que o Q do estágio atual.! Os valores de Q de diversos filtros são

apresentados em tabela no final destemódulo.

al

a2, b2


18/63


c. Projeto de Filtros Passa Baixas (PB)I. Primeira Ordem! Podemos implementar esse filtro usando as configurações de

amplificadores inversores e não inversores. Porém, abordaremos aquiapenas a topologia não inversora.

!

A função transferência é apresentada na Eq. (13) e os valores doscoeficientes são definidos nas Eq. (14) e (15).

!

O projeto do filtro é feito definindo-se f c, o ganho A0 e o capacitor C1.Por fim, resolva as Eq. (16) e (17), obtendo a1 da tabela.

6th order o

2nd order

al , bl

2nd order

a2, b2

a3 b3

2nd order

a3, b3

Figure 16 11 . Cascading Filter Stages for Higher Order Filters

Figure 16-10 demonst ra ted that the h igher the corner f requency o f a

er i ts Q. Therefore, to avoid the saturat ion of the individual stages,

placed in the order of r is ing Q values. The Q values for each f i l ter ord

order) in Sect ion 16-9 , Tab les 16 -4 through 16-10.

1 6 .3 . 1 F i r s t - O r d e r L o w - P a s s F i l t e r

Figures 16 -1 2 and 1 6- 13 show a first-order low-pass filter in the inv

inverting configuration.

a 1

VIN ~~/~/~

-

a

VOUT

Figure 16 12. First Order Noninverting Low Pass Filter

272

Low Pass Filter Design

~ ~ ~ ~ ~ : ~ , ~ . ~ ~ ~ , ~ ~ , ~ , ~ ~ . , ~ . ~ ~ ~ ~ , ~ , , : ~ ~ . . . . . . . - ~ . , , ~ ~ . ~ . . :~ , :~.,..~

V N

01

R1 R ~

V o u T

16 13. Firs t Order Invert ing Low Pass Fil ter

The transfer functions of the circuits are:

R2 R2

1 + R~ and A(s ) = R~

A( s) = 1 + cocR1C lS 1 - - cocR2C lS

The negative sign indicates that the inverting amplifier generates a 180 ~ phase shift from

the filter input to the output.

The coefficient comparison between the two transfer functions and Equation 16- 3 yields:

R2 and Ao = R2

A 0 = 1 + R~ R~

(13)

V N

R1 R ~

Vou T

Figure 16 13. F irst Order Inver ting Low Pass Fil ter


R2 R2

1 + R~ and A( s) = R~

A(s ) = 1 + cocR1ClS 1 - - cocR2C lS

The negative sign indicates that the inverting amplifier gen erates a 180 ~ phase shift fr


The coefficient comparison between the two transfer functions and Equation 16-3 yiel

R2 and Ao = R2

A 0 = 1 + R~ R~

a 1 = cocR1C 1

and

a 1 = e)cR2C 1

To dimension the circuit, specify the corner frequ ency (fc), the dc gain (Ao), and capaci

C1, and then solve for resistors R1 and R2:

al and R2 = al

R1 = 2~fcC1 2~fcC1

a 2

R 2 = R3( A 0 - 1) and R1 = Ao

(14)

(15)


R2 R2

1 + R~ and A(s ) = R~

A(s ) = 1 + cocR1ClS 1 - - cocR2ClS

The negative sign indicates that the inverting amplifier generates a 180 ~ phase shift fr


The coefficient comparison between the two transfer functions and Equation 16-3 yiel

R2 and Ao = R2

A 0 = 1 + R~ R~

a 1 = cocR1C 1

and

a 1 = e)cR2C 1

To dimension the circuit, specify the corner frequency (fc), the dc gain (Ao), and capaci


al and R2 = al

R1 = 2~fcC1 2~fcC1

(16)

1 + R~

A(s ) = 1 + cocR1ClS

The negative sign indicates that the invertin


The coefficient comparison between the two

R2

A 0 = 1 + R~

a 1 = cocR1C 1

To dimension the circuit, specify the corner f


a l

R1 = 2~fcC1

R 2 = R3 (A 0 - 1)

(17)


19/63


c. Projeto de Filtros Passa Baixas (PB)I. Primeira OrdemExemplo: projete um filtro PB com N = 1, ganho unitário e fc = 1kHz.Solução:

Um filtro com N=1 apresenta o mesmo valor de a1 = 1 para qualquer tipo de

otimização. Definimos C1 observando a faixa de frequência citada, a qualsugere um capacitor na faixa de dezenas de nF. Faremos C1 = 47nF.

Os cálculos restantes são:

Com R 2 = 0, R 3 pode ser removido.

A simulação usa um LF347

r Design

1. First Order Unity Gain Low Pass Filter

For a first-order unity-gain low-pass filter with fc = 1 kHz and C1 = 47 nF, R1 calculates

to-

a l

_ 1 _

R1 = 2~fcC1 - 2~. 103 Hz. 47 90- 9F - 3.38 kQ

However, to design the first stage of a third-order unity-gain Bessel low-pass filter, assum-

ing the same value s for fc a nd C1, requires a different val ue for R 1. In this case, obtain

al for a third-order Bessel filter from Table 16-4 in Section 16.9 (Bessel coefficients) to

calculate RI:

a l _ 0.756

R1 = 2~fcC1 - 2~. 103 Hz. 47 90- 9F = 2.56 kQ

When operating at unity gain, the noninverting amplifier reduces to a voltage follower (Fig-

ure 16-14), thus inherently providing a superior gain accuracy. In the case of the inverting

R2 = R3 !(1"1) = 0


20/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda Ordem! Existem duas topologias para filtros de segunda ordem:"

Sallen-Key baseada no amplificador não inversor "

Multiple Feedback (MFB) # baseado no amplificador inversor.

!

Abordaremos aqui apenas a topologia Sallen-Key. O livro apresentadetalhes complementares sobre a topologia MFB.

! Função transferência do filtro:

Low Pass Filte

. . . . . . . . . ~ . . . . . . . . . . . ,~ .

V I N

0 2

C1 3 T "

V O U T

Figure 16 15.

General Sallen Key Low Pass Filter

0 2

VOUT

A0 = 1+ R4

R3

(18)

V I N

0 2

C1 3 T"

V O U T

16 15.

General Sallen Key Low Pass F ilter

0 2

VOUT

16 16. Unity Gain Sallen Key Low Pass Filter

The transfer function of the circuit in Figure 16-15 is:

A(s) =

Ao

For the unity-gain circuit in Figure 16-16 (Ao=I), the transfer function simplifies to:

(19)


21/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda Ordem! Filtros que exijam alta precisão de ganho, ou ganho unitário, ou valor

de Q < 3, não incluem R3 e R4 o AO muda sua topologia para umseguidor de tensão (A0 = 1). A função transferência se torna:

! O procedimento de projeto envolve a definição prévia de C1 e C2. Emseguida, Calculamos R1 e R2 usando Eq. (21). A Eq. (22) deve ser

observada para se obter apenas valores reais na Eq. (21).

Low

. . . . . . . . . ~ . . . . . . . . . . .

V I N

0 2

C1 3 T "

V O U T

Figure 16 15.

General Sallen Key Low Pass Filter

0 2

VOUT

Figure 16 16. Unity Gain Sallen Key Low Pass Filter


A(s) =

Ao

For the unity-gain circuit in Figure 16-16 (Ao=I), the transfer function s

A(s) =

VOUT

Unity Gain Sallen Key Low Pass Filter

e transfer function of the circuit in Figure 16-15 is:

A(s) =

Ao

r the unity-gain circuit in Figure 16-16 (Ao=I), the transfer function simplifies to:

A(s) =

1 + ~ cC l( R 1 + R2) s + e)c2 R1R2C1C2s2

e coefficient comparison between this transfer function and Equation 16-2 yields:

A o = l

a 1 = mcCl (a 1

+ a2)

b 1

=

o)c2R1R2C1C2

ven C1 and C2, the resistor values for R1 and R2 are calculated through:

a1,2

a,C 2 ::F /a ,2C 2 2- 4bl C, C 2

4=fcC 1C2

(20)

Unity Gain Sallen Key Low Pass Filter

A(s) =

Ao

r the unity-gain circuit in Figure 16-16 (Ao=I), the transfer function simplifies to:

A(s) =

1 + ~c Cl (R 1 + R2) s + e)c2 R1R2C1C2s2

coefficient comparison between this transfer function and Equation 16-2 yields:

A o = l

a 1 = mc Cl( a 1 + a2)

b 1 = o)c2R1R2C1C2

en C1 and C2, the resistor values for R1 and R2 are calculated through:

a1,2

a,C 2 ::F /a ,2 C2 2- 4b lC ,C 2

4=fcC 1C2

(21)


In order to obtain real values under the square ro

tion

0 2 > _ 0 1 ~

4b 1

2

a l

Example 16 2. Second Order Unity Gain Tschebyschef

(22)


22/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda OrdemExemplo: projete um filtro PB de 2ª ordem com fc = 1kHz e otimização deTschebyscheff com ripple de 3dB.Solução: da tabela, obtemos os coeficientes a1 =1,0650 e b1 = 1,935.

Podemos arbitrar C1 em um valor comercial da ordem de dezenas de nF, deacordo com f c. Então, façamos C1 = 22nF.

Aplicando C1 na Eq. (22), obtemos:

Calculamos R 1 e R 2 usando Eq. (21)

In order to obtain real values under the square root,

C 2

must satisfy the following condi-

tion

0 2 > _ 0 1 ~

4b 1

2

a l

Example 16 2. Second Order Unity Gain Tschebyscheff Low Pass Filter

The task is to design a second-orde r unity-gain Tschebysche ff low-pass fi lter with a corner

frequency of fc = 3 kHz and a 3-dB passband ripple.

From Table 16-9 (the Tschebyscheff coefficients for 3-dB ripple), obtain the coefficients

al and bl for a second-order fi l ter with al = 1.0650 and bl = 1.9305.

Speci fy ing

C 1 as 22 nF yield s in

a C 2

of:

02 _

c14b1~ = 22 .1 0- 9n F . 4 . 1 . 93 05 _-..

a l 2 i~ 0- ~ - 150 nF

Inserting al and b l into the r esistor e qua tion for R1, 2 results in

1 . 06 5 . 150 . 10 - 9 - , / ( 1 . 0 65 . 1 50 . 1 0 - 9 ) 2 -

4 . 1 . 9 3 0 5 . 2 2 . 1 0 - 9 . 1 5 0 . 1 0 - 9

R 1 = ~ 4~ . 3 . 1 03 . 22 . 10_ 9 . 1 50 . 10_ 9 = 1 . 26 k Q

and

150,1nF # 180nF

R1 =

1,065*180*10!9! 1,065*180*10

!9( )2

! 4 *1,9305*22*10!9*180*10

!9

4 *" *1000* 22*10!9*180*10

!9 # 1632$

R2 =

1,065*180*10!9

+ 1,065*180*10!9( )

2

! 4*1,9305*22*10!9*180*10

!9

4*" *1000* 22*10!9*180*10

!9 # 5431$

R 1 = ~ 4 ~ . 3 . 1 0 3 . 2 2 . 1 0 _ 9 . 1 5 0 . 1 0 _ 9


23/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda OrdemSolução (cont...)

$

Os valores de R 1 E R 2 devem ser obtidos por

meio de potenciômetros$

A s r e s p o s t a s d e

amplitude e fase dofiltro são apresentadas

em comparação ao filtro

de 1ª ordem.$

Observa-se que a taxa

de atenuação é bemmaior, além do ripple deum pico (2ª ordem) na

amplitude de saída.

and

1 . 0 6 5 . 15 0 . 1 0 - 9 + ~ ( 1 . 0 6 5 . 1 5 0 . 1 0 - 9 ) 2 - 4 . 1 . 9 3 0 5 . 2 2 . 1 0 - 9 . 1 5 0 . 1 0

R 2 = 4 g 3 1 0 3 2 2 1 0 _ 9 1 5 0 1 0 _ 9

with the final circuit shown in Figure 16-17.

150n

VouT

Figure 16 17. Second Order Unity Gain Tschebyscheff Low Pass with 3

A special case of the general Sallen-Key topology is the application of

ues an d e qual capa cito r valu es R 1 = R 2 = R and C1 = C2 = C.

276

180nF

1632 5431


24/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda Ordem! A topologia Sallen-Key apresenta um caso especial onde é possível

definir R 1 = R 2 = R e C1 = C2 = C.! Nesta simplificação, a Eq. (19) assume a forma descrita em Eq. (23),

com ganho definido na Eq. (24).

! A partir da Eq. (23) e da definição arbitrária de C, obtemos o valor deR e A0.

! Eq. (26) mostra que a resposta do filtro será unicamente definida peloganho A0, o qual determina o fator de qualidade do filtro.

Low-Pass Filter Design

9 .

The genera l t rans fe r func t ion changes to

A s) = A~

1 + e )cR C( 3 - Ao )s + (eL)cRC)2s 2

a 4

with A o = 1 4-R--~

The coef f ic ien t compar ison wi th Equat ion 16-2 y ie lds :

a 1 = (JL)oRC(3 - Ao )

b 1 = (~c RC ) 2

Given C and solv ing for R and Ao resul ts in

R = ]~ Ao=3

2~ fcC and

al = 3 1

N Q

Thus, Ao depends sole ly on the pole qua l i ty Q and v ice versa; Q, and wi th i t the f i lter type,

(23)


9 .

The general transfer function changes to

A s) = A~

1 + e )cR C(3 - Ao )s + (eL)cRC)2s 2

a 4

wit h A o = 1 4-R--~

The coefficient comparison with Equation 16-2 yields:

a 1 = (JL)oRC(3 - Ao )

b 1 = (~c RC ) 2

Given C and solving for R and Ao results in

R = ]~ Ao=3

2~fcC and

al = 3 1

N Q

(24)


9 .


A s) = A~

1 + e )cRC (3 - Ao )s + (eL)cRC)2s 2

a 4

wi th A o = 1 4-R--~


a 1 = (JL)oRC(3 - Ao )

b 1 = (~ cR C) 2


R =

]~ Ao=3

2~fcC and

a l = 3 1

Q

Thus, Ao depends solely on the pole quality Q and vice versa; Q, and with it the filter type,

is determined by the gain setting of Ao:

Q= 1

3 - A o

(25)


9 .


A s) = A~

1 + e)c RC (3 - Ao) s + (eL)cRC)2s 2

a 4

with A o = 1 4-R--~


a 1 = (JL)oRC(3 - Ao)

b 1 = (~ cR C) 2


R =

]~ Ao=3

2~fcC and

a l = 3 1

N Q

Thus, Ao depends solely on the pole quality Q and vice versa; Q, and with it the filter type,


Q= 1

3 - A o

(26)

s = ~


25/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda Ordem! Podemos usar um potenciômetro no lugar de R4 a ajustar o tipo de

resposta de interesse através da relação R4/R3.Exemplo: Projete e simule três filtros de Tschebycheff com fc = 1kHz e

ripple de 0,5dB, 1dB e 2dB. Use a topologia abaixo

Solução: arbitraremos o valor C = 22nF e R 3 = 1k Ω. Das tabelas temos osseguintes parâmetros:

1 + e )cR C(3 - Ao) s + (eL)cRC)2s 2

with A o =


a 1 = (JL)oRC(3 - Ao )

b 1 = (~c RC ) 2


R = ]~ Ao=3

2~fcC and

al = 3 1

N Q

Thus, Ao depends solely on the pole quality Q and vice versa; Q, an


Q= 1

3 - A o

The circuit in Figure 16-18 allows the filter type to be changed throug

ratios R4/R3.

c

t

V N Vout

Figure 16-18. Adjustable Second-O rder Low-Pass Fi l ter

Ripple bi Q

0,5dB 1,3827 0,861,0dB 1,5515 0,96

2,0dB 1,7775 1,13


26/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda OrdemPara cada Ripple, calculamos o valor de R, R 4 e A0.

As respostas em ganho normalizado e fase dos três filtros são apresentadas aseguir.

R0,5 dB =

1,3827

2!" !

1kHz!22nF

# 8507$ A0 0,5dB = 3! 1

0,86

= 1,837 R

4 = R

3!( A

0 "1)

R4 1,0dB = 1k !(1, 837 "1) = 837#

R1dB

=

1,5515

2 !" !1kHz !22nF # 9011$ A

01,0dB = 3!

1

0,96"1,958 R4 1,0dB = 1k !(1,958 "1) = 958#

R2dB =

1,7775

2 !" !1kHz !22nF

# 9638$ A01,0dB

= 3!1

1,13

" 2,115 R4 1,0dB = 1k !(2,115"1) = 1,115#


27/63


c. Projeto de Filtros Passa Baixas (PB)II. Segunda Ordem

# A T a x a d eatenuação aumenta

com o fator Q dof i l t ro , ou se ja ,aumenta com aa m p l i t u d e d oripple.# O defasamentotambém aumento

com o aumento dofator Q.


28/63


c. Projeto de Filtros Passa Baixas (PB)III. Ordens Superiores! Quanto maior a ordem, mais acentuadas são as características do

filtro. Ordens superiores são obtidas pela serialização de filtros deprimeira e segunda ordens.

!

Com a serialização, o produto das respostas em frequência dosmódulos individuais (estágios) resulta na resposta em frequênciaotimizada e acentuada de todo o filtros.

! Cada estágio apresentará seu próprio ai, bi e Qi obtidos das tabelas edependentes da ordem global do filtro sendo projetado.

! Filtros de ordens impares apresentarão o primeiro estágio como um

filtro de 1ª ordem.! Filtros de ordens pares apresentam apenas filtros de segunda ordem

em cada um de seus estágios.

e as s o e s gn a -or er un y-ga n u erwor ow-pass er w e corner re-


29/63


c. Projeto de Filtros Passa Baixas (PB)III. Ordens SuperioresExemplo: projete e simule um filtro PB de quinta ordem, f c = 50kHz comresposta de Butterworth.Solução.

1.

Obtenha os coeficientes através das tabelas

2. Dimensionar cada filtro parcial usando os coeficientes obtidos2.1 Primeiro estágio 1ª Ordem

Defina C1 = 1nF e calcule R 1.

Filter Co

Table 16 5. Butterworth Coefficients

n i a i bi ki= Qi

f c i / f c

1 1 1.0000 0.0000 1.000

2 1 1.4142 1.0000 1.000 0.71

3 1 1 0000

0.0000 1.000

2 1.0000 1.0000 1.272 1.00

4 1 1.8478 1.0000 0.719 0.54

2 0.7654 1.0000 1.390 1.31

1 1.0000 0.0000 1.000

2 1.6180 1.0000 0.859 0.62

3 0.6180 1.0000 1.448 1.62

1 1.931 9 1.0000 0.676 0.52

2 1.4142 1.0000 1.000 0.71

3 0.5176 1.0000 1.479 1.93

1 1.0000 0.0000 1.000

2 1.801 9 1.0000 0.745 0.55

3 1.2470 1.0000 1.117 0.80

4 0.4450 1.0000 1.499 2.25

Filter Coe

Table 16 5. Butterworth Coefficients

n i a i bi ki= Qi

f c i / f c

1 1 1.0000 0.0000 1.000

2 1 1.4142 1.0000 1.000 0.71

3 1 1 0000

0.0000 1.000

2 1.0000 1.0000 1.272 1.00

4 1 1.8478 1.0000 0.719 0.54

2 0.7654 1.0000 1.390 1.31

1 1.0000 0.0000 1.000

2 1.6180 1.0000 0.859 0.62

quenc y fc = 50 kHz.

First the coefficients for a fifth-order Butterworth filter are obtained from Table 16-5, Sec-

tion 16.9:

ai bi

Filt er I al = 1 bl = 0

Filt er 2 a2 = 1.61 80 b2 = 1

Fi l ter 3 a3 = 0.61 80 b3 = 1

Then dimension each partial f i l ter by specifying the capacitor values and calculating the

required resistor values.

a 1

VIN~ TiT

C

; VouT

. F i r s t OrderUn i ty Ga inLow Pass

With C1 = lnF,

a l

R 1 = 2 ~c C 1

2 ~ . 5 0 . 1 0 3 H z . 1 . 1 0 - 9 F

= 3 . 1 8 k ~

The closest 1 valu e is 3.16 k.Q.

Higher-order low-pass f i lters are required to sharpen a desired f i lte

that purpose, f irst-order and second-order f i lter stages are connect

the product of the individual frequency responses results in the opti

spons e of the overall f i l ter.

In order to simplify the design of the partial filters, the coefficients ai

type are listed in the coeff icient tables (Tables 16-4 through 16-10 i

each table providing sets of coeff icients for the f irst 10 f i lter orders.

Exam ple 16 3. Fifth Orde r Fi l ter

The task is to design a fifth-order unity-gain Butterworth low-pass filte

quency fc = 50 kHz.

First the coefficients for a fifth-order Butterworth filter are obtained fro

tion 16.9:

ai bi

Fil ter I al = 1 bl = 0

Filt er 2 a2 = 1.61 80 b2 = 1

Fi l ter 3 a3 = 0.61 80 b3 = 1

Then dimension each partial f i l ter by specifying the capacitor values

required resistor values.

irst

Fil ter

a 1

VIN~ TiT

C

; VouT

F igu re16 20 . F i rs t Orde rUn i ty Ga inLow Pass

Second i l ter


30/63


c. Projeto de Filtros Passa Baixas (PB)III. Ordens SuperioresSolução (cont...)

2.2 Segundo Estágio! 2ª Ordem

Como o segundo estágio já recebe um sinal pré-filtrado, podemos definir um

capacitor menor, ou seja C1 = 820pF. Calculando C2, temos:

Vamos definir C2 em 1.5nF e calcular R 1 e R 2 deste estágio.

s Filter Design

ilter

VIM

02

II

01 T

VOUT

16 21. Second Order Unity Gain Sallen Key Low Pass Filter

With

C 1 =

820 pF,

4b 2

C 2 >_ C 1- = 82 0.1 0-1 2F. ~

a22

4.1

1.6182

= 1.26 nF

The closest 5% value is 1.5 nE

With C 1 = 820 pF and C 2 = 1.5 nF, calculate the values for R1 and R2 through"

v

2

a2C 2 - a22 0 2 - 4b 20 10 2

4~fcC1C2 and R1 =

•

2

a 2 C 2 H-- a22 C 2 - 4 b 2 0 1 0 2

4~:fcC1C 2

Second i l ter

VIM

02

II

01 T

VOUT

Figure 16 21. Second Order Unity Gain Sallen Key Low Pass Filter

With

C 1 =

820 pF,

4b 2

C 2 >_ C 1 - = 82 0 . 10 - 1 2F . ~

a22

4.1

1 .6182

= 1.26 nF


With

C 1 =

820 pF and

C 2 =

1.5 nF, calculate the values for R1 and R2 thro

v

2

a2C 2 - a22 0 2 - 4 b 20 10 2

R1 = 4~f cC 1C 2 and R1 =

•

2

a 2 C 2 H-- a22 C 2

4~:fcC1C

and obtain

Second i l ter

VIM

02

II

01 T

VOUT

Figure 16 21. Secon d Order Unity Gain Sallen Key Low Pass Filter

With

C 1 =

820 pF,

4b 2

C 2 >_ C 1 - = 82 0 .1 0 - 12 F . ~

a22

4.1

1 .6182

= 1.26 nF


With

C 1 =

820 pF and

C 2 =

1.5 nF, calculate the values for R1 and R2 through"

v

2

a2C 2 - a22 0 2 - 4b 20 10 2

R1 = 4~fc C1C 2 and R1 =

•

2

a 2 C 2 H-- a22 C 2 - 4 b 2 0 1 0 2

4~:fcC1C 2

and obtain

a 1 --

1 . 6 1 8 . 1 . 5 . 1 0 - 9 - V / / (1 . 6 18 . 1 .5 . 1 0- 9 ) 2 - 4 . 1 . 8 2 0 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

4 ~ . 5 0 . 1 03 . 8 2 0 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

= 1 .87 kQ

a 2 -

1 .618 .1 .5 .10 -9 + V / (1 .618 .1 .5 .10 -9 ) 2

_ 4 . 1 . 8 2 0 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

4~ .50 .103 .820 90 - 1 2 . 1 . 5 . 1 0 - 9

= 4 .42 kQ

R1 and R2 are available 1% resistors.

VIM

02

II

01 T

VOUT

Figure 16 21. Second Order Unity Gain Sallen Key Low Pass Filter

With

C 1 =

820 pF,

4b 2

C 2 >_ C 1 - = 82 0 . 10 -1 2F . ~

a22

4.1

1 . 6182

= 1 .26 nF


With

C 1 =

820 pF and

C 2 =

1.5 nF, calculate the values for R1 and R2 through"

v

2

a2C 2 - a22 0 2 - 4b 20 10 2

R1 = 4~fcC 1C2 and R1 =

•

2

a 2 C 2 H-- a22 C 2 - 4 b 2 0 1 0 2

4~:fcC1C 2

and obtain

a 1 --

1 . 6 1 8 . 1 . 5 . 1 0 - 9 - V / / (1 . 6 1 8 .1 . 5 .1 0 - 9 ) 2 - 4 . 1 . 8 2 0 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

4 ~ . 5 0 .1 0 3 . 8 20 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

= 1 .87 kQ

a 2 -

1 . 618 . 1 . 5 . 1 0 -9 + V / (1 . 61 8 . 1 . 5 . 1 0 -9 ) 2

_ 4 . 1 . 8 2 0 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

4~ . 50 . 103 . 820 90 - 1 2 . 1 . 5 . 1 0 - 9

= 4 . 42 kQ


. . . . . . . . . . . . . .

= 1.8


31/63


c. Projeto de Filtros Passa Baixas (PB)III. Ordens SuperioresSolução (cont...)

2.2 Terceiro Estágio! 2ª Ordem

Definimos agora C1 = 330pF e calculamos C2 deste estágio.

Vamos definir C2 = 4.7nF. Com o mesmo procedimento de cálculoapresentado no estágio 2, calculamos R 1 ≅ 1,45k e R 2 ≅ 4,51k .

a 1 --

4 ~ . 5 0 .1 0 3 . 8 20 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

a 2 -

1 . 6 1 8 . 1 . 5 . 1 0 - 9 + V / ( 1 . 6 1 8 . 1 . 5 . 1 0 - 9 ) 2

_ 4 . 1 . 8 2 0 . 1 0 - 1 2 . 1 . 5 . 1 0 - 9

4 ~ .5 0 .1 0 3 .8 2 0 90 - 1 2 . 1 . 5 . 1 0 - 9

= 4.4


Third Fi l ter

The calculation of the third fi l ter is identical to the calculation of the second fi l t

that a 2 and b2 are rep laced by a 3 and b 3, thus resulting in different c apa citor a

values.

Specify C1 as 330 pF, and obtain 0 2 with"

4b 3

C 2 >_ C 1 - - = 3 3 0 . 1 0 - 1 2 F . ~

a32

4.1

0 .6 1 8 2

= 3 . 46 nF

The closest 10% value is 4.7 nF.

280

High Pass Filter Design

With C 1 = 330 pF an d C 2 = 4.7 nF, the valu es for R1 an d R2 are"

9 R1 = 1.45 k.Q, with the closes t 1% valu e bein g 1 .47 k..O.

9 R2 = 4.51 k.Q, with the clo sest 1% value being 4.53 k.O.

Figure 16-22 shows the final fi l ter circuit with its partial fi l ter stages.

3.16k

ViN~

--

1.5n

L

820p - ]

4.7n

II

330p " I

VOUT

Figure 16 22 . Fifth Ord er Unity Gain Butterworth Low Pass Filter


32/63


c. Projeto de Filtros Passa Baixas (PB)III. Ordens SuperioresSimulação

- Fc não é exatamente50kHz dev ido àsaproximações feitasem C2 do segundo eterceiro estágio.- D e v i d o à saproximações, ocorreum pequeno overshoot

na resposta, perto def c.

820p - ]

VOUT


33/63


d. Projeto de Filtros Passa Alta (PA)! A topologia de um filtro PA é muito similar a de um filtro PB. Basta

trocar as posição de resistores e capacitores.

330p " I

Figure 16 22. Fifth OrderUnity G ain Butterworth Low Pass Filter

16.4 High Pass Filter Design

By replacing the resistors of a low-pass filter with capacitors, and its capacitors with resis-

tors, a high-pass filter is created.

0 2

..... c >

c~ T Voo~ > v ~ ~ t t qt

R1 I

_

VOUT

Figure 1 6 23 . Low Pas s o High Pass Transition Through Components Exchange

To plot the gain response of a high-pass filter, mirror the gain response of a low-pass filter

at the corner frequency, .Q=I, thus replacing .0. with 1/.Q and S with 1/S in Equation 16-1.

PB PAigh Pas s Filter Design. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

m

o

I

r

o . , ,

I


34/63


d.

Projeto de Filtros Passa Alta (PA)I. Primeira Ordem! Se fizermos bi = 0 na Eq. (27), obtemos a função transferência de um

filtro PA de 1ª ordem.! O circuito não inversor que implementa o PA é apresentado abaixo em

conjunto com sua função transferência. Versões inversoras podem serconsultadas no livro referência.! O ganho na faixa de passagem é definido pela Eq. (30) e os valores de

R1 e C1 pelas Eqs. (31) e (32).

0.1

1

10

Frequency

Develop ing The Gain Response of a High Pass Filter

general transfer function of a high-pass filter is then:

A s) = A= 16-4)

]-[ ai b,)

i 1 + ~ +

Aoo being the passb and gain.

ce Equation 16-4 represents a cascade of second-order high-pass filters, the transfer

Ai s ) = Aoo 16-5 )

a

+ ~ + ~

h b=0 for all first-order filters, the transfer function of a first-order filter simplifies to:

A(s) - A~ (16-6)

(28)


9 . , - . . . . . . . ~ ~ ~ . . . . . . .

1 6 . 4 . 1 F i r s t O r d e r H i g h P a s s F i l t e r

Figure 1 6-25 and 16-26 show a first-order high-pass filter in the noninverting and the in-

verting configuration.

VIN

01

R1

-- R2

R3

VOUT

Figure 16 25. First Order Noninverting High Pass Filter

Ra

C1 R1

Figure 16 25. First Order Noninverting High Pass Filter

V N

Ra

C1 R1

1

V

Figure 16 26. First Order Inverting High Pass Filter

The trans fer functions of the circuits are:

a 2

1 + ~ and

A s) =

ocR1C 1 s

The negative sign indicates that the inverting amplifi


The coefficient comparison between the two transf

vides two different passband gain factors:

(29)

igure 16 25. First Order Noninverting High Pass Filter

V N

Ra

C1 R1

1

VOUT

igure 16 26. First Order Inverting High Pass Filter


a 2

1 + ~ and

A s ) =

ocR1C 1 s

oJcR1C 1 s

A s) = -

a 2

1 1

The negative sign indicates that the inverting amplifier gene rates a 180 ~ phase shift from


The coeff icient comparison between the two transfer functions and Equation 1 6-6 pro-

vides two different passband gain factors:

R2 and A = R2

A= =I +R 3 = R~

while the term for the coefficient al is the same for both circuits:

a 1

o)cR1C 1

(30)


. . . . . .

To dimension the circuit, specify the corner f requen cy (fc), the dc gain (Aoo), and

(C1), and then solve for R1 and R2:

_ 1

R 1 - 2~ f ca lC 1

R 2 = R3(Aoo -

1)

and

R 2 = -R 1

A oo

1 6 . 4 . 2 S e c o n d - O r d e r H i g h - P a s s F i l t e r

(31)


. . . .

To dimension the circuit, specify the corner frequency (fc), th


_ 1

R 1 - 2~ f ca lC 1

R 2 = R3(Aoo - 1) and R 2


(32)


35/63


d.

Projeto de Filtros Passa Alta (PA)I. Primeira Ordem! Exercício: projete um Filtro PA com f c = 4kHz e ganho unitário.Resolução: Um filtro com N=1 apresenta o mesmo valor de a1 = 1 paraqualquer tipo de resposta (otimização). Definimos C1 observando a faixa de

frequência citada, a qual sugere um capacitor na faixa de dezenas de nF.Faremos C1 = 33nF.

Procedendo ao cálculo de R 1: R1 =1

2 !" ! 4000 !33!10#9

=1205,7 $ 1,2k %

ave c ange .


36/63


d.

Projeto de Filtros Passa Alta (PA)II. Segunda Ordem! Novamente, abordaremos aqui apenas a topologia de Salen-Key e

deixaremos a topologia MFB para estudo no livro.! A topologia do circuito é praticamente igual à do Filtro PB de segunda

ordem, com a diferença da substituição das posições dos resistores ecapacitores.

To dimension the circuit, specify the corner frequency (fc), the dc gain (Aoo)


_ 1

R 1 - 2~ fc a lC 1

R 2 = R3(Aoo -

1)

and

R 2 = -R 1

A oo


High-pas s filters use the same two topologies as the low-pass filters: Sallen

tiple Feedback. The only difference is that the positions of the resistors and

have changed.

16.4.2.1 Sallen Key Topology

The general Sallen-Key topology in Figure 16-27 allows for separate g

Ao = 1+R4/R3.

C1 [ C2

V N III - I I

a 2

k/k/k,

R, - _ m

- - a 4

VOUT

Figure 16 27. GeneralSallen Key High Pass Filter

16.4.2.1 Sallen Key Topology

The general Sallen-Key topology in Figure 16-27 allows for separate gain setting via

Ao = 1+R4/R3.

C1 [ C2

V N III - I I

a 2

k/k/k,

R, - _ m

- - a 4

VOUT

Figure 16 27. Gene ralSallen Key High Pass Filter

A(s) =


cz

1 +

R2(C1+ C2)+ RIC2(1 a 1 1

~ .1 1 - eOc 2 R1 R2 C1C 2 S-

wi th o~ = 1 +

a 4

R3

The unity-gain topology in Figure 16-28 is usually applied in Iow-Q filters with high gain

accuracy.

l

v , . I I -

a 2

~/k/k,

a l T

m

VOUT

(33)

ass filters: Sallen-Key and Mul-

the resistors and the capacitors

s for separate gain setting via

VOUT

h o~ = 1 +

a 4

R3

d in Iow-Q filters with high gain

(34)

~ . ~ ~ ~ . . . ~ ~ ~ ~ ~ ~ ~ . . ~ . ~ . . ~ ~ . ~ ~ ~ ~ ~ . ~ ~ ~ . ~ . ~ ~ ~


37/63


d. Projeto de Filtros Passa Alta (PA)II. Segunda Ordem! Podemos simplificar o projeto do filtro definindo C1 = C2 = C e ganho

unitário (α=1) . A função transferência reduz-se à Eq. (35).

! Os coeficientes A", a1 e b1 são obtidos diretamente de (35).

! E os valores de C será arbitrário. Já os valores de R 1 e R 2 são

calculados pelas Eq. (36) e (37).

High-Pass Filter Design

. ~ ~ ~ ~ ~ ~ ~ . . ~ . ~ . . ~ ~ . ~ ~ ~ ~ ~ . ~ ~ ~ . ~ . ~ ~ ~ ~ . . . .

simplify the circuit design, it is com mon to cho ose unity -gain (c~ = 1), and C1 = 02 = C.

e transfer function of the circuit in Figure 16-28 then simplifies to:

A(s) =

2 1 + 1 .I__

cocalC s eoc2alR2C2 s 2

e coefficient comparison between this transfer function and Equation 16-5 yields:

A o o = l

a 1 --

b 1 =

cocR1C

1

O)c2R1R2 02

(35)


. ~ ~ ~ ~ ~ ~ ~ . . ~ . ~ . . ~ ~ . ~ ~ ~ ~ ~ . ~ ~ ~ . ~ . ~ ~ ~ ~ . . ..

To simplify the circu it design, it is com mon to choose unity -gain (c~ = 1), and C1 = 02 = C.

The transfer function of the circuit in Figure 16-28 then simplifies to:

A(s) =

2 1 + 1 .I__


The coefficient comparison between this transfer function and Equation 16-5 yields:

A o o = l

a 1 --

b 1 =

cocR1C

1

O)c2R1R2 02

Given C, the resistor values for R 1 and R2 are calculated through:


~ . ~ ~ ~ . . . ~ ~ ~ ~ ~ ~ ~ . . ~ . ~ . . ~ ~ . ~ ~ ~ ~ ~ . ~ ~ ~ . ~ . ~ ~ ~ ~ . . . .

To simplify the circuit de sign, it is com mo n to cho ose unity-gain (c~ = 1), and C1 = 02 = C.

The transfer function of the circuit in Figure 16-28 then simplif ies to:

A(s) =

2 1 + 1 .I__


The coeff icient comparison between this t ransfer funct ion an d Equat ion 16- 5 yields:

A o o = l

a 1 --

b 1 =

cocR1C

1

O)c2R1R2 02

Given C, the resistor values for R 1 and R2 are calculate d through:

=

R1 =fcCal

a l

~ . ~ ~ ~ . . . ~ ~ ~ ~ ~ ~ ~ . . ~ . ~ . . ~ ~ . ~ ~ ~ ~ ~ . ~ ~ ~ . ~ . ~ ~ ~

To simplify the circuit design, it is com mon to choos e unity-gain (c~

The transfer function of the circuit in Figure 16-28 then simplif ies

A(s) =

2 1 + 1 .I__

coca lC s eoc2alR2C2 s 2

The coefficient comparison between this transfer function and Eq

Ao o = l

a 1 --

b 1 =

cocR1C

1

O)c2R1R2 02

Given C, the res istor values for R 1 and R2 are calc ulated throug h:

=

R1 =fcCal

a l

R 2 = 4=fcCb 1


. ~ ~ ~ ~ ~ ~ ~ . . ~ . ~ . . ~ ~ . ~ ~ ~ ~ ~ . ~ ~ ~ . ~ . ~ ~ ~ ~ . . . .

To simplify the circuit de sign, it is com mon to choose unity-gain (c~ = 1), and C1 = 02 = C.

The transfer function of the circuit in Figure 16-28 then simplifies to:

A(s) =

2 1+ 1 .I__

cocal C s eoc2alR2C2 s 2

The coefficient comparison between this transfer function and Equation 16-5 yields:

A o o = l

a 1 --

b 1 =

cocR1C

1

O)c2R1R2 02

Given C, the resistor value s for R 1 and R2 are calcu lated through :

=

R1 =fcCal

a l

R 2 = 4=fcCb 1

(36)

To simplify the circuit design, it is common to choose unity-gain (c~

The transfer function of the circuit in Figure 16-28 then simplifies

A(s) =

2 1 + 1 .I__


The coefficient comparison between this transfer function and Eq

A o o = l

a 1 --

b 1 =

cocR1C

1

O)c2R1R2 02

Given C, the resistor values for R 1 and R2 are calcu lated through:

=

R1 =fcCal

a l

R 2 =

4=fcCb 1

16 4 2 2 Multiple Feedback Topology

(37)

b 1 = e)cR1C C2


38/63


d.

Projeto de Filtros Passa Alta (PA)III. Ordens Superiores! Para N > 2, faremos o processo de serialização de filtros PA de ordens

1 e 2 da mesma forma que fizemos para os filtros PB.! Os coeficientes dos filtros serão os mesmos obtidos pelas tabelas

constantes no final deste material, seguindo o mesmo procedimentode projeto apresentados para os filtros PB.! Exercício: Projete um Filtro PA de terceira ordem com f c = 1kHz e

resposta de Bessel.Resolução:

O primeiro passo é obter os coeficientes dos dois estágios que comporão

esse filtro.

Para o primeiro estágio (N = 1), definimos

C = 100nF e calculamos

Given capacitors C and C2, and solving for resistors R 1 and R2:

_ 1 2A=

R1 - 2=fc.C .al

a l

R2 = 2~;fc .blC2 1 - 2Aoo)

The passba nd ga in Aoo) of a MFB high-pass filter can vary significantl

tolerances of the two capacitors C and C2. To keep the gain variation

necessary to use capacitors with tight tolerance values.

16.4 .3 Higher Order High Pass Fi lter

Likewise, as with the low-pass filters, higher-order high-pass filters are

cading first-order and seco nd-order filter stages. Th e filter coefficients

used for the low-pass filter design, and are listed in the coefficient ta

through 16-10 in Section 16.9).

Exam ple 16 4. Third Order High Pass Fi l ter with fc = 1 kHz

The task is to design a third-order unity-gain Bessel high-pass filter w

quency fc = 1 kHz. Obtain the coefficients for a third-order Bessel filte

Section 16.9:

ai bi

Filt er I al = 0.756 bl = 0

Filt er 2 a2 = 0.9996 b2 = 0.4772

and compute each partial filter by specifying the capacitor values and

quired resistor values.

First Filter

Band Pass Filter Design

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

R1 = 1 = 2.1 05 kQ

2 = fca l C 1 2 ~ .1 0 3 H z .0 .7 5 6 .1 0 0 .1 0 - 9 F

Closes t 1 value is 2.1 k~.

an ass e r es gn


39/63


d.

Projeto de Filtros Passa Alta (PA)III. Ordens Superiores

O Segundo estágio (N=2) é projetando arbitrando-se C = 100nF e calculandoR 1 e R 2 usando as Eq. (36) e (37).

O Circuito resultante é apresentado abaixo, usando valores de componentescomerciais:

Faça a simulação !!!!


R1 = 1 = 2.1 05 kQ

2= f ca l C 1 2~ . 103 H z . 0 . 75 6 . 100 . 10 - 9F

Close st 1 value is 2.1 k~.

ilter

With C = 100nF,

_ 1 = 3.18 kQ

R 1 - ~ f cCa l ~ .10 3 .10 0 .10-9 .0 .756


_ a

R 2 _ 1 0.999 6 = 1.67 kQ

4=fcCb 1 4=-103 .100.10 - 9 . 0 . 4 7 7 2

Closest 1 value is 1.65 k ~

Figure 16-30 shows the final fi lter circuit.

V I N

1.65k

1 0 0 n - ~ k A ,

i i_~ l~1761~176 _ ~ T

I ~ II II 1

2.1 0k . . - VouT

3 . 1 6 k

R1 = 1 = 2.1 05 kQ

2= f c a lC 1 2~ . 10 3Hz . 0 . 756 . 10 0 . 10 - 9F

Clos est 1 value is 2.1 k~.

ilter

With C = 100nF,

_ 1 = 3.18 kQ

R 1 - ~ f c Ca l ~ . 103 . 100 . 10 -9 . 0 . 756

Closest 1 value is 3.16 k~.

_ a

R 2 _ 1 0.99 96 = 1.67 kQ

4=fcCb 1 4=-103.100.10 - 9 . 0 . 4 7 7 2


Figure 16- 30 shows the final f i lter circuit.

V I N

1.65k

1 0 0 n - ~ k A ,

i i_~ l~1761~176 _ ~ T

I ~ II II

1

2.1 0k . . - VouT

3 . 1 6 k

w


Second i l ter

With C = 100nF,

_ 1 = 3.18 kQ

R 1 - ~ f cCa l ~ .103 .1 00 .10 -9 .0 .756

Closest 1 value is 3.16 k~.

_ a

R 2 _ 1 0.99 96 = 1.67 kQ

4=fcCb 1 4=-103.100.10 - 9 . 0 . 4 7 7 2


Figure 16-30 shows the final fi lter circuit.

V I N

1.65k

1 0 0 n - ~ k A ,

i i_~ l~1761~176 _ ~ TI ~ II II

1

2.1 0k . . - VouT

3 .1 6 k

w

Figure 16 30. Third Order Unity Gain Bessel High Pass

0 ^ -.~ 0 / -,.


40/63


e. Projeto de Filtros Passa Banda (PBn)! Representa a união de dois filtros, um PB e outro PA, de maneira a

permitir a passagem de uma banda estreita de frequência.! Haverão duas frequências de corte, Ω1 e Ω2, que representam os

pontos de -3dB da faixa de passagem inferior e superior,

respectivamente.! Usamos o filtro PB como base e espelhamos sua resposta a partir desua frequência de corte, conforme figura abaixoand-Pass Filter Design~ . ~ ~ < ` ~ ` ~ ~ ~ ~ < ~ ` ~ ~ < ~ < ` ~ ~ ~ ~ ~ ~ ` ~ / : ~ < ~ .~ . ~ . . . . ~ . ~ . ~ - ~ .~ ~ < . ~ ~ ~ ~ : ~ ~ ~ ~ ~ ~

IAI [dB] ' ~ ~ . ~ _ ~

0 ^ -.~ 0 / -,.

-3

v

-3

v

I

.~ 0 1 ~ ~..0

F igu re 16 -31 . Low-Pa ss to Band-P ass T rans it i on

-3

v

-3

v

I

.~ 0 1 ~ ~..0

ow-P ass to Ban d-Pas s Trans it ion

orner frequency of the low-pass filter transforms to the lower and upper-3 dB fre-

cies of the band-p ass, ~21 and .0-2. The difference betwe en both freq uencies is de-

as the normalized bandwidth A.Q:

ormalized mid frequency, where Q = 1, is

Qm = = Q2.~ 1

alogy to the resonant circuits, the quality factor Q is defined as the ratio of the mid

ency (fm) to the bandw idth (B)

(38)

v v

.~ 0 1 ~ ~..0

Low-P ass t o Band -Pass Transi ti on

corner frequency of the low-pass f i lter transforms to the lower and u pp er -3 dB fre-

ncie s of the band-pass, ~21 and .0-2. The differen ce betw een both fre que ncie s is de-

as the normalized bandwidth A.Q:

normalized mid frequency, where Q = 1, is

Qm = = Q2 .~1

nalogy to the resonant circuits, the quality factor Q is defined as the ratio of the mid

uen cy (fm) to the band width (B)

Q = fm = fm = 1 = 1 (1 6- 8)

B f 2 - f l ~'~2 - ~'~1 A~'-~

simplest design of a band-pass filter is the connection of a high-pass filter and a low-

filter in series, which is commonly done in wide-band filter applications. Thus, a first-

r high-pass and a first-order low-pass provide a second-order band-pass, while a

(39)


41/63


e. Projeto de Filtros Passa Banda (PBn)I. Segunda Ordem! A partir de um filtro PB podemos obter o PBn executando a

transformação abaixo na função transferência do filtro PB.


d Orde r Band Pass F i lte r

develop the frequency response of a second-order band-pass filter, apply the trans-

rmation in Equation 16-7 to a first-order low-pass transfer function"

A0

A(s ) - 1 + s

placing s with A-- -~(s+ 1 )

lds the general transfer function for a second-order band-pass filter:

A~ (16-9)

A(s) = 1 + A ~. s + S 2

hen designing band-pass filters, the parameters of interest are the gain at the mid fre-

ency (Am) and the quality factor (Q), which represents the selectivity of a band-pass

er.

erefore, replace Ao with Am and A.O. with 1/Q (Equation 16-7) and obtain:

Am

A(s) -e-.s

(16-10)

Função Transf.Filtro PB


16 .5 .1 Seco nd Orde r Band Pass F i lt e r

To develop the frequency response of a second-order band-pass filter, apply the trans-

formation in Equation 16-7 to a first-order low-pass transfer function"

A0

A(s ) - 1 + s

Replacing s with A-- -~(s+ 1 )

yields the general transfer function for a second-order band-pass filter:

A~ (16-9)

A(s) = 1 + A~. s + S 2

When designing band-pass filters, the parameters of interest are the gain at the mid fre-

quency (Am) and the quality factor (Q), which represents the selectivity of a band-pass

filter.

Therefore , replace Ao with Am and A.O. with 1/Q (Equation 16- 7) and obtain:

Am

A(s) -e-.s

(16-10)

1 + $.s +

S 2

LJ

Figure 16-32 shows the normalized gain response of a second-order band-pass filter for

Substituir spelo termo acima

16.5.1 Sec ond O rde r Band Pass F i lt e r

To develop the frequency response of a second-order band-

formation in Equation 16-7 to a first-order low-pass transfer

A0

A(s ) - 1 + s


yields the general transfer function for a second-order band-

A~

A(s) = 1 + A ~. s +

S 2

When designing band-pass filters, the parameters of interest

quency (Am) and the quality factor (Q), which represents th

filter.

There fore, replace Ao with Am and A.O. with 1/Q (Equation 1

Am

A(s) -e-.s

1 + $.s + S 2

LJ

Figure 16-32 shows the normalized gain response of a seco

different Qs.

(40)Função Transf. Filtro PBn

n d Or d e r B an d Pass F il te r



A0

A(s ) - 1 + s



A~ (16-9)

A(s) = 1 + A ~ .s + S 2

When designing band-pass filters, the parameters of interest are the gain at the mid fre-


filter.

There fore, replace Ao with Am and A.O. with 1/Q (Equation 16-7) and obtain:

Am

A(s) -e-.s

(16-10)

1 + $.s + S 2

LJ

Figure 16-32 shows the normalized gain response of a second-order band-pass filter for

different Qs.

(41)

1 6 .5 . 1 S e c o n d O r d e r B a n d P a s s F i l te r



A0

A(s) - 1 + s

Replacing s with A-- -~(s+ 1)


A~ (16-9)

A(s) = 1 + A~ .s + S 2

When designing ba nd-pass filters, the parameters of interest are the gain at the mid fre-


filter.

Therefore, replace Ao with Am and A.O. with 1/Q (Equation 16-7) and obtain:

Am

A(s) -e-.s

(16-10)

1 + $.s +

S 2

LJ

Figure 16 -32 shows the normalized gain response of a second-order band-pass filter for

different Qs.

- 5

- 1 0

II1

15

I

c

~ 20

I

~ 25

- 3 0

0=1/

/

J

35 /

0.1

/

/

/

/

r

\

1

Frequency

\

k

\

\

lO

F igure 16 32 . Ga in Response o f a Second Order Band Pass F i lt e r

! Em Eq. (40), Devemos substituir A0 por

Am para representar o ganho na bandamédia e aplicar Eq. (39) para incluir ofator de qualidade. Então:

%-

%-


42/63


e. Projeto de Filtros Passa Banda (PBn)I. Segunda Ordem! Podemos usar a topologia de Sallen-Key para implementar o PBn de 2ª

ordem usando redes RC em PA e PB.! Sua função transferência se torna a Eq. (41), onde:

shows that the frequency response of second-order band-pass filters gets

ith rising Q, thus making the filter more selective.

opology

R

%-

Vow

Key Topology

R

%-

Vow

Sallen KeyBand Pass

e Sallen-K ey band-pass circuit in Figure 1 6-3 3 has the following transfer function:

A(s) = G.RC~m.S

1 + RC e0 m (3 - G ) s + R2C2(.Om2.S2

rough coefficient comparison with Equation 16-10, obtain the following equations:

1

mid-frequency: fm = 2~RC

a 2

inner gain: G = 1 R~

G

gain at fm Am = 3 - G

filter quality: Q =

3 - G

(41)

Figure16 33. Sallen KeyBand Pass

The Sallen-Key band-pass circuit in Figure 16-33 has the foll

A(s) = G.RC~m .S

1 + R Ce 0m (3 - G ) s + R2C2(.Om2.S2

Through coefficient comparison with Equation 16-10, obtain t

1


a 2

inner gain:

G = 1 R~

G


f i l te r q u a l i t y : Q =

3 - G

The Sallen-Key circuit has the advantage that the quality facto

inner gain (G) without modifying the mid frequency (fm). A dra

(42)


The S allen-Key band-pass circuit in Figure 16 -33 has the follo

A(s) = G.RC~m.S

1 + RC e0 m (3 - G ) s + R2C2(.Om2.S2

Through coefficient comparison with Equation 16-10, obtain th

1


a 2

inner gain:

G = 1 R~

G


filter qual ity: Q =

3 - G

The Sallen-Key circuit has the advantage that the quality factor

inner gain (G) without modifying the mid frequency (fm). A dra

and Am cannot be adjusted independently.

(43)

%-


The Sallen-Key band-pass circuit in Figure 16- 33 has the following

A(s) = G.RC~m.S

1 + R C e0 m (3 - G ) s + R2C2(.Om2.S2

Through coefficient comparison with Equation 16-10, obtain the foll

1


a 2

inner gain: G = 1 R~

G

gai n at fm Am = 3 - G

filter quality : Q =

3 - G

The Sallen-Key circuit has the advan tage that the quality factor (Q) c

inner gain (G) without modifying the mid frequency (fm). A drawback


Care mus t be taken when G approaches the value of 3, because then

(44)


The Sallen-Key band-pass circuit in Figure 16-33 has the followi

A(s) = G.RC~m .S

1 + R Ce 0m (3 - G ) s + R2C2(.Om2.S2

Through coefficient comparison with Equation 16-10, obtain the

1


a 2

inner gain:

G = 1 R~

G


f i l t e r qua l i t y : Q =

3 - G

The Sallen-Key circuit has the advantage that the quality factor (Q

inner gain (G) without modifying the m id frequency (fm). A drawb


Care must be taken when G approaches the value of 3, because th

(45)


43/63


e. Projeto de Filtros Passa Banda (PBn)I. Segunda Ordem! Procedimento de projeto:① Definir fm, C e depois calcular② Devido à dependência entre Q e Am podemos calcular R 2 de duas

formas diferentes. A equação escolhida dependerá se o projeto iráseguir algum requisito de Q ou de Am.

③ Use as Eq. (43), (44) e (45) para definir o ganho do filtro.

Exercício: projete um Filtro PBn com frequência central de 1kHz, largurade banda de 200Hz.

f i l te r q u a l i t y : Q =

3 - G

The Sallen-Key circuit has the advantage that the quality factor (Q) can b

inner gain (G) without modifying the mid frequency (fm ). A drawback is, h


Care mus t be taken when G approaches the value of 3, because then A m b

and causes the circuit to oscillate.

To set the mid frequency of the band-pass, specify fm and C and then so

a

2~mC

290

(46)Band Pass Filter Design

Beca use of the dependency between Q and Am, there are two options to solve for R2: ei-

ther to set the gain at mid frequency:

a 2 -

2 A m - 1

1

+ A m

or to design for a specified Q:

R2 = 2 Q - 1

Q

1 2 Multiple Feedback Topology

c

(47.a)

Because of the dependency between Q and Am, there are t

ther to set the gain at mid frequency:

a 2 -

2 A m - 1

1

+ A m

or to design for a specified Q:

R2 = 2 Q - 1

Q

16 5 1 2 Multiple Feedback Topology

a 1

VIN

c

VOUT

(47.b)


44/63


e. Projeto de Filtros Passa Banda (PBn)I. Segunda OrdemResolução.

Com base em f m = 1kHz , podemos definir C = 100nF e calcular R.

Como foi especificada a largura de banda (200Hz), estamos fazendo umprojeto voltado ao fator de qualidade do filtro. Usando Eq. (39), temos Q.

Podemos calcular G e Am usando Eq. (43) e (44).

O Alto valor de Am implicará num valor limitado de V in devido à saturação doamplificador operacional.

R =1

2 !" !103 !100 !10#9 $1592%

Q = f m

B=

1000

200= 5

G =3!Q

"1

Q=

3!5"1

5= 2, 8

Am =

G

3!G=

2,8

3! 2,8=14

Definindo R 1

= 1k e usandoEq. (43), temos:

R2 = R

1! G "1( )

R2 = 1000 ! 2,8 "1( )

R2 = 1,8k #


45/63


e. Projeto de Filtros Passa Banda (PBn)I. Segunda Ordem

Deve-se tomar cuidado com o projeto baseado no Q pelos seguintes motivos:- Grandes valores de Q geram valores bem altos de Am.- Se G se aproxima de 3, teremos Am infinito, o que nos obriga a alterar o valorde Q.

Q _ 10 = 31 8 kQ

R2 - ~frnC =. 1 kHz . 100 nF "


46/63


e. Projeto de Filtros Passa Banda (PBn)II. Quarta Ordem! Filtros PBn de 4ª ordem permitem a especificação do tipo da resposta.

A serialização de dois filtros PBn leva ao filtro de 4ª ordem com afunção transferência abaixo, onde:

"

Ami, fmi e Qi representa o ganho, frequência média e fator de qualidade doenésimo estágio, respectivamente." α e 1/α são os fatores que definem o quão distante estão fm1 e fm2 do f m

de projeto do total do filtro. Filtros com alto valor de Q têm fm1 e fm2 muitopróximos de f m.

!

O fator α é calculado pela Eq. (49) usando o método de aproximaçõessucessivas, com a1 e b1 sendo os coeficiente do filtro PB de 2ª ordem.

R2 _- 31.8 kQ = 7.96 k~

R1 = - 2Am 4

- AmR1 2.7.96 kQ = 80.4

R 3 = 2 Q 2 + A m = 2 0 0 - 2

-Order Band-Pass Filter Staggered Tuning)

gure 16-32 shows that the frequency response of second-order band-pass filters gets

eeper with rising Q. However, there are band-pass applications that require a flat gain

sponse close to the mid frequency as well as a sharp passband-to-stopband transition.

ese tasks can be accomplished by higher-order band-pass filters.

particular interest is the application of the low-pass to band-pass transformation onto

second-order low-pass filter, since it leads to a fourth-order band-pass filter.

eplacing the S term in Equation 16-2 with Equation 16-7 gives the general transfer func-

n of a fourth-order band-pass:

s2.A0(AQ) 2

A(s ) = b, (16-1 1)

a, [ (Aa)2 ]-S2 4- a,

1 +~ /k Q. S+ 2+ b~ _1 ~A Q' S3 + s4

milar to the low-pass filters, the fourth-order transfer function is split into two second-or-

r band-pass terms. Further mathematical modifications yield:

Ami Ami s

A(s) =

[os ] [ ]

1

q u a t io n 1 6 - 1 2 r e p r e s e n t s t h e c o n n e c t i o n o f t w o s e c o n d - o r d e r b a n d - p a s s f il te rs in s e -

es where

(16-12)

(48)

Band Pass Filter De

9 - . .̂ . . 9 , . .. ~ , ~ . ~ \ . . ~ ~ . . ~ ~

9 Ami is the gain at the mid frequency, fmi, of each partial filter

9 Qi is the pole qual ity of each filter

9 o~ and 1/0~ are the factors by which th e mid fr eque ncies of the individ ual filters,

and fm2, derive from the mid frequency, fm, of the overall bandpass.

In a fourth-order band-pass filter with high Q, the mid frequencies of the two partial filt

differ only slightly from the overall mid frequency. This method is called staggered tuni

Factor ~ needs to be determined through successive approximation, using equat

16-13:

b 1

= 0 (16-

with al and bl being the second-order low-pass coefficients of the desired filter type

(49)

,

16-13:


47/63


e. Projeto de Filtros Passa Banda (PBn)II. Quarta Ordem! Para facilitar a tarefa de projeto, a tabela abaixo sumariza as três

principais respostas com os respectivos valores de α, ai, bi eQ=1;10;100.

!

Através deα calculamos:

Table 16 2.

b 1

= 0 (16-1 3)

with al and bl being the second-order low-pass coefficients of the desired filter type.

To simplify the filter design, Table 16 -2 lists those coefficients, and provides the o~ values

for three different quality factors, Q = 1, Q = 10, and Q = 100.

Values of o~ Fo r Different Filter Types and Differen t Q s

al

b l

O

X

Bessel

1.3617

100

0 01

1.0032

0.6180

10 1

0.1 1

1.0324 1.438

Butterworth

al 1.4142

bl 1.0000

Q 100 10

A~ 0 01 0 1

1.0035 1.036

1

1

1.4426

Tschebyschef f

al 1 0650

bl 1.9305

O 100 10

A~ 0 01 0 1

o~ 1.00 33 1.03 38

1

1

1.39

ign

o~ has been determined, all quantities of the partial filters can be calcu lated using the

ing equations:

id frequency of filter 1 is:

fm (16-14)

fml = -E

id frequency of filter 2 is:

(50)


Afte r o~ has been determined, all quantities of the partial filters can be calcu lated using

following equations:

The mid frequency of filter 1 is:

fm (16-

fml = -E

the mid frequency of filter 2 is:

fm2 = fm'~ (16 -

with fm being the mid frequency of the overall forth-order band-pass filter.

(51)


After o~ has been determined, all quantities of the partial filters can be calcu lated



fm

fml = -E


fm2 = fm'~

with fm being the mid frequency of the overall forth-order band-pass filter.

The individual pole quality, Qi, is the same for both filters:

Qi = Q (1 + (~2)b 1

a-a 1

with Q being the quality factor of the overall filter.

(52)

Afte r o~ has been determined, all quantities of the p



fm

fml = -E


fm2 = fm'~

with fm being the mid frequency of the overall fort

The individual pole quality, Qi, is the same for bot

Qi = Q (1 + (~2)b 1

a-a 1

with Q being the quality factor of the overall filter.

The individual gain (Ami) at the partial mid frequen

filters:

Am = e i ll///Am

with Am being the gain at mid frequency, fm, of th

(53)


48/63


e. Projeto de Filtros Passa Banda (PBn)II. Quarta Ordem! Exercício. Projete um Filtro PBn com freq. central de 10kHz, largura de

banda de 1kHz, ganho unitário e resposta maximamente plana na faixade passagem e resposta de ButterWorth.

Resolução.Da tabela, obtemos os coeficientes: a1 =1.4141, b1 = 1 eα=1.036, porque

Q =10kHz/1kHz = 10, Eq. (39).

Usando as Eq. (50) e (51) calculamos as freq. médias de cada estágio.

O fator de qualidade e o ganho dos dois estágio são sempre iguais. Seuscálculos são feitos abaixo:


================= =============== . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . : : : : : . . . . . . . . ~ . . . . . . . . . . . . . . . . : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

In accordance with Equations 16-14 and 16-15, the mid frequencies for the partial filters

are:

fmi -- 10 kH z = 9.6 53 kH z and

1.036

fro2 = 10 kHz .1.0 36 = 10.36 kHz

The overall Q is defined as Q = frn/B, and for this example results in Q = 10.

Using Equation 16-16, the Qi of both filters is:

Qi = 10 (1 + 1 0362)'1

1.6-36:1~1~,2 = 14.15


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . : : : : : . . . . . . . . ~ . . . . . . . . . . . . . . . . : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

In accordance with Equations 16-14 and 16-15, the mid frequencies for the partial filters

are:

fmi -- 10 kH z = 9.6 53 kHz and

1.036

fro2 = 10 kHz .1.0 36 = 10.36 kHz

The o verall Q is defined as Q = frn/B, and for this examp le results in Q = 10.

Using Equation 16-16, the Qi of both filters is:

Qi = 10 (1 + 1 0362)'1

1.6-36:1~1~,2 = 14.15


= = = = = = = = = == = = = = = = = = = = = = == = . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . : : : : : . . . . . . . . ~ . . . . . . . . . . . . . . . . :

In accordance with Equations 16-14 and 16-15, the mid frequencies for the partial fi l ters

are:

fmi -- 10 kH z = 9.6 53 kH z and

1.036

fro2 = 10 kHz .1 .03 6 = 10.36 kHz

The overall Q is defined as Q = frn/B, and for this example results in Q = 10.

Using Equation 16-16, the Qi of both fi lters is:

Qi = 10 (1 + 1 0362)'1

1.6-36:1~1~,2 = 14.15

With Equation 16- 17, the passb and gain of the partial fi lters at

Documents

3. Filtros Ativos