High Pass Filters, 2nd Order Filters, Active Filters,Resonances.pdf

Lecture 20Lecture 20High Pass Filters, 2nd Order g ,

Filters, Active Filters, RResonances

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Series Resonance

fjRfLjfZs π 12)( −+=

resonanceForfC

jfjfs π

11:

2)(

2

For resonance the reactance of the inductor and the capacitor cancel:

f

LCf

CfLf

πππ

1)2(1

212 2

20

00 =→=


LCf

π20 =

Series ResonanceQuality factor QS

Resistance resonance at inductance of ReactanceQs ≡

RLf

Resistance02π

=

LCffrom

CfLSubstitute

R

020

2 21

)()2(1

ππ==

CRfQ

LCCf

s

0

21

2)()2(

π

ππ

=


CRf02π

Series Resonance−+=

fCjfLjRfZs 2

12)(π

π

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

fRCRfLjR

2121

ππ

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⎥⎦⎢⎣ ⎠⎝

LCffff

RLf

jR

f

20

)2(12

1π

==

⎥⎦⎢⎣⎟⎠

⎜⎝

RLf

QandfSubstitute

LCfffR

s0

0

02

0

21

)2(

π

π

⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

−+=ffjQRfZ

RQ

LCf s

0

0

1)(

2π


⎥⎥⎦⎢

⎢⎣

⎟⎟⎠

⎜⎜⎝

+ff

jQRfZ ss0

1)(

Series Resonance

⎟⎟⎠

⎞⎜⎜⎝

⎛−=∠⎟⎟

⎠

⎞⎜⎜⎝

⎛−+= −

ff

ffQTanZ

ff

ffQ

RZ

sss01

2021


⎟⎠

⎜⎝

⎟⎠

⎜⎝ ffffR 00

Series Resonant Band-Pass Filter

Rss /VVI

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

==

ff

ffjQ

fZs

s

s

s

0

01

)(I

⎟⎟⎞

⎜⎜⎛

−+=→

⎟⎟⎞

⎜⎜⎛

−+==

⎠⎝

ffjQffjQR

s

RsR

00 1

1

1VVV

IV


⎟⎟⎠

⎜⎜⎝

−+⎟⎟⎠

⎜⎜⎝

−+ff

jQff

jQ ss00

11


1V( )ffffjQss

R

0011

−+=

VV





fffB 0

sLH Q

fffB 0=−=

sQ

For Q >>1

B

For QS>>1

20BffH +≅

20BffL −≅


22

Example 6.6

HzFxHLC

f 1000)1015920)(159260(2

12

160 ===

−ππ

HHzR

LfQ

FxH

s 10100

)15926.0)(1000(22)101592.0)(15926.0(2

0 =Ω

==πππ

BB

HzHzQf

Bs

10010

10000 ===


HzHzHzBffHzHzHzBff LH 9505010002

10505010002 00 =−=−≈=+=+≈

Example 6.6

:resonanceAt

Ω

Ω===

1000111000)1592.0)(1000(22 0L

jjjZ

jHHzjLfjZ ππ

Ω=++=

Ω−=−=−=−

100

1000)101592.0)(1000(22 6

0

CLs

C

ZZRZ

jFxHz

jCf

jZππ


CLs

Example 6.6

oo

001001∠=

∠== SV

I

oo 01)0010)(100(

001.0100

∠=∠==

∠===

R

Z

R

s

IV

I

oo 9010)001.0)(1000(

01)001.0)(100(

∠=∠==

∠=∠==

jZ

R

LL

R

IV

IV


oo 9010)001.0)(1000( −∠=∠−== jZCC IV

Example 6.6

L Q VV = ssL Q VV

ssC Q VV =


Parallel Resonance

Z 1= ( ) ( )fLjfCjR

Z p ππ 2121 −+=

At resonance Z is purely resistive:At resonance ZP is purely resistive:

( ) fLfjCfj ππ 1212 000 =→=


( )LC

ffjfjπ2

000

Parallel ResonanceQuality factor QP

ResistanceQP ⎟⎟⎞

⎜⎜⎛

≡

Rresonance at inductance of Reactance

QP

=

⎟⎟⎠

⎜⎜⎝

≡

ffromLSubstitute

Lf0

112π

=

CRfQLC

ffromCf

LSubstitute

P 0

020

2

22)()2(

πππ

=

==


fQP 0

Parallel Resonance

( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=−+

=

fLRjfRCj

RfLjfCjR

Z P

221

21211

ππ

ππ

⎟⎞

⎜⎛

=⎟⎞

⎜⎛

=

⎟⎠

⎜⎝

f

R

f

RfL

11

2π

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟

⎟⎠

⎞⎜⎜⎝

⎛−+

RfLCff

fjQfLCff

fRCfj P0

200

20

0 )2(11

)2(121

πππ

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

ff

ffjQ

R

P0

01


⎠⎝

Parallel Resonance

⎟⎟⎞

⎜⎜⎛

+==

ffjQ

RZPout01

IIV Vout for constant current, varying the frequency


⎟⎟⎠

⎜⎜⎝

−+ff

jQP0

1

Parallel Resonance

Fi d th h t I I d I f

2.159,2.159,10,010 3 ==Ω=∠= − pFCHLkR μoI

Find the phasor currents IR, IC and IL for:

10)2.159)(101(2

102

1012

16

4

0

60 =====

HHzxLfRQHzx

LCf P μπππ


Parallel Resonanceoo 010)10(010 43 ∠=∠== −Rout IV

oo

010010

)(

344 ∠=

∠=== −

Routout

R

out

VVI

oo

90100101010

2

44

∠=∠

=== −

R

outout

R

VVI

o010

9010102 3

0

∠

−∠====jLfjZ L

L π

VV

I

o901010

010

2

23 ∠=

−∠

=−

== −

jCfjZ

out

C

outC

VVI


2 0Cfπ

II PC Q= II PC Q=

II PL Q=


Ideal Filters


Noise Rejectionj


Second Order Low-Pass Filter

−− jj

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=−+

=++

=

LCffff

RLfj

fRC

fCjfLjR

fCZZZ

Zininin

CLR

Cout

00

0

2121

2

22

2

ππ

π

ππ

π VVVV

⎟⎞

⎜⎛

−=

⎟⎞

⎜⎛

−

==

⎠⎝

ffffjQ

fLffRC

j

fH S

in

out

0

0

0

00

)/(12

2)(π

πVV


⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

ff

ffjQ

LCffff

RLfj S

in 0

000

0 12

121π

πV

Second Order Low Pass FilterSecond-Order Low-Pass Filter

( )ffjQV( ) ( )( )00

0

1 ffffjQffjQ

fHs

s

−+−

==in

outVV

( )( ) ( )122

0

1

90

ffffQTffffQ

ffQs

∠

−∠=

−

o

( ) ( )

( ) ( )0

0012

0021

ffQfH

ffffQTanffffQ

s

sS −∠−+ −

( ) ( )( )200

20

1 ffffQfH

S

s

−+=


Second Order Low-Pass Filter


Second Order High-Pass Filterg

At low frequency the capacitor i i iis an open circuit

At high frequency the capacitor is a short and the inductor is open


Second Order Band-Pass Filter

At low frequency the capacitor i i iis an open circuit

At high frequency the inductor is an open circuit


Second Order Band-Reject Filterj

At low frequency the i icapacitor is an open

circuit

At high frequency the inductor is an open circuit


Example 6.8

Design a filter that passes frequency components higher than 1 kHz and rejects components lower than 1 kHz. Chose L=50 mHChose L 50 mH

FxxLf

CLC

kHzf μπππ

507.0)1050()101()2(

1)2(1

211 3232220 ===→==

−


xxLfLC πππ )1050()101()2()2(2 0

Example 6.8

To avoid amplifying the signal at f0 choose Qs=1


Example 6.8

Ω−

1314)1050)(1(222 300 HxkHzLf

RLf

Q πππΩ===→= 1.314

1))((00

Qf

RRf

Qs


Exercise 6.20

Design a filter with QS=1 that passes frequency components lower than 5 kHz and rejects components higher than 5 kHz. Chose L=5 mHChose L 5 mH

FxxLf

CLC

kHzf μπππ

2026.0)105()105()2(

1)2(1

215 32322

020 ===→==

−


xxLfLC πππ )105()105()2()2(2 0

Exercise 6.20

Ω−

1157)105)(5(222 300 HxkHzLf

RLf

Q πππΩ===→= 1.157

1))((00

Qf

RRf

Qs


Exercise 6.21

Design a filter that passes frequency components between fL=45 kHz and fH=55 kHz. Chose L=1 mH

nFxxLf

CLC

kHzf 13.10)101()1050()2(

1)2(1

2150 32322

020 ===→==

−πππ


Exercise 6.21=−=

55010

0 kHzfQ

kHzffB LH

=

===

2

510

0

0

LfR

kHzBf

Q

π

Ω==−

83.62)101)(50(2 3 HxkHz

QR

πΩ83.62

5


Active Filters

Ideally, an active filter circuit should:Ideally, an active filter circuit should:

1. Contain few components. Co ta ew co po e ts

2. Have a transfer function that is insensitive to . ave a t a s e u ct o t at s se s t ve to component tolerances


3. Place modest demands on the op amp’s gain–p p gbandwidth product, output impedance, slew rate, and other specifications

4. Be easily adjusted

5. Require a small spread of component values

6. Allow a wide range of useful transfer functions to be realized


Active First-Order Low-Pass Filter

ff

i

f

i

o

CfRjZZ

VV

fH

π21111

)( −==

f

ff

f

f

ff

R

RCfRj

RfCj

RZπ

π

21

21111

+=+=

ff

ff

ff

RZ

CfRjR

Zπ

1

21

⎟⎞

⎜⎛

+=

f

ffi

f

i

f

R

CfRjRZfH

π

1

211)(

⎥⎤

⎢⎡⎟⎟⎞

⎜⎜⎛−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

ffB

Bi

CRf

ffjR

π21

)/(1

=

⎥⎦

⎢⎣ +⎟⎟⎠

⎜⎜⎝


ff

A low-pass filter with a dc gain of -Rf/Ri

t

( ) ( )dttvRC

tvt

o in1∫−=


RC 0∫

Active First-Order High-Pass Filterg

i

f

i

o

ZZ

vv

fH )( −==

ff

ffi

ii

RZfH

RZCfj

RZπ

)(

21

==

=+=

ifiif

ii

i

CRfjRCRfjCfj

RZfH

πππ

222

1)(

⎟⎞

⎜⎛

+−=−=

Bf

ii

if

i

i

ii

if

ffjffj

RR

CRfjRCRfj ππ

)/(1)/(

2121

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

+⎟⎟⎠

⎜⎜⎝−=

+−=

iiB

Bi

CRf

ffjR

π21

)/(1

=

⎥⎦

⎢⎣ +⎟⎠

⎜⎝


A high-pass filter with a high frequency gain of -Rf/Ri

Differentiator Circuit

d( )dt

dvRCtvoin−=


dt

Higher Order Active Filtersg

n fHfHfHfH ∗∗∗= )()()()( 21n

B

n

i

fn

ffjRR

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

)/(11)1(


Butterworth Transfer FunctionButterworth filters are characterized by having a particularly flat pass-band.

( )( ) nff

HfH2

0

1=


( ) nBff 21+

Sallen Key CircuitsSallen–Key Circuits

RCfB π2

1=

RCπ2


Sallen–Key Circuitsy

vR

v f

vv

vRKR

v

i

off

f

=

−+=− )1(

KR

RKA

v

vv

fv

o

i

=−

+==

= −

)1(1

Rv fv

i

N i ti lifi ith i f K


Non-inverting amplifier with a gain of K

K Values for Low-Pass and High-Pass Butterworth Filters of Various Orders


Low-Pass Active Filter Designg

Design a fourth order low pass Butterworth


Design a fourth-order low-pass Butterworth filter having a frequency cut-off of 100 Hz


= F0 1CChoose μ

==− HzxCf

RB )100)(1010(2

12

1F0.1C Choose

6ππ

μ

Ω= kHzxCfB

92.15)100)(101.0(22 ππ

From the table a fourth order filter requires K values of 1.152 and 2.235. The DC gain is (1.152)(2.235) = 2.575




Low-pass Active Filter Designg


Low-pass Active Filter Designg


Review for Next Lecture!

• Magnetic fields (B flux density H magneticMagnetic fields (B flux density, H magnetic field intensity, magnetic permeability μ)

• Right hand rule• Right hand rule• Forces on charges and current carrying wires

i i i fi ldmoving in a magnetic field• Faraday’s Law• Lenz’s Law• Ampere’s Law


Ampere s Law


Equal Component Sallen-Key Low-Pass Active Filter

1RC

fB π21

=



va

0=−

+−

+− −

Rvv

Zvv

Rvv aoaina


RZR C


0−−− −vvvvvv aoaina

1

0=++

RRZR

a

C

oaina

1)1(

=−+

=− vK

vRKR

Rv oo

ff

f

0/

=−

+−

+−

RKvv

Zvv

Rvv oa

C

oaina

0/

)( =−

+−+−

RKvv

CjvvR

vv oaoa

ina

C

ω


RR


va

/

0

−

=+−−

vKvZv

Rvv

C

aa


0/

=+ CjvR

vKva

ao ω

High Pass Filters, 2nd Order Filters, Active Filters,Resonances.pdf

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