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Lecture 20

Lecture 20

High Pass Filters, 2

g

,

nd

Order

Filters, Active Filters,

R

(2)

Series Resonance

f

j

R

fL

j

f

Z

s

(

)

=

2

π

+

1

resonance

For

fC

j

f

j

f

s

π

1

1

:

2

)

(

2

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

f

LC

f

C

f

L

f

π

π

π

1

)

2

(

1

2

1

2

2

2

0

0

0

=

=

LC

f

π

2

0

=

(3)

Series Resonance

Quality factor Q

S

Resistance

resonance

at

inductance

of

Reactance

Q

s

R

L

f

Resistance

0

2

π

=

LC

f

from

C

f

L

Substitute

R

0

2

0

2

2

1

)

(

)

2

(

1

π

π

=

=

CR

f

Q

LC

C

f

s

0

2

1

2

)

(

)

2

(

π

π

π

=

CR

f

0

2

π

(4)

Series Resonance

+

=

fC

j

fL

j

R

f

Z

s

2

1

2

)

(

π

π

⎟⎟

⎜⎜

+

=

fRC

R

fL

j

R

2

1

2

1

π

π

+

=

⎥⎦

⎢⎣

LC

ff

f

f

R

L

f

j

R

f

2

0

)

2

(

1

2

1

π

=

=

⎥⎦

⎢⎣

R

L

f

Q

and

f

Substitute

LC

ff

f

R

s

0

0

0

2

0

2

1

)

2

(

π

π

⎟⎟

⎜⎜

+

=

R

jQ

f

f

f

Z

R

Q

LC

f

s

0

0

1

)

(

2

π

+

jQ

⎜⎜

f

f

⎟⎟

R

f

Z

s

s

0

1

)

(

(5)

Series Resonance

⎟⎟

⎜⎜

=

⎟⎟

⎜⎜

+

=

f

f

f

f

Q

Tan

Z

f

f

f

f

Q

R

Z

s

s

s

0

1

2

0

2

1

f

f

f

f

R

0

0

(6)

Series Resonant Band-Pass Filter

R

s

s

V

/

V

I

⎟⎟

⎜⎜

+

=

=

f

f

f

f

jQ

f

Z

s

s

s

s

0

0

1

)

(

I

⎟⎟

⎜⎜

+

=

⎟⎟

⎜⎜

+

=

=

f

f

jQ

f

f

jQ

R

s

R

s

R

0

0

1

1

1

V

V

V

I

V

⎟⎟

⎜⎜

+

⎟⎟

⎜⎜

+

f

f

jQ

f

f

jQ

s

1

s

1

(7)

Series Resonant Band-Pass Filter

1

V

(

f

f

f

f

)

jQ

s

s

R

0

0

1

1

+

=

V

V

(8)
(9)

Series Resonant Band-Pass Filter

f

f

f

B

0

s

L

H

Q

f

f

f

B

=

=

0

s

Q

For Q >>1

B

For Q

S

>>1

2

0

B

f

f

H

+

2

0

B

f

f

L

2

2

(10)

Example 6.6

Hz

F

x

H

LC

f

1000

)

10

1592

0

)(

15926

0

(

2

1

2

1

6

0

=

=

=

π

π

H

Hz

R

L

f

Q

F

x

H

s

10

100

)

15926

.

0

)(

1000

(

2

2

)

10

1592

.

0

)(

15926

.

0

(

2

0

=

Ω

=

=

π

π

π

B

B

Hz

Hz

Q

f

B

s

100

10

1000

0

=

=

=

Hz

Hz

Hz

B

f

f

Hz

Hz

Hz

B

f

f

H

0

+

=

1000

+

50

=

1050

L

0

=

1000

50

=

950

(11)

Example 6.6

:

resonance

At

Ω

Ω

=

=

=

1000

1

1

1000

)

1592

.

0

)(

1000

(

2

2

0

L

j

j

j

Z

j

H

Hz

j

L

f

j

Z

π

π

Ω

=

+

+

=

Ω

=

=

=

100

1000

)

10

1592

.

0

)(

1000

(

2

2

0

6

C

L

s

C

Z

Z

R

Z

j

F

x

Hz

j

C

f

j

Z

π

π

C

L

s

(12)

Example 6.6

o

o

0

01

0

0

1

=

=

=

V

S

I

o

o

)

1

0

0

01

0

)(

100

(

0

01

.

0

100

=

=

=

=

=

=

R

Z

R

s

I

V

I

o

o

)

10

90

0

01

.

0

)(

1000

(

0

1

)

0

01

.

0

)(

100

(

=

=

=

=

=

=

j

Z

R

L

L

R

I

V

I

V

o

o

)

10

90

0

01

.

0

)(

1000

(

=

=

=

Z

I

j

V

(13)

Example 6.6

L

Q V

V

L

=

Q V

s

s

V

s

s

C

Q V

V

=

(14)

Parallel Resonance

Z

=

1

( )

R

j

fC

j

(

fL

)

Z

p

π

π

1

2

2

1

+

=

At resonance Z is purely resistive:

At resonance Z

P

is purely resistive:

(

f

L

)

f

j

C

f

j

2

π

0

=

(

1

2

π

0

)

0

=

1

LC

f

f

j

f

j

π

2

0

0

0

(15)

Parallel Resonance

Quality factor Q

P

Resistance

Q

P

⎜⎜

⎟⎟

R

resonance

at

inductance

of

Reactance

Q

P

=

⎟⎟

⎜⎜

f

from

L

Substitute

L

f

0

1

1

2

π

=

CR

f

Q

LC

f

from

C

f

L

Substitute

P

0

0

2

0

2

2

2

)

(

)

2

(

π

π

π

=

=

=

f

Q

P

0

(16)

Parallel Resonance

( )

(

)

⎟⎟

⎜⎜

+

=

+

=

fL

R

j

fRC

j

R

fL

j

fC

j

R

Z

P

2

2

1

2

1

2

1

1

π

π

π

π

=

=

f

R

f

R

fL

1

1

2

π

+

+

R

fLC

f

f

f

jQ

fLC

f

f

f

RC

f

j

P

0

2

0

0

2

0

0

)

2

(

1

1

)

2

(

1

2

1

π

π

π

⎟⎟

⎜⎜

+

=

f

f

f

f

jQ

R

P

0

0

1

(17)

Parallel Resonance

⎟⎟

⎜⎜

+

=

=

f

f

jQ

R

Z

P

out

0

1

I

I

V

V

out

for constant current,

varying the frequency

⎟⎟

⎜⎜

+

f

f

jQ

P

0

1

(18)

Parallel Resonance

Fi d th

h

t I

I

d I f

2

.

159

,

2

.

159

,

10

,

0

10

3

=

Ω

=

=

=

o

R

k

L

μ

H

C

pF

I

Find the phasor currents I

R

, I

C

and I

L

for:

10

)

2

.

159

)(

10

1

(

2

10

2

10

1

2

1

6

4

0

6

0

=

=

=

=

=

H

Hz

x

L

f

R

Q

Hz

x

LC

f

P

μ

π

π

π

(19)

Parallel Resonance

o

o

(

10

)

10

0

0

10

3

4

=

=

=

R

out

I

V

o

o

0

10

0

10

)

(

3

4

4

=

=

=

=

R

out

out

R

out

V

V

I

o

o

90

10

0

10

10

10

2

4

4

=

=

=

=

R

out

out

R

V

V

I

o

0

10

90

10

10

2

0

3

=

=

=

=

j

L

f

j

Z

L

L

π

V

V

I

o

90

10

10

0

10

2

2

3

=

=

=

=

j

C

f

j

Z

out

C

out

C

V

V

I

2

π

f

0

C

(20)

I

I

C

=

Q

P

I

I

C

=

Q

P

I

I

L

=

Q

P

(21)
(22)
(23)

Second Order Low-Pass Filter

j

j

⎟⎟

⎜⎜

+

=

+

=

+

+

=

LC

ff

f

f

R

L

f

j

fRC

fC

j

fL

j

R

fC

Z

Z

Z

Z

in

in

in

C

L

R

C

out

0

0

0

2

1

2

1

2

2

2

2

π

π

π

π

π

π

V

V

V

V

=

=

=

f

f

f

f

jQ

f

L

f

fRC

j

f

H

S

in

out

0

0

0

0

0

)

/

(

1

2

2

)

(

π

π

V

V

⎟⎟

⎜⎜

+

⎟⎟

⎜⎜

+

f

f

f

f

jQ

LC

ff

f

f

R

L

f

j

S

in

0

0

0

0

0

1

2

1

2

1

π

π

V

(24)

Second Order Low Pass Filter

Second-Order Low-Pass Filter

(

f

f

)

jQ

V

( )

(

(

)

)

0

0

0

1

jQ

f

f

f

f

f

f

jQ

f

H

s

s

+

=

=

o u t

i n

V

V

(

)

(

)

2

1

(

)

2

0

1

90

f

f

f

f

Q

T

f

f

f

f

Q

f

f

Q

s

=

o

(

)

(

)

( )

(

0

)

0

0

1

2

0

0

2

1

f

f

Q

f

H

f

f

f

f

Q

Tan

f

f

f

f

Q

s

s

S

+

( )

(

)

(

)

2

0

0

2

0

1

Q

f

f

f

f

f

H

S

s

+

=

(25)
(26)

Second Order High-Pass Filter

g

At low frequency the capacitor

i

i

i

is an open circuit

At high frequency the

capacitor is a short and the

inductor is open

(27)

Second Order Band-Pass Filter

At low frequency the capacitor

i

i

i

is an open circuit

At high frequency the inductor

is an open circuit

(28)

Second Order Band-Reject Filter

j

At low frequency the

i

i

capacitor is an open

circuit

At high frequency the

inductor is an open

circuit

(29)

Example 6.8

Design a filter that passes frequency components higher

than 1 kHz and rejects components lower than 1 kHz.

Chose L=50 mH

Chose L 50 mH

F

x

x

L

f

C

LC

kHz

f

μ

π

π

π

(

2

)

(

1

10

)

(

50

10

)

0

.

507

1

)

2

(

1

2

1

1

3

2

3

2

2

2

0

=

=

=

=

=

x

x

L

f

LC

π

π

π

(

2

)

(

2

)

(

1

10

)

(

50

10

)

2

0

(30)

Example 6.8

To avoid amplifying the

signal at f

0

choose Q

s

=1

(31)

Example 6.8

Ω

1

314

)

10

50

)(

1

(

2

2

2

0

f

0

L

kHz

x

3

H

R

L

f

Q

=

π

=

π

=

π

=

314

.

1

Ω

1

)

)(

(

0

0

Q

f

R

R

f

Q

s

(32)

Exercise 6.20

Design a filter with Q

S

=1 that passes frequency components

lower than 5 kHz and rejects components higher than 5 kHz.

Chose L=5 mH

Chose L 5 mH

F

x

x

L

f

C

LC

kHz

f

μ

π

π

π

(

2

)

(

5

10

)

(

5

10

)

0

.

2026

1

)

2

(

1

2

1

5

3

2

3

2

2

0

2

0

=

=

=

=

=

x

x

L

f

LC

π

π

π

(

2

)

(

2

)

(

5

10

)

(

5

10

)

2

0

(33)

Exercise 6.20

Ω

1

157

)

10

5

)(

5

(

2

2

2

0

f

0

L

kHz

x

3

H

R

L

f

Q

=

π

=

π

=

π

=

157

.

1

Ω

1

)

)(

(

0

0

Q

f

R

R

f

Q

s

(34)

Exercise 6.21

Design a filter that passes frequency components between

f

L

=45 kHz and f

H

=55 kHz. Chose L=1 mH

nF

x

x

L

f

C

LC

kHz

f

10

.

13

)

10

1

(

)

10

50

(

)

2

(

1

)

2

(

1

2

1

50

2

2

3

2

3

0

2

0

=

=

=

=

=

π

π

π

(35)

Exercise 6.21

=

=

5

50

10

0

kHz

f

Q

kHz

f

f

B

H

L

=

=

=

=

2

5

10

0

0

L

f

R

kHz

B

f

Q

π

Ω

=

=

2

(

50

)(

1

10

)

62

.

83

3

H

x

kHz

Q

R

π

Ω

83

.

62

5

(36)

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

(37)

3. Place modest demands on the op amp’s gain–

p

p

g

bandwidth 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

(38)

Active First-Order Low-Pass Filter

f

f

i

f

i

o

C

fR

j

Z

Z

V

V

f

H

π

2

1

1

1

1

)

(

=

=

f

f

f

f

f

f

f

R

R

C

fR

j

R

fC

j

R

Z

π

π

2

1

2

1

1

1

1

=

+

=

+

f

f

f

f

f

f

R

Z

C

fR

j

R

Z

π

1

2

1

+

=

f

f

f

i

f

i

f

R

C

fR

j

R

Z

f

H

π

1

2

1

1

)

(

⎟⎟

⎜⎜

=

+

⎟⎟

⎜⎜

=

=

f

f

B

B

i

C

R

f

f

f

j

R

π

2

1

)

/

(

1

=

⎣ +

⎟⎟

⎜⎜

f

f

(39)

t

( )

v

( )

t

dt

RC

t

v

t

o

in

1

=

RC

0

(40)

Active First-Order High-Pass Filter

g

i

f

i

o

Z

Z

v

v

f

H

(

)

=

=

f

f

f

f

i

i

i

R

Z

f

H

R

Z

C

f

j

R

Z

π

)

(

2

1

=

=

=

+

=

i

f

i

i

f

i

i

i

C

R

f

j

R

C

R

f

j

C

f

j

R

Z

f

H

π

π

π

2

2

2

1

)

(

+

=

=

B

f

i

i

i

f

i

i

i

i

i

f

f

f

j

f

f

j

R

R

C

R

f

j

R

C

R

f

j

π

π

)

/

(

1

)

/

(

2

1

2

1

+

⎟⎟

⎜⎜

=

+

⎟⎟

⎜⎜

=

+

=

i

i

B

B

i

C

R

f

f

f

j

R

π

2

1

)

/

(

1

=

⎣ +

(41)

Differentiator Circuit

d

( )

dt

dv

RC

t

v

o

=

in

dt

(42)

Higher Order Active Filters

g

n

f

H

f

H

f

H

f

H

(

)

=

1

(

)

2

(

)

(

)

n

B

n

i

f

n

f

f

j

R

R

+

⎟⎟

⎜⎜

=

)

/

(

1

1

)

1

(

(43)

Butterworth Transfer Function

Butterworth filters are

characterized by

having a particularly

flat pass-band.

( )

(

)

n

f

f

H

f

H

2

0

1

=

(

)

n

B

f

f

2

1

+

(44)

Sallen Key Circuits

Sallen–Key Circuits

RC

f

B

π

2

1

=

RC

π

2

(45)

Sallen–Key Circuits

y

v

R

v

f

v

v

v

R

K

R

v

i

o

f

f

f

=

+

=

)

1

(

K

R

R

K

A

v

v

v

f

v

o

i

=

+

=

=

=

)

1

(

1

R

v

i

v

f

N

i

ti

lifi

ith

i

f K

(46)

K Values for Low-Pass and High-Pass

Butterworth Filters of Various Orders

(47)

Low-Pass Active Filter Design

g

Design a fourth order low pass Butterworth

Design a fourth-order low-pass Butterworth

filter having a frequency cut-off of 100 Hz

(48)

Low-Pass Active Filter Design

g

=

0 1

F

C

Choose

μ

=

=

Hz

x

Cf

R

B

2

(

0

1

10

)(

100

)

1

2

1

F

0.1

C

Choose

6

π

π

μ

Ω

=

k

Hz

x

Cf

B

92

.

15

)

100

)(

10

1

.

0

(

2

2

π

π

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

(49)
(50)
(51)
(52)

Review for Next Lecture!

• Magnetic fields (B flux density H magnetic

Magnetic 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 ld

moving in a magnetic field

• Faraday’s Law

• Lenz’s Law

(53)
(54)

Equal Component Sallen-Key

Low-Pass Active Filter

1

RC

f

B

π

2

1

=

(55)

Equal Component Sallen-Key

Low-Pass Active Filter

v

a

0

=

+

+

R

v

v

Z

v

v

R

v

v

a

in

a

o

a

R

Z

R

C

(56)

Equal Component Sallen-Key

Low-Pass Active Filter

0

v

v

v

v

v

v

a

in

a

o

a

1

0

=

+

+

R

R

Z

R

a

C

o

a

in

a

1

)

1

(

=

+

=

v

K

v

R

K

R

R

v

o

o

f

f

f

0

/

=

+

+

R

K

v

v

Z

v

v

R

v

v

a

o

C

o

a

in

a

0

/

)

(

+

=

+

R

K

v

v

C

j

v

v

R

v

v

a

o

o

a

in

a

C

ω

R

R

(57)

Equal Component Sallen-Key

Low-Pass Active Filter

v

a

/

0

=

+

v

K

v

Z

v

R

v

v

C

a

a

0

/

=

+

v

j

C

R

v

K

v

a

a

o

ω

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