ECE 550 LINEAR SYSTEM THEORY LECTURE NOTES

115

Passive Filters

Up to now, we have driven a circuit with a dc or a fixed frequency ac signal and looked at the effects of changing the magnitudes and phase angles of thevoltages, currents and impedances. In some systems, you really want toknow the response of the system when the frequency is varied, i.e., thefrequency response. Often, filters weaken (attenuate) the input signal outside a particular frequency band. For example, you might have a graphic equalizer an your stereo which is a collection of filter circuits designed to allow you to amplify the sound (or audible frequency kHzHz 3~300 ) in a particular frequency range and attenuate frequencies outside that band.

116

Transfer function:

)( ωjV

)( ωjI

System

)()()(

ωωω

jIjVjZ = transfer impedance

)()(

)( 0

ωω

ωjVjV

jGi

V = voltage gain

)()()(

ωωωjVjIjY = transfer admittance

)()(

)( 0

ωω

ωjIjI

jGi

I = current gain

)( ωjZ , )( ωjY , )( ωjGV , and )( ωjGI are called the ‘transfer functions’

which describe the system characteristics. We generally represent it as )( ωjH

System

)( ωjI i

)( ωjVi

)(0 ωjI

)(0 ωjV

117

Consider a linear system (current proportional to the input voltage, output response proportional to the input)

)( ωjX )( ωjY)( ωjH

Linear System

The Linear System can be characterized by a Transfer Fct.: )()()(

ωωωjXjYjH =

Let us consider an example:

RI

+_ Ljω

+

_ 0ViV

The transfer function,

LjRLj

jVjV

jGjHi

V ωω

ωω

ωω+

===)()(

)()( 0

By Voltage Divider

LjRLjVV i ω

ω+

=0

118

We show voltages and currents as functions of “ ωj ”indicating that we are interested in the frequency characteristics or response of a circuit /system. In ENGR 203 you will see this developed into Fourier Transform Analysis.

We can write )( ωjH as: (multiply by L

L1

1 to isolate jω)

LRj

jjH+

=ω

ωω)( and: (recall, for a complex number c=a+jb, |c|=sqrt(a2+b2)

( )( ωjH is a complex function, so it can be represented in polar form, i.e. using mag. and phase, )(|)(|)( ωθωω jjHjH ∠= , and you can give the “phase angle plot” of )( ωθ j vs. freq. ω)

119

2

222 11

1)(

ωω

ωω

⎟⎠⎞

⎜⎝⎛+

=

⎟⎠⎞

⎜⎝⎛+

=

LR

LR

jH

0)(0=

=ωωjH ; 1)( =

∞=ωωjH

Magnitude Plot ( ωω ..)( freqvsjH )

ω

)( ωjH

707.02

1=

cω

1.0

This network behaves like a “High-Pass” filter. As you can see for lowfrequencies )( ωjH is smaller. “ωc” gives an indication of the frequencies that can be passed by the filter. Itis called the “cut-off frequency” or the “half-power frequency” at which theaverage power delivered to the load is half of the maximum power.

120

For this example, when

LR

=ω

21

2)( =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

LR

LR

jH ω LR

c =⇒ω

Consider another example:

R

I

+_

Ljω

+

_iV

0V

By simple inspection, V0=Vi at dc since jωL=0 V0=0 at ∞=ω , since ∞=Ljω (open circuit) This implies that this is a ‘low-pass’ filter!

2)( maxH

jH c =ω

121

LRj

LR

jH+

=ω

ω)( ⇒ ( )22

)(

LR

LR

jH+

=ω

ω

ω

)( ωjH

707.02

1=

cω

1.0

Low pass behavior

122

Some other filter examples:

RI

+_

+

_iV

0VC

R

I

+_

+

_iV

0V

C

0)(1)0(=∞=

HH } Low-pass

RCc1

=ω

RCc1

=ω

RCc1

=ω

RCc1

=ω

RCjRCjH

1

1)(

+=

ωω

} High-pass1)(0)0(=∞=

HH

RCjjjH

1)(

+=

ωωω

123

Parallel Resonant Circuits: Consider:

+

_

outVinI C R L

Let us first analyze the circuit by inspection:

0=outV @ dc since ⇒= 0Ljω inductor shorts the output.

0=outV @ ∞=ω since ⇒= 01Cjω

capacitor shorts the output.

124

The transfer function of the circuit can be given as (Transfer Impedance)

⎥⎦⎤

⎢⎣⎡ −+

==

LCj

RjIjV

jHi

out

ωωω

ωω

111

)()(

)( =Zeq = ...)111(

1

321 ZZZ++

or 22 11

1)(

⎥⎦⎤

⎢⎣⎡ −+⎟

⎠⎞

⎜⎝⎛

=

LC

R

jH

ωω

ω ,

when the capacitor and inductor are in resonance (exchange energy) 01

=−L

Cω

ω , )1(L

Cω

ω = , ω2= LC1 ,

LC1

=ω , then ω = ωo (resonant frequency)

125

RHjH == max)( ω , when 0ωω = , such that

01

00 =−

LC

ωω

or: LC1

0 =ω , again is called the “resonant frequency”

or the center frequency of the “bandpass” filter

ω1 and ω2 are called the “half-power frequencies” where 2

)( maxHjH =ω

(or lower and upper cutoff frequencies) We can calculate ω1 and ω2 to be

LCRCRC1

21

21 2

2,1 +⎟⎠⎞

⎜⎝⎛++=ω

We can also show that 210 ωωω = 12 ωω −=B is called the bandwidth of the filter. (defines the pass band

– the range of frequencies passed with appreciable output)

126

Quality Factor Quality Factor (Q Factor), or “selectivity factor”, indicates the quality of the filter circuit

Q = 12

00

ωωω

βω

−= (dimensionless)

Where: High Q – good selectivity (narrow bandwidth passed) Low Q – poor selectivity (broad bandwidth passed)

127

Series Resonance:

R

+_

+

_

iV 0V

C

L

By inspection: ( 0=ω ) V0=Vi @ dc since ⇒∞==

CjZc ω

1 capacitor is dominant, like an open circuit

iVV

H 01)0( ==⇒

AND V0=Vi @ ∞=ω , since ⇒∞== LjZ ω inductor is dominant, like an open circuit

1)( =∞⇒ H

128

At 0ωω = whenCjLj0

0 ωω = ; 0)(0 00 =⇒= ωHV (Zeq = 0

00 =

−+

CjLj

ωω )

This implies that the circuit exhibits “band-stop” characteristic.

⎟⎠⎞

⎜⎝⎛ −+

⎟⎠⎞

⎜⎝⎛ −

==

CLjR

CLj

VV

jHi

ωω

ωω

ω1

1

)( 0

22 1

1

)(

⎟⎠⎞

⎜⎝⎛ −+

⎟⎠⎞

⎜⎝⎛ −

=

CLR

CL

jH

ωω

ωω

ω

0)( =ωjH , when C

L0

01

ωω = , ⇒

LC1

0 =ω resonant frequency

129

Bandstop Characteristic.

ω1 and ω2 can be determined by equating 2

)( maxHjH =ω

Where,

LCLR

LR 1

22

2

2,1 +⎟⎠⎞

⎜⎝⎛++=ω

130

Multi-Section Filter Design References:

• ‘Microwave Filters, Impedance-Matching Networks and Coupling Structures’ by Matthaei, Young and Jones

• “Micro Engineering’ by David M. Pozar.

maxH

)( ωjH

ωcωLow-pass

maxH

)( ωjH

ωcωHigh-pass

131

maxH

)( ωjH

ω0ω

Band-pass

maxH

)( ωjH

ωBandstop

0ω

N-SectionFilter

R(load)

Rs

Vs Pinc

Zin

Pload

132

In this filter design procedure, we characterize the performance of a filter using a transfer function called Power Transfer Ratio or Power Loss Ratio, PLR defined as

load

incLR P

PloadtodeliveredPowersourcefromavailablePower

P ==

It can be expressed in terms of the input impedance inZ as Eqn. A

)(21

*

2

inin

inLR ZZ

ZP

+

+=

133

Design Procedure:

Filter Specs.

Low-passprototype

design

Scaling andconversion

Final Circuit

Type of filter No. of sections

csR ω, for Low Pass / High Pass

BRs ,0, ω etc., Bandpass / Bandstop.

134

Lowpass Prototype Design:

R0=g0=1

+_ C1=g1

Zin

L2=g2

C3=g3

L4=g4

C5=g5 gn+1

Ladder Circuit (a)

g0=1

+_

Zin

L1=g1

C2=g2

L3=g3

C4=g4 gn+1

L5=g5

Ladder Circuit (b)

135

In order to have a maximally flat response in the pass band, with a cut-off frequency of 1=cω and a source resistance of Ω= 10R , The power loss ratio, PLR for the networks shown above should be Eqn. B Where N is the No. of sections (counting only L’s and C’s). We can find out g1, g2 … gn+1 for the networks (a) and (b) by equating the two equations A and B and comparing the coefficients of Kω

N

LRP 21 ω+=

136

For example: Consider the network with N=1 (network (b))

Zin

Ω1 L1=g1

g2=R

According to A, )(2

1)(2

1 2

*

2

LjRLjRLjR

ZZZ

Pinin

inLR ωω

ω−++

++=

+

+=

( )R

LRR

LR4)1(

4)()1( 222

222 ωω ++

=++

=

LjRZin ω+=

137

And this should be equal to 22 11 ωω +=+ N

Equating 0ω th coefficient, 0)1(14

)1( 22

=−⇒=+ RR

R or R=1

14

2

=R

L or L=2

21 =∴ g and 12 =g

138

Design table for Low-pass filter prototype

Impedance Scaling: If source resistance is Ω0R instead of Ω1 , the modified filter components are given by:

0' RRS = ,

' '0

0

, CL R L CR

= = LL RRR 0' =

139

For example: (For a source resistance of 100Ω)

Ω1 L=2

R=1

Ω100 200H

Ω100R=

Ω1

R=1

Ω100

Ω100R=2 0.02F

140

Frequency Scaling: To change the cutoff frequency from 1=cω to “ cω ”, New L & C’s are given by: No change in R’s.

c

kk

LL

ω='

c

kk

CC

ω='

141

Impedance and frequency scaling: (will show how we use this in an example) New filter elements are given by Filter to Filter Transformations: The main advantage of using this method of low-pass filter prototype is to be able to come up with other types of filter designs such as high-pass, bandpass and bandstop from the original low-pass prototype itself.

0' RRS = , LL RRR 0

' =

c

kok

LRL

ω=' ,

oc

kk R

CC

ω='

142

Low-pass to High-pass Transformation:

g0=1 L1=g1

C2=g2

L3=g3

Low pass

R0

High pass

C1' C3'

L2' R0

143

Transformation: Eqn. C Examples: Design a 5-section maximally flat low-pass filter with a cut-off frequency of 100kHz. The source resistance is 1kΩ. From the design table, for N=5, g1=0.618; g2=1.618; g3=2; g4=1.618; g5=0.618; g6=1 R0=1000; sec/10283.6101002 53 radc ×=××= πω

kck LR

Cω0

' 1=

kck C

RL

ω0' =

144

Considering the network (a) for the low-pass prototype.

nFR

gCc

9836.010283.61000

618.05

0

1'1 =

××==

ω

mHgR

Lc

6.220'2 ==

ω nF

Rg

Cc

183.30

3'3 ==

ω

mHgR

Lc

6.240'4 ==

ω nF

Rg

Cc

9836.00

5'5 ==

ω

The final network will be

1000

0.984nF

2.6mH

3.183nF

2.6mH

0.984nF 1000

Ω

Ω

Note: You can use the network (b) (starting with an inductor) and do the design. Both designs should give you identical response.

145

Example 2: Design a 5-section maximally flat high-pass filter with a cut-off frequency of 100kHz. Source resistance is 1kΩ. Again from the design table: g1=0.618; g2=1.618; g3=2; g4=1.618 and g5=0.618. R0=1000Ω; sec/10283.6 5 radc ×=ω Considering network (a) (capacitance first in the low-pass prototype), applying equations given in C,

mHg

RL

c

6.21

01 ==

ω nF

gRC

c

984.01

202 ==

ω

mHg

RL

c

796.03

03 ==

ω nF

gRC

c

984.01

404 ==

ω

mHg

RL

c

6.25

05 ==

ω

146

The high-pass filter is

1000Ω

1000Ω

0.984nF 0.984nF

2.6mH 0.796mH 2.6mH

147

Low-pass to Band-pass Transformation:

Transforms toL L' 'C

In a low-pass prototype where Eqn. D1

Δ=

0

0'

ωLR

L

oLRC

0

'

ωΔ

=

ω0 center frequency of the bandpass filter

0

12

ωωω −

=Δ is called the fractional

bandwidth.

148

Transforms toC L" "C

In a low-pass prototype Eqn. D2

CR

L0

0"

ωΔ

=

oRCCΔ

=0

"

ω

149

Low-pass to Band-stop Transformation:

LTransforms to

L

'

'

CIn a low-pass prototype Where Eqn. E1

0

0'

ωLR

LΔ

=

oLRC

Δ=

0

' 1ω

ω0 center frequency

0

12

ωωω −

=Δ fractional bandwidth.

150

Transforms toL"

"C

C

Where Eqn. E2

CR

LΔ

=0

0"'

ω

oRCC

0

"

ωΔ

=

151

Example for Bandpass filter design: Design a maximally flat bandpass filter with N=2; center frequency is100kHz; bandwidth is 10%; Source resistance is 50Ω. From the design table, g1=1.4142, g2=1.4142 Let us use the low-pass prototype network given in (b) (First element: inductor)

sec/10283.6101002 530 rad×=××= πω , %)10(1.0=Δ

152

From Eqn. D1, mHRg

L 1.11.010283.6

504142.15

0

011 =

×××

=Δ

=ω

nFRg

C 25.2504142.110283.6

1.05

0101 =

×××=

Δ=ω

From Eqn. D2, uHgR

L 63.520

02 =

Δ=ω

nFR

gC 15.450

501.010283.64142.1

500

22 =

×××=

Δ=ω

153

The bandpass circuit will be,

50 L1=1.1mH C1=2.25nF

L2=5.63uH C2=450.15nF 50

Ω

Ω

154

Example for Bandstop filter design: Design a maximally flat bandstop filter with N=2; center frequency is100kHz; bandwidth is 10%; Source resistance is 50Ω. Again, start with low-pass prototype table: g1=1.4142, g2=1.4142 Let us start with the low-pass prototype given in (a) (First element: capacitor)

From Eqn. E2, mHg

RL 563.0

4142.11.010283.650

510

01 =

×××=

Δ=ω

nFRg

C 5.45010283.6

4142.11.05

00

11 =

×××

=Δ

=ω

155

From Eqn. E1, uHRg

L 25.1110283.6

504142.11.05

0

022 =

×××

=Δ

=ω

nF

RgC 1.225

504142.11.010283.611

5020

2 =××××

=Δ

=ω

The bandstop circuit will be,

Ω

Ω