Design of RF and Microwave Filters II

7/28/2019 Design of RF and Microwave Filters II

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Oct. , 2011 Fall , KAIST1

Design of RF and Microwave Filters II

EE542 Microwave Engineering Class

S.-O. Park

EE542 Microwave Engineering,




Outline

1. Lumped Filter Implementation

2. Filter Implementation

3. Stepped-Impedance Low-Pass Filters

4. Coupled Line Filters

5. Filters Using Coupled Resonators





Low-pass Filter

~ L Z 2V C 1

V

R

GV

G Z

~GV GV Filter 2V

GV

G Z R

C L Z 2V

1 2 3 4

Cascading four ABCD-networks.

1 1 01 1 0

1 10 1 10 1

11

11

G

L

G G L

L

L

R A B R

RC D j C

R R j C R R R

j C R





RF Filter Parameter

GV

G Z R

C L Z 2V

1 2 3 4

Cascading four ABCD-networks.

1 1 01 1 0

1 10 1 10 1

11

11

G

L

G G L

L

L

R A B R

RC D j C

R R j C R R R

j C R





Low-Pass Filter Frequency Response

Frequency Response from the ABCD

Definitions:

2

1

2 0i

V A

V

So the Transfer Function is Simply:

1 1

1 G

H

A j R R C

Corresponding Phase is:

Group Delay:

1Im

tanRe

H

H

g

d t

d





High-Pass Filter

~

G Z

GV

R

L L Z

(a) High-pass filter with load resistance (b) Network and input/output voltages

G

Z

GV

R

L L Z 2V

1 01 01 1

11110 10 1

1 11

1 11

G

L

G G L

L

L

R A B R

C D R j L

R R R R j L R

j L R



http://slidepdf.com/reader/full/design-of-rf-and-microwave-filters-ii 7/69Oct. , 2011 Fall , KAIST7

High-Pass Filter Frequency Response

Frequency Response from the ABCD

Definitions:

2

1

2 0i

V A

V

So the Transfer Function is simply:

1 1

1 11 G

L

H A

R R j L R

For :

2 1

1

L

GG L G

L

V R

R RV R R R R

Inductive Influence can be neglected




FILTER IMPLEMENTATION

Richard’s Transformation

tan tan ,

tan ,

tan ,

1 tan ,

p

L

L

l l v

jX j L jL l

jX j C jC l

l

Figure 8.34 (p. 407)

Richard’s transformation.(a) For an inductor to a short-circuited stub.

(b) For a capacitor to an open-circuited stub.




4l

Distributed Lines as series and shunt Resonators;

Short circuited line. <parallel resonance>4

00 0

0

tantan

tan

Lin

L

Z jZ l Z Z jZ l

Z jZ l

Now assume that at and0

let 0

Then, for TEM line,

0

02 2 p p

l l

l v v

0 0 0

0

0 00 0 0 0

0 0 0 0

cot cot cot2 2

tan

2 2 4 4

in

e

Y j Y l jY l jY

jY jY jY jY

The input admittance of

a shunt resonant circuit is,

02inY j C

4

0 , Z

l

inY

Short-Circuited Line4




2 2

0 002 2

0

2 20 0 0

0

1 21 2

2 2

inY j C j C j C j C j C LC

let

00

0 0

2 ,

2 4

Y C Y C

The capacitance of the equivalent circuit as,

The inductance of the equivalent circuit can be found as

2

0

1 L

C

4

0, Z

l

inY

inY





in Z

at0

4l

0

0

2

in

Z Z

l j

l inZ

(6.29)

The input impedance can be written as

0

20 0 0

1

, ,4

Z

R C Ll Z C

0 0

0 0

0 04 4 2

C Z Q RC

G l Z l

2

1 1

12

in

in

in

Y G j C

Z Y

j C R

( 2l

at 1st resonance)

0 00

0 0

1 2

2 2in

j l Z Z Z

l j l j

0 0

2

0 0 0 0

4, ,

1 2 4in

Z Z R l Z R C L

j RC l Z C




The open-circuited Line

0

0 0

0

cot tantan

tan

in in

in eq eq

Z Z jZ l Y jY l

j l Y

B C C l

0

0

0

tan( )

( ) tan

L

in

L

Z jZ l V l Z Z I l Z jZ l

inY 0Y

in inY jB

Open

l

Open-circuited Transmission Line




The equivalence between them is satisfied by ensuring that the susceptance

slope parameter B

02

B B

for a each circuit is equal.

0

0021 2

2 2 B C C C

L

Likewise, the susceptance slope parameter of the corresponding distributed circuit is

0 0

2

cot

2 4l

Y l Y l B

l

For open-circuited line,4

2

2

2

2

1cot csc

sin

1tan sec

cos

z z z z

z z z z




00 0 0 0

0 0 0

2

cot tan2 2 4

11

in

in

Z jZ l jZ jZ jZ

j Z jx j L j L

C C

reactance slope parameters of the circuit are equal.

The quantity is defined by

0

0

02

2

1

2

X X

X L L

C

its reactance slope parameters is

0 0

2

cot

2 2 4l

Z l Z l X

l




Waveguide as a Resonator

Low losses => tanh

Close to the frequency at which:

=>

Now, , and

l l

f

f

f f f l l g )/tan(tan:2/

0( / )in Z Z l j f f

'

'

0C

L Z ''' /)2/( C L R l C Ll r

''

22

0 0

tanh tantanh( ) 1 tan tanh

in

l j l Z Z j l d Z j l l

in Z

L

R C

l in Z




Series, R L C 2in Z R j L

00 2

0 0

1, ,

2

Z R Z l L C

L

0 0

0

22 2

L Z Q R Z l l

( at 1st resonance)l

Smaller Larger.Q (6.27)

((6.25))

in Z

L R C

0 , , Z

l in Z




Waveguide as a Resonator

We get: (waveguide)

(series resonant circuit near resonance)

-Relation:

-Quality factor:

f l jLl R

Z In ''

2

f L j R Z In 2

2/'l R R

2/'l L L

LC 2/1

2'

'

R

L

R

LQ

shorted 4/ g

open 2/ g parallel resonator

open 4/ g

shorted 2/ g series resonator

23 EE542 Microwave Engineering,



tanin X l

6

4

2

0

2

4

6

2

3

2

2

5

2

2 l l l v

Equivalentlumped-circuit

behavior of Xin

Figure. Normalized input reactance versus for a short-circuited, lossless transmission line.l

Normalized input reactance versus l





0

X

r

L

C

L C

(a)

L

C

L C

(b)

0

B

r

Figure. Frequency variation of reactance and susceptance for series and parallel resonant

circuits(resonant frequency ).1 LC

X and B for series and parallel resonant circuits at Resonant





2, 4

Transmission Line Resonators

0

0

tanh

tanh tan

1 tan tanh

in Z Z j l

l j l Z

j l l


l in

Z

02l

lossless : 00 , tanin Z jZ l

Short-circuited lossy transmission line

low loss : 0

tan

1 tanin

l j l Z Z

j l l

1, I

at

we discuss the use of transmission lines to realize the RLC resonator

for a resonator, we are interested in Q and therefore,we need to consider lossy transmission lines

tanh( jx )=jtan(x )

note that tanh(A+B)=(tanh A + tanh B)/(1+ tanh A tanh B),





0

0 0

tanh( )

sinh cos cosh sin tanh tan

cosh cos sinh sin 1 tan tanh

in Z Z j l

l l j l l l j l Z Z

l l j l l j l l

tanh l l 0let

0 , p p p

l l l l

v v v

02

pvl

0

l

0 0 0

tan tan tanl

cosh( ) cosh cos sinh sin

sinh( ) sinh cos cosh sin

x iy x y i x y

x iy x y i x y





0

0 0

0 01in l j Z Z Z l j

j l

0since 1l 2 ,in Z R j L

0 R Z l 0

02

Z L

2

0

1C

L

0

02 2

LQ

R l

0

tanh( )in

Z Z j l

2 N

0 0

tanh tantanh( )

1 tan tanhin

l j l Z Z j l d Z

j l l

tanh l l

0 0 0

tan tan tanl


in Z

L R C






LC G

in Z

at0

4l

0

02

in

Z Z

l j

0 , , Z

l in

Z

(6.29)

The input impedance can be written as

0

20 0 0

1

, ,4

Z

R C Ll Z C

0 00 0

0 04 4 2

C Z Q RC

G l Z l

2

1 1

12

in

in

in

Y G j C

Z Y

j C R

( 2

l

at 1st resonance)

0 00

0 0

1 2

2 2in

j l Z Z Z

l j l j

0 0

2

0 0 0 0

4, ,

1 2 4in

Z Z R l Z R C L

j RC l Z C





Open-Circuited Line2

at0

2l

0

0

in

Z Z

l j

in Z

l in Z

(6.33)

for an open-circuited line

the input impedance for the open-circuited l/2 line

can be rewritten as:

0

1 tan tanh

tanh tanin

j l l

Z Z l j l

0 0cot cothin Z jZ l Z j l

0 00

0 0

00 2

0 0 0 0

1

1 2

21, ,

2

in

in

j l Z Z Z

l j l j R

Z j RC

Z R Z l C L

Z C





l

1n

2n

V

0

00

2 2

C Q RC

G l

02

0 0 0

1, ,2

Z R C Ll Z C

( at 1st resonance)l

2

1 1

12

in

in

in

Y G j C

Z Y

j C R

l in

Z





0

X

r

L

C

L C

(a)

L

C

L C

(b)

0

B

r

Figure. Frequency variation of reactance and susceptance for series and parallel resonant

circuits(resonant frequency ).1 LC

X and B for series and parallel resonant circuits at Resonant





The open-circuited Line

0

0 0

0

cot tantan

tan

in in

in eq eq

Z Z jZ l Y jY l

j l Y

B C C l

0

0

0

tan( )( ) tan

L

in

L

Z jZ l V l Z Z I l Z jZ l

inY 0Y

in inY jB

Open

l

Open-circuited Transmission Line





TABLE 8.7 The Four Kuroda Identities

( )a

( )b

( )c

( )d

2

1

Z 1 Z

2 Z

1 Z

2 Z 1 Z

1 Z

2

1

Z

2

2

Z

n

1

2

Z n

21n Z 2

2

1n Z

2

2

Z

n1

2

Z n

21: n

21n Z

2 :1n2

2

1

n Z

22 11where n Z Z





The two circuit of identity (a) in Table 8.7 can

be redrawn as shown in Figure 8.35; we will

show that these two networks are equivalent

by showing that their ABCD matrices are

identical. From Table 4.1, the ABCD matrix

of a length l of transmission line with charac-

teristic impedance Z 1 is

1

1

1

2

1

cos sin

sin cos

11

1 1

l jZ l A B

jC D l l

Z

Bj Z

j

Z

where . Now the open-circuited

shunt stub in the first circuit in Figure 8.35

has an impedance of

so the ABCD matrix of the entire circuit is

tan l

2 2cot jZ l jZ

1

2

2 1

1

2 12

1 2 2

1 0 11

11 1

11 .1 1 11

L

j Z A B

j j

C D Z Z

j Z

Z j

Z Z Z





The short-circuited series stub in the second

circuit in Figure 8.35 has an impedance of

so the ABCD

matrix of the entire circuit is 2 2

1 1tan , j Z n l j Z n

212

2

2 2

2

11 22

222 1

2 2

11 1

11 0 1

11 .

11

R

Z j j Z

A B nn

j nC D

Z

Z Z Z

n

j n Z

Z Z

The result in (8.80a) and (8.80b) are identical

if we choose

The other identities in Table 8.7 can be proved

in the same way.

22 11 .n Z Z

Figure 8.35 (p. 408)

Equivalent circuits illustrating Kuroda

identity

(a) in Table 8.7.





EXAMPLE 8.6 Low-Pass Filter Design Using Stubs

Design a low-pass filter for fabrication using

microstrip lines. The specifications are: cutoff

frequency of 4GHz , third order, impedance of

50 , and a 3dB equal-ripple characteristic.

Solution

From Table 8.4, the normalized low-pass

prototype element values are

1 1

2 2

3.3487 ,0.7117

g L g C

Figure 8.36a (p. 409) Filter design procedure

for Example 8.5.

(a) Lumped-element low-pass filter prototype.

(b) Using Richard’s transformations to

convert inductors and capacitors to series and

shunt stubs.

(c ) Adding unit elements at ends of

filter.





3 3

4

3.3487 ,

1.0000 L

g L

g R

Figure 8.36b (p. 410)

(d ) Applying the second Kuroda identity.

(e) After impedance and frequency scaling.

(f ) Microstrip fabrication of final filter.

with the lumped-element circuit shown inFigure 8.36a.

The next step is to use Richard’s transformations

to convert series inductors to series stubs, and

shunt capacitors to shunt stubs, as shown in

Figure 8.36b.

According to (8.78), the characteristic impedance

of a series stub (inductor) is L, and the characteristicimpedance of a shunt stub (capacitor) is 1/C . For

commensurate line synthesis, all stubs are /8 long

at . (It is usually most convenient to work

with normalized quantities until the last step in the

design.)

The series stubs of Figure 8.36b would be verydifficult to implement in microstrip form, so we will

use one of the Kuroda identities to convert these to

shunt stubs. First, we must add unit elements at

either end of the filter, as shown in Figure 8.36c.

These redundant elements do not affect filter

performance since

c





they are matched to the source and load

(Z 0 =1) Then we can apply Kuroda identity

(b) from Table 8.7 to both ends of the filter.

In both cases we have that

2 2

1

11 1 1.2993.3487

Z n

Z

16GHz , as a result of the periodic nature of

Richard’s transformation.

Impedance and Admittance Inverters

As we have seen, it is often desirable to use

only shunt, elements which implementing a

filter with a particular type of transmission line.

The Kuroda identity can be used for conversions

of this form, but another possibility is to use

impedance or admittance ( j ) inverters[1],[4],[7].

such inverters are especially useful for bandpass

or bandstop filters with narrow(<10%)bandwidths.

The conceptual operation of impedance andadmittance is illustrated in Figure 8.38; since

these inverters essentially form the inverse of the

load impedance or admittance, they can be used

to transform series-connected elements to shunt-

connected elements, or vice versa. This procedure

The result is shown in Figure 8.36d.

Finally, we impedance and frequency

scale the circuit, which simply involves

multiplying the normalized characteristic

impedances by 50 and choosing the

line and stub lengths to be /8 at 4GHz .

The final circuit is shown in Figure 8.36e,

with a microstrip layout in Figure 8.36f.The calculated amplitude response of the

lumped-element version. Note that the

passband characteristics are very similar

up to 4 GHz , but the distributed-element

filter has a response which repeats every





Figure 8.37 (p. 411) Amplitude responses

of lumped-element and distributed-element

low-pas filter of Example 8.5.

Figure 8.38a/b (p. 412) Impedance and

admittance inverters.

(a) Operation of impedance and admittance

inverters.

(b) Implementation as quarter-wave transformers.

will be illustrated in later sections for bandpass

and bandstop filters.

In its simplest form, a j or K inverter can be

constructed using a quarter-wave transformer

of the appropriate characteristic impedance, as

shown in Figure 8.38b. This implementation also

allows the ABCD matrix of the inverter to be

easily found from the ABCD parameters for a

length of transmission line given in Table 4.1.

Many other types of circuits can also be used as





Figure 8.38c/d (p. 412) Impedance and

admittance inverters.

(c ) Implementation using transmission lines

and reactive elements.

(d ) Implementation using capacitor networks.

J or K inverters, with one such alternative

being shown in Figure 8.38c. Inverters of

this form turn out to be useful for modelingthe coupled resonator filters of Section 8.8.

The lengths, /2 , of the transmission line

sections are generally required to be negative

for this type of inverter, but this poses no

problem if these lines can be absorbed into

connecting transmission lines on either side.





STEPPED-IMPEDANCE LOW-PASS FILTERS

Approximate Equivalent Circuits for Short

Transmission Line Sections

Figure 8.39 (p. 413) Approximate equivalent

circuits for short sections of transmission lines.(a) T -equivalent circuit for a transmission line

section having

(b) Equivalent circuit for small and large Z 0.

(c ) Equivalent circuit for small and small Z 0.

11 22 0

12 21 0

11 12 0 0

cot ,

1 cos .

cos 1tan ,

sin 2

A Z Z jZ l C

Z Z jZ l

C l l

Z Z jZ jZ l

Now assume a short length of line

(say < /4)l

0

0

0

0

0

0

tan ,2 2

1 sin .

,

0,

0,

( ),

( ),

inductor

capac itor

h

l

l X Z

B l Z

X Z l

LR B l

Z

X

CZ B Y l l

R





EXAMPLE 8.7 Stopped-Impedance Filter Design

Design a stepped-impedance low-pass filter having a maximally flat response and a cutoff

frequency of 2.5GHz . It is necessary to have

more than 20dB insertion loss at 4.0GHz .

The filter impedance is 50 ; the highest

practical line impedance is 150 , and the

lowest is 10 .

To use Figure 8.26, we calculate

4.01 1 0.6,2.5c

then the figure indicates N=6 should give thenecessary attenuation at 4.0 GHz . Table 8.3

gives the low-pass prototype values as

g1 = 0.517 = C1,

g2 = 1.414 = L2,

g3 = 1.932 = C3,

g4 = 1.932 = L4,

g5 = 1.414 = C5,

g4 = 0.517 = L6,

1 1

0

02 2

3 3

0

04 4

5 5

0

06 6

5.9 ,

27.0 ,

22.1 ,

36.9 ,

16.2 ,

l

h

l

h

l

h

Z l g R

Rl g

Z

Z l g

R

Rl g Z

Z l g

R

Rl g

Z





Figure 8.40 (p. 414) Filter design for Example 8.6.

(a) Low-pass filter prototype circuit. (b) Stepped-impedance implementation. (c ) Microstrip

layout of final filter.





The final filter circuit is shown in Figure 8.40b,

where

Note that in all cases. A layout in

microstrip is shown in Figure 8.40c.

Figure 8.41 shows the calculated amplitude

response, compared with the response of the

corresponding lumped-element filter. The

passband characteristics are very similar, butthe lumped-element circuit gives more

attenuation at higher frequencies. This is

because the stepped-impedance filter

elements depart significantly from the lumped-

element values at the higher frequencies. The

stepped-impedance filter may have other

passbands at higher frequencies, but the

response will not be perfectly periodic because

the lines are not commensurate.

Figure 8.41 (p. 415)

Amplitude response of the stepped-impedance

low-pass filter of Example 8.6, with (dotted line)and without (solid line) losses. The response of

the corresponding lumped-element filter is also

shown.





Figure 8.42 (p. 417)Definitions pertaining to coupled line filter section.

(a) A parallel coupled line section with port voltage

and current definitions. (b) A parallel coupled line

section with even- and odd-mode current sources.

(c ) A two-port coupled line section having a

bandpass response.

1 1 2

2 1 2

3 3 4

4 3 4

1 1 2 2 1 2

3 3 4 1 4 3

,

,,

.

1 1, ,2 2

1 1, .2 2

I i i

I i i I i i

I i i

i I I i I I

i I I i I I





1 0 1 0 2 0 1 0 2

0 3 0 4 0 4 0 3

11 22 33 44 0 0

12 21 34 43 0 0

13 31 24 42 0 0

14 41 23 32 0 0

cot2

csc .2

cot ,2

cot ,2

csc ,2

csc .2

e e o o

e e o o

e o

e o

e o

e o

jV Z I Z I Z I Z I

j Z I Z I Z I Z I

j Z Z Z Z Z Z

j Z Z Z Z Z Z

j Z Z Z Z Z Z

j Z Z Z Z Z Z





0 0

j z j l V z V e V e

l 0 z

0

,

,

V z I z

Z

L

V L Z

L I

0

z 0 cote

in e Z jZ l

1 1

1 1

1

0 0 1

1

0 1

3 3

, 2 cos

0 0 2 cos

cot

2 cos sin

cos

sin

j z l j z l a b e

e

e

a b e in

e e

e

a b e

j l j l

a b e e

v z v z V e e

z l V l z

v v V l i Z

jZ l Z i

V i jl l

l z v z v z jZ i

l

v z v z V e e V

At

2 cos z

l





Similarly, the voltages due to current sources

driving the line in the even mode are3i

First consider the line as being driven in the

even mode by the current sources. If the

other ports are open-circuited, the impedance

seen at port 1 or 2 is

1i

0 cot . 8.88ein e Z jZ l

The voltage on either conductor can be

expressed as

1 1

2 cos . 8.89

j z l j z l a b e

e

v z v z V e e

V l z

so the voltage at port 1 or 2 is

1 110 0 2 cos .e

a b e inv v V l i Z

This result and (8.88) can be used to rewrite

(8.89) in terms of as1i

1 10 1

cos.

sin8.90a b e

l z v z v z jZ i

l

3 30 3cos .

sin8.91a b e z v z v z jZ i

l

Now consider the line as being driven in the

odd mode by current If the other ports are

open-circuited, the impedance seen at 1 or 2 is

2.i

0 cot . 8.92oin o Z jZ l

The voltage on either conductor can be

expressed as

2 2

2 cos . 8.93

j z l j z l a b e

e

v z v z V e e

V l z





Then the voltage at port 1 or port 2 is

2 2

0 2

2 20 2

4 40 4

1 2 3 41

0 1 0 2 0 3 0 4

1 11 1 13 3, 3 31 1 33 3,

2 11 13

11

0 0 2 cos .cos 1

.sin

cos.

sin

0 0 0 0

cot csc .

o

a b in

a b o

a b o

a a a a

e o e o

i

v v V l i Z z

v z v z jZ il

z v z v z jZ i

l

V v v v v

j Z i Z i j Z i Z i

V Z I Z I V Z I Z I

Z Z

Z Z

2

33

2 22 2

0 0 0 0

0 0

1 csc cot .2

1 ,2

e o e o

i e o

Z

Z Z Z Z

Z Z Z

0 01 2

0 0

11 33 0 0112

13 0 013

cos cos .

cos cos ,

e o

e o

e o

e o

Z Z Z Z

Z Z Z Z Z Z Z Z Z

which shows is real for

where

1 2 1

1 0 0 0 0cos e o e o Z Z Z Z

Figure 8.43 (p. 420)

The real part of the image impedance of

the bandpass network of Figure 8.42c .





3 3

a bv l v l

3 0

0 33 0

3 3

0 3

2 2

0 2

4 4

0 4

2 cos cot

cot2 cos sin

cos

sin

cos

sincos

sin

e e

e in in e

ee e

a b e

a b o

a b o

V l i Z Z jZ l

jZ l iV i jZ l l

z v z v z jZ i

l

l z v z v z jZ i

l z

v z v z jZ il

Total voltage at port 1 is

1 2 3 4

1

0 1 0 2 0 3 0 4

0 0 0 0

cot csc

a a a a

e o e o

V v v v v

j Z i Z i j Z i Z i





Figure 8.44 (p. 421)

Equivalent circuit of the coupled line section

of Figure 8.42c .

0 0

0 0

22

0 0

0

2 202

0

cos sin cos sin0

sin sin0cos cos

cos1 sin cos sin

1 sin cos

jZ jZ A B B j J

j jC D j J

Z Z

JZ j JZ JZ J

j J JZ JZ

0

2 2 2

0

2 2 20

20

0

0

1 sin cos

sin 1 cos ,1 sin cos

2 : .

1cos sin cos

i

i

JZ

JZ J B Z C JZ J

Z JZ

A JZ JZ





20 0 0

0 00

0 0 0

2

0 0 0 0

2

0 0 0 0

1 ,2

1 ,

1 ,

1 .

e o

e o

e o

e

o

Z Z JZ

Z Z JZ Z Z JZ

Z Z JZ JZ

Z Z JZ JZ





TABLE 8.8 Ten Canonical Coupled Line Circuits

Circuit Image Impedance Response

1i Z

1i Z

1i Z

1i Z

1i Z 2i Z

1Re i Z

02 3

2

1Re i Z

02 3

2

1Re i Z

02 3

2

0

12 2

20 0 0 0

0 02

1

2 cos

cos

oe oi

e o e o

e oi

i

Z Z Z

Z Z Z Z

Z Z Z

Z

0

12 2 2

0 0 0 0

2 sin

cos

oe oi

e o e o

Z Z Z

Z Z Z Z

2 2

20 0 0 0

1

cos

2sin

e o e o

i

Z Z Z Z Z







All pass

All pass

1i Z 1i Z

1i Z 1i Z

1i Z

2i Z

1Re i Z

02 3

2

2 220 0 0 0 0

1

0

0 02

1

cos

sin

oe o e o e o

i

oe o

e oi

i

Z Z Z Z Z Z Z

Z Z

Z Z Z

Z

0 01

2e o

i

Z Z Z

0 01

0 0

2 e oi

e o

Z Z Z

Z Z







All pass

All stop

All stop

All stop

1i Z

1i Z

1i Z

1i Z

1i Z

1i Z

1i Z

2i Z

1 0i oe o Z Z Z

0 01

0 0

0 02

1

2cote o

i

e o

e oi

i

Z Z Z j

Z Z

Z Z Z Z

1 0 tani oe o Z j Z Z

1 0 coti oe o Z j Z Z





Figure 8.45(a-c) (p. 422)

Development of an equivalent circuit for

derivation of design equations for a coupled line

bandass filter.

(a) Layout of an N + 1 section coupled line

bandpass filter.

(b) Using equivalent circuit of Figure 8.44 for

each coupled line section.

(c ) Equivalent circuit for transmission lines of

length 2

Figure 8.45 (d-f) (p. 422)

Development of an equivalent circuit for

derivation of design equations for a coupled

line bandass filter.(d ) Equivalent circuit for the admittance

inverters. (e) Using results of (c ) and (d ) for

the N = 2 case. (f ) Lumped-element circuit for

a bandpass filter for N = 2.





2 2 2 211 11 12 11 11 12

12 12 12 12

11 11

12 12 12 12

1 0

0 11 1

8.109

Z Z Z Z Z Z A B Z Z Z Z

C D Z Z Z Z Z Z

Equating this result to the ABCD parameters for a transmission line of length 2 and chacacteristic

impedance gives the parameters of the equivalent circuit as

0 Z

0

12

11 22 12 0

1

,sin2cot 2 .

8.110a8.110b

jZ

Z C Z Z Z A jZ

Then the series arm impedance is

11 12 0 0cos2 1 cot

sin28.111 Z Z jZ jZ





20 0 0 0

12

0 0 0

0

20 0 00

00

0

0

. ,sin 1 2

2 1,2

0 01 0

00 0

jZ jZ jL Z Z

Z L C Z L

jZ jZ

A B N N j jN C D N Z

Z





12

0

sin2

~ 22

jz Z

at center frequeny

2 l

2 2 2 p

l l l z v

0

0

2

0

0

0

4

1

2

pv

0 0 0 0

12

0

0

sin2sin 1

jZ jZ jZ Z

2

0

11 12 0

0

2cos 1 1cos 2 1

sin 2 2sin cos

cot

jZ Z Z jZ

jZ





20 0 0 0

12

0 0 0

0

20 0 00

00

0

0

. ,sin 1 2

2 1,2

0 01 0

00 0

jZ jZ jL Z Z

Z L C Z L

jZ jZ

A B N N j jN C D N Z

Z





With reference to Figure 8.45e, the admittance just to the right of the J 2 inverter is

2 2022 0 3 0 3

1 2 0

1 ,C

j C Z J j Z J j L L

since the transformer scales the load admittance by the square of the ratio. Then the admittance

seen at the input of the filter is

22

12 2211 0

2 2 0 0 0 3

201 2

2 221 01 0

2 2 0 0 0 3

1 1

1 . 8.116

J Y j C

j L J Z j C L Z J

C J j

L J Z j C L Z J

These results also use the fact, from (8.114), that LnC n=1/ for all LC resonators. Now the

admittance seen looking into the circuit of Figure 8.45f is

20

1

1 2 2 0

01

1 02 2 0 0 0

1 11

1 8.117

Y j C j L j L j C Z

C j

L j L C Z





which is identical in form to (8.116). Thus,

the two circuits will be equivalent if the

following conditions are met :

1 1

2 21 11 0

2 21 0 2 2

22 22

2 2 21 0 3

022

1 ,

,

8.118a

8.118b

8.118c

C C L L J Z

J Z C C L L J

J Z J Z

J

We know Ln and C n from (8.114);L’ n and

C’ n are determined form the element

values of a lumpes-element low-pass

prototype which has been impedance

scaled and frequency transformed to a

bandpass filter. Using the results in Table

8.6 and the impedance scaling formulas

of (8.64) allows the L’ n and C’ n values to

be written as

01

0 1

1

110 0

2 0

2

0

21

0 2 0

,

,

,

,

8.119a

8.119b

8.119c

8.119d

Z L

g

g

C Z

g Z L

C g Z

where is the fractionalbandwidth of the filter. Then (8.118) can be

solved for the inverter constants with the

following results ( for N=2 ):

2 1 0

1 4

1 11 0

1 1 1

1 4

2 2 22 0 1 0

2 2 1 2

23 0

1 2

,

2

,2

.2

8.120a

8.120b

8.120c

C L J Z

L C g

C C J Z J Z

L L g g

J J Z

J g





After the J ns are found, Z 0e and Z 0o for each

coupled line section can be calculated form

(8.108).

The above results were derived for the

special case of N=2 (three coupled line

sections), but more general results can be

derived for any number of sections, and for

the case where Z L Z 0 (or g n+1 1, as in the

case of an equal-ripple response with N even). Thus, the design equations for a

bandpass filter with N+1 coupled line

sections are

0 1

1

0

0 1

1

,

2

, 2,3,..., .2 1

.2

8.121a

for 8.121b

8.121c

n

n n

N

N N

Z J

g

Z J n N g g

Z J g g

The even and odd mode characteristic

impedances for each section are then found form

(8.108).

EXAMPLE 8.8 Coupled Line Bandpass Filter

Design

Design a coupled line band pass filter with N=3

and a 0.5 dB equal-ripple response. The center

frequency is 2.0GHz , the bandwidth is 10%,and Z0= 50 . What is the attenuation at 1.8 GHz ?

Solution

The fractional bandwidth is =1. We can use

Figure 8.27a to obtain the attenuation is 1.8 GHz ,

but first we must use (8.71) to convert thisfrequency to the normalized low-pass form

1 :C

0

0

1.8 2.01 1 2.11.0.1 2.0 1.8





Then the value on the horizontal scale of Figure

8.27a is

1 2.11 1 1.11C

Which indicates an attenuation of about 20 dB for

N=3.

The low-pass prototype values, gn , are given in

Table 8.4; then (8.121) can be used to calculate

the admittance inverter constants, Jn. Finally, the

even-and odd-mode characteristic impedances

can be found from (8.108). These results are

summarized in the following table:

ngn Z0Jn

Z0e

( ) Z0o

( )

1 1.5963 0.3137 70.61 39.24

2 1.0967 0.1187 56.64 44.77

3 1.5963 0.1187 56.64 44.77

4 1.0000 0.3137 70.61 39.24

Note that the filter sections are symmetric

about the midpoint. The calculated response of

this filter is shown in Figure 8.46; passbands

also occur at 6 GHz, 10 GHz , etc.Many other types of filters can be constructed

using coupled line sections; most of these are

of the bandpass or bandstop variety. One

particularly compact design is the interdigitated

filter, which can be obtained from a coupled line

filter by folding the lines at their midpoints; see

[1] and [3] for details.


l.millitron 38GHz system



Oct. , 2011 Fall , KAIST60EE542 Microwave Engineering,




HUGEHES 14GHz Ku-Band Earth Station


FILTERS USING COUPLED RESONATORS




FILTERS USING COUPLED RESONATORS

Bandstop and Bandpass Filters Using Quarter-

Wave Resonators

Figure 8.47 (p. 427)

Bandstop and bandpass filters using shunt

transmission line resonators ( = /2 at the

center frequency).

(a) Bandstop filter. (b) Bandpass filter.

0 cot ,n Z jZ

Figure 8.48 (p. 428)

Equivalent circuit for the bandstop filter of Figure

8.47a. (a) Equivalent circuit of open-circuited

stub for near /2. (b) Equivalent filter circuit

using resonators and admittance inverters. (c )

Equivalent lumped-element bandstop filter.


Bandpass Filters Using Capacitively Coupled Resonators




p g p y p

Figure 8.50a/b (p. 431)Development of the equivalence of a capacitive-gap coupled resonator bandpass filter to the

coupled line bandpass filter of Figure 8.45.

(a) The capacitive-gap coupled resonator bandpass filter.

(b) Transmission line model.


Bandpass Filters Using Capacitively Coupled Resonators




Figure 8.50cd (p. 431)

Development of the equivalence of a capacitive-

gap coupled resonator bandpass filter to the

coupled line bandpass filter of Figure 8.45.(c ) Transmission line model with negative-length

sections forming admittance inverters ( i/2 <0).

(d ) Equivalent circuit using inverters and /2

resonators ( = at 0). This circuit is now

identical in form with the coupled line bandpass

filter equivalent circuit in Figure 8.45b.

p g p y p

10 2

0

1 10 0 1

1 1 1, 1,2,..., ,

2 2tan 2 . .

1

1 tan 2 tan 2 .2

for i i i

ii i i

i

i i i

i N

J Z B B

Z J

Z B Z B





EXMAPLE 8.10 Coupled Resonator Bandpass

Filter Design

Design a bandpass filter using capacitive

coupled resonators, with a 0.5dB equal-ripple

passband characteristic. The center frequency is

2.0GHz , the bandwidth is 10%, and the

impedance is 50 . At least 20dB of attenuation

is required at 2.2GHz .

Solution

We first determine the order of the filter to satisfy

the attenuation specification at 2.2GHz . Using

(8.71) to convert to normalized frequency gives

0

0

2.2 2.01 1 1.91.0.1 2.0 2.2

1 1.91 1.0 0.91.C

Then,

From Figure 8.27a, we see that N=3 should

satisfy the attenuation specification at 2.2GHz.

The low-pass prototype values are given in

Table 8.4, from which the inverter constants

can be calculated using (8.121). Then the

coupling susceptances can be found from

(8.134), and the coupling capacitor values as

0.nn BC





Finally, the resonator lengths can be calculated from (8.135). The following table summarizes

these results.

n g n Z 0 J n Bn C n n

1 1.5963 0.3137 6.96 10-3 0.554pF 155.8

2 1.0967 0.1187 2.41 10-3 0.192pF 166.5

3 1.5963 0.1187 2.41 10-3 0.192pF 155.8

4 1.0000 0.3187 6.96 10-3

0.554pF _

The calculated amplitude response is plotted in Figure 8.51. The specifications of this filter are

the same as the coupled line bandpass filter of Example 8.8, and comparison of the results in

Figures 8.51 and 8.46 shows that the responses are identical near the passband region.





Figure 8.51 (p. 433)

Amplitude response for the capacitive-gap coupled series resonator bandpass filter of

Example 8.10.


TABLE 4.1 The ABCD Parameters of Some Useful Two-Port Circuits.




TABLE 4.1 The ABCD Parameters of Some Useful Two Port Circuits.

Circuit ABCD Parameters

Z

Y

0 Z

l

:1 N

A= 1 B= Z

C= 0 D= 1

A= 1 B= 0

C= Y D= 1

A= N B= 0

C= 0 D=

A= B=

C= D=

cos l 0 sin jZ l

0 sin jY l cos l

1 N


TABLE 4.1 The ABCD Parameters of Some Useful Two-Port Circuits.



TABLE 4.1 The ABCD Parameters of Some Useful Two Port Circuits.

Circuit ABCD Parameters

1Y 2Y

3Y

1 Z 2 Z

3 Z

A= B=

C= D=

2

3

1Y Y

3

1Y

1 21 2

3

YY Y Y

Y 1

3

1Y Y

A= B=

C= D=

1

3

1 Z Z

3

1 Z

1 21 2

3

Z Z Z Z Z

2

3

1Z Z

Documents

Design of RF and Microwave Filters II