Microwave Filter DesignBy

Professor Syed Idris Syed Hassan

Sch of Elect. & Electron Eng

Engineering Campus USM

Nibong Tebal 14300

SPS Penang

Contents

2

1. Composite filter2. LC ladder filter3. Microwave filter

Composite filter

3

m=0.6 m=0.6m-derivedm<0.6

constantkT

π2

1 π2

1

Matchingsection

Matchingsection

High-fcutoff

Sharpcutoff

Z iTZ iT Z iT

Z oZ o

m<0.6 for m-derived section is to place the pole near the cutoff frequency(ωc)

oZZZZZ =+ 2121 '4/'1''

iTZZZZZ =+ 2121 '4/'1/''

For 1/2 π matching network , we choose the Z’1 and Z’2 of the circuit so that

Image method

DC

BAZ i1 Z i2

I 1 I 2

+V 1

-

+V 2

-

Z in1 Z in2

221

221

DICVI

BIAVV

+=+=

Let’s say we have image impedance for the network Zi1 and Zi2

Where Zi1= input impedance at port 1 when port 2 is terminated with Zi2

Zi2= input impedance at port 2 when port 1 is terminated with Zi1

Then

4@

Where Zi2= V2 / I2

and V1 = - Zi1 I1

ABCD for T and π network

5

Z 1 /2 Z 1 /2

Z 2

Z 1

2Z 2 2Z 2

T-network π -network

++

+

2

12

2

1

2

12

1

21

4

12

1

Z

Z

Z

Z

Z

ZZ

Z

+

++

2

1

2

2

21

12

1

21

142

1

Z

Z

Z

Z

ZZ

Z

Z

Image impedance in T and π network

6

Z 1 /2 Z 1 /2

Z 2

Z 1

2Z 2 2Z 2

T-network π -network

2121 4/1 ZZZZZiT +=

( ) ( )22

212121 4//2/1 ZZZZZZe +++=γ

iTi ZZZZZZZZ /4/1/ 212121 =+=π

( ) ( )22

212121 4//2/1 ZZZZZZe +++=γ

Image impedance Image impedance

Propagation constant Propagation constant

Substitute ABCD in terms of Z1 and Z2 Substitute ABCD in terms of Z1 and Z2

Composite filter

7

m=0.6 m=0.6m-derivedm<0.6

constantkT

π2

1 π2

1

Matchingsection

Matchingsection

High-fcutoff

Sharpcutoff

Z iTZ iT Z iT

Z oZ o

Constant-k section for Low-pass filter using T-network

8

L/2

C

L/2

414/1

2

2121LC

C

LZZZZZiT

ω−=+=

LjZ ω=1

CjZ ω/12 =

If we define a cutoff frequency LC

c2=ω

And nominal characteristic impedanceC

LZo =

Thenc

oiT ZZ2

2

1ωω−= Zi T= Zo when ω=0

continue

9

Propagation constant (from page 11), we have

( ) ( ) 122

14//2/12

2

2

222

212121 −+−=+++=

ccc

ZZZZZZeωω

ωω

ωωγ

Two regions can be considered

∀ω<ωc : passband of filter --> Zit become real and γ is imaginary (γ= jβ )since ω2/ωc

2-1<1

∀ω>ωc : stopband of filter_--> Zit become imaginary and γ is real (γ= α ) since ω2/ωc

2-1<1

ωc

ω

Mag

ωcα,β

ω

πβ

αpassband stopband

Constant-k section for Low-pass filter using π-network

10

LjZ ω=1

CjZ ω/12 =

−

=

−

==

2

2

2

2

2

21

11

/

c

o

co

oiTi

Z

Z

ZZZZZ

ωω

ωω

π

( ) ( ) 122

14//2/12

2

2

222

212121 −+−=+++=

ccc

ZZZZZZeωω

ωω

ωωγ

Zi π= Zo when ω=0

Propagation constant is the same as T-network

C/2

L

C/2

Constant-k section for high-pass filter using T-network

11

LCC

LZZZZZiT 22121

4

114/1

ω−=+=

CjZ ω/11 =

LjZ ω=2

If we define a cutoff frequency LC

c2

1=ω

And nominal characteristic impedanceC

LZo =

Then2

2

1ωωc

oiT ZZ −= Zi T= Zo when ω = ∞

2C

L

2C

Constant-k section for high-pass filter using π-network

12

CjZ ω/11 =

LjZ ω=2

−

=

−

==

2

2

2

2

2

21

11

/

c

c

o

co

oiTi

Z

Z

ZZZZZ

ωω

ωω

π

( ) ( ) 122

14//2/12

2

2

222

212121 −+−=+++=

ωω

ωω

ωωγ cccZZZZZZe

Zi π= Zo when ω=

Propagation constant is the same for both T and π-network

∞

2L

C

2L

Composite filter

13

m=0.6 m=0.6m-derivedm<0.6

constantkT

π2

1 π2

1

Matchingsection

Matchingsection

High-fcutoff

Sharpcutoff

Z iTZ iT Z iT

Z oZ o

m-derived filter T-section

14

Z 1 /2 Z 1 /2

Z 2

Z' 1 /2 Z' 1 /2

Z' 2

mZ 1 /2 mZ 1 /2

Z 2 /m

1

2

4

1Z

m

m−

Constant-k section suffers from very slow attenuation rate and non-constant image impedance . Thus we replace Z1 and Z2 to Z’1 and Z’2 respectively.

Let’s Z’1 = m Z1 and Z’2 to obtain the same ZiT as in constant-k section.

4'

4

'''

4

21

2

21

21

21

21

21Zm

ZmZZ

ZZZ

ZZZiT +=+=+=

4'

4

21

2

21

21

21Zm

ZmZZ

ZZ +=+

Solving for Z’2, we have

( )m

Zm

m

ZZ

4

1'

21

22

2−+=

Low -pass m-derived T-section

15

Lm

m

4

1 2−

mC

mL/2mL/2

LjZ ω=1

CjZ ω/12 =

For constant-k section

LmjZ ω=1'( )

Ljm

m

CmjZ ω

ω 4

11'

2

2−+=and

( ) ( )22

212121 '4/''/''2/'1 ZZZZZZe +++=γ

( ) ( )( )

( )( ) 22

2

22

1

/11

/2

4/1/1'

'

c

c

m

m

mmLjCmj

Lmj

Z

Z

ωωωω

ωωω

−−−=

−+=

( )( )( ) 22

2

2

1

/11

/1

'4

'1

c

c

mZ

Z

ωωωω

−−−=+

Propagation constant

LCc

2

1=ωwhere

continue

16

( )( )2

2

2

1

/1

/1

'4

'1

op

c

Z

Z

ωωωω

−−=+( )

( )2

2

2

1

/1

/2

'

'

op

cm

Z

Z

ωωωω

−−=

If we restrict 0 < m < 1 and 21 m

cop

−= ωω

Thus, both equation reduces to

( )( )

( )( )

( )( )

−−

−−+

−−+=

2

2

2

2

2

2

/1

/1

/1

/2

/1

/21

op

c

op

c

op

c mme

ωωωω

ωωωω

ωωωωγ

Then

When ω < ωc, eγ is imaginary. Then the wave is propagated in the network. When ωc<ω <ωop, eγ is positive and the wave will be attenuated. When ω = ωop, eγ becomes infinity which implies infinity attenuation. When ω>ωop, then eγ become positif but decreasing.,which meant decreasing in attenuation.

Comparison between m-derived section and constant-k section

17

Typical attenuation

0

5

10

15

0 2 4ω c

att

en

ua

tio

n

m-derived

const-k

composite

ωop

M-derived section attenuates rapidly but after ω>ωop , the attenuation reduces back . By combining the m-derived section and the constant-k will form so called composite filter.This is because the image impedances are nonconstant.

High -pass m-derived T-section

18

2C/m

L/m

2C/m

Cm

m21

4

−

CjmZ ω/'1 =

( )Cmj

m

m

LjZ

ωω

4

1'

2

2−+=

and

( ) ( )22

212121 '4/''/''2/'1 ZZZZZZe +++=γ

( ) ( )( )

( )( ) 22

2

22

1

/11

/2

4/1/

/

'

'

ωωωω

ωωω

c

c

m

m

CmjmmLj

Cjm

Z

Z

−−−=

−+=

( )( )( ) 22

2

2

1

/11

/1

'4

'1

ωωωω

c

c

mZ

Z

−−−=+

Propagation constant

LCc

2

1=ωwhere

continue

19

( )( )2

2

2

1

/1

/1

'4

'1

ωωωω

op

c

Z

Z

−−=+( )

( )2

2

2

1

/1

/2

'

'

ωωωω

op

c m

Z

Z

−−=

If we restrict 0 < m < 1 and cop m ωω 21−=

Thus, both equation reduces to

( )( )

( )( )

( )( )

−−

−−+

−−+=

2

2

2

2

2

2

/1

/1

/1

/2

/1

/21

ωωωω

ωωωω

ωωωωγ

op

c

op

c

op

c mme

Then

When ω < ωop , eγ is positive. Then the wave is gradually attenuated in the networ as function of frequency. When ω = ωop, eγ becomes infinity which implies infinity attenuation. When ωχ>ω >ωop, eγ is becoming negative and the wave will be propagted.

Thus ωop< ωc

continue

20

α

ωωop ωc

M-derived section seem to be resonated at ω=ωop due to serial LC circuit. By combining the m-derived section and the constant-k will form composite filter which will act as proper highpass filter.

m-derived filter π-section

21

mZ 1

m

Z22

m

Z22

( )m

Zm

4

12 12−( )

m

Zm

4

12 12−

( )( ) 2

22121

21/1

4/1/''

co

iTiZ

mZZZZZZZ

ωωπ

−

−+==

11' mZZ =

( )m

Zm

m

ZZ

4

1'

21

22

2−+=

Note that

The image impedance is

Low -pass m-derived π-section

22

mL

2

mC

2

mC

( )m

Lm

4

12 2−( )m

Lm

4

12 2−LjZ ω=1

CjZ ω/12 =

For constant-k section

221 / oZCLZZ == ( ) 22222

1 /4 coZLZ ωωω −=−=Then

and

Therefore, the image impedance reduces to

( )( )( ) o

c

ci Z

mZ

2

22

/1

/11

ωω

ωωπ

−

−−=

The best result for m is 0.6which give a good constant Ziπ . This type of m-derived section can be used at input and output of the filter to provide constant impedance matching to or from Zo .

Composite filter

23

m=0.6 m=0.6m-derivedm<0.6

constantkT

π2

1 π2

1

Matchingsection

Matchingsection

High-fcutoff

Sharpcutoff

Z iTZ iT Z iT

Z oZ o

Matching between constant-k and m-derived

24

πiiT ZZ ≠The image impedance ZiT does not match Ziπ, I.e

The matching can be done by using half- π section as shown below and the image impedance should be Zi1= ZiT and Zi2=Ziπ

Z' 1 / 2

2Z' 2Z i2 =Z i πZ i1 =Z iT

+

1'2

12

'

'4

'1

2

1

2

1

Z

Z

Z

Z

12121 '4/'1'' iiT ZZZZZZ =+=

22121 '4/'1/'' ii ZZZZZZ =+=π

It can be shown that

11' mZZ =

( )m

Zm

m

ZZ

4

1'

21

22

2−+=

Note that

Example #1

25

Design a low-pass composite filter with cutoff frequency of 2GHz and impedance of 75Ω . Place the infinite attenuation pole at 2.05GHz, and plot the frequency response from 0 to 4GHz.

SolutionFor high f- cutoff constant -k T - section

C

L/2 L/2

LCc

2=ωC

LZo =

LC

c

122

=

ω

2oZ

LC = 2

oCZL =or

CL

c

122

=

ω

Rearrange for ωc and substituting, we have

nHZL co 94.11)1022/()752(/2 9 =×××== πω

pFZC co 122.2)10275/(2/2 9 =××== πω

continue

26

cop m ωω 21−=

( ) ( ) 2195.01005.2/1021/12992 =××−=−= opcm ωω

For m-derived T section sharp cutoff

nHnHmL

31.12

94.112195.0

2=×=

pFpFmC 4658.0122.22195.0 =×=

nHnHLm

m94.1294.11

2195.04

2195.01

4

1 22

=×

−=−

Lm

m

4

1 2−

mC

mL/2mL/2

continue

27

For matching sectionmL/2

mC/2mC/2

( )m

Lm

2

1 2−( )m

Lm

2

1 2−

mL/2

Z iT

Z oZ o

m=0.6

nHnHmL

582.32

94.116.0

2=×=

pFpFmC

6365.02

122.26.0

2=×=

nHnHLm

m368.694.11

6.02

6.01

2

1 22

=×

−=−

continue

28

3.582nH 5.97nH 1.31nH

6.368nH

0.6365pF

2.122pF

12.94nH

0.4658pF

3.582nH

6.368nH

0.6365pF

1.31nH5.97nH

Can be addedtogether

Can be addedtogether

Can be addedtogether

A full circuit of the filter

Simplified circuit

12.94nH9.552nH

6.368nH7.28nH 4.892nH

0.6365pF 0.6365pF0.4658pF

2.122pF

6.368nH

continue

30

Freq response of low-pass filter

-60

-40

-20

0

0 1 2 3 4

Frequency (GHz)

S11

Pole due to m=0.2195

section

Pole due to m=0.6section

N-section LC ladder circuit(low-pass filter prototypes)

31

g o =G og 1

g 2

g 3

g 4

g n+1

g o =R o

g 1

g 2

g 3

g 4

g n+1

Prototype beginning with serial element

Prototype beginning with shunt element

Type of responses for n-section prototype filter

32

•Maximally flat or Butterworth•Equal ripple or Chebyshev•Elliptic function•Linear phase

Maximally flat Equal ripple Elliptic Linear phase

Maximally flat or Butterworth filter

33

( )12

21

−

+=

n

c

CHωωω

For low -pass power ratio response

( )

−=

n

kgk 2

12sin2

π

g0 = gn+1 = 1

( )( )cA

nωω /log2

110log

110

10/10 −= co

kk Z

gC

ω=

c

kok

gZL

ω=

where

C=1 for -3dB cutoff pointn= order of filter ωc= cutoff frequency

No of order (or no of elements)

Where A is the attenuation at ω1 point and ω1>ωc

Prototype elements

k= 1,2,3…….n

Series element

Shunt element

Series R=Zo

Shunt G=1/Zo

Example #2

34

Calculate the inductance and capacitance values for a maximally-flat low-pass filter that has a 3dB bandwidth of 400MHz. The filter is to be connected to 50 ohm source and load impedance.The filter must has a high attenuation of 20 dB at 1 GHz.

( )( )cA

nωω /log2

110log

110

10/10 −=

( )1

32

12sin21 =

×−= π

g

g0 = g 3+1 = 1First , determine the number of elements

Solution

( )( ) 51.2

400/1000log2

110log

10

10/2010 >−=

c

Thus choose an integer value , I.e n=3

Prototype values

( )2

32

122sin22 =

×−×= π

g

( )1

32

132sin23 =

×−×= π

g

continue

35

nHgZ

LLc

o 9.19104002

1506

113 =

××××===

πω

pFZ

gC

co

9.1510400250

26

22 =

××××==

πω

15.9pF

19.9nH

50 ohm

50 ohm 19.9nH

or

36

nHgZ

Lc

o 8.39104002

2506

22 =

××××==

πω

pFZ

gCC

co

95.710400250

16

113 =

××××===

πω

7.95pF

39.8nH

50 ohm

50 ohm

7.95pF

Equi-ripple filter

37

( )1

21

−

+=

cnoCFH

ωωω

For low -pass power ratio response

110 10/ −= LroF

where

Cn(x)=Chebyshev polinomial for n order and argument of x n= order of filter ωc= cutoff frequencyFo=constant related to passband ripple

Chebyshev polinomial

Where Lr is the ripple attenuation in pass-band

(x)(x)-CCx(x)C n-n-n 212=

x(x)C =1

cn ei)(C ωω == .11

1=(x)Co

Continue

38

Prototype elements

=

372.17cothln

4

11

LrF

( )

=+ evennforF

oddnforgn

121 coth

1

ckk

kkk bb

aag

1

1

−

−=

2

11 F

ag =

where

=n

FF 1

22

sinh

( )nk

n

kak ,....2,1

2

1sin2 =

−= π

nkn

kFbk ,....2,1

2sin22

2 =

+= π

c

kok

gZL

ω=

co

kk Z

gC

ω=

Series element

Shunt element

Example #3

39

Design a 3 section Chebyshev low-pass filter that has a ripple of 0.05dB and cutoff frequency of 1 GHz.

From the formula given we have

g2= 1.1132

g1 = g3 = 0.8794

F1=1.4626 F2= 1.1371

a1=1.0 a2=2.0

b1=2.043

nHLL 7102

8794.050931 =

××==π

pFC 543.310250

1132.192 =

××=

π

3.543pF

7nH

50 ohm

50 ohm 7nH

Transformation from low-pass to high-pass

40

•Series inductor Lk must be replaced by capacitor C’k

•Shunts capacitor Ck must be replaced by inductor L’k

ck

ok g

ZL

ω=

ckok gZC

ω1=

ωω

ωω c

c

−→

g o =R o

g 1

g 2

g 3

g 4

g n+1

Transformation from low-pass to band-pass

41

•Thus , series inductor Lk must be replaced by serial Lsk and Csk

o

ksk

LL

ωΩ=

kosk LC

ωΩ=

−

Ω→

ωω

ωω

ωω o

oc

1where

oωωω 12 −=Ω 21 ωωω =oand

skskk

ok

ok

o

o C

jLjLjLjLjjX

''

111

ωω

ωω

ωω

ωω

ωω −=

Ω−

Ω=

−

Ω=

Now we consider the series inductor

kok gZL =

Impedance= series

normalized

continue

42

•Shunts capacitor Ck must be replaced by parallel Lpk and Cpk

kopk CL

ωΩ=

o

kpk

CC

ωΩ=

pkpkk

ok

ok

o

ok L

jCjCjCjCjjB

''

111

ωω

ωω

ωω

ωω

ωω −=

Ω−

Ω=

−

Ω=

Now we consider the shunt capacitor

o

kk Z

gC =

Admittance= parallel

Transformation from low-pass to band-stop

43

•Thus , series inductor Lk must be replaced by parallel Lpk and Cskp

o

kpk

LL

ωΩ=

kopk L

CΩ

=ω

1

11

−

−

Ω→

ωω

ωω

ωω o

ocwhere

oωωω 12 −=Ω 21 ωωω =oand

pkpk

k

o

ko

o

okk L

jCj

Lj

Lj

Lj

Xj

''

1111

ωω

ωω

ωω

ωω

ωω −=

Ω−

Ω=

−

Ω=

Now we consider the series inductor --convert to admittance

kok gZL =

admittance = parallel

Continue

44

•Shunts capacitor Ck must be replaced by parallel Lpk and Cpk

kosk CL

ωΩ= 1

o

kpk

CC

ωΩ=

sksk

k

o

ko

o

okk C

jLj

Cj

Cj

Cj

Bj

''

1111

ωω

ωω

ωω

ωω

ωω −=

Ω−

Ω=

−

Ω=

Now we consider the shunt capacitor --> convert to impedance

o

kk Z

gC =

Example #4

45

Design a band-pass filter having a 0.5 dB ripple response, with N=3. The center frequency is 1GHz, the bandwidth is 10%, and the impedance is 50Ω.

Solution

From table 8.4 Pozar pg 452.

go=1 , g1=1.5963, g2=1.0967, g3= 1.5963, g4= 1.000

Let’s first and third elements are equivalent to series inductance and g1=g3, thus

nHgZ

LLo

oss 127

1021.0

5963.1509

131 =

×××=

Ω==

πω

pFgZ

CCoo

ss 199.05963.150102

1.09

131 =

×××=Ω==

πω

kok gZL =

continue

46

Second element is equivalent to parallel capacitance, thus

nHg

ZL

o

op 726.0

0967.1102

501.09

22 =

×××=Ω=

πω

pFZ

gC

oop 91.34

1021.050

0967.19

22 =

×××=

Ω=

πω

o

kk Z

gC =

50 Ω 127nH 0.199pF

0.726nH 34.91pF

127nH 0.199pF

50 Ω

Implementation in microstripline

47

Equivalent circuitA short transmission line can be equated to T and π circuit of lumped circuit. Thus from ABCD parameter( refer to Fooks and Zakareviius ‘Microwave Engineering using microstrip circuits” pg 31-34), we have

j ω L=jZ o sin( β d)

j ω C/2=jY o ta n( β d)/2 j ω C/2=jY o ta n( β d/2)

j ω L/2=jZ o tan( β d/2)j ω L/2=jZ o ta n( β d/2)

j ω C=jY o si n( β d)

Model for series inductor with fringing capacitors

Model for shunt capacitor with fringing inductors

48

d

Z o

L

Z oL

Z o

=

d

oCfC

dZL

λπ

ωtan

=

doLfL

d

ZC

λπ

ωtan

1

π-model with C as fringing capacitance

Τ-model with L as fringing inductance

ZoL should be high impedanceZoC should be low impedance

d

Z oZ oCC Z o

= −

oL

d

Z

Ld

ωπ

λ 1sin2

( )oCd CZd ωπ

λ 1sin2

−=

Example #5

49

From example #3, we have the solution for low-pass Chebyshev of ripple 0.5dB at 1GHz, Design a filter using in microstrip on FR4 (εr=4.5 h=1.5mm)

nHLL 731 == pFC 543.32 =

Let’s choose ZoL=100Ω and ZoC =20 Ω.

mmZ

Ld

oL

d 25.10100

107102sin

2

1414.0sin

2

9911

3,1 =

×××=

=

−−− π

πω

πλ

cmf

c

rd 14.14

5.410

1039

8

=×==ε

λ

pFd

ZC

doLfL 369.0

1414.0

01025.0tan

102100

1tan

19

=

×

××=

=

λπ

πλπ

ω

Note: For more accurate calculate for difference Zo

continue

50

( ) ( ) mmCZd oCd 38.102010543.3102sin

2

1414.0sin

212911

2 =××××== −−− ππ

ωπ

λ

nHdZ

Ld

oCfC 75.0

1414.0

01038.tan

102

20tan

9=

×

×=

=

λπ

πλπ

ω

pFC 543.32 =

The new values for L1=L3= 7nH-0.75nH= 6.25nH and C2=3.543pF-0.369pF=3.174pF

Thus the corrected value for d1,d2 and d3 are

mmd 08.9100

1025.6102sin

2

1414.0 991

3,1 =

×××=−

− ππ

( ) mmd 22.9201017.3102sin2

1414.0 12912 =××××= −− π

π

More may be needed to obtain sufficiently stable solutions

51

mmmmhZ

wroL

31.05.157.15.4100

37757.1

377100 =

−=

−=

ε

mmmmhZ

wroL

97.105.157.15.420

37757.1

37720 =

−=

−=

ε

−

=57.1

377

hw

Z

r

o

ε

Now we calculate the microstrip width using this formula (approximation)

mmmmhZ

wroL

97.25.157.15.450

37757.1

37750 =

−=

−=

ε

10.97mm

2.97mm

0.31mm

9.08mm

9.22mm

9.08mm

2.97mm

0.31mm

Implementation using stub

52

Richard’s transformation

βξ tanjLLjjX L == βξ tanjCCjjBc ==

At cutoff unity frequency,we have ξ=1. Then

1tan =β 8

λ=

L

C

jX L

jB c

λ /8

S.C

O.C

Z o =L

Z o =1/C

jX L

jB c

λ /8

The length of the stub will be the same with length equal to λ/8. The Zo will be difference with short circuit for L and open circuit for C.These lines are called commensurate lines.

Kuroda identity

53

It is difficult to implement a series stub in microstripline. Using Kuroda identity, we would be able to transform S.C series stub to O.C shunt stub

d

d d d

S.C seriesstub

O.C shuntstub

Z 1Z 2 /n 2

n 2 =1+Z 2 /Z 1

Z 1 /n 2

Z 2

d=λ/8

Example #6

54

Design a low-pass filter for fabrication using micro strip lines .The specification: cutoff frequency of 4GHz , third order, impedance 50 Ω, and a 3 dB equal-ripple characteristic.

Protype Chebyshev low-pass filter element values are

g1=g3= 3.3487 = L1= L3 , g2 = 0.7117 = C2 , g4=1=RL

1

1 3.3487

0.7117

3.3487

Using Richard’s transform we have

ZoL= L=3.3487 Zoc=1/ C=1/0.7117=1.405and

1λ/ 8

1λ/ 8

λ/ 8

λ/ 8

λ/ 8

Z oc =1.405

Z oL =3.3487Z oL =3.3487

Zo Zo

Using Kuroda identity to convert S.C series stub to O.C shunt stub.

299.13487.3

111

1

22 =+=+=Z

Zn

3487.3

1

1

2 =Z

Z3487.3/ 2

1 == oLZnZ 1/ 22 == oZnZ

thus

We haveand

Substitute again, we have

35.43487.3299.121 =×== oLZnZ 299.1299.112

2 =×== nZZ oand

55

d d d

S.C seriesstub

O.C shuntstub

Z 1Z 2 /n 2 =Z o

n 2 =1+Z 2 /Z 1

Z 1 /n 2 =Z oL

Z 2

50 Ω217.5 Ω

64.9 Ω 70.3 Ω

λ /8

64.9 Ωλ /8

λ /8

217.5 Ω50 Ω

56

λ /8

λ /8λ /8 λ /8

λ /8

Z o =50 Ω

Z 2 =4.35x50=217.5 Ω

Z 1 =1.299x50=64.9 Ω

Zoc=1.405x50=70.3 Ω

Z L =50 Ω

Z 1 =1.299x50=64.9 Ω

Z 2 =4.35x50=217.5 Ω

Band-pass filter from λ/2 parallel coupled lines

57

Input

λ /2 resonator

λ /2 resonator

Output

J' 0 1+ π /2rad

J' 23+ π /2rad

J' 12+ π /2rad

λ /4 λ /4λ /4

Microstrip layout

Equivalent admittance inverter

Equivalent LC resonator

Required admittance inverter parameters

58

21

1001 2'

Ω=

ggJ

π

1,...2,11

2'

11, −=×Ω=

++ nkfor

ggJ

kkkk

π

tionsofnongg

Jnn

nn sec.2

'21

11, =

Ω=

++

π

oωωω 12 −=Ω

The normalized admittance inverter is given by

[ ]21,1,1, ''1, +++ ++= kkkkokkoe JJZZ

[ ]21,1,1,, ''1 +++ +−= kkkkokkoo JJZZ

okkkk ZJJ 1,1,' ++ =where

where A

B

C

D

E

Example #7

59

Design a coupled line bandpass filter with n=3 and a 0.5dB equi-ripple response on substrate er=10 and h=1mm. The center frequency is 2 GHz, the bandwidth is 10% and Zo=50Ω.

We have g0=1 , g1=1.5963, g2=1.0967, g3=1.5963, g4= 1 and Ω=0.1

3137.05963.112

1.0

2'

21

21

1001 =

×××=

Ω= ππ

ggJ

[ ] Ω=++== 61.703137.03137.0150,, 24,31,0 oeoe ZZ

[ ] Ω=+−== 24.393137.03137.0150 24,3,1,0, oooo ZZ

3137.015963.12

1.0

2'

21

21

434,3 =

×××=

Ω= ππ

ggJ

A

C

D

E

60

1187.00967.15963.1

1

2

1.01

2'

212,1 =

×××=×Ω= ππ

ggJ

1187.05963.10967.1

1

2

1.01

2'

323,2 =

×××=×Ω= ππ

ggJB

B

[ ] Ω=++== 64.561187.01187.0150,, 23,22,1 oeoe ZZ

[ ] Ω=+−== 77.441187.01187.0150 23,2,2,1, oooo ZZ

D

E

Using the graph Fig 7.30 in Pozar pg388 we would be able to determine the required s/h and w/h of microstripline with εr=10. For others use other means.

mf r

r 01767.0101024

103

2

1034/

9

88

=×××=×=

ελThe required resonator

61

Thus we have

For sections 1 and 4 s/h=0.45 --> s=0.45mm and w/h=0.7--> w=0.7mm

For sections 2 and 3 s/h=1.3 --> s=1.3mm and w/h=0.95--> w=0.95mm

50 Ω

50 Ω

0.7mm

0.45mm

0.95mm

1.3mm

0.95mm

1.3mm

0.45mm

0.7mm

17.67mm 17.67mm 17.67mm 17.67mm

Band-pass and band-stop filter using quarter-wave stubs

62

n

oon g

ZZ

4

Ω= π

n

oon g

ZZ

Ω=

π4

Band-pass

Band-stop

....Z 01

Z 02 Z on-1Z on

Z oZ oZ oZ o

Z o

λ /4

λ /4λ /4λ /4λ /4

λ /4

....Z 01

Z 02Z on-1 Z on

Z oZ oZ oZ o

Z o

λ /4

λ /4λ /4λ /4λ /4

λ /4

Example #8

63

Design a band-stop filter using three quarter-wave open-circuit stubs . The center frequency is 2GHz , the bandwidth is 15%, and the impedance is 50W. Use an equi-ripple response, with a 0.5dB ripple level.

We have g0=1 , g1=1.5963, g2=1.0967, g3=1.5963, g4= 1 and Ω=0.1

n

oon g

ZZnote

Ω=

π4

:Ω=

×××== 9.265

5963.115.0

504031 πZZo

Ω=××

×= 3870967.115.0

5042 πoZ

50 Ω

λ /4

265.9

Ω

387Ω

265.9

Ω λ /4

λ/4

λ/4

λ/4

Note that: It is difficult to impliment on microstripline or stripline for characteristic > 150Ω.

Capacitive coupled resonator band-pass filter

64

Z o Z oZ oZ o ....B 2B 1

θ 2θ 1

B n+1

Z o

θ n

21

1001 2'

Ω=

ggJ

π

1,...2,11

2'

11, −=×Ω=

++ nkfor

ggJ

kkkk

π

tionsofnongg

Jnn

nn sec.2

'21

11, =

Ω=

++

π

oωωω 12 −=Ωwhere

( ) 21 io

ii

JZ

JB

−=

( )[ ] ( )[ ]111 2tan

2

12tan

2

1+

−− ++= ioioi BZBZπθ

i=1,2,3….n

Example #9

65

Design a band-pass filter using capacitive coupled resonators , with a 0.5dB equal-ripple pass-band characteristic . The center frequency is 2GHz, the bandwidth is 10%, and the impedance 50W. At least 20dB attenuation is required at 2.2GHz.

First , determine the order of filter, thus calculate

91.12.2

2

2

2.2

1.0

11 =

−=

−

Ω ωω

ωω o

o

91.0191.11 =−=−cω

ω

From Pozar ,Fig 8.27 pg 453 , we have N=3

prototype

n gn ZoJn Bn Cn θn

1 1.5963 0.3137 6.96x10-3 0.554pF 155.8o

2 1.0967 0.1187 2.41x10-3 0.192pF 166.5o

3 1.0967 0.1187 2.41x10-3 0.192pF 155.8o

4 1.0000 0.3137 6.96x10-3 0.554pF -

Other shapes of microstripline filter

66

Rectangular resonator filter

U type filter

λ /4

In

Outλ /4

In Out

Interdigital filterλ /2

inout

Wiggly coupled line

ϕ 1

ϕ 2

67

ϕ1= π/2

ϕ2= π/4

The design is similar to conventional edge coupled line but the layout is modified to reduce space.

ϕ 1

Modified Wiggly coupled line to improve 2nd and 3rd harmonic rejection. λ/8 stubs are added.