Filter Design1

Microwave Filter DesignBy

Professor Syed Idris Syed HassanSch of Elect. & Electron EngEngineering Campus USM

Nibong Tebal 14300SPS Penang

Contents

2

1. Composite filter2. LC ladder filter3. Microwave filter

Composite filter

3

m=0.6 m=0.6m-derivedm<0.6

constantkT

21

21

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(c)

oZZZZZ 2121 '4/'1''

iTZZZZZ 2121 '4/'1/''

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

Image method

DCBA

Z i1 Z i2

I1 I2

+V 1-

+V 2-

Z in1 Z in2

221

221

DICVIBIAVV

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 network

5

Z 1/2 Z 1/2

Z 2

Z 1

2Z 2 2Z 2

T-network -network

2

12

2

1

2

12

1

21

41

21

ZZ

ZZ

Z

ZZZ

2

1

2

2

21

12

1

211

421

ZZ

Z

ZZZ

ZZ

Image impedance in T and network

6

Z 1/2 Z 1/2

Z 2

Z 1

2Z 2 2Z 2

T-network -network

2121 4/1 ZZZZZiT

22

212121 4//2/1 ZZZZZZe

iTi ZZZZZZZZ /4/1/ 212121

22

212121 4//2/1 ZZZZZZe

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

21

21

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

CLZZZZZiT

LjZ 1

CjZ /12

If we define a cutoff frequency LCc2

And nominal characteristic impedanceCLZo

Thenc

oiT ZZ 2

2

1

Zi T= Zo when =0

continue

9

Propagation constant (from page 11), we have

12214//2/1 2

2

2

222

212121

ccc

ZZZZZZe

Two regions can be considered

<c : passband of filter --> Zit become real and is imaginary (= j )since 2/c

2-1<1

>c : stopband of filter_--> Zit become imaginary and is real (= ) since 2/c

2-1<1

c

Mag

c

passband stopband

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

10

LjZ 1

CjZ /12

2

2

2

2

2

21

11

/

c

o

co

oiTi

Z

Z

ZZZZZ

12214//2/1 2

2

2

222

212121

ccc

ZZZZZZe

Zi= Zo when =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

LCCLZZZZZiT 22121 4

114/1

CjZ /11

LjZ 2

If we define a cutoff frequency LCc 21

And nominal characteristic impedanceCLZo

Then2

2

1c

oiT ZZ Zi T= Zo when =

2C

L

2C

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

12

CjZ /11

LjZ 2

2

2

2

2

2

21

11

/

c

c

o

co

oiTi

Z

Z

ZZZZZ

12214//2/1 2

2

2

222

212121

cccZZZZZZe

Zi= Zo when =

Propagation constant is the same for both T and -network

2L

C

2L

Composite filter

13

m=0.6 m=0.6m-derivedm<0.6

constantkT

21

21

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

mZ1/2 mZ 1/2

Z2 /m

1

2

41 Z

mm

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

21ZmZmZZZZZZZZiT

4'

4

21

2

21

21

21ZmZmZZZZ

Solving for Z’2, we have

m

ZmmZZ

41'

21

22

2

Low -pass m-derived T-section

15

Lmm

41 2

mC

mL/2mL/2

LjZ 1

CjZ /12

For constant-k section

LmjZ 1' Lj

mm

CmjZ

411'

2

2

and

22

212121 '4/''/''2/'1 ZZZZZZe

22

2

22

1

/11/2

4/1/1''

c

c

mm

mmLjCmjLmj

ZZ

22

2

2

1

/11/1

'4'1

c

c

mZZ

Propagation constant

LCc 21

where

continue

16

2

2

2

1

/1/1

'4'1

op

c

ZZ

2

2

2

1

/1/2

''

op

cmZZ

If we restrict 0 < m < 1 and 21 m

cop

Thus, both equation reduces to

2

2

2

2

2

2

/1/1

/1/2

/1/21

op

c

op

c

op

c mme

Then

When <c, e is imaginary. Then the wave is propagated in the network. When c<<op, eis positive and the wave will be attenuated. When = op, e becomes infinity which implies infinity attenuation. When >op, then ebecome 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 4 c

atte

nuat

ion

m-derived

const-k

composite

op

M-derived section attenuates rapidly but after >op , 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

Cmm

214

CjmZ /'1

Cmj

mm

LjZ

41'

2

2

and

22

212121 '4/''/''2/'1 ZZZZZZe

22

2

22

1

/11/2

4/1//

''

c

c

mm

CmjmmLjCjm

ZZ

22

2

2

1

/11/1

'4'1

c

c

mZZ

Propagation constant

LCc 21

where

continue

19

2

2

2

1

/1/1

'4'1

op

c

ZZ

2

2

2

1

/1/2

''

op

c mZZ

If we restrict 0 < m < 1 and cop m 21

Thus, both equation reduces to

2

2

2

2

2

2

/1/1

/1/2

/1/21

op

c

op

c

op

c mme

Then

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

Thus op< c

continue

20

op c

M-derived section seem to be resonated at =op 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 -section

21

mZ 1

mZ22

mZ22

m

Zm4

12 12

mZm

412 1

2

2

22121

21/1

4/1/''co

iTiZ

mZZZZZZZ

11' mZZ

m

ZmmZZ

41'

21

22

2

Note that

The image impedance is

Low -pass m-derived -section

22

mL

2mC

2mC

m

Lm4

12 2 m

Lm4

12 2LjZ 1

CjZ /12

For constant-k section

221 / oZCLZZ 22222

1 /4 coZLZ Then

and

Therefore, the image impedance reduces to

o

c

ci ZmZ

2

22

/1

/11

The best result for m is 0.6which give a good constant ZiThis 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

21

21

Matchingsection

Matchingsection

High-fcutoff

Sharpcutoff

Z iTZ iT Z iT

Z oZ o

Matching between constant-k and m-derived

24

iiT ZZ The image impedance ZiT does not match Zi, I.eThe matching can be done by using half- section as shown below and the image impedance should be Zi1= ZiT and Zi2=Zi

Z' 1 / 2

2Z' 2Z i2=Z iZ i1=Z iT

1'2

12'

'4'1

2

1

2

1

Z

ZZZ

12121 '4/'1'' iiT ZZZZZZ

22121 '4/'1/'' ii ZZZZZZ

It can be shown that

11' mZZ

m

ZmmZZ

41'

21

22

2

Note that

Example #1

25

Design a low-pass composite filter with cutoff frequency of 2GHz and impedance of 75 . 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

LCc2

CLZo

LC

c

122

2oZLC 2

oCZL or

CL

c

122

Rearrange for c and substituting, we have

nHZL co 94.11)1022/()752(/2 9

pFZC co 122.2)10275/(2/2 9

continue

26

cop m 21

2195.01005.2/1021/12992 opcm

For m-derived T section sharp cutoff

nHnHmL 31.12

94.112195.02

pFpFmC 4658.0122.22195.0

nHnHLmm 94.1294.11

2195.042195.01

41 22

Lmm

41 2

mC

mL/2mL/2

continue

27

For matching sectionmL/2

mC/2mC/2

m

Lm2

1 2 m

Lm2

1 2

mL/2

Z iT

Z oZ o

m=0.6

nHnHmL 582.32

94.116.02

pFpFmC 6365.02122.26.0

2

nHnHLmm 368.694.11

6.026.01

21 22

continue

28

3.582nH 5.97nH 1.31nH

6.368nH

0.6365pF

2.122pF12.94nH

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

00 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

go=G og1

g2

g3

g 4

g n+1

go=R o

g1

g2

g3

g4

gn+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

cCH

For low -pass power ratio response

n

kgk 212sin2

g0 = gn+1 = 1

c

A

n /log2

110log

110

10/10

co

kk Z

gC

c

kok

gZL

whereC=1 for -3dB cutoff pointn= order of filter c= cutoff frequency

No of order (or no of elements)

Where A is the attenuation at point and 1>c

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.

c

A

n /log2

110log

110

10/10

1

3212sin21

g

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

Solution

51.2

400/1000log2110log

10

10/2010

c

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

Prototype values

232122sin22

g

132132sin23

g

continue

35

nHgZLLc

o 9.19104002

1506

113

pFZgC

co9.15

104002502

62

2

15.9pF

19.9nH

50 ohm

50 ohm 19.9nH

or

36

nHgZLc

o 8.39104002

2506

22

pFZgCC

co95.7

104002501

61

13

7.95pF

39.8nH

50 ohm

50 ohm

7.95pF

Equi-ripple filter

37

1

21

cnoCFH

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 c= 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 .11

1(x)Co

Continue

38

Prototype elements

372.17cothln

41

1LrF

evennforF

oddnforgn

121 coth

1

ckk

kkk bb

aag1

1

2

11 F

ag

where

nFF 1

22sinh

nkn

kak ,....2,12

1sin2

nkn

kFbk ,....2,12

sin222

c

kok

gZL

co

kk Z

gC

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 71028794.050

931

pFC 543.310250

1132.192

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

cko

k gZC

1

c

c

go=R o

g 1

g 2

g 3

g4

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

ko

sk LC

o

oc

1where

o 12 21 oand

skskk

ok

ok

o

o CjLjLjLjLjjX'

'111

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 C

L

o

kpk

CC

pkpkk

ok

ok

o

ok L

jCjCjCjCjjB'

'111

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

ko

pk LC

1

11

o

ocwhere

o 12 21 oand

pkpk

k

o

ko

o

okk LjCj

Lj

Lj

Lj

Xj

''1111

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 C

L

1

o

kpk

CC

sksk

k

o

ko

o

okk CjLj

Cj

Cj

Cj

Bj

''1111

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 50.

SolutionFrom 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

nHgZLLo

oss 127

1021.05963.150

91

31

pFgZ

CCoo

ss 199.05963.150102

1.09

131

kok gZL

continue

46

Second element is equivalent to parallel capacitance, thus

nHgZL

o

op 726.0

0967.1102501.0

92

2

pFZ

gCoo

p 91.341021.050

0967.19

22

o

kk Z

gC

50 127nH 0.199pF

0.726nH 34.91pF

127nH 0.199pF

50

Implementation in microstripline

47

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

jL=jZ osin( d)

jC/2=jY o ta n(d)/2 jC/2=jY ota n(d/2)

jL/2=jZ otan( d/2)jL/2=jZ ota n(d/2)

jC=jY osi n(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

tan

doLfL

dZ

C

tan1

-model with C as fringing capacitance

-model with L as fringing inductance

ZoL should be high impedance ZoC should be low impedance

d

Z oZ oCC Z o

oL

d

ZLd

1sin2

oCd CZd 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 (r=4.5 h=1.5mm)

nHLL 731 pFC 543.32

Let’s choose ZoL=100 and ZoC =20 .

mmZ

LdoL

d 25.10100

107102sin21414.0sin

2

9911

3,1

cmf

c

rd 14.14

5.410103

9

8

pFdZ

CdoL

fL 369.01414.001025.0tan

1021001tan1

9

Note: For more accurate calculate for difference Zo

continue

50

mmCZd oCd 38.102010543.3102sin

21414.0sin

212911

2

nHdZLd

oCfC 75.0

1414.001038.tan

10220tan 9

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

1025.6102sin21414.0 99

13,1

mmd 22.9201017.3102sin21414.0 1291

2

More may be needed to obtain sufficiently stable solutions

51

mmmmhZ

wroL

31.05.157.15.4100

37757.1377100

mmmmhZ

wroL

97.105.157.15.420

37757.137720

57.1

377

hw

Zr

o

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

mmmmhZ

wroL

97.25.157.15.450

37757.137750

10.97mm

2.97mm

0.31mm

9.08mm

9.22mm

9.08mm

2.97mm

0.31mm

Implementation using stub

52

Richard’s transformation

tanjLLjjX L tanjCCjjBc

At cutoff unity frequency,we have =1. Then

1tan 8

L

C

jX L

jB c

/8

S.C

O.C

Z o=L

Z o=1/C

jX L

jB c

/8

The length of the stub will be the same with length equal to/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

dd d d

S.C seriesstub

O.C shuntstub

Z 1Z 2/n 2

n2=1+Z 2/Z 1

Z 1/n 2

Z 2

d=/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 , 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

18

18

8

8

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.

299.13487.3111

1

22 ZZn

3487.31

1

2 ZZ

3487.3/ 21 oLZnZ 1/ 2

2 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

50217.5

64.9 70.3

/8

64.9 /8

/8

217.5 50

56

/8

/8/8

/8

/8

Z o=50

Z 2=4.35x50=217.5

Z 1=1.299x50=64.9

Zoc=1.405x50=70.3

Z L=50

Z 1=1.299x50=64.9

Z 2=4.35x50=217.5

Band-pass filter from /2 parallel coupled lines

57

Input/2 resonator

/2 resonatorOutput

J' 01+/2rad

J' 23+/2rad

J' 12+/2rad

/4 /4/4

Microstrip layout

Equivalent admittance inverter

Equivalent LC resonator

Required admittance inverter parameters

58

21

1001 2'

ggJ

1,...2,112

'1

1,

nkforgg

Jkk

kk

tionsofnongg

Jnn

nn sec.2

'21

11,

o 12

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=50.

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

3137.05963.1121.0

2'

21

21

1001

gg

J

61.703137.03137.0150,, 24,31,0 oeoe ZZ

24.393137.03137.0150 24,3,1,0, oooo ZZ

3137.015963.12

1.02

'21

21

434,3

gg

J

A

C

D

E

60

1187.00967.15963.1

12

1.012

'21

2,1

ggJ

1187.05963.10967.1

12

1.012

'32

3,2

ggJB

B

64.561187.01187.0150,, 23,22,1 oeoe ZZ

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 r=10. For others use other means.

mf r

r 01767.0101024

1032

1034/ 9

88

The 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

50

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

ZZ4

n

oon g

ZZ

4

Band-pass

Band-stop

....Z01

Z02 Zon-1Zon

Zo ZoZoZo Zo

/4

/4/4/4/4

/4

....Z01

Z02Zon-1 Zon

Zo ZoZoZo Zo

/4

/4/4/4/4

/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 =0.1

n

oon g

ZZnote

4:

9.2655963.115.0

504031

ZZo

3870967.115.0

5042 oZ

50

/4

265.

9

387

265.

9 /4

/4

/4

/4

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

Capacitive coupled resonator band-pass filter

64

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

21

B n+1

Z o

n

21

1001 2'

ggJ

1,...2,112

'1

1,

nkforgg

Jkk

kk

tionsofnongg

Jnn

nn sec.2

'21

11,

o 12 where

21 io

ii JZ

JB

111 2tan

212tan

21

ioioi BZBZ

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

222.2

1.011

o

o91.0191.11

c

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

prototype

n gn ZoJn Bn Cn n

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

/4

In

Out/4

In Out

Interdigital filter/2

in out

Wiggly coupled line

1

2

67

1= /2

2= /4

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

1

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