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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
2
1 2
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(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
DC
BAZ i1 Z i2
I1 I2
+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 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
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 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
2
1 2
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
LjZ 1
CjZ /12
If we define a cutoff frequency LC
c2
And nominal characteristic impedanceC
LZo
Thenc
oiT ZZ2
2
1
Zi T= Zo when =0
continue
9
Propagation constant (from page 11), we have
122
14//2/12
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
122
14//2/12
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
LCC
LZZZZZiT 22121
4
114/1
CjZ /11
LjZ 2
If we define a cutoff frequency LC
c2
1
And nominal characteristic impedanceC
LZo
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
122
14//2/12
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
2
1 2
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
mZ1/2 mZ 1/2
Z2 /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 1
CjZ /12
For constant-k section
LmjZ 1'
Ljm
m
CmjZ
4
11'
2
2
and
22
212121 '4/''/''2/'1 ZZZZZZe
22
2
22
1
/11
/2
4/1/1'
'
c
c
m
m
mmLjCmj
Lmj
Z
Z
22
2
2
1
/11
/1
'4
'1
c
c
mZ
Z
Propagation constant
LCc
2
1where
continue
16
2
2
2
1
/1
/1
'4
'1
op
c
Z
Z
2
2
2
1
/1
/2
'
'
op
cm
Z
Z
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
att
en
ua
tio
n
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
Cm
m21
4
CjmZ /'1
Cmj
m
m
LjZ
4
1'
2
2
and
22
212121 '4/''/''2/'1 ZZZZZZe
22
2
22
1
/11
/2
4/1/
/
'
'
c
c
m
m
CmjmmLj
Cjm
Z
Z
22
2
2
1
/11
/1
'4
'1
c
c
mZ
Z
Propagation constant
LCc
2
1where
continue
19
2
2
2
1
/1
/1
'4
'1
op
c
Z
Z
2
2
2
1
/1
/2
'
'
op
c m
Z
Z
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
m
Z22
m
Z22
m
Zm
4
12 12
m
Zm
4
12 12
2
22121
21/1
4/1/''
co
iTiZ
mZZZZZZZ
11' mZZ
m
Zm
m
ZZ
4
1'
21
22
2
Note that
The image impedance is
Low -pass m-derived -section
22
mL
2
mC
2
mC
m
Lm
4
12 2 m
Lm
4
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 Z
mZ
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
2
1 2
1
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.e
The 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
Z
Z
Z
12121 '4/'1'' iiT ZZZZZZ
22121 '4/'1/'' ii ZZZZZZ
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 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
LCc
2
C
LZo
LC
c
122
2oZ
LC 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
94.112195.0
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
go=G og1
g2
g3
g4
gn+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
c
CH
For low -pass power ratio response
n
kgk 2
12sin2
g0 = gn+1 = 1
c
A
n /log2
110log
110
10/10
co
kk Z
gC
c
kok
gZL
where
C=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
132
12sin21
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
g
1
32
132sin23
g
continue
35
nHgZ
LLc
o 9.19104002
1506
113
pFZ
gC
co
9.1510400250
26
22
15.9pF
19.9nH
50 ohm
50 ohm 19.9nH
or
36
nHgZ
Lc
o 8.39104002
2506
22
pFZ
gCC
co
95.710400250
16
113
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
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
nkn
kFbk ,....2,1
2sin22
2
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 7102
8794.050931
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
ckok gZ
C
1
c
c
go=R o
g1
g2
g3
g4
gn+1
Transformation from low-pass to band-pass
41
•Thus , series inductor Lk must be replaced by serial Lsk and Csk
o
ksk
LL
kosk L
C
o
oc
1where
o 12 21 oand
skskk
ok
ok
o
o C
jLjLjLjLjjX
''
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 L
jCj
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 C
jLj
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.
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
pFgZ
CCoo
ss 199.05963.150102
1.09
131
kok gZL
continue
46
Second element is equivalent to parallel capacitance, thus
nHg
ZL
o
op 726.0
0967.1102
501.09
22
pFZ
gC
oop 91.34
1021.050
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 ota 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
d
ZC
tan
1
-model with C as fringing capacitance
-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
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
Ld
oL
d 25.10100
107102sin
2
1414.0sin
2
9911
3,1
cmf
c
rd 14.14
5.410
1039
8
pFd
ZC
doLfL 369.0
1414.0
01025.0tan
102100
1tan
19
Note: For more accurate calculate for difference Zo
continue
50
mmCZd oCd 38.102010543.3102sin
2
1414.0sin
212911
2
nHdZ
Ld
oCfC 75.0
1414.0
01038.tan
102
20tan
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
1025.6102sin
2
1414.0 991
3,1
mmd 22.9201017.3102sin2
1414.0 12912
More may be needed to obtain sufficiently stable solutions
51
mmmmhZ
wroL
31.05.157.15.4100
37757.1
377100
mmmmhZ
wroL
97.105.157.15.420
37757.1
37720
57.1
377
hw
Z
r
o
Now we calculate the microstrip width using this formula (approximation)
mmmmhZ
wroL
97.25.157.15.450
37757.1
37750
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
d
d 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.
299.13487.3
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
n2=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 resonator
Output
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,...2,11
2'
11,
nkfor
ggJ
kkkk
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
3137.05963.112
1.0
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
1.0
2'
21
21
434,3
gg
J
A
C
D
E
60
1187.00967.15963.1
1
2
1.01
2'
212,1
gg
J
1187.05963.10967.1
1
2
1.01
2'
323,2
gg
JB
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
103
2
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
ZZ
4
n
oon g
ZZ
4
Band-pass
Band-stop
....Z01
Z02 Zon-1Zon
ZoZoZoZo
Zo
/4
/4/4/4/4
/4
....Z01
Z02Zon-1 Zon
ZoZoZoZo
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,...2,11
2'
11,
nkfor
ggJ
kkkk
tionsofnongg
Jnn
nn sec.2
'21
11,
o 12 where
21 io
ii
JZ
JB
111 2tan
2
12tan
2
1
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
2
2
2.2
1.0
11
o
o
91.0191.11 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
inout
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.
