Transcript Microwave Engineering - USM :: Universiti Sains Malaysia

Microwave Filter Design
By
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
Contents
1. Composite filter
2. LC ladder filter
3. Microwave filter
2
Composite filter
Matching
section
High-f
cutoff
m=0.6
Zo
1
2
constant
k
T

ZiT
Matching
section
Sharp
cutoff
m=0.6
mderived
m<0.6
ZiT
1
2

Zo
ZiT
m<0.6 for m-derived section is to place the pole near the cutoff frequency(wc)
For 1/2  matching network , we choose the Z’1 and Z’2 of the circuit so that
Z '1 Z '2 1  Z '1 / 4Z '2  Zo
Z '1 Z '2 / 1  Z '1 / 4Z '2  ZiT
3
Image method
I1
I2
+
Zi1
A B
C D 


V1
Zin1
+
V2
Zi2
Zin2
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
V1  AV2  BI2
I1  CV2  DI2
@
Where Zi2= V2 / I2
and V1 = - Zi1 I1
4
ABCD for T and  network
Z1/2
Z1/2
Z2
T-network

Z1
1 
 2Z 2
 1
 Z
2

2
Z1 
Z1 

4Z 2 
Z1 
1
2Z 2 
Z1
2Z2
2Z2
-network
Z1

 1  2Z
2

 1  Z1
 Z 2 4 Z 2 2



Z1 
1
2Z 2 
Z1
5
Image impedance in T and  network
Z1/2
Z1/2
Z2
T-network
Z1
2Z2
2Z2
-network
Substitute ABCD in terms of Z1 and Z2 Substitute ABCD in terms of Z1 and Z2
Image impedance
Image impedance
ZiT  Z1Z2 1  Z1 / 4Z2
Zi  Z1Z2 / 1 Z1 / 4Z2  Z1Z2 / ZiT
Propagation constant
e  1  Z1 / 2 Z 2 
Z1 / Z 2   Z12 / 4Z 22 
Propagation constant
e  1  Z1 / 2 Z 2 
Z1 / Z 2   Z12 / 4Z 22 
6
Composite filter
Matching
section
High-f
cutoff
m=0.6
Zo
1
2
constant
k
T

ZiT
Matching
section
Sharp
cutoff
m=0.6
mderived
m<0.6
ZiT
1
2

Zo
ZiT
7
Constant-k section for Low-pass filter
using T-network
L/2
L/2
C
Z1  jw L
Z 2  1 / jw C
L
w 2 LC
ZiT  Z1Z 2 1  Z1 / 4Z 2 
1
C
4
2
w

If we define a cutoff frequency
c
LC
L
And nominal characteristic impedance Z o 
C
w2
Zi T= Zo when w=0
Then
Z iT  Z o 1  2
w c
8
continue
Propagation constant (from page 11), we have

e  1  Z1 / 2Z 2 
Z1 / Z 2   
Z12
/ 4Z 22
  1 w
2w 2
2
c

2w w 2
wc
w c2
1
Two regions can be considered
w<wc : passband of filter --> Zit become real and  is imaginary (= jb )
since w2/wc2-1<1
w>wc : stopband of filter_--> Zit become imaginary and  is real (= a )
since w2/wc2-1<1
Mag
a,b
wc

w
passband
b
wc
stopband
a
w 9
Constant-k section for Low-pass filter
using -network
L
Z1  jw L
C/2
Z i  Z1Z 2 / Z iT 

e  1  Z1 / 2Z 2 
C/2
Z o2
2 

w
 Z 1

o
2

w c 

Z1 / Z 2   
Z12

Z 2  1 / jw C
Zo
2 

w
 1

2

w c 

/ 4Z 22
  1 w
Zi = Zo when w=0
2w 2
2
c

2w w 2
wc
w c2
1
Propagation constant is the same as T-network
10
Constant-k section for high-pass filter
using T-network
2C
2C
L
Z1  1 / jw C
Z 2  jw L
L
1
1 2
C
4w LC
1
w

c
If we define a cutoff frequency
2 LC
L
And nominal characteristic impedance Z o 
C
2
w
Zi T= Zo when w = 
Then
Z iT  Z o 1  c2
w
ZiT  Z1Z 2 1  Z1 / 4Z 2 
11
Constant-k section for high-pass filter
using -network
C
Z1  1 / jw C
2L
2L
Z 2  jw L
Zi  Z1Z 2 / ZiT 

e  1  Z1 / 2Z 2 
Z o2
2 

w
 Z 1 c 
2 
 o
w


Z1 / Z 2   
Z12

Zo
2 

w
 1 c 
2 

w
c 

/ 4Z 22
  1
2w c 2
w
2
Zi = Zo when w= 

2w c
w
wc 2
1
2
w
Propagation constant is the same for both T and -network
12
Composite filter
Matching
section
High-f
cutoff
m=0.6
Zo
1
2
constant
k
T

ZiT
Matching
section
Sharp
cutoff
m=0.6
mderived
m<0.6
ZiT
1
2

Zo
ZiT
13
m-derived filter T-section
mZ1/2
Z1/2
Z1/2
Z'1/2
mZ1/2
Z'1/2
Z2 /m
Z2
Z'2
1  m2
Z1
4m
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.
Z12
Z1 '2
m2 Z12
Z iT  Z1Z 2 
 Z1 ' Z 2 '
 m Z1Z '2 
4
4
4
Solving for Z’2, we have
2
2
Z1
m2 Z1
Z1Z 2 
 m Z1Z '2 
4
4


Z 2 1  m2 Z1
Z '2 

m
4m
2
14
Low -pass m-derived T-section
mL/2
mL/2
mC
1  m2
L
4m
Z '1  jw Lm
Propagation constant
and
For constant-k
section
Z1  jw L
Z 2  1 / jw C


1
1  m2
Z '2 

jw L
jw Cm
4m
Z '1 / Z '2   Z '12 / 4Z '22 
 2w m / w c 2
Z '1
jw Lm


2
Z '2 1 / jw Cm   jw L1  m / 4m 1  1  m 2 w / w c 2
1  w / w c 2
Z '1
1
1

where
w

c
4Z '2 1  1  m 2 w / w c 2
2 LC
e  1  Z '1 / 2 Z '2 
15
continue
If we restrict 0 < m < 1 and
Thus, both equation reduces to
Z '1  2w m / w c 

Z '2 1  w / w op 2
2

Then
e  1 

 2w m / w c 2

1  w / w op
2
wop 
wc
1  m2
1  w / w c 2
Z '1
1

4Z '2 1  w / w op 2


  2w m / w 2  1  w / w 2 
c
c


 
2
2
 1  w / w op  1  w / w op 







When w < wc, e is imaginary. Then the wave is propagated in the
network. When wc<w <wop, e is positive and the wave will be attenuated.
When w = wop, e becomes infinity which implies infinity attenuation.
When w>wop, then e become positif but decreasing.,which meant
16
decreasing in attenuation.
Comparison between m-derived section
and constant-k section
attenuation
Typical attenuation
15
m-derived
10
const-k
5
composite
0
0
w c wop 2
4
M-derived section attenuates rapidly but after w>wop , 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.
17
High -pass m-derived T-section
2C/m
2C/m
Z '1  m / jw C
L/m
and
4m
C
2
1 m


jw L 1  m2
Z '2 

m
j 4mw C
Propagation constant
e  1  Z '1 / 2 Z '2 
Z '1 / Z '2   Z '12 / 4Z '22 
 2w c m / w 2
Z '1
m / jw C


2
Z '2  jw L / m  1  m / j 4mw C 1  1  m 2 w c / w 2


1  w c / w 2
Z '1
1

4Z '2 1  1  m 2 w c / w 2




where w c 
1
2 LC
18
continue
If we restrict 0 < m < 1 and
Thus, both equation reduces to
Z '1  2w c m / w 

Z '2 1  w op / w 2
2

Then
e  1 

 2w c m / w 2

1  w op / w
2
Thus wop< wc
w op  1 m 2 w c
1  w c / w 2
Z '1
1

4Z '2 1  w op / w 2


  2w m / w 2  1  w / w 2 
c
c


 
2
2
 1  w op / w  1  w op / w 







When w < wop , e is positive. Then the wave is gradually attenuated in
the networ as function of frequency. When w = wop, e becomes infinity
which implies infinity attenuation. When wc>w >wop, e is becoming
19
negative and the wave will be propagted.
continue
a
wop
wc
w
M-derived section seem to be resonated at w=wop 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.
20
m-derived filter -section
mZ1
Note that
2Z 2
m


2 1  m Z1
4m
2
2Z 2
m

Z '1  mZ1


Z 2 1  m2 Z1
Z '2 

m
4m

2 1  m Z1
4m
2
2
The image impedance is
Z i  Z1 ' Z 2 ' / Z iT 


Z1Z 2  Z12 1  m 2 / 4
Z o 1  w / w c 
2
21
Low -pass m-derived -section
mL



2 1 m L
4m
2
For constant-k
section
mC
2
mC
2

2 1 m L
4m
Then
Z1Z2  L / C  Zo2
and
2
Z1  jw L
Z 2  1 / jw C
Z12  w 2 L2  4Z o2 w / wc 2
Therefore, the image impedance reduces to
Z i 


1  1  m 2 w / w c 
2
1  w / w c 
2
Zo
The best result for m is 0.6which give a good constant Zi . 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 .
22
Composite filter
Matching
section
High-f
cutoff
m=0.6
Zo
1
2
constant
k
T

ZiT
Matching
section
Sharp
cutoff
m=0.6
mderived
m<0.6
ZiT
1
2

Zo
ZiT
23
Matching between constant-k and m-derived
ZiT  Zi
The image impedance ZiT does not match Zi,
I.e matching can be done by using half-  section as shown below and the
The
image impedance should be Zi1= ZiT and Zi2=Zi
Z' 1 / 2
It can be shown that
Z '1

1  4 Z '
2

 1
 2 Z '2
Z '1 
2 

1 

Zi1=ZiT
ZiT  Z '1 Z '2 1  Z '1 / 4Z '2  Zi1
Zi  Z '1 Z '2 / 1 Z '1 / 4Z '2  Zi 2
Zi2=Zi
2Z'2
Note that
Z '1  mZ1


Z 2 1  m2 Z1
Z '2 

m
4m
2
24
Example #1
Design a low-pass composite filter with cutoff frequency of 2GHz and
impedance of 75W . Place the infinite attenuation pole at 2.05GHz, and plot
the frequency response from 0 to 4GHz.
L/2
L/2
Solution
For high f- cutoff constant -k T - section
C
L
Zo 
C
L
Zo
2
or
Rearrange for wc and substituting, we have
2
 2  1
L   
 wc  C
L  CZo
wc 
2
C
2
LC
L  2Zo / wc  (2  75) /(2  2 109 )  11.94nH
2
 2  1
C   
 wc  L
C  2 / Zowc  2 /(75 2 109 )  2.122pF
25
continue
mL/2
mL/2
mC
For m-derived T section sharp cutoff
1  m2
L
4m
w op  1 m 2 w c

m  1  wc / wop
2 

1  2 109 / 2.05109

2
 0.2195
mL 0.2195 11.94 nH

 1.31nH
2
2
mC  0.2195 2.122pF  0.4658pF
1  m2
1  0.21952
L
11.94nH  12.94nH
4m
4  0.2195
26
continue
mL/2
mL/2
For matching section
mC/2
Zo
mC/2
1  m L
Zo
1  m  L
2
2
2m
2m
m=0.6
mL 0.6 11.94 nH

 3.582 nH
2
2
ZiT
mC 0.6  2.122 pF

 0.6365 pF
2
2
1  m2
1  0.62
L
11.94nH  6.368nH
2m
2  0.6
27
continue
A full circuit of the filter
Can be added
together
3.582nH
5.97nH
Can be added
together
5.97nH
6.368nH
1.31nH
12.94nH
Can be added
together
1.31nH
3.582nH
6.368nH
2.122pF
0.6365pF
0.4658pF
0.6365pF
28
Simplified circuit
9.552nH
6.368nH
7.28nH
4.892nH
12.94nH
6.368nH
2.122pF
0.6365pF
0.4658pF
0.6365pF
continue
Freq response of low-pass filter
Pole due to
m=0.2195
section
0
S11
0
1
2
3
4
-20
-40
-60
Frequency (GHz)
Pole due to
m=0.6
section
30
N-section LC ladder circuit
(low-pass filter prototypes)
Prototype beginning with serial element
go=Ro
g2
g4
g1
g3
gn+1
Prototype beginning with shunt element
g2
g4
go=Go
g1
g3
gn+1
31
Type of responses for n-section prototype filter
•Maximally flat or Butterworth
•Equal ripple or Chebyshev
•Elliptic function
•Linear phase
Maximally flat
Equal ripple
Elliptic
Linear phase
32
Maximally flat or Butterworth filter
For low -pass power ratio response
2n 

2 w  

H w   1  C  

wc  



1
where
C=1 for -3dB cutoff point
n= order of filter
wc= cutoff frequency
No of order (or no of elements)


log10 10
1
n
2 log10 w1 / wc 
A / 10
Prototype elements
g0 = gn+1 = 1
Series R=Zo
Shunt G=1/Zo
 2k  1 
g k  2 sin 

 2n 
Lk 
Zo gk
wc
gk
Ck 
Z ow c
Where A is the attenuation at w1 point and w1>wc
Series element
Shunt element
k= 1,2,3…….n
33
Example #2
Calculate the inductance and capacitance values for a maximally-flat lowpass 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.
Solution
Prototype values
First , determine the number of elements


log10 10A / 10  1
n
2 log10 w1 / wc 


log10 1020 / 10  1

> 2.51
2 log10 1000/ 400c 
Thus choose an integer value , I.e n=3
g0 = g 3+1 = 1
 2  1 
g1  2 sin 
1

 23 
 2  2  1 
g 2  2 sin 
2

 23 
 2  3  1 
g3  2 sin 
1

 23 
34
continue
L3  L1 
Z o g1
wc
501

 19.9nH
6
2    40010
g2
2
C2 

 15.9 pF
6
Z ow c 50 2    40010
50 ohm 19.9nH 19.9nH
15.9pF
50 ohm
35
or
g1
1
C3  C1 

 7.95 pF
6
Z ow c 50 2    40010
L2 
Zo g2
wc
50 2

 39.8nH
6
2    40010
50 ohm
39.8nH
7.95pF
7.95pF
50 ohm
36
Equi-ripple filter
For low -pass power ratio response

2 w 

H w   1  FoCn   
 wc  

1
where
Cn(x)=Chebyshev polinomial for n order
and argument of x
n= order of filter
wc= cutoff frequency
Fo=constant related to passband ripple
Fo  10Lr /10 1
Chebyshev polinomial
Co(x)  1
Cn( 1 )  1
i.e w  wc
C1(x)  x
Cn(x)  2 x Cn-1(x)-Cn-2(x)
Where Lr is the ripple attenuation in pass-band
37
Continue
Prototype elements
a1
g1 
F2
1 
 Lr 
F1  lncoth

4 
 17.372
ak 1ak
gk 
bk 1bk c
for n odd
1
g n1  
2
F1  for n even
coth

Lk 
Zo gk
wc
gk
Ck 
Z ow c
where
Series element
 2 F1 
F2  sinh

 n 
 k  1 
ak  2 sin 

2
n


bk 
F22
2  k

 sin  
 2n 
k  1,2,....n
k  1,2,....n
Shunt element
38
Example #3
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
F1=1.4626
F2= 1.1371
a1=1.0
a2=2.0
b1=2.043
50 0.8794
L1  L3 
 7 nH
9
2 10
1.1132
C2 
 3.543pF
9
50 2 10
50 ohm
7nH
7nH
g1 = g3 = 0.8794
g2= 1.1132
3.543pF
50 ohm
39
Transformation from low-pass to high-pass
wc
w

 
wc
w
•Series inductor Lk must be replaced by capacitor C’k
•Shunts capacitor Ck must be replaced by inductor L’k
Zo
Lk 
g kw c
1
Ck 
Z o g kw c
go=Ro
g2
g1
g4
g3
gn+1
40
Transformation from low-pass to band-pass
w
1  w wo 



 
wc
W  wo w 
where
w 2  w1
W
wo
and
wo  w1 w2
Now we consider the series inductor
1  w wo 
1 w
1 wo
j


jX  j 

Lk  j
Lk  j
Lk  jwL'sk 

W  wo w 
W wo
W w
w C 'sk
normalized
Lk
W
Lsk 
Csk 
Lk  Z o g k
Ww o
w o Lk
•Thus , series inductor Lk must be replaced by serial Lsk and Csk
Impedance= series
41
continue
Now we consider the shunt capacitor
1  w wo 
1 w
1 wo
j
Ck  j
jBk  j 

Ck  j
Ck  jwC ' pk 
W  wo w 
W wo
W w
w L' pk
L pk
W

w o Ck
C pk
Ck

Ww o
gk
Ck 
Zo
•Shunts capacitor Ck must be replaced by parallel Lpk and Cpk
Admittance= parallel
42
Transformation from low-pass to band-stop
w
1  w wo 





wc
W  wo w 
1
where
w 2  w1
W
wo
and
wo  w1 w2
Now we consider the series inductor --convert to admittance
1
1
j
 j
Xk
WLk
L pk 
WLk
wo
 w wo 
1 w
1 wo
j

  j

j
 jwC ' pk 
W w o Lk
W wLk
w L' pk
 wo w 
C pk
1

w o WLk
Lk  Z o g k
•Thus , series inductor Lk must be replaced by parallel Lpk and Cskp
admittance = parallel
43
Continue
Now we consider the shunt capacitor --> convert to impedance
1
1
j
j
Bk
WCk
Lsk 
 w wo 
1 w
1 wo
j

   j
j
 jwL'sk 
W w o Ck
W wCk
w C 'sk
 wo w 
1
Ww oCk
C pk 
WCk
wo
gk
Ck 
Zo
•Shunts capacitor Ck must be replaced by parallel Lpk and Cpk
44
Example #4
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 50W.
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
Lk  Z o g k
Z o g1
501.5963
Ls1  Ls 3 

 127nH
9
Ww o 0.1 2 10
W
0.1
Cs1  Cs3 

 0.199pF
9
wo Z o g1 2 10  501.5963
45
continue
Second element is equivalent to parallel capacitance, thus
Lp2
WZ o
0.1 50


 0.726nH
9
wo g 2 2 10 1.0967
C p2
g2
1.0967


 34.91pF
9
Z o Ww o 50 0.1 2 10
50W
127nH
0.199pF
0.726nH
127nH
gk
Ck 
Zo
0.199pF
34.91pF
50W
46
Implementation in microstripline
Equivalent circuit
A 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
jwL=jZosin(bd)
Model for series inductor
with fringing capacitors
jwC/2=jYotan(bd)/2
jwL/2=jZotan(bd/2)
Model for shunt capacitor
with fringing inductors
jwC/2=jYotan(bd/2)
jwL/2=jZotan(bd/2)
jwC=jYosin(bd)
47
L
d
d
Zo
ZoL
Zo
C
Zo
ZoC
-model with C as fringing
capacitance
T-model with L as fringing
inductance
d
1  w L 

d
sin 
2
 Z oL 
d
d
sin 1 w CZ oC 
2
C fL
 d 

tan  
Z oLw
 d 
1
ZoL should be high impedance
L fC
Zo
 d 

tan  
w
 d 
Z oC
ZoC should be low
impedance
48
Example #5
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 (er=4.5 h=1.5mm)
L1  L3  7 nH
C2  3.543pF
Let’s choose ZoL=100W and ZoC =20 W.
3 108
d 
 9
 14.14cm
f e r 10 4.5
c
Note: For more accurate
calculate for difference Zo
d
0.1414 1  2 109  7 109 
1  w L 
  10.25m m
 
d1,3 
sin 
sin 

2
2
100
 Z oL 


C fL
 d 
1
   0.01025

tan  
tan

  0.369pF
9
Z oLw
  0.1414 
 d  100 2 10
1
49
continue
C2  3.543pF


d
0.1414
d2 
sin 1 w CZ oC  
sin 1 2 10 9  3.543 10 12  20  10.38mm
2
2
L fC
 d 
20
   .01038

tan  
tan

  0.75nH
9
w
  0.1414 
 d  2 10
Z oC
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
d1,3
0.1414 1  2 109  6.25109 
  9.08m m

sin 

2
100


d2 


0.1414
sin 1 2 10 9  3.17 10 12  20  9.22 mm
2
More may be needed to obtain sufficiently stable solutions
50
Now we calculate the microstrip width using this formula
377
(approximation)
Zo 
w

 1.57
h

er 
 377

 377



w100 
 1.57 h  
 1.571.5m m  0.31m m
Z

 100 4.5

 oL e r

 377

 377

w20  
 1.57h  
 1.571.5m m  10.97m m
Z

 20 4.5

 oL e r

 377

 377


w50 
 1.57h  
 1.571.5m m  2.97m m
Z

 50 4.5

 oL e r

9.22mm
10.97mm
0.31mm
0.31mm
2.97mm
2.97mm
9.08mm
9.08mm
51
Implementation using stub
Richard’s transformation
jBc  j C  jC tan b 
jX L  j L  jL tan b 
At cutoff unity frequency,we have =1. Then
tan b   1
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.


8
/8
jXL
L
jXL
S.C
Zo=L
/8
jBc
C
jBc
O.C
Zo=1/C
52
Kuroda identity
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
O.C shunt
stub
d
d
Z1
S.C series
stub
d
Z1/n 2
d
Z2/n 2
Z2
n2=1+Z2/Z 1
d=/8
53
Example #6
Design a low-pass filter for fabrication using micro strip lines .The
specification: cutoff frequency of 4GHz , third order, impedance 50 W, 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
3.3487
0.7117
ZoL=3.3487
3.3487
1
ZoL=3.3487
8
8
1
8
8
Zo
Zo
1
Using Richard’s transform we have
Zoc=1.405
ZoL= L=3.3487 and
8
Zoc=1/ C=1/0.7117=1.405
54
Using Kuroda identity to convert S.C series stub to O.C shunt stub.
We have
2
2
and
Z
/
n
 Zo  1
2
Z1 / n  ZoL  3.3487
Z2
1

Z1 3.3487
Z2
1
n  1
 1
 1.299
Z1
3.3487
2
thus
Substitute again, we have
Z1  n ZoL  1.299 3.3487 4.35
2
O.C shunt
stub
d
Z1
and Z2  Zon2  11.299  1.299
S.C series
stub
d
Z1/n 2=ZoL d
Z2/n 2=Zo
Z2
n2=1+Z2/Z 1
55
Z1=1.299x50
=64.9W
Zoc=1.405x50
=70.3W
Z1=1.299x50
=64.9W
Zo=50W
/8
/8
Z2=4.35x50
=217.5W
/8
Z2=4.35x50
ZL=50W
/8 =217.5W /8
50W
50W
217.5W
/8
/8
64.9W
217.5W
/8
70.3W
64.9W
56
Band-pass filter from /2 parallel coupled lines
Output
/2 resonator
/2 resonator
Microstrip
layout
Input
/4
/4
/4
J'01
+/2
rad
J'12
+/2
rad
J'23
+/2
rad
Equivalent
admittance
inverter
Equivalent
LC resonator
57
Required admittance inverter parameters
The normalized admittance inverter is given by
1
2
 W
J '01  

2
g
g
 0 1
where W  w 2  w1
A
for k  1,2,...n  1
B
wo
W
1
J 'k ,k 1 

2
g k g k 1
1
2
 W
J 'n,n1  

2
g
g
 n n1 

Z oe , k ,k 1  Z o 1  J 'k ,k 1  J 'k ,k 12

Z oo,k ,k 1  Z o 1  J k ,k 1 ' J 'k ,k 12
n  no. of sec tions


C
D
where J 'k ,k 1  J k ,k 1Zo
E
58
Example #7
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=50W.
We have g0=1 , g1=1.5963, g2=1.0967, g3=1.5963, g4= 1 and W=0.1
A
1
2
1
2
 W
   0.1
J '01  



  0.3137
 2 11.5963
 2 g 0 g1 
1
2
1
2
C
 W
   0.1
J '3,4  



  0.3137
 2 1.59631
 2 g3 g 4 
D
Zoe ,0,1  Zoe ,3,4  501  0.3137 0.31372  70.61W
E
Zoo,0,1  Zoo,3,4  501  0.3137 0.31372  39.24W




59
W
B
1
  0.1
1
J '1,2 



 0.1187
2
2
g1g 2
1.59631.0967
B
J ' 2, 3 
D
Zoe ,1,2  Zoe ,2,3  501  0.1187 0.11872  56.64W
E
Zoo,1,2  Zoo,2,3  501  0.1187 0.11872  44.77W
W
The required resonator
2

1
  0.1
1


 0.1187
2
g 2 g3
1.09671.5963




3 108
3 108
r / 4 

 0.01767 m
9
2 f e r 4  2 10 10
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 er=10. For others use other means. 60
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
0.95mm
0.45mm
50W
0.95mm
0.7mm
0.7mm
1.3mm
50W
1.3mm
0.45mm
17.67mm
17.67mm
17.67mm
17.67mm
61
Band-pass and band-stop filter using quarter-wave stubs
/4
Zo
Zo
Band-pass
Z01
/4
Z02
/4
....
Zo
Zo
/4
Z01
/4
/4
Zo
Zo
Band-stop
Z02
4gn
/4
/4
....
Zo
Z on 
 Zo W
Zon
Zon-1
/4
Zo
Zo
Zon-1
/4
Zo
4 Zo
Z on 
 Wg n
Zon
/4
/4
62
Example #8
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 W=0.1
4  50
Z o1  Z 03 
 265.9W
  0.151.5963
4  50
Zo2 
 387W
  0.151.0967
/4
265.9W
/4
/4
265.9W
/4
/4
387W
50 W
4 Zo
note: Z on 
 Wg n
Note that: It is difficult to
impliment on microstripline
or stripline for characteristic
> 150W.
63
Capacitive coupled resonator band-pass filter
Zo
1
2
1
2
Zo
Zo
B1
Zo
Zo
Bn+1
B2
 W
J '01  

2
g
g
 0 1
W
1
J 'k ,k 1 

2
g k g k 1
w 2  w1
where W 
wo
for k  1,2,...n  1
1
2
 W
J 'n,n1  

2
g
g
 n n1 
Ji
Bi 
2
1  Z o J i 

....
n
n  no. of sec tions
i=1,2,3….n
 

1
1
1
 i    tan 2Z o Bi   tan 1 2Z o Bi 1 
2
2
64
Example #9
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
prototype w
1  w w o  1  2.2 2 

  

 1  1.91 1  0.91

  1.91
W  w o w  0.1  2 2.2 
wc
From Pozar ,Fig 8.27 pg 453 , we have N=3
n
1
2
3
4
gn
1.5963
1.0967
1.0967
1.0000
ZoJn
0.3137
0.1187
0.1187
0.3137
Bn
6.96x10-3
2.41x10-3
2.41x10-3
6.96x10-3
Cn
0.554pF
0.192pF
0.192pF
0.554pF
n
155.8o
166.5o
155.8o
-
65
Other shapes of microstripline filter
Rectangular resonator filter
/2
Interdigital filter
in
out
U type filter
/4
Out
In
Out
/4
In
66
Wiggly coupled line
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.
1
1= /2
2= /4
2
67