design of all-pole microwave filters - intranet...
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Design of all-pole microwave filters
Giuseppe MacchiarellaPolytechnic of Milan, Italy
Electronic and Information Department
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In-line filters with all-equal resonators
K23Kn,n+1
Leq, f0
K12K01
R0 R0
From the general design equations:
Leq, f0 Leq, f0 Leq, f0
, 1 , 1 0q q q q eqK k L 001 , 1 1
eqn n
E
LK K R
Q
with:
, 10 1
1q q
q q
Bkf g g
1
0E
gQB f
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/2 waveguide resonators coupled with inductive reactances
Leq, f0
Kq,q+1
Zc=1
0/2
Zc=1 Zc=1jxq,q+1
q,q+1
20
002
geq CL Z
q,q+1
, 1, 1 2
, 1
1, 1 , 1
1
tan 2
q q cq q
q q
C
q q q q
K Zx
KZ
x
….
L1 L2 Ln
x01 x12 x23 xn-1,n xn,n+1
1, , 1
0
12
q q q qq
g
L
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Inductive reactances in rectangular waveguide
Iris with finite thickness
Aperture W Zc=1 Zc=1jxs
Equivalent circuit (reference sections: symmetry axis)
11 12 11 12, 1 tan
2 2
e oj je o
os o e
S S e S S e
x
S11, S12: scattering parameters (computed numerically with EM simulators)
a
b
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Evaluation of iris equivalent parameters
• With the previous formulas, the curves of x and vs. W are drawn (see the graph above).
• For the required value X the aperture W is first obtained through the blue curve. The corresponding electrical length is derived from the red curve
10 12.5 15 17.5 20 22.5 25W (mm)
0
0.1
0.2
0.3
0.4
x
0.5
0.875
1.25
1.625
2
delta
(de
g)
Reactance x
ElectricalLength
X
W
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The separation of the iris are obtained from the lengths Lq corrected with the irises parameters q,q+1:
1, , 10
1q q q q q q
g
L L
Filter dimensioning
….
L’1 L’2 L’n
W01 W12 W23 Wn-1,n Wn,n+1
Top view
a
The apertures Wi,i+1 are evaluated as explained in the previous slide from the Ki,i+1.
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Example of design
Specifications:f0=4 GHz, B=40 MHzReturn Loss in passband ≥26 dBAs1 ≥ 45 dB for f>4.05 GHzAs2 ≥ 60 dB for f>3.9 GHzChebycheff characteristic
Being fs1 and fs2 not geometrically symmetric, the requested order n of the filter is obtained by selecting the larger between the following ones:
1
11
6 5.5420log 6
s
s
A RLn
2
22
6 4.5820log 6
s
s
A RLn
n=6
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Step 1: Evaluation of equivalent circuit param.
Prototype normalized coefficients:gq={0.7919, 1.3649, 1.7002, 1.5379, 1.5089, 0.7163}
Selected waveguide: a=50mm, b=25mm
Waveguide parameters:Cutoff frequency: fc=v/2a=3 GHz
0
0
2
02
00
113.3 , 2.2861
gg
c
mmf f
Coupling reactances and resonator lengths:xq,q+1={0.223, 0.0346, 0.0236, 0.022, 0.0236,0.0346,0.223}Lq ={52.25, 55.61, 55.83, 55.83, 55.61, 52.25}
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Filter response (ideal inductive reactance)
3.95 3.96 3.97 3.98 3.99 4 4.01 4.02 4.03 4.04 4.05Frequency (GHz)
-60
-50
-40
-30
-20
-10
0
Ideal ChebycheffIdeal waveguide
The reactances are assumed to vary with frequency as g/g0
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Step 2: Irises dimensioning
Wq,q+1={21.27, 10.86, 9.426, 9.219, 9.426, 10.86, 21.27} mmq,q+1 ={1.609, 0.807, 0.683, 0.665, 0.683, 0.807, 1.609}°
Corrected lengths of resonators:L’q={51.49, 55.14, 55.41, 55.41, 55.14, 51.49} mm
Example of fabricated device (4 resonators sample)
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Computed filter response (Mode Matching)
3.95 3.96 3.97 3.98 3.99 4 4.01 4.02 4.03 4.04 4.05Frequency (GHz)
-60
-50
-40
-30
-20
-10
0
Iris (MM)
Ideal reactances
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Broad frequency range response
3.8 4 4.2 4.4 4.6 4.8 5 5.2 5.4 5.5Frequency (GHz)
-250
-200
-150
-100
-50
0
Waveguide filter
Ideal Chebycheff
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Effect of filter bandwidth on accuracy
3.8 3.85 3.9 3.95 4 4.05 4.1 4.15 4.2Frequency (GHz)
Normalized Bandwidth: 2.5%
-60
-50
-40
-30
-20
-10
0
3.6 3.7 3.8 3.9 4 4.1 4.2 4.3 4.4Frequency (GHz)
Normalized Bandwidth: 5%
-60
-50
-40
-30
-20
-10
0
3.5 3.6 3.7 3.8 3.9 4 4.1 4.2 4.3 4.4 4.5Frequency (GHz)
Normalized Bandwidth: 10%
-60
-50
-40
-30
-20
-10
0
3.95 3.96 3.97 3.98 3.99 4 4.01 4.02 4.03 4.04 4.05Frequency (GHz)
Normalized Bandwidth: 1%
-60
-50
-40
-30
-20
-10
0
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In-line filters with assigned inverters
G0 G0Ceq,1, f0
J1
Ceq,2, f0 Ceq,3, f0 Ceq,n, f0
Each resonator is loaded with two conductances with value J1.The loaded Q of q-th resonator is obtained from the general design equations:
0 ,
1 0
12 2
eq q qq
C gQ
J B f
J0J1 J0
0 0 1J G J
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Inverters with 0/4 line sections
….YC0 YC1 YC1 YC0
0/4 0/4 0/4 0/4
Ceq,1, f0 Ceq,2, f0 Ceq,3, f0 Ceq,n, f0
G0 G0
1 1
0 0 1
0 , 1 10
1) Assign
2) Compute
3) Compute 2
c
c c
qeq q c q c
Y J
Y G Yg
C Y Q YB f
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Resonators with shorted transmission lines
Ycs,q
0/4
Ceq,q, f0
, 10
4 qcs q c
gY Y
B f
0 , ,4eq q cs qC Y
NOTE: The choice of Yc1 affects the computed values of Ycs,q. These latter must be physically realizable with the chosen fabrication technology of the filter
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Example: filter in microstrip technology
Let assume the following specifications:f0=1 GHz, B/f0=0.1, RL=15 dB, n=5
Assuming the Chebycheff characteristic:g={1.2327 1.3592 2.0599 1.3592 1.2327}
Now suppose that the range of realizable characteristic impedance is 20-150 Ohm. Then we assume Yc1=1/150=0.0067. Using the previous formula:
Ycs,q={0.1046 0.1154 0.1748 0.1154 0.1046}
1/Ycs,q={9.5570 8.6676 5.7192 8.6676 9.5570}
The computed impedances are not realizable with microstrip technology!
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Possible solution to allow the feasibility
Increase the imposed return loss (not too much effective) Use two stubs in parallel with twice Ycs,q
Choice a different implementation of resonators (for obtaining smaller Ycs,q for the same equivalent capacitance). For instance, using open-circuited, /2 stubs the characteristic impedances are doubled.Another choice is the use of capacity-loaded resonators:
short Ycs
L<0/4
Cs
0 24CS
eqY
C
, 10
42
qcs q c
gY Y
B f
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Implementation with double /2 resonators
Substrate: Duroid (r=2.16, H=1.2 mm, t=50)
1/Ycs,q={38.23, 34.67, 22.88, 34.67, 38.23}
0 0 1 00.0115 (1 86.6)c c cY G Y Y Yc1=1/150=0.0067
Layout:
Discontinuities!
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Computed response
0.8 0.9 1 1.1 1.2Frequency (GHz)
-60
-50
-40
-30
-20
-10
0Ideal Chebycheff
0.8 0.9 1 1.1 1.2Frequency (GHz)
-60
-50
-40
-30
-20
-10
0Ideal TransmissionLines
0.8 0.9 1 1.1 1.2Frequency (GHz)
-60
-50
-40
-30
-20
-10
0MicrostripLines
0.8 0.9 1 1.1 1.2Frequency (GHz)
-60
-50
-40
-30
-20
-10
0
Microstrip(DiscontinuitiesCompensated)
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Attenuation produced by losses
An estimate of the overall attenuation at center frequency f0 can be obtained with the following formula:
1
1, 0
210 log 1 q
dBq n
QA
Q
Where Qq is expressed as function of the prototype parameters gq:
0
12
gQ
B f
Note: Narrow band filters require high Q0resonators to reduce passband attenuation
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Coupled-line filters
Physical structure: Z0
Ze,01 Zo,01
0/4
Ze,12 Zo,12
0/4
Z0
Ze,23 Zo,23
0/4
di,i+1
Basic Block:
,( , 1) ,( , 1) ,( , 1)
,( , 1) ,( , 1) ,( , 1)( , 1) ,( , 1)
0
12
, sin 2
m i i e i i o i i
i i e i i o i ii i i i
Z Z Z
Z Z ZK Z
L
Ze,i,i+1 Zo,i,i+1
0/4
di,i+1
Open
Open
wi,i+1
wi,i+1
Ki,i+1
Zm,(i,i+1)Zm,(i,i+1)
0/40/4
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Typical dependance of Zm and Z on d/w
0.4 0.9 1.4 1.9 2.4 2.9 3.4 3.9d/w
0
10
20
30
40
50
60
70
80
90
100
Zm
Zdelta
For d/w >1 Zm is practically independent on d for assigned wand practically coincides with the characteristic impedance of the isolated line with the same w.
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Equivalent circuit (inner blocks)
Ki-1,i
Zm,(i-1,i)
Ki,i+1
Zm,(i,i+1)Zm,(i,i+1)Zm,(i-1,i)
Leq, f0 0 , ,( 1, ) ,( , 1)
4 2eq i m i i m i i cL Z Z Z
Design Equations:
, 1 ,( , 1)
, 1 , 1,( , 1)0 , 0 , 1
2 2q q i iq q i i
m i ieq q eq q
K Zk m
ZL L
i-th resonator
, 1 , 12i i i im k
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First (last) block
2
0101 ,(0,1)
0 0
cot2m
K fX ZR f
K0,1
Zm,(0,1)Zm,(0,1)
0/40/4
R0 K0,1
Zm,(0,1)
R0
jX01
R0>>X01
X1
2
011 01 ,(0,1) ,(1,2)
0 0 0
cot 2 cot2 2m m c
Kf fX X Z Z Zf R f
0 ,1 0 1,20 ,1 0,1 1,2
0 0
22, ,
1 14 4
eq Eceq
c c c
E E
L R Q kZL m mZ Z Z
R Q R Q
Zm,(1,2) Zm,(1,2)
0 ,120,1 0
eqE
LQ
K R
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Dimensioning of the filter
1. The dimension w of the lines is first assigned 2. The values of mi,i+1 are computed with the previous
formulas3. Using a graph of m vs. d (like the one shown below) the
distances di,i+1 are evaluated
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1d (mm)
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
m
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Example of microstrip filter design
Specifications:f0=1 GHz, B=200 MHz, RL=26 dB, n=7g={0.807, 1.397, 1.758, 1.634, 1.758, 1.397, 0.807}
Assigned substrate parameters:r=10, H=1.2mm, t=10
Assigned width of lines (for Zc≈85 Ohm): w=0.25 mm Computed mi,i+1:
m={0.415, 0.258, 0.201, 0.186 , 0.186, 0.201, 0.258, 0.415}
Computed distances di,i+1 (with the previous graph):d={0.272, 0.649, 0.861, 0.93, 0.93, 0.861, 0.649, 0.272}
Evaluation of length of blocks (about 0/4 for a not coupled line): L=30.44mm
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Computed filter response (circuit model)
0.7 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3Frequency (GHz)
-80
-70
-60
-50
-40
-30
-20
-10
0
ChebycheffIdeal
Microstrip(including discontinuities)
0.7 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3Frequency (GHz)
-80
-70
-60
-50
-40
-30
-20
-10
0
Microstripoptimized
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Filters with an array of coupled-lines
Interdigital:
Comb:
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Basic block: array of coupled lines
1 2 3 n….
1 2 3 n
n+1 n+2 n+3 2n
1 2 3 n
n+1 n+2 n+3 2n
AdmittanceMatrix (2nx2n)
The admittance matrix Y (2nx2n) can be evaluated from the static capacitance matrix p.u.l C (dimension n x n):
, ,tan( ) sin( )v v
j L j L
Y Y C CY Y YY Y
….
….
L
is the phase velocity
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Matrix Y of Interdigital Array
1 3
n+2 2n
AdmittanceAdmittanceMatrix (2nx2n)Matrix (2nx2n)
……..
……..1 3
n+2 2n
AdmittanceAdmittanceMatrix (2nx2n)Matrix (2nx2n)
……..
……..2 n1 3
AdmittanceMatrix YI (nxn)
….
Let assume that Ci,j=0 for j>i+1 (no coupling between non-adjacent lines). Elements of matrix YI are then given by:
,
, 1
tan( )
( , ) 1sin( )
0 1
i i
i iI
v Ci j
j Lv C
y i j j ij L
j i
Matrix YI is tri-diagonal
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Condition on the Y matrix for representing a filter
Y2,3 Y3,4 Yn-1,nY1,2Y1,1 Y2,2Y3,3 Y4,4 Yn,n
2 n1 3 4
Equivalent parameters (Yi,j=jBi,j):
0
, 0, 0 0
0
, 1, 1 , 1 0 , 1
0
,0 , 0 ,
0 tan( ) 2 4
sin( )
12 4
i ii i
i ii i i i i i
i ieq i i i
v CB L L
Lv C
J B CL
BC C
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Coupling with external loads
J0,1R0 B1,1
1
20,1
1,1 1,10
JY jB
G
0
0 0
1,10
,0 ,120,1 0 1,1 1,1
12 4Re Re
i ieq
E
BCC
QJ G Y Y
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Equations for interdigital filters design
0
1,1, 1 , 1
, 11,10 , , , 1, 1
4 4, Re
i i i ii i E
eq i eq j i i i i
CJ Ck Q
YC C C C
Typically the coupled lines are assumed of equal width (assigned a priori). The unknowns are then represented by the distances di,i+1 between the lines and by the dimensional parameter (de) of the external coupling structure. The solution is obtained by solving the system of non-linear equations in the variables di,i+1 and de :
, 1 , 1 1,24 ( ) 0, , 0
4i i i i E ek F d Q U d d
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Matrix Y of Comb Array
2 n1 3
AdmittanceMatrix YC (nxn)
….1 2 3 n
AdmittanceMatrix (2nx2n)
….
….
,,
, 1
tan( )
( , ) 1tan( )
0 1
i is i
i iC
v Cj C i j
Lv C
y i j j ij L
j i
With no coupling between non-adjacent lines:
To get yc(i,i+1)≠0:
2L
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Equations for comb filters design (l=/4)
0
0
, 1 , 1 , 1 0 , ,
1,1, 1
, 11,1, 1, 1
2, 4
24 4,
2 Re
i i i i i i eq i i i
i ii i E
i i i i
J B C C C
CCk Q
YC C
As in the case of interdigital filters, assigning all the lines with the same width, the solution is obtained by solving the system of non-linear equations in the variables di,i+1and de :
, 1 , 1 1,24 2( ) 0, , 0
2 4i i i i E ek F d Q U d d
The tuning capacitances are obtained by the resonance condition: ,
0 , ,tan( 4)i i
s i i i
v CC v C
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Approximated evaluation of Capacitance Matrix
Two coupled lines:Single line:
C0/2 C0/2C0/2
Ca/2
C0/2
Ca/2
C
0
0
2
22
aeven
aodd
C CC
C CC C
, even even odd oddY C Y C
02
2
a even
odd even
C C CC C
C
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Array of coupled lines: evaluation of C
,(2,3) 2
aC ,(2,3) 2
aC ,(3,4) 2
aC ,(3,4) 2
aC,(1,2) 2
aC ,(1,2) 2
aC
,(1,2) C ,(2,3) C ,(3,4) C
C0/2C0/2
,( 1, ) ,( , 1)
, 1, , 1
,( 1, ) ,( 1, ) ,( , 1) ,( , 1)0
2
2 2
a i i a i ii i i i i i
even i i odd i i even i i odd i i
C CC C C
C C C CC
,(1,2) 0 ,(1,2) ,(1,2)
1,1 1,22 2a odd evenC C C C
C C
,( , 1) ,( , 1), 1 ,( , 1) 2
even i i odd i ii i i i
C CC C
, 0i iC C
If the lines are not too close:
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Design equations
The design equations for comb and interdigital filters become:
0
,( , 1) ,( , 1) 0, 1
0 1,1
1 , 2 Re
even i i odd i ii i E
Y Y Yk Q M
M Y Y
with:2 (Comb) (Interdigital)
4 4M M
,( , 1) ,( , 1) ,( , 1) ,( , 1)
2 2even i i odd i i even i i odd i iC C Y Y
Being:
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Example: Comb Filters in slab-line
L
One-side short circuited array of coupled rods
Tuning capacitances
….1 2 3 n
si,i+1
d
h….1 2 3 n
si,i+1
d
h
Array of coupled rods(coupled slab-lines)
C=Capacitance matrix p.u.l
Inline comb filter configuration