specifications microwave filter design · case study: parallel coupled-line combline filter. part b...
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MICROWAVE AND RF DESIGNMICROWAVE AND RF DESIGN
Based on material in Microwave and RF Design: A Systems Approach, 2nd
Edition, by Michael Steer. SciTech Publishing, 2014.
Presentation copyright Michael Steer
Case Study:Parallel Coupled-Line Combline Filter
Presented by Michael Steer
Reading: § 16.1–16.4
Index: CS_PCL_Filter
Case Study: Parallel Coupled-Line Combline Filter
Copyright 2013 M. Steer and IETADA
C b C bC 1 C 3
50
50 t2COutput
1 2
Input
40
11 S 21
60
20
10
30
S50
S21
(dB
)
0
4
8
12
16
20
11S
(dB
)
240.5 1.0 1.5
0
Frequency (GHz)
1
Specifications● Bandpass filter● Center frequency: 1 GHz● 10% Bandwidth● Steep filter skirts
– requires Chebyshev response, choose a ripple factor of 0.1
● Low loss in passband– Microstrip technology
(Also low fabrication cost and very good performance.)
2
Microwave filter design
● Combination of Art and Science● Art: knowing the structures that intrinsically
have the desired response.● Science: knowing how to use mathematics in a
synthesis process to obtain the required tailoring of the response.
3
Art: choice of topologyParallel microstrip lines in combline configuration.
0
dB
12111 30-80
-60
-40
-20
21S
Frequency (GHz)
4
50
50 Output
1 2
Input
Science: synthesis procedure
50
50
Output
1 2
Input
5
Vg
L 11L 21
C 21
C31
L 31
11C
11
How to go from
to
using mathematical synthesis,while maintaining desired electrical characteristics.
(perhaps more complicated)
Optimization: an alternative to synthesis
● Optimization given a final structure that almost has the right response, use optimization to get the exact final response.
● E.G. Adjust line widths and lengths; number of microstrip lines, capacitor values.
● Works if the starting solution is very close.
● Does not provide insight or lead to new solutions.
● Even then, optimization with more than 6 variables is a problem.
6
50
50 Output
1 2
InputSummary
● Filter design, as with most RF design, is a combination of art and science.
● The art is identifying the structures that intrinsically have the desired response.
● The science is developing the mathematical procedure to go from the mathematical specification of the desired response to the final microstrip realization.
● Choose topology (art), use synthesis procedure (science), use optimization to almost perfect design, use fabrication and test to perfect design.
7
Case Study: Parallel Coupled-Line Combline Filter. Part B
8
Filter design is based on circuit transformations.
Vg
L11L 21
C21
C31
L31
11C
11
Outline● Begin with a lumped element filter.
● Consider circuit model of coupled lines● Work out the steps to go from the lumped
element circuit to a transmission line-based circuit.
9
Z0
Vg
L12 C12
L22 C21
L32 C32
Z0Vg
L11L 21
C21
C31
L31
11C
11 or
Third-order filter● The lumped element filter has three
LC resonators:
● So (perhaps) the transmission line equivalent has three resonators:
● Consider as two pairs of coupled lines:
Vg
L 11L 21
C 21
C31
L 31
11C
11
1
2
2
1
10
Network model of a pair of coupled lines
1
2
4
3
1: n
4
3
:1n
Z
1
2
02
01
Z
The equivalent circuit of a pair of coupled lines
Is obtained by equating symbolic ABCD equations
11
Combline section and network models
1
2
4
3
Combline section
1: n n
4
3
:1
1
201Z
02Z
1: n n
4
3
:1
1
201Z
02Z
12
Combline section and network models
1
2
4
3
Combline section
1: n n
4
3
:1
1
201Z
02Z
2
1: n
Z 02
Z 012
1
( fr = f0 ) Z1: n
02
Z011 2
Z
Z
012
1 2Z 011 022
13
Translation of a circuit with stubs to coupled lines
1
2
4
3
Z
Z
012
1 2Z 011 022
So if the following structure is seen in a circuit (a Pi arrangement of shorted stubs)
Then it can be replaced by a combline section
14
Comparison of lumped element filter and Pi arrangement of stubs
Z
Z
012
1 2Z 011 022
Pi arrangement of shorted stubs.
Three connected resonators.Vg
L 11L 21
C 21
C31
L 31
11C
11
Each stub is a resonator.
A shorted stub corresponding to a parallel LC resonator.
An open circuit stub corresponds to a series LC resonator.
15
So the conversion from a lumped element filter to Pi network of shorted stubs is not direct.
But The Idea is Starting to Come Through
Summary● The key idea is to begin with a lumped element filter
prototype and put the circuit in the form of a collection of shorted stubs in a PI configuration.
16
Z
Z
012
1 2Z 011 022
Want basic circuit structure to be
But cannot start from here (3rd order BPF)
(Model of two PCL in combline configuration.)
Case Study: Parallel Coupled-Line Combline Filter. Part C, Step 1:
Develop Lowpass Prototype Filter
17
L21
C11 C31Vg
11
Outline● Begin with a lumped element filter.
● Calculate element values.
18
L21
C11 C31Vg
11
19
TRANSMISSION2 REFLECTION2
This is the 6th order Chebyshev response
ε is called the ripple factor.
Passband ripple,PBR = (1+ ε2)
Ripple in dB,RdB = 10 log(PBR)
Steeper filter skirt for• Higher order• Larger ripple
19
Third-order Chebyshev filter
Coefficients of a Chebyshev lowpass prototype filter normalized to a radian corner frequency of ω0 = 1 rad/s and a 1 Ω system impedance (i.e., g0 = 1 = gn+1).
The ripple factor is ε. ε = 0.1 is a ripple of 0.0432 dB.
ω0 is the radian frequency at which the transmission response of a Chebyshev filter is down by the ripple. Here ω0 = 1 radian/s.
20 21
L21
C11 C31Vg
11
A third-order Chebyshev lowpass filter prototype
g0 = 1g1 = 0.85158g2 = 1.10316g3 = 0.85158g4 = 1
C11 = 0.85158 FL21 = 1.10316 HC31 = 0.85158 F
ω0 = 1 rad/s
21
22
Chebyshev filter coefficients from recursive formula
22
Summary● Step 1: developed 3rd order Chebychev lowpass
prototype filter.
23
L21
C11 C31Vg
11
C11 = 0.85158 FL21 = 1.10316 HC31 = 0.85158 F
ω0 = 1 rad/s
Case Study: Parallel Coupled-Line Combline Filter. Part D, Step 2:
Remove Series Inductor
24
L21
C11 C31Vg
11
C11 C21 C31
111 1
Outline
● Use an inverter(s) to replace series inductor.
● An inverter can be implemented using transmission lines.
● Where there are transmission lines it may be possible to equate them to an inverter (if they are /4 long).
25
Inverters
K
ZL
Zin
K
= K0Z
= K0ZinZ
LZImpedanceinverter
A /4 long line is an inverter
2
inL
KZZ
An inverter can be realized using a transmission line.
Wherever there are /4 long transmission lines an impedance inverter can be realized (probably).
Lossless telegrapher’s equation:
0in 0
0
tan( )tan( )
L
L
Z jZZ ZZ jZ
/ 2; tan( )
0in 0
tan( )tan( )L
jZZ ZjZ
20
inL
ZZZ
26
Consider combline section and network model
1
2
4
3
Combline section
1:n n
4
3
:1
1
201Z
02Z
A combline section of coupled lines /4 long inherently presents two impedance inverters.
27
Replacement of a series inductor by a shunt capacitor plus inverters
L
C
-1:1
KK
Equivalence is demonstrated using ABCD parameters.
1T
0 1L
sL
K
L
2
0T
/ 0jK
j K
1
1 0T
1sC
C
-1:1
3
1 0T
0 1
For the cascade2
CASCADE 2 1 2 3
0 1 0 0 1 0 1T T T T T
/ 0 1 / 0 0 1 0 1jK jK sCK
j K sC j K
2L CK
sL
28
Equivalence of a series inductor and a shunt capacitor plus inverters
L
C
-1:1
KK 2L CK
29
C KK
Drop negative unity transformer as it only affects phase and not filter response.
Inverter form of lowpass prototype filter
L21
C11 C31Vg
11
C11 C21 C31
111 1
C11 = 0.85158 FL21 = 1.10316 HC31 = 0.85158 F
ω0 = 1 rad/s C11 = 0.85158 FC21 = 1.10316 FC31 = 0.85158 F
ω0 = 1 rad/s
30 31
+1
( odd)
( even)
g2 g4
g1 g3
gn
g0
Vg
gn n
gn n
1
Vg
g1 g2 g3 g4
1 1 1 1 1
-1:1 -1:1 -1:1
1
Vg
g1 g2 g3 g4
1 1
g5
1 1 1
Ladder prototype filters using impedance inverters
31
Summary: Inverter form of lowpassprototype filter
C11 C21 C31
111 1
C11 = 0.85158 FC21 = 1.10316 FC31 = 0.85158 F
ω0 = 1 rad/s
32
Case Study: Parallel Coupled-Line Combline Filter. Part E, Step 3:
Bandpass Transformation
33
C11 C21 C31
111 1
' ' ' ' ' '
Vg
111 1
LC C L C L1 1 2 32 3
34
First lumped element transformation to BPF, 1 GHz
L21
C11 C31Vg
11
Z 0
Vg
L12 C12
L 22 C 21
L32 C 32
Z 0
34
+11(rad/s)
(s)T 2
(rad/s)1
T s( ) 2
0 2
LPF HPF
35
BPF and center frequency transformation
1 11 22 2
1 10 1 22 2
0 1 2
1112 12
0 11 0
950MHz, 1050MHz
1000 MHzfractional bandwidth,
1, CC LC
R
dB
This assumes that the LPF corner frequency is 1 radian/s.
L21
C11 C31Vg
11
Z0
Vg
L12 C12
L22 C21
L32 C32
Z0
36
1
Vg
L12 C12
L22 C22
L32 C32
1
Transformation to BPF, 1 GHz
C12 = 1.35533 nF = C31L12 = 18.6894 pH = L32C22 = 14.4271 pFL22 = 1.75573 nH
10
20
30
40
60
50
S214
8
12
16
24
20
11S(d
B)
21S(d
B) S11
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5
0
Frequency (GHz)
0
36
' ' ' ' ' '
Vg
111 1
LC C L C L1 1 2 32 3
37
BPF and center frequency transformation
C11 C21 C31
111 1
1 11 22 2
1 10 1 22 2
0 1 2
/ /111 1
0 11 0
950MHz, 1050MHz
1000 MHzfractional bandwidth,
1, CC LC
R
dB
This assumes that the LPF corner frequency is 1 radian/s.
' ' ' ' ' '
Vg
111 1
LC C L C L1 1 2 32 3
38
Prototype BPF and center frequency transformation
C1 = 1355.33 pF = C3
L1 = 0.0186894 nH = L3
C2 = 1755.73 pF
L2 = 0.0144271 nH
/
/
/
/
/
/
Z
Z
012
1 2Z 011 022
Recall: desired basic circuit structure
(Model of two PCL in combline configuration.)
' ' ' ' ' '
Vg
111 1
LC C L C L1 1 2 32 3
39
Summary prototype BPF
C1 = 1355.33 pF = C3
L1 = 0.0186894 nH = L3
C2 = 1755.73 pF
L2 = 0.0144271 nH
/
/
/
/
/
/
Case Study: Parallel Coupled-Line Combline Filter. Part F, Step 4:
Impedance Scaling
40
' ' ' ' ' '
Vg
111 1
LC C L C L1 1 2 32 3
||
Vg
505050 50
LC C L C L1 1 2 32 3|| || || || ||
Principle of impedance scaling
● Every impedance in the circuit is scaled by the same amount
● So to go from 1 to 50 – The value of a resistor is increased by a factor of 50.– The value of an inductor is increased by a factor of 50.– The value of a capacitor is reduced by a factor of 50.– The value of an impedance inverter is increased by a factor
of 50.– The value of an admittance inverter is reduced by a factor of
50.
41
Summary, Step 4: BPF scaled to 50 .
' ' ' ' ' '
Vg
111 1
LC C L C L1 1 2 32 3
||
Vg
505050 50
LC C L C L1 1 2 32 3|| || || || ||
C1 = 1355.33 pF = C3
L1 = 0.0186894 nH = L3
C2 = 1755.73 pF
L2 = 0.0144271 nH
/
/
/
/
/
/
C1 = 27.1066 pF = C3
L1 = 0.934468 nH = L3
C2 = 35.1147 pF
L2 = 0.721359 nH
||
||
||
||
42
Case Study: Parallel Coupled-Line Combline Filter. Part G, Step 5:Conversion of Lumped-Element
Resonators
43
LC C0 Z 1Z 01
,
Z 1
Z 01
Outline
● Central idea: Obtain a broadband realization of the LC resonators in the BPF without using inductors.
● Realize the LC resonant circuit by a circuit with C and a stub.
● Equate admittances and the derivatives of admittances
44
LC C0 Z 1Z 01
45
LC Z 1Z 01
Narrowband resonator equivalence at ω0
Z01 is the characteristic impedance of the transmission line and Z1is the input impedance of the shorted transmission line.
2 degrees of freedom. 2 degrees of freedom.
Can only match admittance at one frequency.
45
46
LC C0 Z 1Z 01
,
Z 1
Z 01
Broadband resonator equivalence at ω0Z01 is the characteristic impedance of the line and Z1 is the input impedance of the shorted line.2 degrees of freedom. 3 degrees of freedom.
0
inin 0
at
and at .YY
Broadband match is obtained
by matching
Yin Yin/
r is, the radian resonant frequency of the stub (i.e. the frequency at which it is /4 long).
46 47
LC C0 Z 1Z 01
,
Z 1
Z 01
Broadband resonator equivalence at ω0
2 degrees of freedom. 3 degrees of freedom.
0
inin 0
at
and at .YY
Broadband match is obtained
by matching
Yin Yin/
Specific design choice r 0 (most common).
The admittance of the networks are equivalent(at 0) when:
Also (at 0)Z1 = jZ01
,
Z 1
Z 01The derivatives of the admittance of the networks are equivalent(at 0) when:
47
Z01 is the characteristic impedance of the line and Z1 is the input impedance of the shorted line.
48
The transmission line stubs present impedances Z1 = jZ01, Z2 = jZ02, and Z3 = jZ03 since the resonant frequencies of the stubs are twice that of the design center frequency.
Step 5. Bandpass combline filter with broadband realization of lumped-
element inverters/ / / /
1 3/ /2
01 03
02
21.0881 pF
27.3181 pF7.54713
5.82598
C CCZ ZZ
/ // // / Z 01 Z 02 Z 032C1C C 3
50 50
Convert LC resonators to hybrid C‐stub resonators.
• The commensurate frequency, fr, of the design is the resonant frequency of the stubs.
• By default all the stubs have the same fr.
• The design choice here is that fr =2f0. f0 is the center frequency of the design.
48 49
Summary, Step 5
/ // // / Z 01 Z 02 Z 032C1C C 3
50 50
• Broadband, but stubs have different characteristic impedances.
• Really want them to be the same as they will be realized by microstrip lines and we want them to have the same width.(Kind of, this is a little imprecise as the inverters are yet to be realized.)
49
/ / / /1 3/ /2
01 03
02
21.0881 pF
27.3181 pF7.54713
5.82598
C CCZ ZZ
Case Study: Parallel Coupled-Line Combline Filter. Part H
Step 6: Equalize Stub Impedances
50
/ / // / // / Z 01 Z 02 Z 032C1C C 3
56.9084 56.9084
Outline● Result of Step 5 (previous):
● Broadband, but want stubs to have the same characteristic impedances.
● Result of this step (Step 6):
/ / / /1 3/ /2
01 03
02
21.0881 pF
27.3181 pF7.54713
5.82598
C CCZ ZZ
/ // // / Z 01 Z 02 Z 032C1C C 3
50 50
/ / // / // / Z 01 Z 02 Z 032C1C C 3
56.9084 56.9084
/ / / / / / /1 3 2
/01 03 02
21.0881 pF
7.54713
C C CZ Z Z
51
52
12
Target combline filter physical layout
/ / // / // / Z 01 Z 02 Z 032C1C C 3
56.9084 56.9084
Very approximately
52
Compare prototype with comblinenetwork model
53
• The impedances of the shunt stubs are mostly determined by the impedances of the individual microstrip lines.
• For manfacturability reasons we would like the microstrip lines to have the same width.
• Therefore we want the shunt stubs to have the same characteristic impedance.
• The impedances of the series stubs are mostly determined by the coupling of the individual microstrip lines.
/ / // / // / Z 01 Z 02 Z 032C1C C 3
56.9084 56.9084
Procedure/ / / /
1 3/ /2
01 03
02
21.0881 pF
27.3181 pF7.54713
5.82598
C CCZ ZZ
/ // // / Z 01 Z 02 Z 032C1C C 3
50 50
Want Z02 scaled so that new Z02 = Z01.
00
0y
2 31
J3J1 J J
0
Better to use admittance now as the analysis is based on building a nodal admittance matrix.
54 55
Element values are impedances except for y and y1, which are admittances.
x xy1
y = yx1
2
0 00
J3J1 J J
1 3
y J3J1
0
2 31
j/Jj/J
-j/J
j/J
-j/J
j/J
y = yx1
y1 J3J1
0
1 2 3
j/dj/d
-j/d -j/d
j/dj/d
Inverter impedance scaling
00
0y
2 31
J3J1 J J
0
Admittances are the same
if
Scaled original network
Original network
Procedure is:(A) Develop nodal admittance matrix of original network.(B) Develop nodal admittance matrix of scaled network.Then equate to find required parameters.
A
B
55
Jd = x
56
Realization of a series inductor as a shunt capacitor with 10 inverters.
1 nH
IMPEDANCE OR ADMITTANCEINVERTERS
y 11
1 nF
C
IMPEDANCE INVERTERS
10 10 1C
10 pF
xx
y = yx1
ADMITTANCE INVERTERS
1C= 0.1 S= 0.1 S
JJ
Example
Note the impact on the size of the capacitor!
56 57
Summary, step 6
/ / // / // / Z 01 Z 02 Z 032C1C C 3
56.9084 56.9084
/ / / / / / /1 3 2
/01 03 02
21.0881 pF
7.54713
C C CZ Z Z
• The stubs now have the same impedance, and the capacitances are the same.
• After scaling so that Z01 = Z02:
57
Case Study: Parallel Coupled-Line Combline Filter. Part I
Step 7: Inverter Realization
58
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
Outline● The prototype filter from Step 6 is
● Realize inverters using stubs.
● Combine adjacent stubs.
● Result of this step (Step 7):
/ / // / // / Z 01 Z 02 Z 032C1C C 3
56.9084 56.9084
59
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
60
Realization using short‐circuited stubs resonant at twice the passband center frequency.
56.9084
-j -j
j
56.910256.9084
56.9084
j
j
j
Inverter realization using stubs
Impedance inverter
Realization as alumped‐element circuit
Equivalence was established using ABCD parameters.
60
Inverter translation
/ / // / // / Z 01 Z 02 Z 032C1C C 3
56.9102 56.9102
/ / / / / / /1 3 2
/01 03 02
21.0881 pF
7.54713
C C CZ Z Z
j
j
j
j
-j
Z 01= 56.9102Z x
7.54713 56.9102
= 7.54713 Stubs can be combined.
61
62
j
-j
Z 01= 56.9084Z x
7.54713 56.9084
= 7.54713
-jj 7.54713 56.9084
j 8.70106
j
Z 01= 8.70106
8.70106
Combining stubs
Represent impedance as one stub.
Represent parallel stubs as parallel impedances.
Convert to a single impedance.
62 63
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
Bandpass filter prototype without inverters
/ / / / / / /1 3 2
/01 03
012 023/ /02
21.0881 pF
8.70106 56.9084
10.2715
C C CZ ZZ ZZ
fr = 2f0f0 is the center frequency of the design.
Note that in many designs fr = 2f0.This is simply assumed sometimes. But fr could have another relationship.
63
Compare prototype with comblinenetwork model
1
2
4
3
Combline section
Z
Z
012
1 2Z 011 022
64
Model:
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
Z
Z
012
1 2Z 011 022
Z
Z
012
1 2Z 011 022
C 1 C32C
Output
1 2
InputWith capacitors:
An issue with resonant frequency
● So the f0’s are different!
● What do we do?
● We need to re-examine the development the lead to the assignment of fr .
1
2
4
3
Combline section
( fr = f0 )
Z
Z
012
1 2Z 011 022
Model:
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
( fr = 2f0 )
Here f0 is the center frequency of the match.
Here f0 is the center frequency of the bandpass filter.
65
Consider exact network model of combline section
1
2
4
3
Combline section
66
There is nothing here that depends on the relationship of fr and f0 .
Z 0 oo21: 2
1: 21: 2
1: 2
V1
V2
2 :1
2 :1 2 :1
2 :1
V4
V3
VX
WV
ZV
YVWI YI
ZIIX
I4
I3I1
I2
Z 0 e e2
Exact model:
Reconsider network models of comblinesection1
2
4
3
Combline section
67
There is nothing here that depends on the relationship of fr and f0 .
Z 0 oo21: 2
1: 21: 2
1: 2
V1
V2
2 :1
2 :1 2 :1
2 :1
V4
V3
VX
WV
ZV
YVWI YI
ZIIX
I4
I3I1
I2
Z 0 e e2
Exact model:
1: n n
4
3
:1
1
201Z
02Z
Approximate model:This is believed to be most accurate when fr = f0 .That is, when the lines are /4 long at the operating frequency.But it is a reasonably good model all frequencies, even when it is /8 long .
Here f0 is the operating frequency
Reconsider simplified network model of combline section
1
2
4
3
Combline section
1: n n
4
3
:1
1
201Z
02Z
2
1: n
Z 02
Z 012
1
( fr = f0 ) Z1: n
02
Z011 2
Z
Z
012
1 2Z 011 022
68
Pretty good model even when fr = f0 (e.g when it is /8 long).
Compare prototype with comblinenetwork model
1
2
4
3
Combline section
Z
Z
012
1 2Z 011 022
69
Model:
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
Z
Z
012
1 2Z 011 022
Z
Z
012
1 2Z 011 022
C 1 C3
50
50 2COutput
1 2
Input
With capacitors:
These lines are /8 long at f0.
( fr = 2f0 )
70
Summary, Step 7
70
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
/ / / / / / /1 3 2
/01 03
012 023/ /02
21.0881 pF
8.70106 56.9084
10.2715
C C CZ ZZ ZZ
Case Study: Parallel Coupled-Line Combline Filter. Part J
Step 8:Scaling Characteristic Impedances of Stubs
71
t tt
tt
2C
Z0 23Z0 12
C 30Z t 1 0Z t 2 0Z t 3C 1
Outline● From Step 7
● Want the characteristic impedances of the shunt stubs to be between 30 and 80 .
/ / / / / / /1 3 2
/01 03
012 023/ /02
21.0881 pF
8.70106 56.9084
10.2715
C C CZ ZZ ZZ
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
72
Desired stub impedances
73
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
• The impedances of the shunt stubs are mostly determined by the impedances of the individual microstrip lines.
• For manfacturability reasons we would like the microstriplines to have reasonable width.
• On Alumina (r around 10) that means that we want the characteristic impedances of the stubs to be between 30 and 80 .
• The impedances of the series stubs are mostly determined by the coupling of the individual microstrip lines.
Scale Impedances
/ / / / / / /1 3 2
/01 03
012 023/ /02
21.0881 pF
8.70106 56.9084
10.2715
C C CZ ZZ ZZ
// / // // // / / Z 03C 3Z 022C1C Z 01
Z 012 Z 023
Scale to 80 .
Multiply impedances by a factor of 80/10.2715.
74
Want characteristic impedances of the stubs to be between 30 and 80 .
Scale Impedances/ / / / / / /
1 3 2/
01 03
012 023/ /02
21.0881 pF
8.70106 56.9084
10.2715
C C CZ ZZ ZZ
// / // // // / / Z03C3Z022C1C Z01
Z012 Z023
Multiply impedances by a factor of 80/10.2715.
t tt
tt
2C
Z0 23Z0 12
C 30Z t 1 0Z t 2 0Z t 3C 1
1 3 2/
0 1 03
0 12 0 23
0 2
2.70759 pF
67.7683 443.232
80
t t t
t
t t
t
C C CZ ZZ ZZ
75
76
Summary, Step 8
76
t tt
tt
2C
Z0 23Z0 12
C 30Z t 1 0Z t 2 0Z t 3C 1
1 3 2/
0 1 03
0 12 0 23
0 2
2.70759 pF
67.7683 443.232
80
t t t
t
t t
t
C C CZ ZZ ZZ
Case Study: Parallel Coupled-Line Combline Filter. Part K
Step 9: 50 Match
77
ttt
tt
139.404
139.404
Z0 23Z0 12
C 30Z t 1 0Z t 22C 0Z t 3C 1
78
Use Impedance Inverters
t tt
tt
2C
Z0 23Z0 12
C 30Z t 1 0Z t 2 0Z t 3C 1
Result of Step 8:
78
ttt
tt
139.404
139.404
Z0 23Z0 12
C 30Z t 1 0Z t 22C 0Z t 3C 1
79
Summary, Step 9
79
1 3 2/
0 1 03
0 12 0 23
0 2
2.70759 pF
67.7683 443.232
80
t t t
t
t t
t
C C CZ ZZ ZZ
ttt
tt
139.404
139.404
Z0 23Z0 12
C 30Z t 1 0Z t 22C 0Z t 3C 1
Case Study: Parallel Coupled-Line Combline Filter. Part L
Step 10: Implementing the Input/Output Inverters
80
t
tt Z0 23Z0 12
1C 3C
CbCb
0Z t 1 2C 0Z t 2 0Z t 3
Outline● From Step 9
● Implement input and output inverters.
81
ttt
tt
139.404
139.404
Z0 23Z0 12
C 30Z t 1 0Z t 22C 0Z t 3C 1
RL
Yin
C
C
a
b
K RL
Yin
An inverter as a capacitor network
K C
C
a
b
These are equivalent but only for resistive loads.
Equate admittances, note that Yin of inverter is real.
This is not the same as general matching which works with complex conjugate impedances.
It is the same as matching If input and output are resistances.
82
An inverter as a capacitor network
K C
C
a
b
These are equivalent but only for resistive loads.
This is can be shown by using a complex load and calculating the input impedance of the capacitive network with a complex load.
83
Derivation capacitor network (at 1 GHz)
84
in 22
50 1 1389.426 1139.540
La
b L
RY sCK sC R
1.06484 pF1.22170 pF
a
b
CC
C
C
a
b
389.426 50
inYLR
85
External inverters as capacitive networks
ttt
tt
139.404
139.404
Z0 23Z0 12
C 30Z t 1 0Z t 22C 0Z t 3C 1
Ca
Cb
Ca
Cb
85
Note that Ca and C1 are in parallel.
86
t
tt Z0 23Z0 12
1C 3C
CbCb
0Z t 1 2C 0Z t 2 0Z t 3
Summary, Step 10
1 1 3
2
0 1 0 3
0 12 0 23
0 2
1.64276 pF
2.70759 pF
1.22170 pF
67.7683 443.232
80
a t
t
b
t t
t t
t
C C C CCCZ ZZ ZZ
Case Study: Parallel Coupled-Line Combline Filter. Part M
Physical Design of Combline Filter
87
Cb CbC1 C3
50
50 t 2COutput
1 2
Input
Outline● From Step 10
● Capacitors stay as lumped-element capacitors● Implement the following in microstrip:
88
t
tt Z0 23Z0 12
1C 3C
CbCb
0Z t 1 2C 0Z t 2 0Z t 3
tt Z0 23Z0 12
0Z t 1 0Z t 2 0Z t 3
Key Concept
89
1
22
1
2
1
• Can treat three coupled lines as two pairsof coupled lines with the center line shared.
• Error is small.
• One transmission path that is missing is direct coupling of the first line to the third line. This coupling is very small.
90
w2w1
L
s1
w3w2
L
s2
Cb C1 C3t 2C Cb
w3w2w1
L
s s21
tt Z0 23Z0 12
0Z t 1 0Z t 2 0Z t 3
tZ0 12
0Z t 1 0Z t 2
t
0Z t 3
Z0 23
0Z t 2
Physical design of the three coupled lines
90 91
w2w1
L
s1
tZ0 12
0Z t 1 0Z t 2
Implement one pair at a time.
91
Equivalent circuits for a combline section.
Z011 Z022
Z012
21
1: n
Z 02
Z01
1
2
1
2
4
3
92
Derivation of parameters
92
Equivalent circuits for a combline section:
Z011 Z022
Z012
21
1: n
Z02
Z01
1
2
012
011
011 02202 012 01
011 022 012
11 7.540 0.1326
3342 and 69.17
Zn KZ n
Z ZZ nZ Z nZ Z Z
From model theory:
93
Derivation of parameters
93
Equivalent circuits for a combline section.1: n
Z02
Z01
1
2
1
2
4
3
Two estimates of coupled line system impedance:2
20 ,1 01 0 ,2 02 2
1 68.56 and 55.80 1
S SKZ Z K Z Z
K
This happened because the shunt stubs in the Pi arrangement of stubs is not symmetrical. So take mean:
0 0 ,1 0 ,2 63.8 S S SZ Z Z
94
Derivation of parameters
94
Equivalent circuits for a combline section.
1: n
Z02
Z01
1
2
1
2
4
3
0
0
0 0
20 0
63.8
1 55.8 1
/ 72.9 S
S
o S
e o
Z
nZ Zn
Z Z Z
Dimensions of microstriplines determined using tables or iteratively solving coupled line equations.
95
Cb C1 C3t 2C Cb
w3w2w1
L
s s21
Physical design of the three coupled lines
95
rh w
Conductor pattern
Stript
0
0
0
63.8 55.8 72.9
S
o
e
ZZZ
Choose alumina substrate with r = 10, h = 635 m.
Use lookup table for a 50 system impedance.
1 2 3
1 2
591 m (600 m rounded)635 m (650 m rounded)
7.24, 5.95
Take 6.56ee eo
e ee eo
w w ws s
96
Physical design
Cb C1 C3t 2C Cb
w3w2w1
L
s s21
r = 10, h = 635 mw1 = w2 = w3 = 600 ms1 = s2 = 650 mL = g/8 = 14.65 mm
w1
w2
w3
L
via
s
s1
2
Layout (to scale)
Capacitor values are unchanged (e.g. implement using surface‐mount capacitors).
96
0
0
30 cm @ 1 GHz
/g e
(recall fr = 2f0 )
97
Revisit Assumptions
Cb C1 C3t 2C Cb
w3w2w1
L
s s21
r = 10, h = 635 mw1 = w2 = w3 = 600 ms1 = s2 = 650 mL = 14.65 mm
These values were derived looking up a table for a50 system impedance.
However Z0S = 63.8
97
98
Revisit Assumptions, wCb C1 C3
t 2C Cb
w3w2w1
L
s s21
r = 10, h = 635 mw1 = w2 = w3 = 600 ms1 = s2 = 650 mL = 14.65 mmThese values were derived looking up a table for a50 system impedance.
Have three system impedances:
20 ,1 01
2
0 ,2 02 2
0 0 ,1 0 ,2
1 68.56
55.80 1
63.8
S
S
S S S
Z Z K
KZ ZK
Z Z Z
This choice mostly affects w1, w2, and w3.
Could optimize in EM simulation, but better to get closer now.
Choose Z0S = 55.8 .98 99
Update w
Cb C1 C3t 2C Cb
w3w2w1
L
s s21
r = 10, h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 650 mL = 14.65 mm
Use Z0S = 55.8 .
99
100
Revisit Assumption, LCb C1 C3
t 2C Cb
w3w2w1
L
s s21
r = 10, h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 1150 mL = 14.65 mm
For e , used geometric mean of even and odd mode effective permittivity (affects L).
100
LC C0 Z 1Z 01
Instead of adjusting L in EM‐based optimization (to get Z1 right) we can tune capacitor (C0) .
101
Summary, physical design
1 3
2
1.64276 pF
2.70759 pF
1.22170 pFt
b
C CCC
101
Cb C1 C3t 2C Cb
w3w2w1
L
s s21
Alumina (r = 10), h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 650 mL = 14.65 mm
Case Study: Parallel Coupled-Line Combline Filter. Part N
Microwave Circuit Simulation
102
Cb CbC1 C3
50
50 t2COutput
1 2
Input
Outline● Physical Design
● First developed lumped-element BPF reference● Microwave circuit simulation
– Use microstrip coupled line element (MCLIN)– Compare and interpret response– Optimize
103
Cb C1 C3t 2C Cb
w3w2w1
L
s s21
w1
w2
w3
L
via
s
s1
2
104
Z0 = 50 ΩC13 = C33 = 27.107 pFL13 = L33 = 934.47 pHC23 = 288.54 fFL23 = 87.787 nH
Z0
Vg
L13 C13
L23 C23
L33 C33
Z0
Lumped-Element BPF for Reference
10
20
30
40
60
50
S214
8
12
16
24
20
11S(d
B)
21S(d
B) S11
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5
0
Frequency (GHz)
0
104 105
0
Frequency (GHz)
60 24
20
10
30
40
50
4
8
12
16
20
11S(d
B)
S 21(d
B)
0.5 6.53.5
11S
S21
8.5
0
Wideband Response
Z0
Vg
L11 C11
L21 C21
L31 C31
Z0
105
106
0.80 GHz
1.20 GHz
0.90 GHz
0.94 GHz
0.92 GHz
1.10 GHz1.08 GHz
1.06 GHz1.02 GHz
0.98 GHz
S11 of the lumped-element BPF
Z0
Vg
L11 C11
L21 C21
L31 C31
Z0
10
20
30
40
60
50
S214
8
12
16
24
20
11S(d
B)
21S(d
B) S11
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5
0
Frequency (GHz)
0
106 107
The zeros of the S11response, and hence the poles of the S21response, are at 0.96, 1.00, and 1.04 GHz.
0.80 GHz
1.20 GHz
0.90 GHz
0.94 GHz
0.92 GHz
1.10 GHz
1.08 GHz
1.06 GHz1.02 GHz
0.98 GHz
0.96 GHz1.00 GHz1.04 GHz
S11 of the lumped-element BPF
Z0
Vg
L11 C11
L21 C21
L31 C31
Z0
107
zeros
108
0.80 GHz
1.20 GHz
0.90 GHz
0.94 GHz
0.92 GHz
1.10 GHz
1.08 GHz
1.06 GHz1.02 GHz
0.98 GHz
0.96 GHz1.00 GHz1.04 GHz
S11 of the lumped-element BPF
108
S214
8
12
16
24
20
11S(d
B)
S11
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5Frequency (GHz)
0
zeros
zeros109
Details: 6 μm gold metallization
Cb CbC1 C3
50
50 t2COutput
1 2
InputCb CbC1 C3
50
50 t 2COutput
1 2
MCLIN
Input
1 2 3
Circuit model using MCLIN element
1 3
2
1.64276 pF
2.70759 pF
1.22170 pFt
b
C CCC
Alumina (r = 10), h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 650 mL = 14.65 mm
109
110
Response with MCLIN element
10
20
30
40
60
50
S214
8
12
16
24
20
11S(d
B)
21S(d
B)
S11
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5
0
Frequency (GHz)
0
Response of lumped‐element BPF
110
(a) s1 = s2 = 650 m (b) s1 = s2 = 1150 m
20
10
30
40
50
4
8
12
16
20
11S(d
B)
S 21(d
B)
1.060 24
0
0.5 1.5
0
Frequency (GHz)
S21(a)
S21(b)
S11(b)
1.06 GHz
0.88 GHz
1.20 GHz
1.12 GHz
1.08 GHz
1.10 GHz
1.04 GHz
1.00 GHz0.98 GHz
0.96 GHz
0.94 GHz 0.92 GHz
0.90 GHz
1.02 GHz
0.80 GHz
111
S11 response with MCLIN element
Locus gets close to origin twice.
11S(d
B)
4
8
12
16
20
S11
0
0.5 1.0 1.5Frequency (GHz)
24
Frequency (GHz)
40
11 , MCLIN
S 21, Lumped
S21 , MCLIN60
20
10
30
S50
S21
(dB
)
0
4
8
12
16
20
11S
(dB
)
240.5 1.0 1.5
0
112
1 3
2
1.22170 pF
1.64276 pF
2.70759 pF
b
t
CC CC
Alumina (r = 10), h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 1150 mL = 14.65 mm
1 3
2
1.9076 pF
2.8748 pFt
C CC
Optimized S11 response with MCLIN element
Values for optimized response. Account for error in L.
Cb CbC1 C3
50
50 t 2COutput
1 2
Input
L
Frequency (GHz)
40
11 , MCLIN
S 21, Lumped
S21 , MCLIN60
20
10
30
S50
S21
(dB
)
0
4
8
12
16
20
11S
(dB
)
240.5 1.0 1.5
0S11 response with optimized MCLIN element
Cb CbC1 C3
50
50 t 2COutput
1 2
Input
Path 1Path 2
Path 2 not considered in synthesis.
At 1.2 GHz Path 1 and Path 2 cancel.
At 0.8 GHz Path 1 and Path 2 reinforce.
There is partial reinforcement below 0.9 GHz.
There is partial cancellation above 1.1 GHz.
113
1.06 GHz
0.94 GHz
0.92 GHz
0.90 GHz0.80 GHz
1.20 GHz1.04 GHz
1.02 GHz
1.00 GHz0.98 GHz
0.96 GHz
1.10 GHz1.08 GHz
114
S11 response with optimized MCLIN element
1.06 GHz
0.94 GHz
0.92 GHz0.90 GHz
0.80 GHz
1.20 GHz1.04 GHz1.02 GHz
1.00 GHz0.98 GHz
0.96 GHz
1.10 GHz1.08 GHz
115
Comparison of S11 response
0.80 GHz
1.20 GHz
0.90 GHz
0.94 GHz
0.92 GHz
1.10 GHz
1.08 GHz
1.06 GHz1.02 GHz
0.98 GHz
Lumped BPF BPF with optimized MCLIN
1.06 GHz
0.94 GHz
0.92 GHz0.90 GHz
0.80 GHz
1.20 GHz1.04 GHz1.02 GHz
1.00 GHz0.98 GHz
0.96 GHz
1.10 GHz1.08 GHz
116
S11 response with with optimized MCLIN
40
11 , MCLIN S21 , MCLIN60
20
10
30
S50
S21
(dB
)
0
4
8
12
16
20
11S
(dB
)
240.5 1.0 1.5
0
During manual tuning look at both rectangular S11 plot and Smith chart plot.
117
0
Frequency (GHz)
60 24
20
10
30
40
50
4
8
12
16
20
11S(d
B)
S 21(d
B)
0.5 6.53.5
11S
S21
8.5
0
Wideband response of lumped-element BPF
Z0
Vg
L11 C11
L21 C21
L31 C31
Z0
117
10
30
40
50
4
8
12
16
20
11S
(dB
)
S21
(dB
)
0.5 6.53.5
S 21
0
8.5
0
Frequency (GHz)
11S
60 24
20
118
Optimized S11 response with MCLIN element
Cb CbC1 C3
50
50 t 2COutput
1 2
Input
Recall that fr = 2f0.So transmission lines look the same atfr , 3fr , 5fr …, i.e. f0 , 6f0 , 10f0 …
BUT impedance of capacitance is not the same, hence the spurious passbands are shifted.
Spurious basebands at f0 , 4f0 , 7.5f0 , …
fr = 2f010
30
40
50
4
8
12
16
20
11S
(dB
)
S21
(dB
)
0.5 6.53.5
S 21
0
8.5
0
Frequency (GHz)
11S
60 24
20
119
Effect of higher fr ?
Cb CbC1 C3
50
50 t 2COutput
1 2
InputWhat if fr = 3f0?
Spurious passbandswould be shifted higher in frequency. BUT diminishing returns, design becomes more sensitive.
fr = 2f0
Frequency (GHz)
40
11 , MCLIN
S 21, Lumped
S21 , MCLIN60
20
10
30
S50
S21
(dB
)
0
4
8
12
16
20
11S
(dB
)
240.5 1.0 1.5
0
120
Alumina (r = 10), h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 1150 mL = 14.65 mm
1 3
2
1.22170 pF
1.9076 pF
2.8748 pF
b
t
CC CC
Summary, optimized physical design
Cb CbC1 C3
50
50 t 2COutput
1 2
Input
LOnly adjusted C1, Ct2, and C3.
Case Study: Parallel Coupled-Line Combline Filter. Part O
EM Simulation
121
Cb CbC1 C3
50
50 t2COutput
1 2
Input
400 μm× 400 μm tantalum vias6 μm gold metallization. EM enclosure has perfect conducting walls withXDIM = 22 mm, YDIM = 20 mm and height = 5.635 mm.
Cb CbC1 C3
50
50 t2COutput
1 2
Input
Cb CbC1 C3
50
50 t 2COutputInput
EM Subcircuit
231
1
3
2
w1
w2
w3
XDIM
YDIM
L
via
Enclosure
s
s1
2
122
Alumina (r = 10), h = 635 mw1 = w2 = w3 = 500 ms1 = s2 = 1150 mL = 14.65 mm
1 3 2
1.22170 pF
1.9076 pF , 2.8748 pFb
t
CC C C
Use optimized MCLIN‐based BPF values
S 21
(dB
)
10
20
30
40
60
50
0
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5
4
8
12
16
24
20S11, EM
S21, MCLIN
S21, EM 11S(d
B)
Frequency (GHz)
0
123
Comparison of responses
Could further optimize . . .
S 21
(dB
)
10
20
30
40
60
50
0
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5
4
8
12
16
24
20S11, EM
S21, MCLIN
S21, EM 11S(d
B)
Frequency (GHz)
0
124
Comparison of responses
• Bandwidth is smaller• Indicates lower overall coupling
• Notch above passband has shifted lower.
• Overall response is almost the same as with MCLIN‐based analysis
• Perhaps slight mismatch at center of passband.
• Use MCLIN‐based analysis to optimize design.
• Gridding in EM analysis (50 m used here could have resulted in EM analysis differences).
• Some subtle effects are captured in EM Simulation not in MCLIN analysis
• E.G. via coupling.
S11 response on a Smith chart
EM Analysis
0.80 GHz
1.00 GHz
0.98 GHz
0.96 GHz
0.94 GHz
0.92 GHz
1.08 GHz
0.90 GHz
1.06 GHz
1.20 GHz
1.04 GHz
1.02 GHz
1.10 GHz
1.06 GHz
0.94 GHz
0.92 GHz
0.90 GHz0.80 GHz
1.20 GHz1.04 GHz
1.02 GHz
1.00 GHz0.98 GHz
0.96 GHz
1.10 GHz1.08 GHz
126
S11 response (optimized)
MCLIN Analysis
S11 response (optimized)
MCLIN Analysis
EM Analysis
127
128
S11 response on a Smith chart
S 21
(dB
)
10
20
30
40
60
50
0
1.00.90.80.70.60.5 1.1 1.31.2 1.4 1.5
4
8
12
16
24
20S11, EM
S21, MCLIN
S21, EM 11S(d
B)
Frequency (GHz)
0
129
Wideband response
Frequency (GHz)
60 24
20
10
30
40
50
4
8
12
16
20
11S(d
B)
S 21(d
B)
0.5 6.53.5
S21
11S
0
8.5
0
Same as with MCLIN‐based analysis
130
Manufactured filter considerations
Cb CbC1 C3
50
50 t 2COutput
1 2
Input
L
w1
w2
w3
XDIM
YDIM
L
via
Enclosure
s
s1
2
It will be necessary to tune every filter manufactured.
Fabrication tolerances are about 1%.
Greater accuracy than that is required.
Tuning done by adjusting capacitor values.
131
Summary, Parallel Coupled-Line Combline Filter
Cb CbC1 C3
50
50 t2COutput
1 2
Input
L
Filter synthesized using a methodical process.
Microwave simulation required to optimize design.
EM simulation as a check as there are coupling mechanisms that cannot be captured otherwise.
Every filter manufactured will require tuning.