Case Study: Parallel Coupled- Line Combline Filter

MICROWAVE AND RF DESIGN MICROWAVE AND RF DESIGN Case Study: Parallel Coupled- Line Combline Filter Presented by Michael Steer Reading: 6. 6.4 Index: CS_PCL_Filter Based on material in Microwave and RF Design: A Systems Approach, 2 nd Edition, by Michael Steer. SciTech Publishing, 204. Presentation copyright Michael Steer

Case Study: Parallel Coupled-Line Combline Filter 0 0 Input 50 C C b C t2 C 3 C b 2 Output 50 S 2 (db) 0 20 30 40 4 8 2 6 S (db) 50 S S 2 20 ADA Copyright 203 M. Steer and IET 60 0.5.0.5 Frequency (GHz) 24

Bandpass filter Specifications Center frequency: GHz 0% Bandwidth Steep filter skirts requires Chebyshev response, choose a ripple factor of 0. 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 topology Parallel microstrip lines in combline configuration. Input 50 Output 0 2 50 db -20-40 S 2-60 -80 2 30 Frequency (GHz) 4

Science: synthesis procedure How to go from C 2 V g L L 2 C L 3 C 3 to Input 50 Output 2 50 using mathematical synthesis, while maintaining desired electrical characteristics. (perhaps more complicated) 5

Optimization: an alternative to synthesis Input 50 Output 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. 2 50 6

Summary 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 Filter design is based on circuit transformations. C 2 V g L L 2 C L 3 C 3 8

Outline Begin with a lumped element filter. C 2 0 L 22 C 2 V g L L 2 C L 3 C 3 or V g 0 L 2 C 2 Consider circuit model of coupled lines L 32 C 32 Work out the steps to go from the lumped element circuit to a transmission line-based circuit. 9

Third-order filter The lumped element filter has three LC resonators: So (perhaps) the transmission line equivalent has three resonators: C 2 V g L L 2 C L 3 C 3 Consider as two pairs of coupled lines: 2 2 0

Network model of a pair of coupled lines The equivalent circuit of a pair of coupled lines 4 2 3 Is obtained by equating symbolic ABCD equations : n 02 n : 4 2 0 3

Combline section and network models : n n: 4 02 4 2 3 Combline section 2 0 3 : n n: 02 4 2 0 3 2

Combline section and network models : n n: 4 02 4 2 3 Combline section 2 0 3 ( f r = f 0 ) 02 : n 2 0 2 : n 02 2 0 02 2 0 022 3

Translation of a circuit with stubs to coupled lines So if the following structure is seen in a circuit (a Pi arrangement of shorted stubs) 02 2 0 022 Then it can be replaced by a combline section 2 4 3 4

Comparison of lumped element filter and Pi arrangement of stubs Pi arrangement of shorted stubs. 02 2 0 022 C 2 Three connected resonators. V g L L 2 C L 3 C 3 Each stub is a resonator. A shorted stub corresponding to a parallel LC resonator. An open circuit stub corresponds to a series LC resonator. 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 5

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. Want basic circuit structure to be 02 But cannot start from here (3rd order BPF) 2 0 022 (Model of two PCL in combline configuration.) 6

Case Study: Parallel Coupled-Line Combline Filter. Part C, Step : Develop Lowpass Prototype Filter L 2 V g C C 3 7

Outline Begin with a lumped element filter. L 2 V g C C 3 Calculate element values. 8

This is the 6 th order Chebyshev response ε is called the ripple factor. Passband ripple, PBR = (+ ε 2 ) Ripple in db, R db = 0 log(pbr) TRANSMISSION 2 REFLECTION 2 Steeper filter skirt for Higher order Larger ripple 9

Third-order Chebyshev filter Coefficients of a Chebyshev lowpass prototype filter normalized to a radian corner frequency of ω 0 = rad/s and a Ω system impedance (i.e., g 0 = = g n+ ). The ripple factor is ε. ε = 0. is a ripple of 0.0432 db. ω 0 is the radian frequency at which the transmission response of a Chebyshev filter is down by the ripple. Here ω 0 = radian/s. 20

A third-order Chebyshev lowpass filter prototype L 2 V g C C 3 g 0 = g = 0.8558 g 2 =.036 g 3 = 0.8558 g 4 = C = 0.8558 F L 2 =.036 H C 3 = 0.8558 F ω 0 = rad/s 2

Chebyshev filter coefficients from recursive formula 22

Summary Step : developed 3rd order Chebychev lowpass prototype filter. V g L 2 C C 3 ω 0 = rad/s C = 0.8558 F L 2 =.036 H C 3 = 0.8558 F 23

Case Study: Parallel Coupled-Line Combline Filter. Part D, Step 2: Remove Series Inductor L 2 Vg C C 3 C C 2 C 3 24

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

Impedance inverter K A /4 long line is an inverter in Inverters K L in K 2 L in 0 = K Lossless telegrapher s equation: L j0tan( ) in 0 /2; tan( ) j tan( ) 0 in 0 j 0 j L tan( ) tan( ) L in 2 0 L L 0= K An inverter can be realized using a transmission line. Wherever there are /4 long transmission lines an impedance inverter can be realized (probably). 26

Consider combline section and network model :n n: 4 02 4 2 Combline section 3 2 0 3 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 -: K C K L CK 2 Equivalence is demonstrated using ABCD parameters. L T L sl 0 C T 0 sc T K T For the cascade CASCADE 2 2 3 2 0 jk j/ K 0 -: T 3 0 0 sl 2 0 jk 0 0 jk 0 sck T T T T j/ K 0 sc j/ K 0 0 0 28

Equivalence of a series inductor and a shunt capacitor plus inverters L -: K C K L CK 2 K C K Drop negative unity transformer as it only affects phase and not filter response. 29

Inverter form of lowpass prototype filter L 2 V g C C 3 C C 2 C 3 ω 0 = rad/s ω 0 = rad/s C = 0.8558 F L 2 =.036 H C 3 = 0.8558 F C = 0.8558 F C 2 =.036 F C 3 = 0.8558 F 30

Ladder prototype filters using impedance inverters g 0 g g 3 g n ( n odd) g V g g 2 g 4 gn ( neven) n+ -: -: -: Vg g g 2 g 3 g 4 Vg g g 2 g 3 g 4 g 5 3

Summary: Inverter form of lowpass prototype filter C C 2 C 3 ω 0 = rad/s C = 0.8558 F C 2 =.036 F C 3 = 0.8558 F 32

Case Study: Parallel Coupled-Line Combline Filter. Part E, Step 3: Bandpass Transformation C C 2 C 3 V g C' L' C' L' C' L' 2 2 3 3 33

First lumped element transformation to BPF, GHz L 2 0 L 22 C 2 Vg C C 3 V g 0 L 2 C 2 L 32 C 32 T(s) 2 T ( s) 2 LPF HPF (rad/s) + 0 2 (rad/s) 34

BPF and center frequency transformation L 2 0 L 22 C 2 Vg C C 3 V g 0 L 2 C 2 L 32 C 32 950MHz, 2 2 2 2 0 2 2 fractional bandwidth, C 0 2 000 MHz C, L 2 2 0 C 0 050MHz R db This assumes that the LPF corner frequency is radian/s. 35

Transformation to BPF, GHz V g L 22 C 22 C 2 =.35533 nf = C 3 L 2 = 8.6894 ph = L 32 C 22 = 4.427 pf L 22 =.75573 nh L 2 C 2 L 32 C 32 0 0 S 2 (db) 0 20 30 S S 2 4 8 2 S (db) 40 6 50 20 60 24 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) 36

BPF and center frequency transformation V g C C 2 C 3 C' L' C' L' C' L' 2 2 3 3 950MHz, 2 2 2 050MHz R db 2 0 2 2 fractional bandwidth, C 0 2 000 MHz C, L / / 0 C 0 This assumes that the LPF corner frequency is radian/s. 37

V g Prototype BPF and center frequency transformation C' L' C' L' C' L' 2 2 3 3 C = 355.33 pf = C 3 / / L = 0.086894 nh = L 3 / C 2 = 755.73 pf / L 2 = 0.04427 nh Recall: desired basic circuit structure 02 / / 2 0 022 (Model of two PCL in combline configuration.) 38

Summary prototype BPF V g C' L' C' L' C' L' 2 2 3 3 C = 355.33 pf = C 3 / L = 0.086894 nh = L 3 C 2 = 755.73 pf / / / L 2 = 0.04427 nh / / 39

Case Study: Parallel Coupled-Line Combline Filter. Part F, Step 4: Impedance Scaling V g C' L' C' L' C' L' 2 2 3 3 V g 50 50 50 50 C L C L C L 2 2 3 3 40

Principle of impedance scaling Every impedance in the circuit is scaled by the same amount So to go from 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. 4

Summary, Step 4: BPF scaled to 50. / C = 355.33 pf = C 3 / V g C' L' C' L' C' L' 2 2 3 3 / L = 0.086894 nh = L 3 C 2 = 755.73 pf / / L 2 = 0.04427 nh / V g 50 50 50 50 C L C L C L 2 2 3 3 C = 27.066 pf = C 3 L = 0.934468 nh = L 3 C 2 = 35.47 pf L 2 = 0.72359 nh 42

Case Study: Parallel Coupled-Line Combline Filter. Part G, Step 5: Conversion of Lumped-Element Resonators 0, L C C 0 0 43

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. L C C 0 0 Equate admittances and the derivatives of admittances 44

Narrowband resonator equivalence at ω 0 L C 0 2 degrees of freedom. 2 degrees of freedom. Can only match admittance at one frequency. 0 is the characteristic impedance of the transmission line and is the input impedance of the shorted transmission line. 45

Broadband resonator equivalence at ω 0 Y in C L C 0 0 0 2 degrees of freedom. 3 degrees of freedom. / Y in, 0 is the characteristic impedance of the line and is the input impedance of the shorted line. Broadband match at is obtained by matching Y in in 0 0 Y and at. r is, the radian resonant frequency of the stub (i.e. the frequency at which it is /4 long). 46

Broadband resonator equivalence at ω 0 Y in C L C 0 0 0 2 degrees of freedom. 3 degrees of freedom. / Y in, 0 is the characteristic impedance of the line and is the input impedance of the shorted line. Broadband match at is obtained by matching Y in in 0 0 Y and at. Specific design choice r 0 (most common). The admittance of the networks are equivalent (at 0 ) when: The derivatives of the admittance of the networks are equivalent (at 0 ) when: Also (at 0 ) = j 0 0, 47

Step 5. Bandpass combline filter with broadband realization of lumpedelement inverters Convert LC resonators to hybrid C stub resonators. 50 50 / / / / / / 0 C 2 02 03 C C 3 C C // // 3 // 2 0 03 02 2.088 pf 27.38 pf 7.5473 5.82598 C The commensurate frequency, f r, of the design is the resonant frequency of the stubs. By default all the stubs have the same f r. The transmission line stubs present impedances = j 0, 2 = j 02, and 3 = j 03 since the resonant frequencies of the stubs are twice that of the design center frequency. The design choice here is that f r = 2f 0. f 0 is the center frequency of the design. 48

Summary, Step 5 50 50 / / / / / / 0 C 2 02 03 C C 3 Broadband, but stubs have different characteristic impedances. C C // // 3 // 2 27.38 pf 0 03 02 2.088 pf 7.5473 5.82598 C 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

Case Study: Parallel Coupled-Line Combline Filter. Part H Step 6: Equalize Stub Impedances 56.9084 56.9084 / / / / / / / / 0 C 2 02 03 C C 3 50

Outline Result of Step 5 (previous): 50 50 / / / / / / 0 C 2 02 03 C C 3 C C // // 3 // 2 0 03 02 2.088 pf C 27.38 pf 7.5473 5.82598 Broadband, but want stubs to have the same characteristic impedances. Result of this step (Step 6): 56.9084 56.9084 / / / / / / / / 0 C 2 02 03 C C 3 C 2.088 pf C C // // /// 3 2 7.5473 / 0 03 02 5

Target combline filter physical layout 56.9084 56.9084 / / / / / / / / 0 C 2 02 03 C C 3 Very approximately 2 52

Compare prototype with combline network model 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. 56.9084 56.9084 / / / / / / / / 0 C 2 02 03 C C 3 53

Want 02 scaled so that new 02 = 0. Procedure 50 50 / / / / / / 0 C 2 02 03 C C 3 C C // // 3 // 2 0 03 02 2.088 pf 27.38 pf 7.5473 5.82598 C 2 3 J J y J 0 J 3 0 0 0 Better to use admittance now as the analysis is based on building a nodal admittance matrix. 54

Inverter impedance scaling A 2 3 -j/j -j/j 2 3 J J y J 0 J 3 J j/j j/j y j/j J 3 j/j 0 0 0 Original network 2 3 B -j/d 0 -j/d 2 3 Admittances are the same if d = J x J J x J Element values are impedances except for y and y, which are admittances. y 0 0 0 y = yx Scaled original network x J 3 J j/d j/d y j/d J 3 j/d 0 y = yx Procedure is: (A) Develop nodal admittance matrix of original network. (B) Develop nodal admittance matrix of scaled network. Then equate to find required parameters. 55

Example Realization of a series inductor as a shunt capacitor with 0 inverters. nh IMPEDANCE OR ADMITTANCE INVERTERS y C nf ADMITTANCE INVERTERS J x = 0. S C y = yx J x = 0. S IMPEDANCE INVERTERS Note the impact on the size of the capacitor! 0 C 0 pf 0 56

Summary, step 6 After scaling so that 0 = 02 : 56.9084 56.9084 / / / / / / / / 0 C 2 02 03 C C 3 C 2.088 pf C C // // /// 3 2 7.5473 / 0 03 02 The stubs now have the same impedance, and the capacitances are the same. 57

Case Study: Parallel Coupled-Line Combline Filter. Part I Step 7: Inverter Realization 02 023 / / / / / / / / / / / C 0 C 2 02 C 3 03 58

Outline The prototype filter from Step 6 is 56.9084 56.9084 / / / / / / / / 0 C 2 02 03 C C 3 Realize inverters using stubs. Combine adjacent stubs. Result of this step (Step 7): 02 023 / / / / / / / / / / / C 0 C 2 02 C 3 03 59

Inverter realization using stubs j 56.9084 j 56.9084 -j56.9084 -j 56.902 j j Impedance inverter Realization as a lumped element circuit Realization using shortcircuited stubs resonant at twice the passband center frequency. Equivalence was established using ABCD parameters. 60

Inverter translation 56.902 56.902 / / / / / / / / 0 C 2 02 03 C C 3 C 2.088 pf C C // // /// 3 2 7.5473 / 0 03 02 j j j j7.5473 56.902 -j 0 x = 7.5473 = 56.902 Stubs can be combined. 6

Combining stubs j7.5473 56.9084 -j 0 x = 7.5473 = 56.9084 j 7.5473 -j56.9084 Represent parallel stubs as parallel impedances. j 8.7006 Convert to a single impedance. j 8.7006 0 = 8.7006 Represent impedance as one stub. 62

Bandpass filter prototype without inverters 02 023 / / / / / / / / / / / C 0 C 2 02 C 3 03 C 2.088 pf C C // // /// 3 2 / 0 03 02 023 // 02 8.7006 56.9084 0.275 f r = 2f 0 f 0 is the center frequency of the design. Note that in many designs f r = 2f 0. This is simply assumed sometimes. But f r could have another relationship. 63

Compare prototype with combline network model 4 02 023 2 3 Combline section / / / / / / / / / / / C 0 C 2 02 C 3 03 02 02 Model: 02 2 0 022 2 0 022 2 0 022 With capacitors: Input C C 2 C 3 2 Output 64

An issue with resonant frequency 4 02 023 2 3 Combline section / / / / / / / / / / / C 0 ( f r = 2f 0 ) C 2 02 C 3 03 Model: 02 Here f 0 is the center frequency of the bandpass filter. 2 ( f r = f 0 ) 0 022 Here f 0 is the center frequency of the match. So the f 0 s are different! What do we do? We need to re-examine the development the lead to the assignment of f r. 65

Consider exact network model of combline section 4 Exact model: 2 Combline section 3 I V : 2 : 2 IW V W 2 0 e e V Y IY 2: 2: V 3 I 3 I 2 V 2 : 2 : 2 I X V X 2 0 o o V I 2: 2: V 4 I4 There is nothing here that depends on the relationship of f r and f 0. 66

2 Reconsider network models of combline Combline section Here f 0 is the operating frequency There is nothing here that depends on the relationship of f r and f 0. 4 3 V section Exact model: I I 2 V 2 : 2 : 2 : 2 : 2 IW V W I X V X 2 0 e e 2 0 o o V Y V IY I 2: 2: 2: 2: V 3 V 4 I 3 I4 This is believed to be most accurate when f r = f 0. 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. Approximate model: : n n: 02 0 2 4 3 67

Reconsider simplified network model of combline section : n n: 4 02 4 2 3 Combline section 2 0 3 ( f r = f 0 ) 02 : n 2 0 2 : n 02 2 0 02 2 Pretty good model even when f r = f 0 0 022 (e.g when it is /8 long). 68

Compare prototype with combline network model 4 02 023 2 3 Combline section / / / / / / / / / / / C 0 C 2 02 C 3 03 02 02 Model: 02 2 0 022 2 0 022 2 0 022 With capacitors: Input Output 50 C C 2 C 3 2 50 ( f r = 2f 0 ) These lines are /8 long at f 0. 69

Summary, Step 7 02 023 / / / / / / / / / / / C 0 C 2 02 C 3 03 C 2.088 pf C C // // /// 3 2 / 0 03 02 023 // 02 8.7006 56.9084 0.275 70

Case Study: Parallel Coupled-Line Combline Filter. Part J Step 8:Scaling Characteristic Impedances of Stubs 0 t2 0 t23 Ct 0 t Ct 2 0 t 2 Ct3 0 t 3 7

Outline From Step 7 02 023 / / / / / / / / / / / C 0 C 2 02 C 3 03 C 2.088 pf C C // // /// 3 2 / 0 03 02 023 // 02 8.7006 56.9084 0.275 Want the characteristic impedances of the shunt stubs to be between 30 and 80. 72

Desired stub impedances The impedances of the shunt stubs are mostly determined by the impedances of the individual microstrip lines. 02 023 For manfacturability reasons we would like the microstrip lines to have reasonable width. / / / / / / / / / / / C 0 C 2 02 C 3 03 On Alumina ( r around 0) 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. 73

Scale Impedances 02 023 / / / / / / / / / / / C 0 C 2 02 C 3 03 C 2.088 pf C C // // /// 3 2 / 0 03 02 023 // 02 8.7006 56.9084 0.275 Want characteristic impedances of the stubs to be between 30 and 80. Scale to 80. Multiply impedances by a factor of 80/0.275. 74

Scale Impedances C / / / / / / / / / / / 0 02 C 2 02 023 Multiply impedances by a factor of 80/0.275. C 3 03 C 2.088 pf C C // // /// 3 2 / 0 03 02 023 // 02 8.7006 56.9084 0.275 0 t2 0 t23 Ct 0 t Ct 2 0t 2 Ct 3 0t 3 C 2.70759 pf C C t t3 t2 / 0 t 03 02 t 023 t 02 t 67.7683 443.232 80 75

Summary, Step 8 0 t2 0 t23 Ct 0 t Ct 2 0t 2 Ct 3 0t 3 C 2.70759 pf C C t t3 t2 / 0 t 03 02 t 023 t 02 t 67.7683 443.232 80 76

Case Study: Parallel Coupled-Line Combline Filter. Part K Step 9: 50 Match 0 t2 0 t23 39.404 39.404 Ct 0 t Ct 2 0t 2 Ct 3 0 t 3 77

Use Impedance Inverters Result of Step 8: 0 t2 0 t23 Ct 0 t Ct 2 0t 2 Ct 3 0t 3 0 t2 0 t23 39.404 39.404 Ct 0 t Ct 2 0t 2 Ct 3 0 t 3 78

Summary, Step 9 0 t2 0 t23 39.404 39.404 Ct 0 t Ct 2 0t 2 Ct 3 0 t 3 C 2.70759 pf C C t t3 t2 / 0 t 03 02 t 023 t 02 t 67.7683 443.232 80 79

Case Study: Parallel Coupled-Line Combline Filter. Part L Step 0: Implementing the Input/Output Inverters C b 0 t2 0 t23 C b C 0 t Ct 2 0 t 2 3 0t 3 C 80

Outline From Step 9 0 t2 0 t23 39.404 39.404 Ct 0 t Ct 2 0t 2 Ct 3 0 t 3 Implement input and output inverters. 8

An inverter as a capacitor network C b K C a These are equivalent but only for resistive loads. Equate admittances, note that Y in of inverter is real. C b K R L C a R L Y in Y in 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 C b K C a 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 GHz) 389.426 50 C a C b R L Y in Y R 50 sc K sc R L in 2 2 a 39.540 389.426 b L C C a b.06484 pf.2270 pf 84

External inverters as capacitive networks 0 t2 0 t23 39.404 39.404 Ct 0 t Ct 2 0t 2 Ct 3 0 t 3 C b C b Ca Ca Note that C a and C are in parallel. 85

Summary, Step 0 C b 0 t2 0 t23 C b C 0 t Ct 2 0 t 2 3 0t 3 C C C C.64276 pf C C C a t 3 t 2 b 0 t 03 t 02 t 023 t 02 t 2.70759 pf.2270 pf 67.7683 443.232 80 86

Case Study: Parallel Coupled-Line Combline Filter. Part M Physical Design of Combline Filter Input Output 50 C Ct b C 2 C C 3 b 2 50 87

Outline From Step 0 C b 0 t2 0 t23 C b C 0 t Ct 2 0 t 2 3 0t 3 C Capacitors stay as lumped-element capacitors Implement the following in microstrip: 0 t2 0 t23 0 t 0t 2 0 t 3 88

Key Concept 2 2 Can treat three coupled lines as two pairs of coupled lines with the center line shared. 2 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. 89

Physical design of the three coupled 0 t2 0 t23 lines C Ct 2 b C C 3 C b L 0 t 0t 2 0 t 3 s s 2 w w 2 w 3 0 t2 0 t23 L L 0 t 0t 2 s 0 t 2 0 t 3 s 2 w w 2 w 2 w 3 90

Implement one pair at a time. 0 t2 L 0 t 0t 2 s w w 2 Equivalent circuits for a combline section. : n 02 02 4 0 022 2 2 0 2 3 9

Derivation of parameters Equivalent circuits for a combline section: : n 02 02 0 022 2 2 0 From model theory: n K n 02 7.540 0.326 0 0 022 02 n02 3342 and 0 n 69.7 0 022 02 92

Derivation of parameters Equivalent circuits for a combline section. : n 02 4 2 0 2 3 Two estimates of coupled line system impedance: 2 2 K 0 S,0 K 68.56 and 0 S,202 55.80 2 K This happened because the shunt stubs in the Pi arrangement of stubs is not symmetrical. So take mean: 0S 0 S, 0 S,2 63.8 93

Derivation of parameters Equivalent circuits for a combline section. : n 02 4 2 0 2 3 0S 63.8 n 55.8 n 0o 0S / 72.9 2 0e 0o 0 S Dimensions of microstrip lines determined using tables or iteratively solving coupled line equations. 94

Physical design of the three coupled lines C C t 2 b C C 3 C b Conductor pattern L t Strip s s 2 h w r w w 2 w 3 Choose alumina substrate with r = 0, h = 635 m. 0S 0o 0e 63.8 55.8 72.9 Use lookup table for a 50 system impedance. w w2 w3 59 m (600 m rounded) s s2 635 m (650 m rounded) 7.24, 5.95 ee eo Take 6.56 e ee eo 95

Physical design C Ct 2 b C C 3 C b 30 cm @ GHz 0 / g 0 e L s s 2 r = 0, h = 635 m w = w 2 = w 3 = 600 m s = s 2 = 650 m L = g /8 = 4.65 mm (recall f r = 2f 0 ) w w 2 w 3 Capacitor values are unchanged (e.g. implement using surface mount capacitors). Layout (to scale) L via w 3 s2 w 2 s w 96

Revisit Assumptions L C Ct 2 b C C 3 C b r = 0, h = 635 m w = w 2 = w 3 = 600 m s = s 2 = 650 m L = 4.65 mm w s s 2 w 2 w 3 These values were derived looking up a table for a 50 system impedance. However 0S = 63.8 97

Revisit Assumptions, w L C Ct 2 b C C 3 w s s 2 w 2 Have three system impedances: 0 S, 0 0 S,2 02 0S 0 S, 0 S,2 w 3 C b 2 K 68.56 2 K 55.80 2 K 63.8 r = 0, h = 635 m w = w 2 = w 3 = 600 m s = s 2 = 650 m L = 4.65 mm These values were derived looking up a table for a 50 system impedance. This choice mostly affects w, w 2, and w 3. Could optimize in EM simulation, but better to get closer now. Choose 0S = 55.8. 98

Update w L C Ct 2 b C C 3 s s 2 C b Use 0S = 55.8. r = 0, h = 635 m w = w 2 = w 3 = 500 m s = s 2 = 650 m L = 4.65 mm w w 2 w 3 99

Revisit Assumption, L L C Ct 2 b C C 3 C b r = 0, h = 635 m w = w 2 = w 3 = 500 m s = s 2 = 50 m L = 4.65 mm w s s 2 w 2 w 3 For e, used geometric mean of even and odd mode effective permittivity (affects L). optimization (to get right) we can tune capacitor (C 0 ). Instead of adjusting L in EM based C L C 0 0 00

Summary, physical design C.64276 pf C 3 C Ct 2 b C C 3 C b C t 2 2.70759 pf C b.2270 pf L s s 2 Alumina ( r = 0), h = 635 m w = w 2 = w 3 = 500 m s = s 2 = 650 m L = 4.65 mm w w 2 w 3 0

Case Study: Parallel Coupled-Line Combline Filter. Part N Microwave Circuit Simulation Input Output 50 C C b C t2 C C 3 b 2 50 02

Physical Design L Outline w 3 s2 w 2 s w L C Ct 2 b C C 3 s s 2 C b First developed lumped-element BPF reference Microwave circuit simulation w w 2 Use microstrip coupled line element (MCLIN) Compare and interpret response Optimize via w 3 03

Lumped-Element BPF for Reference 0 = 50 Ω C 3 = C 33 = 27.07 pf L 3 = L 33 = 934.47 ph C 23 = 288.54 ff L 23 = 87.787 nh S 2 (db) 0 0 20 30 40 50 V g 0 L 3 C 3 S L 23 C 23 L 33 C 33 0 S 2 4 8 2 6 20 S (db) 0 60 24 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) 04

Wideband 0 L 2 C 2 Response V g 0 L C L 3 C 3 0 0 0 4 S 2 (db) 20 30 40 S 2 8 2 6 S (db) 50 20 S 60 24 0.5 3.5 6.5 8.5 Frequency (GHz) 05

S of the lumpedelement BPF V g 0 L 2 C 2 0 0.80 GHz 0.90 GHz 0.92 GHz 0.94 GHz 0.98 GHz.02 GHz.06 GHz L C L 3 C 3.20 GHz 0 0.0 GHz.08 GHz S 2 (db) 0 20 30 S S 2 4 8 2 S (db) 40 6 50 20 60 24 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) 06

S of the lumped-element 0.90 GHz 0.92 GHz BPF 0.80 GHz The zeros of the S response, and hence the poles of the S 2 response, are at 0.96,.00, and.04 GHz. 0.98 GHz.02 GHz 0.96 GHz.00 GHz.04 GHz 0.94 GHz.06 GHz 0 L 2 C 2.20 GHz.08 GHz V g 0.0 GHz zeros L C L 3 C 3 07

S of the 0.90 GHz 0.92 GHz lumped-element BPF 0.80 GHz 0.98 GHz 0.96 GHz.00 GHz.04 GHz 0.94 GHz.02 GHz.06 GHz 0 S S 2 4 24 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) zeros 8 2 6 20 S (db).20 GHz.0 GHz.08 GHz zeros 08

Circuit model using MCLIN element Input Output 50 Cb Ct2 C C 3 Cb 2 50 Input Output 50 Cb Ct C 2 C C 3 b 2 50 2 3 C C C 3 t 2 b.64276 pf 2.70759 pf.2270 pf C Alumina ( r = 0), h = 635 m w = w 2 = w 3 = 500 m s = s 2 = 650 m L = 4.65 mm Details: 6 μm gold metallization MCLIN 09

Response with MCLIN element S 2 (db) 0 0 20 30 0 4 8 2 S (db) (a) s = s 2 = 650 m (b) s = s 2 = 50 m 40 50 (b) S (b) S 2 (a) S 2 6 20 60 0.5.0 24.5 Frequency (GHz) Response of lumpedelement BPF S 2 (db) 0 0 20 30 40 50 S 0 S 2 4 8 2 6 20 S (db) 60 24 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) 0

S response with MCLIN element.0 GHz.2 GHz.20 GHz Locus gets close to origin twice..08 GHz.06 GHz 0.98 GHz.00 GHz.04 GHz.02 GHz 0 4 0.96 GHz S 8 2 6 S (db) 0.94 GHz 0.92 GHz 0.90 GHz 0.88 GHz 0.80 GHz 20 24 0.5.0.5 Frequency (GHz)

Optimized S response with MCLIN element Alumina ( r = 0), h = 635 m w = w 2 = w 3 = 500 m s = s 2 = 50 m L = 4.65 mm C C C b S 2 (db) 0 0 20 30 40 50 3 t 2.2270 pf.64276 pf 2.70759 pf Input Output 50 C Ct b C 2 C C 3 b 2 50 L S, MCLIN S 2, Lumped S 2, MCLIN 60 0.5.0.5 Frequency (GHz) C C C 3 t 2.9076 pf 2.8748 pf C Values for optimized response. Account for error in L. 0 4 8 2 6 20 24 S (db) 2

S response with optimized MCLIN element Path 2 not considered in synthesis. At.2 GHz Path and Path 2 cancel. At 0.8 GHz Path and Path 2 reinforce. S 2 (db) There is partial reinforcement below 0.9 GHz. There is partial cancellation above. GHz. 0 0 20 30 40 50 S, MCLIN S 2, Lumped S 2, MCLIN 60 0.5.0.5 Frequency (GHz) Input Output 50 C Ct b C 2 C C 3 b 2 50 Path 2 Path 0 4 8 2 6 20 24 S (db) 3

S response with optimized MCLIN element.08 GHz.0 GHz.06 GHz 0.96 GHz.00 GHz.02 GHz.04 GHz 0.98 GHz.20 GHz 0.94 GHz 0.92 GHz 0.90 GHz 0.80 GHz 4

Comparison of S response Lumped BPF BPF with optimized MCLIN 0.90 GHz 0.92 GHz.08 GHz.0 GHz 0.80 GHz 0.98 GHz.02 GHz 0.94 GHz.06 GHz.06 GHz 0.96 GHz.00 GHz.02 GHz.04 GHz 0.98 GHz.20 GHz 0.94 GHz.20 GHz.0 GHz.08 GHz 0.92 GHz 0.90 GHz 0.80 GHz 5

S response with with optimized.08 GHz.0 GHz MCLIN.06 GHz 0.96 GHz.00 GHz.02 GHz.04 GHz 0.98 GHz.20 GHz 0.94 GHz 0 0 0 4 0.92 GHz 0.90 GHz 0.80 GHz S 2 (db) 20 30 40 50 S, MCLIN S 2, MCLIN 60 0.5.0.5 8 2 6 20 24 S (db) During manual tuning look at both rectangular S plot and Smith chart plot. 6

Wideband response 0 L 2 C 2 of lumped-element V g 0 BPF L C L 3 C 3 0 0 0 4 S 2 (db) 20 30 40 S 2 8 2 6 S (db) 50 20 S 60 24 0.5 3.5 6.5 8.5 Frequency (GHz) 7

Optimized S response with MCLIN element S 2 (db) 0 0 20 30 40 50 f r = 2f 0 S 2 S 0 4 8 2 6 20 S (db) Spurious basebands at f 0,4f 0, 7.5f 0, Input Output 50 C Ct b C 2 C C 3 b 2 50 60 0.5 3.5 6.5 24 8.5 Frequency (GHz) Recall that f r = 2f 0. So transmission lines look the same at f r,3f r, 5f r, i.e. f 0,6f 0, 0f 0 BUT impedance of capacitance is not the same, hence the spurious passbands are shifted. 8

0 0 Effect of higher f r? S 2 (db) 0 20 30 40 f r = 2f 0 S 2 4 8 2 6 S (db) 50 S 20 60 0.5 3.5 6.5 24 8.5 Frequency (GHz) Input Output 50 C Ct b C 2 C C 3 b 2 50 What if f r = 3f 0? Spurious passbands would be shifted higher in frequency. BUT diminishing returns, design becomes more sensitive. 9

Summary, optimized physical design Alumina ( r = 0), h = 635 m w = w 2 = w 3 = 500 m s = s 2 = 50 m L = 4.65 mm S 2 (db) Input Output 50 C Ct b C 2 C C 3 b 2 50 0 0 20 30 40 50 S, MCLIN S 2, Lumped S 2, MCLIN 60 0.5.0.5 Frequency (GHz) C C C b 3 t 2.2270 pf.9076 pf 2.8748 pf C 0 4 8 2 6 20 24 S (db) L Only adjusted C, C t2, and C 3. 20

Case Study: Parallel Coupled-Line Combline Filter. Part O EM Simulation Input Output 50 C C b C t2 C C 3 b 2 50 2

Input Output 50 C C b C t2 C C 3 b 2 50 Input 50 C Ct 2 b C C 3 2 3 C b Output 50 EM Subcircuit Use optimized MCLINbased BPF values C.2270 pf b C.9076 pf C, C 2.8748 pf 3 t 2 3 2 Enclosure Alumina ( r = 0), h = 635 m w = w 2 = w 3 = 500 m s = s 2 = 50 m X DIM L = 4.65 mm 400 μm 400 μm tantalum vias 6 μm gold metallization. EM enclosure has perfect conducting walls with X DIM = 22 mm, Y DIM = 20 mm and height = 5.635 mm. L via Y DIM w 3 s2 w 2 s w 22

Comparison of responses 0 0 0 4 S 2 (db) 20 30 S 2, MCLIN S 2, EM 8 2 S (db) 40 6 50 S, EM 20 60 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) 24 Could further optimize... 23

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. S 2 (db) 0 0 20 30 40 50 S, EM S 2, MCLIN S 2, EM 60 24 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) 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. 0 4 8 2 6 20 S (db) 24

S response on a Smith chart.06 GHz.04 GHz.08 GHz.0 GHz EM Analysis 0.94 GHz.02 GHz.00 GHz 0.98 GHz 0.92 GHz 0.96 GHz.20 GHz 0.90 GHz 0.80 GHz

S response (optimized).08 GHz.0 GHz MCLIN Analysis.06 GHz 0.96 GHz.00 GHz.02 GHz.04 GHz 0.98 GHz.20 GHz 0.94 GHz 0.92 GHz 0.90 GHz 0.80 GHz 26

S response (optimized) MCLIN Analysis EM Analysis 27

S response on a Smith chart 0 0 0 4 S 2 (db) 20 30 40 50 S, EM S 2, MCLIN S 2, EM 8 2 6 20 S (db) 60 24 0.5 0.6 0.7 0.8 0.9.0..2.3.4.5 Frequency (GHz) 28

Wideband response 0 0 0 4 S 2 (db) 20 30 S 2 8 2 S (db) 40 6 50 S 20 60 0.5 3.5 6.5 24 8.5 Frequency (GHz) Same as with MCLIN based analysis 29

Manufactured filter considerations Input Output 50 C Ct b C 2 C C 3 b 2 50 L It will be necessary to tune every filter manufactured. Fabrication tolerances are about %. Enclosure L Y DIM Greater accuracy than that is required. w 3 s2 w 2 s Tuning done by adjusting capacitor values. X DIM via w 30

Summary, Parallel Coupled-Line Combline Filter Filter synthesized using a methodical process. Microwave simulation required to optimize design. Input Output 50 C C b C t2 C C 3 b 2 50 EM simulation as a check as there are coupling mechanisms that cannot be captured otherwise. Every filter manufactured will require tuning. L 3