Design of all-pole microwave filters - Intranet...

Design of all-pole microwave filters

Giuseppe MacchiarellaPolytechnic of Milan, Italy

Electronic and Information Department

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

/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

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

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

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.

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

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}

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

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)

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

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

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

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

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

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

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!

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

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!

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)

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

qq

gQ

B f

Note: Narrow band filters require high Q0resonators to reduce passband attenuation

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

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.

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

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

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

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

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

Filters with an array of coupled-lines

Interdigital:

Comb:

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

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

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

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

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

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

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

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

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:

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:

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