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426 Chapter 8: Microwave Filters

1 dB at 2 GHz. The response of the corresponding lumped-element filter is also

shown in Figure 8.41. The passband characteristic is similar to that of the stepped

impedance filter, but the lumped-element filter gives more attenuation at higher

frequencies. This is because the stepped-impedance filter elements depart sig-

nificantly from the lumped-element values at higher frequencies. The stepped-

impedance filter may have other passbands at higher frequencies, but the responsewill not be perfectly periodic because the lines are not commensurate.

8.7COUPLED LINE FILTERS

The parallel coupled transmission lines discussed in Section 7.6 (for directional couplers)

can be used to construct many types of filters. Fabrication of multisection bandpass or

bandstop coupled line filters is particularly easy in microstrip or stripline form for band-

widths less than about 20%. Wider bandwidth filters generally require very tightly coupled

lines, which are difficult to fabricate. We will first study the filter characteristics of a single

quarter-wave coupled line section, and then show how these sections can be used to designa bandpass filter [7]. Other filter designs using coupled lines can be found in reference [1].

Filter Properties of a Coupled Line Section

A parallel coupled line section is shown in Figure 8.42a, with port voltage and current

definitions. We will derive the open-circuit impedance matrix for this four-port network by

considering the superposition of even- and odd-mode excitations [8], which are shown in

Figure 8.42b. Thus, the current sourcesi 1 andi 3 drive the line in the even mode, while i 2andi4drive the line in the odd mode. By superposition, we see that the total port currents,

Ii , can be expressed in terms of the even- and odd-mode currents as

I1= i1 + i2, (8.87a)I2= i1 i2, (8.87b)I3= i3 i4, (8.87c)I4= i3 + i4. (8.87d)

First consider the line as being driven in the even mode by the i1 current sources. If

the other ports are open-circuited, the impedance seen at port 1 or 2 is

Z

e

in= j Z0ecot . (8.88)The voltage on either conductor can be expressed as

v1a (z) = v1b (z) = V+e[ej(z) + ej(z)]= 2V+e cos ( z), (8.89)

so the voltage at port 1 or 2 is

v1a (0) = v1b (0) = 2V+e cos = i1Zein.

This result and (8.88) can be used to rewrite (8.89) in terms ofi1as

v1a (z) = v1b (z) = j Z0ecos ( z)

sin i1. (8.90)


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This result and (8.92) can be used to rewrite (8.93) in terms ofi2as

v2a (z) = v2b (z) = j Z0ocos ( z)

sin i2. (8.94)

Similarly, the voltages due to current i4driving the line in the odd mode are

v4a (z) = v4b (z) = j Z0ocos z

sin i4. (8.95)

The total voltage at port 1 is

V1= v1a (0) + v2a (0) + v3a (0) + v4a (0)= j (Z0ei1 + Z0oi2) cot j (Z0ei3 + Z0oi4) csc , (8.96)

where the results of (8.90), (8.91), (8.94), and (8.95) were used, and = . Next, wesolve (8.87) for thei j in terms of the Is:

i1=1

2(I1 + I2), (8.97a)

i2=1

2(I1 I2), (8.97b)

i3=1

2(I3 + I4), (8.97c)

i4=1

2(I4 I3), (8.97d)

and use these results in (8.96):

V1=j

2(Z0eI1 + Z0eI2 + Z0oI1 Z0oI2) cot

j2

(Z0eI3 + Z0eI4 + Z0oI4 Z0oI3) csc . (8.98)

This result yields the top row of the open-circuit impedance matrix [Z] that describes the

coupled line section. From symmetry, all other matrix elements can be found once the first

row is known. The matrix elements are then

Z11

= Z22

= Z33

=Z44

=

j

2

(Z0e

+Z0o) cot (8.99a)

Z12= Z21= Z34= Z43=j

2(Z0e Z0o) cot (8.99b)

Z13= Z31= Z24= Z42=j

2(Z0e Z0o) csc (8.99c)

Z14= Z41= Z23= Z32=j

2(Z0e + Z0o) csc (8.99d)

A two-port network can be formed from a coupled line section by terminating two

of the four ports with either open or short circuits, or by connecting two ends; there are

10 possible combinations, as illustrated in Table 8.8. As indicated in the table, the various

circuits have different frequency responses, including low-pass, bandpass, all pass, and all

stop. For bandpass filters, we are most interested in the case shown in Figure 8.42c, as open

circuits are easier to fabricate in microstrip than are short circuits. In this case, I2= I4= 0,


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8.7 Coupled Line Filters 429

TABLE 8.8 Ten Canonical Coupled Line Circuits

Circuit Image Impedance Response

Re(Zi1)

0Zi1 Zi2

Zi1

Zi1

Zi1

Zi2

Zi1

Zi1

Zi1

Zi1

Zi1 Zi1

Zi1

Zi2

Zi1

Zi1

Zi1 Zi1

Zi1

Zi1

2

3

2Low-pass

Re(Zi1)

0

2

3

2Bandpass

Re(Zi1)

0

2

3

2Bandpass

Re(Zi1)

0

2

3

2Bandpass

Zi1=

Zi1Zi2=

2Z0eZ

0ocos

(Z0e+Z0o)2cos 2 (Z0eZ0o)

2

Z0eZ0o

Zi1=2Z0eZ0osin

(Z0eZ0o)2 (Z0e+Z0o)

2 cos2

Zi1=2 sin

(Z0eZ0o)2 (Z0e+Z0o)

2 cos2

Zi1=

Z0eZ0o

(Z0eZ0o)2 (Z0e+Z0o)

2 cos2

(Z0e+Z0o)sin

Z0eZ0o

Zi1Zi2=

Zi1= j

Z0eZ0o

Zi1

Zi2=Z0eZ0o

2Zi1=

Z0e +Z0o

Zi1=

Zi1=

Z0eZ0oZi1=j

2Z0eZ0o

Z0e +Z0o

2Z0eZ0o

Z0e +Z0ocot

tan

Z0eZ0oZi1= j cot

All pass

All pass

All pass

All stop

All stop

All stop


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00

1 2

2 2

3

Z0e Z0o

2

Re(Zi)

FIGURE 8.43 The real part of the image impedance of the bandpass network of Figure 8.42c.

so the four-port impedance matrix equations reduce to

V1= Z11I1 + Z13I3, (8.100a)V

3= Z

31I

1 +Z

33I

3, (8.100b)

whereZi j is given in (8.99).

We can analyze the filter characteristics of this circuit by calculating the image imped-

ance (which is the same at ports 1 and 3), and the propagation constant. From Table 8.1,

the image impedance in terms of the impedance parameters is

Zi=

Z211 Z11Z

213

Z33

= 12

(Z0e Z0o)2 csc2 (Z0e + Z0o)2 cot2 . (8.101)

When the coupled line section is /4 long (= /2), the image impedance reduces to

Zi=1

2(Z0e Z0o), (8.102)

which is real and positive since Z0e > Z0o. However, when 0 or , Zi j,indicating a stopband. The real part of the image impedance is sketched in Figure 8.43,

where the cutoff frequencies can be found from (8.101) as

cos 1= cos 2=Z0e Z0oZ0e + Z0o

.

The propagation constant can also be calculated from the results of Table 8.1 as

cos =

Z11Z33

Z213

= Z11Z13

= Z0e + Z0oZ0e Z0o

cos , (8.103)

which shows is real for 1 < < 2= 1, where cos 1= (Z0e Z0o)/(Z0e +Z0o).

Design of Coupled Line Bandpass Filters

Narrowband bandpass filters can be made with cascaded coupled line sections of the form

shown in Figure 8.42c. To derive the design equations for filters of this type, we first showthat a single coupled line section can be approximately modeled by the equivalent circuit

shown in Figure 8.44. We will do this by calculating the image impedance and propagation

constant of the equivalent circuit and showing that they are approximately equal to those


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J

90Z0 Z0

FIGURE 8.44 Equivalent circuit of the coupled line section of Figure 8.42c.

of the coupled line section for = /2, which will correspond to the center frequency ofthe bandpass response.

The ABCD parameters of the equivalent circuit can be computed using the ABCD

matrices for transmission lines from Table 4.1:

A B

C D

=

cos j Z0sin jsin

Z0cos

0 j/Jj J 0

cos j Z0sin jsin Z0

cos

=

J Z0 + 1J Z0

sin cos j

J Z20sin

2 cos2

J

j

1

J Z20

sin2 Jcos2

J Z0 +1

J Z0

sin cos

. (8.104)

The ABCD parameters of the admittance inverter were obtained by considering it as a

quarter-wave length of transmission of characteristic impedance, 1/J. From (8.27) the

image impedance of the equivalent circuit is

Zi= B

C =J Z20sin

2

(1/J) cos2

(1/J Z20 ) sin2 Jcos2 , (8.105)

which reduces to the following value at the center frequency,= /2:Zi= J Z20 . (8.106)

From (8.31) the propagation constant is

cos = A =

J Z0 +1

J Z0

sin cos . (8.107)

Equating the image impedances in (8.102) and (8.106), and the propagation constants of

(8.103) and (8.107), yields the following equations:

1

2(Z0e Z0o) = J Z20 ,

Z0e + Z0oZ0e Z0o

= J Z0 +1

J Z0,

where we have assumed sin 1 for near /2. These equations can be solved for theeven- and odd-mode line impedances to give

Z0e= Z0[1 + J Z0 + (J Z0)2], (8.108a)

Z0o= Z0[1 J Z0 + (J Z0)2

]. (8.108b)Now consider a bandpass filter composed of a cascade ofN+ 1 coupled line sections,

as shown in Figure 8.45a. The sections are numbered from left to right, with the load on the


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FIGURE 8.45 Development of an equivalent circuit for derivation of design equations for a cou-

pled line bandpass filter. (a) Layout of an (N+ 1)-section coupled line bandpassfilter. (b) Using the equivalent circuit of Figure 8.44 for each coupled line section.

(c) Equivalent circuit for transmission lines of length 2. (d) Equivalent circuit

of the admittance inverters. (e) Using results of (c) and (d) for the N

=2 case.

(f) Lumped-element circuit for a bandpass filter for N= 2.


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right, but the filter can be reversed without affecting the response. Since each coupled line

section has an equivalent circuit of the form shown in Figure 8.44, the equivalent circuit of

the cascade is as shown in Figure 8.45b. Between any two consecutive inverters we have

a transmission line section that is effectively 2 in length. This line is approximately /2

long in the vicinity of the bandpass region of the filter, and has an approximate equivalent

circuit that consists of a shunt parallel L Cresonator, as in Figure 8.45c.The first step in establishing this equivalence is to find the parameters for the T-

equivalent and ideal transformer circuit of Figure 8.45c (an exact equivalent). The ABCD

matrix for this circuit can be calculated using the results in Table 4.1 for a T-circuit and an

ideal transformer:

A B

C D

=

Z11

Z12

Z211 Z212Z12

1

Z12

Z11

Z12

1 0

0 1=

Z11Z12

Z212 Z211Z12

1Z12

Z11Z12

. (8.109)

Equating this result to theABCDparameters for a transmission line of length 2and char-

acteristic impedance Z0gives the parameters of the equivalent circuit as

Z12=1C

= j Z0sin2

, (8.110a)

Z11= Z22= Z12A = j Z0cot 2. (8.110b)Then the series arm impedance is

Z11 Z12= j Z0cos 2+ 1

sin2= j Z0cot . (8.111)

The 1:

1 transformer provides a 180

phase shift, which cannot be obtained with the

T-network alone; since this does not affect the amplitude response of the filter, it can be

discarded. For /2 the series arm impedances of (8.111) are near zero and can also beignored. The shunt impedance Z12, however, looks like the impedance of a parallel reso-

nant circuit for /2. If we let = 0 + , where= /2 at the center frequency0, then we have 2= = /vp= (0 + )/0= (1 + /0), so (8.110a)can be written for small as

Z12=j Z0

sin (1 + /0) j Z00

( 0). (8.112)

From Section 6.1 the impedance near resonance of a parallel LCcircuit is

Z= j L20

2( 0), (8.113)

with20= 1/LC. Equating this to (8.112) gives the equivalent inductor and capacitor val-ues as

L= 2Z0 0

, (8.114a)

C= 120L

= 2Z00

. (8.114b)

The end sections of the circuit of Figure 8.45b require a different treatment. The lines

of length on either end of the filter are matched to Z0 and so can be ignored. The end

inverters, J1and JN+1, can each be represented as a transformer followed by a /4 section


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of line, as shown in Figure 8.45d. The ABCDmatrix of a transformer with a turns ratio N

in cascade with a quarter-wave line is

A B

C D = 1

N0

0 N

0 j Z0j

Z0

0

=

0j Z0

Nj NZ0

0

. (8.115)

Comparing this to the ABCDmatrix of an admittance inverter [part of (8.104)] shows that

the necessary turns ratio is N= J Z0. The/4 line merely produces a phase shift and socan be ignored.

Using these results for the interior and end sections allows the circuit of Figure 8.45b

to be transformed into the circuit of Figure 8.45e, which is specialized to the N= 2 case.We see that each pair of coupled line sections leads to an equivalent shunt LC resonator,

and an admittance inverter occurs between each pair of LC resonators. Next, we show

that the admittance inverters have the effect of transforming a shunt LC resonator into

a series LCresonator, leading to the final equivalent circuit of Figure 8.45f (shown for

N= 2). This will then allow the admittance inverter constants, Jn , to be determined fromthe element values of a low-pass prototype. We will demonstrate this for the N= 2 case.With reference to Figure 8.45e, the admittance just to the right of the J2inverter is

j C2 +1

j L2+ Z0J23= j

C2

L 2

0 0

+ Z0J23 ,

since the transformer scales the load admittance by the square of the turns ratio. Then the

admittance seen at the input of the filter is

Y

=

1

J2

1Z2

0j C1 +

1

j L 1 +

J22

jC2/L2[(/0) (0/)] + Z0J2

3

= 1J21Z

20

j

C1

L 1

0 0

+ J

22

j

C2/L2[(/0) (0/)] + Z0J23

. (8.116)

These results also use the fact, from (8.114), that L nCn= 1/20for all L Cresonators.Now the admittance seen looking into the circuit of Figure 8.45f is

Y= j C1 +1

j L 1+ 1

j L 2 + 1/j C2 + Z0

= jC1L 1

0

0

+ 1

j

L 2/C

2[(/0) (0/)] + Z0

, (8.117)

which is identical in form to (8.116). Thus, the two circuits will be equivalent if the fol-

lowing conditions are met:

1

J21Z20

C1

L 1=

C1L 1

, (8.118a)

J21Z20

J2

2

C2

L 2=

L 2C

2

, (8.118b)

J21Z30J

23

J22

= Z0. (8.118c)


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We knowL n andCn from (8.114); Ln andC

n are determined from the element values

of a lumped-element low-pass prototype that has been impedance scaled and frequency

transformed to a bandpass filter. Using the results in Table 8.6 and the impedance scaling

formulas of (8.64) allows the L n andCn values to be written as

L 1= Z00g1, (8.119a)

C1=g1

0Z0, (8.119b)

L 2=g2Z0

0, (8.119c)

C2=

0g2Z0, (8.119d)

where =

(2

1)/0 is the fractional bandwidth of the filter. Then (8.118) can be

solved for the inverter constants with the following results (for N= 2):

J1Z0=

C1L1

L 1C1

1/4=

2g1, (8.120a)

J2Z0= J1Z20

C2C2

L2L2

1/4=

2

g1g2, (8.120b)

J3Z0=J2

J1=

2g2. (8.120c)

After the Jn are found, Z0e and Z0o for each coupled line section can be calculated from

(8.108).

The above results were derived for the special case ofN= 2 (three coupled line sec-tions), but more general results can be derived for any number of sections, and for the case

where ZL= Z0 (or gN+1= 1, as in the case of an equal-ripple response with N even).Thus, the design equations for a bandpass filter with N+ 1 coupled line sections are

Z0J1=

2g1, (8.121a)

Z0Jn=

2

gn1gnforn= 2, 3, . . . ,N, (8.121b)

Z0JN+1=

2gNgN+1. (8.121c)

The even- and odd-mode characteristic impedances for each section are found from (8.108).

EXAMPLE 8.7 COUPLED LINE BANDPASS FILTER DESIGN

Design a coupled line bandpass filter with N= 3 and a 0.5 dB equal-ripple re-sponse. The center frequency is 2.0 GHz, the bandwidth is 10%, and Z0= 50.What is the attenuation at 1.8 GHz?


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