Microwave Filters (8) - ntut.edu.tjuiching/microwave3.pdf · Microwave Filters (8) ... Use Table...

Microwave Circuits 1

Microwave Filters (8)Periodic Structures

where the ABCD matrix represents the cascade ofa transmission line section of length , a shunt

susceptance , and another transmission linesection of length .

where .


Assume the port voltages and currents satisfywave equation

Then,

or

For a nontrivial solution to exist, the following mustbe satisfied

That is.

The characteristic impedance of this wave isdefined as

Also

Then


For symmetrical unit cells, , therefore

The solutions correspond to the positively andnegatively traveling waves, respectively.

Due to reciprocity, , the above equationbecomes

.

Let , then

1. .a. Nonattenuating, propagating wave.b. Passband.

c.

Has multiple solutions only when

d. is pure real.2. .

a. Attenuating, nonpropagating wave.b. Stopband.c.


Has two solutions only when

d. is pure imaginary.

Terminated Periodic Structures

if .

Example 8.1

, slow wave.


Filter Design by The Image ParameterMethod(8.2)

= input impedance at port 1 when port isterminated with .

= input impedance at port 1 when port isterminated with .From ABCD matrix, we have

Solving and gives

and . If symmetric, and .

If port 2 is terminated in , Voltage ratio equals


Current ratio

Define propagation factor as

or

Constant-k Filter Sections

Low pass filters

From Table 8.1, the image impedance equals

where

and


1. : Passband. is real. is imaginary..

2. : Stopband. is imaginary. is real.. Attenuation rate for is 40

dB/decade.

Disadvantages:1. Slow cutoff.2. Image impedance is a function of frequency.

High Pass Filters


m-Derived Filter Sections

Let

In order to keep the image impedance the same

For a low pass filter, and . Then

where


If , is real, and for . Stopband.

When , , implying infinite attenuation.

Disadvantages:1. Slow attenuation when .2. Image impedance is a function of frequency.


Erecta:1. p. 434, in Table 8.1, the propagation

constant should be .

2. p. 434, Eq. 8.36, add absolute value to theterm .

3. p. 437, Eq. 8.43, add absolute value to the

term .


m-derived - section

Goals: 1. Provide the same form of the propagation

factor of an m-derived T-section.2. Provide a less frequency dependent image

impedance.

Procedures:1. Use Table 8.1, to find out the equivalent

- section that give the same thepropagation constants as the T-section.

2. Compute the image impedance.

Choose m=0.6 to minimize the variation.


3. Now . Split the - section to twoparts to match . It can be proved that theimage impedances equals to and respectively.

4. Let the propagation constant of the original- section be and the split - section

. It is obviously.

Composite Filters1. From the given impedance and the cut-off

frequency , determine the values of theinductance and the capacitance by

.

Then, the constant-k T section can beimplemented.2. Determine the m value from the infinite

attenuation pole by


Then m-derived T section can beimplemented.3. Choose m=0.6 as the m value of the bisected

- section.

Example 8.2: LOW-PASS COMPOSITE FILTERDESIGN


From Table 8.21. Constant-k T section

2. m-derived T section

3. m=0.6 matching section


Filter Design by the Insertion Loss Method (8.3)

Benefits: can synthesize a desired response systematically.Disadvantage: trade-off between the ideal and real responses.

Define: insertion loss or power loss ratio, by

Since is an even function,

Maximally flat:

1. Order: N2. Low pass, binomial or Butterworth response.3. Flattest passband.4. Cutoff frequency: 5. Passband: .

6. Power loss ratio at : . For , 3-dB loss.

7. When , .

8. Attenuation rate: 9. The first (2N-1) derivatives are zero at

Equal ripple:

1. Order: N


2. Low pass.3. is the Chebyshev polynomial of order N.

4. Equal ripple of amplitude .

5. When , .

6. Sharper cutoff than maximally flat types by a factor .

7. Attenuation rate: .

Linear Phase:

Then the group delay

is maximally flat.


Maximally Flat Low-Pass Filter Prototype

Consider the above circuit,

Since

We have

In order to fit the above equation to a maximally flat low-passfilter with , , and , that is,

,It is required that

,, and

.

Solving for , and , we have


.

Procedures:1. Referring to Fig. 8.26, determine the order from the

required attenuation characteristics.2. By looking up Table 8.3, determine the normalized

component values by the following rules:a. is the generator resistance if the ladder circuit

starts with a shunt capacitance, or generatorconductance if starts with a series inductors.

b. is the inductance for series inductors, orcapacitance for shunt capacitors.

c. the load resistance if is a shunt capacitor, orload conductance if is a series inductor.

3. Impedance and frequency scaling. Let and be thereal generator resistance and cutoff frequencyrespectively, then


Example 8.3Design a maximal flat low-pass filter with a cutoff frequency of2 GHz, impedance of , and at least 15 dB insertion loss at3 GHz


Equal Ripple Low-Pass Filter PrototypeProcedures:1. Referring to Fig. 8.27, determine the order from the

required attenuation characteristics and choose thedesired ripple level.

2. Decide the component values by Table 8.4 and follow thesame procedure in the maximal flat prototype. Note whenN is even, the generator resistance and the loadresistance are not equal.

3. Impedance and frequency scaling. Use the sameprocedure in the maximal flat prototype.


Linear Phase Low-Pass Filter PrototypesProcedure: Same as previous prototypes except using Table8.5.

Filter Transformation (8.4)

Goals:1. Low pass high pass.2. Low pass band pass.3. Low pass band stop.

Low Pass High Pass

Let , then the low pass response becomes a high pass


response.Procedure:1. Convert the inductances in the low pass prototypes to

capacitances as follows

2. Convert the capacitances in the low pass prototypes toinductances as follows

3. Include the effect of impedance scaling

Low Pass Band Pass

Let , where and . Then,

which is a band pass response with cutoff at and .

An inductance in the low pass prototype will converts to aseries inductance and a series capacitance as follow


An capacitance in the low pass prototype will converts to aparallel inductance and a parallel capacitance as follow

Notice that both pairs of inductances and capacitance have thesame resonant frequency .

Low Pass Band Stop

Similarly, Let .

Convert an inductance in the low pass prototype to a parallelinductance and a parallel capacitance as follow

Convert an capacitance in the low pass prototype to a seriesinductance and a series capacitance as follow


To sum up


Example 8.4: BANDPASS FILTER DESIGN

N=3, 0.5 dB equal ripple, , , .


Filter Implementation (8.5)

Richard’s Transformation

Choose at such that and .A zero occur at . Kuroda’s identities• Physically separate transmission line stubs.• Transform series stubs into shunt stubs, or vice versa.• Change impractical characteristic impedance into more

realizable ones.


Consider Table 8.7(a),


where .

Since the left and the right ABCD matrix must be the same, wehave

Example 8.5 LOW-PASS FILTER DESIGN USING STUBS, 3 dB equal ripple.


Impedance and Admittance Inverters


Stepped-Impedance Low-Pass Filters (8.6)

Convert the ABCD matrix of a short transmission line to Z-parameters, we have

Using T equivalent circuit, we have

If is large

.

If is small

To sum up,


Inductor:

Capacitor:

Example 8.7 STEPPED IMPEDANCE FILTER DESIGN, 20 dB attenuation at 4 GHz. Maximal flat.

. Substrate:


Filters Using Coupled Resonators (8.8)

Bandstop and Bandpass Filters Using Quarter-waveResonators

For a open-circuit transmission line,

where for . Let , where . Assume

ideal transmission line, then . Now

For a series LC circuit near resonance

Thus


For the circuit


Which should equal to the bandstop prototype

To make the two values the same, we need

Solving for the above equations, we have

Then,

where .

To sum up,

bandstrop filter: open stub with ,


bandpass filter: short stub with

Example 8.8 BANDSTOP FILTER DESIGN

, N=3, 0.5 dB equal ripple.

n

1 1.5963 265.9

2 1.0967 387.0

3 1.5963 265.9


Coupled Line Filters

Even-Odd Mode AnalysisDenote the left side as port 1 and the right side port 2.Let be the even-odd mode voltagesand currents.Goal: find out the impedance matrices1. Even mode:


2. Odd mode:

To sum up,

Since

We have


Let port 2 and 4 open, we have

Computing the image impedance

If , which is real and positive since

.


.Cutoff frequency

Propagation constant

Passband is real for

The image impedance of the above is

At , . The propagation constant is

Comparing to the original equations, we have

with . Solve for the even and odd mode impedance togive


The shunt impedance

For a parallel LC circuit near resonance

Matching to the bandpass prototype, we have

In general:


Example 8.7 COUPLED LINE BANDPASS FILTER DESIGN


n

1 1.5963 0.3137 70.61 39.24

2 1.0967 0.1187 56.64 44.77

3 1.5963 0.1187 56.64 44.77

4 1.0000 0.3137 70.61 39.24


Bandpass Filters Using Capacitive Coupled Resonators


Example 8.10 CAPACITIVELY COUPLED SERIESRESONATOR BANDPASS FILTER DESIGN


n

1 1.5963 0.3137 0.554 155.8

2 1.0967 0.1187 0.192 166.5

3 1.5963 0.1187 0.192 155.8

4 1.0000 0.3137 0.554


Bandpass Filters Using Capacitively Coupled ShuntResonators


Similar to 8.8 bandstop filter,

From Fig. 8.38


Example 8.10 CAPACITIVELY COUPLED SHUNTRESONATOR BANDPASS FILTER DESIGN

N=3, 0.5 dB equal-ripple, , , .

n

1 1.5963 0.2218 0.2896 -0.3652 -0.04565 73.6

2 1.0967 0.0594 0.0756 -0.1512 -0.0189 83.2

3 1.5963 0.0594 0.0756 -0.3652 -0.04565 73.6

4 1.0000 0.2218 0.2896

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