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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 .
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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
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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.
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Has two solutions only when
d. is pure imaginary.
Terminated Periodic Structures
if .
Example 8.1
, slow wave.
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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
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Current ratio
Define propagation factor as
or
Constant-k Filter Sections
Low pass filters
From Table 8.1, the image impedance equals
where
and
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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
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m-Derived Filter Sections
Let
In order to keep the image impedance the same
For a low pass filter, and . Then
where
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If , is real, and for . Stopband.
When , , implying infinite attenuation.
Disadvantages:1. Slow attenuation when .2. Image impedance is a function of frequency.
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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 .
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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.
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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
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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
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From Table 8.21. Constant-k T section
2. m-derived T section
3. m=0.6 matching section
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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
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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.
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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
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.
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
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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
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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.
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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
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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
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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
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To sum up
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Example 8.4: BANDPASS FILTER DESIGN
N=3, 0.5 dB equal ripple, , , .
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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.
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Consider Table 8.7(a),
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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.
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Impedance and Admittance Inverters
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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,
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Inductor:
Capacitor:
Example 8.7 STEPPED IMPEDANCE FILTER DESIGN, 20 dB attenuation at 4 GHz. Maximal flat.
. Substrate:
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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
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For the circuit
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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 ,
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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
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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:
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2. Odd mode:
To sum up,
Since
We have
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Let port 2 and 4 open, we have
Computing the image impedance
If , which is real and positive since
.
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.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
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The shunt impedance
For a parallel LC circuit near resonance
Matching to the bandpass prototype, we have
In general:
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Example 8.7 COUPLED LINE BANDPASS FILTER DESIGN
, N=3, 0.5 dB equal ripple.
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
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Bandpass Filters Using Capacitive Coupled Resonators
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Example 8.10 CAPACITIVELY COUPLED SERIESRESONATOR BANDPASS FILTER DESIGN
, N=3, 0.5 dB equal ripple.
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
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Bandpass Filters Using Capacitively Coupled ShuntResonators
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Similar to 8.8 bandstop filter,
From Fig. 8.38
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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