Microwave Filters - Theoretical Information

8/12/2019 Microwave Filters - Theoretical Information

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THEORETICAL INFORMATION

The insertion method can be used to characterise a filter response in microwave. It is

defined as the ratio of power available from source to power delivered to load. In thisprogram two common types of filter characteristics are used: maximally flat and equalripple (or Chebyshev) filters.

MAXIMALLY FLAT FILTER:

It provides the flattest possible pass band region for a given filter complexity or order (N).it is specified by

(1)

Where wc is cut off frequency and N is the order of filter. At cut off it gives 3 dB powerloss. For wwc the attenuation increases monotically as shown in figure 1. Note that L(w) is

10logPLR where PLRis power loss.

Figure 1:Low-pass filter response for maximally flat filter.Frequency vs power loss

EQUAL RIPPLE (CHEBYSHEV):

It provides a sharper cut off , however the pass band has ripples of amplitude 1+k2. It is

specified by

(2)

TN(x) oscillates between +1 and 1. k2 determines the pass band ripple level.


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Figure 2: Low-pass filter response for equal ripple filter. Frequency vs power loss

LOW-PASS FILTER DESIGN

Ladder circuits, as shown below, gives the power loss characteristics of those filtertypes. The lumped circuit elements are shunt capacitors and series inductors as shown in

figure 3. For maximally flat or equal ripple response in the pass band, the ladder network issymmetric for odd number of elements. If we let Z in be the input impedance seen from Rs,then reflection coefficient is

(3)

Then, power loss ratio becomes

(4)

At w=0, all capacitors open and all inductors short circuit , and hence Zin=1. P LRis zero for

maximally flat and equal ripple filter when N is odd. For equal ripple with N is evenPLR=1+k

2at w=0.

Figure 3:Low-pass ladder prototype filter networks

For maximally flat filter with a power loss ratio


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PLR=1+w2N (6)

the element values can be calculated by with R=1,

k=1,2,,N (7) where gk values are the value of inductance in

henries or the value of capacitance in farads.

Some numerical values of gk for maximally flat filter listed up to N =5 in table 1.

N

K 2 3 4 5

1 1.414 1.00 0.76 0.62

2 1.414 2.00 1.85 1.62

3 1.00 1.85 2.00

4 0.76 1.62

5 0.62

table 1: Values of gk for maximally flat filter for N=2,..,5

Equal ripple low pass filter element values can be calculated by

N even (8)

gN+1=1 N odd

when element gN is a capacitor gN+1=R, but when gNis an inductor gN+1=1/R.


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(9)

Some numerical values of gk for equal ripple filter N up to 5 are listed in table 2.

N

K 2 3 4 5

1 0.84 1.03 1.11 1.15

2 0.84 1.15 1.35 1.37

3 1.03 1.77 1.97

4 0.82 1.37

5 1.15

table 2: Values of gk for equal ripple filter with k2=0.023 (0.1 dB ripple)

FILTER TRANSFORMATION:

The low pass prototype element values which are calculated for R=1 and wc=1 can be

modified for arbitrary source impedance and cut off frequency. In addition, by frequency

transformation, band pass and high pass filter element values can be obtained or viceversa.


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Impedance and Frequency Transformation: Element values calculated for wc=1 and

R=1 can be transformed for arbitrary cut off frequency wc and source impedance R0byusing following simple equations L =R0L/wc (10)

C= C/(R0wc)Low pass to High pass Transformation: The frequency substitution wc/w yields highpass filter. The point at w=0 is mapped to infinity, meaning that pass band of a low pass

filter is at infinity for a high pass filter. Then new element values can be found by

(11)

which shows that the series inductors must be replaced with capacitors Ck and the shuntcapacitors must be replaced with inductors Lkgiven by

(12)

Then new ladder network is shown in figure 4.

figure 4: high pass filterladder network


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Low pass to Band pass Transformation:

where wo=(w1w2)1/2center frequency.

w=0 is mapped to w=wo, meaning that pass band of low pass filter is now centred around

wo. Cut off frequency 1 mapped to the edge frequencies w=w1 and w=w2 as desired for aband pass filter. Similarly, new element values are determined by

(13)

Above equations show that the series inductor Lkis transformed to a series LC circuit with

element values,

(14)

and the shunt capacitor Ckis transformed to a shunt LC circuit with element values,

(15)

Then new circuit for band pass filter is as shown in figure 5


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figure 5: band pass filter circuit.

STEP-IMPEDANCE LOW-PASS AND HIGH-PASS FILTER DESIGN

Low pass filter and high pass filter can be realised by using alternating sections of veryhigh and very low characteristic impedance lines. They can be implemented either in

microstrip or stripline. Impedance values are limited by fabrication. Typical value is 150W

for Zh (highest impedance) and 10W for Zl (lowest impedance). Because of theapproximations involved the electrical performances are not perfect but they are easier todesign and take less place compared to other designs.

Approximate Equivalent Circuits for short section transmission line:

Z-parameters for transmission line having Zo characteristic impedance given by,

(16)

the series elements of T equivalent circuit are given by

(17)

and shunt element value is Z12. The equivalent circuit (see figure 6) has the followingelement values:

X/2=Zotan(b l/2) (18)

B=sin(b l )/Zo

Assuming length is short and the characteristic impedance is large then,X ~ Zob l, (19)

B~0,

The result implies that it is a series inductor. Similarly, assuming very small characteristic

impedance results in a shunt capacitor.(see figure 6)

X ~ 0 (20)

B~Yob l,


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figure 6: Approximate equivalent circuits fortransmission lines

a) T equivalent circuit for a TLb) Equivalentcircuit for smallbl

and large Zo c) Equivalent circuit for small b l

and small Zo

From (19) b l=LR0/ Zh for inductor, and from Eq 20 b l=CZl/R0for capacitor can beobtained. The required lengths of the lines are determined from

l = vpgkR0/(Zh wc) for inductor,

l = vpgkZh/(R0wc) for inductor, where vPis the phase velocity of thepropagating wave on the line.

A typical to view of the stepped impedance low pass filter can be seen in figure 2. For high

pass filter it is just replacing high impedance lines with low impedance transmission linesection and element values transformed.

figure 7: Microstrip layout of low pass filter

COUPLED LINE FILTERS


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Coupled transmission lines can be used to implement various filter circuit.

figure 8: a)a parallel coupled line b) even and odd mode currents

Filter characteristics of a single quarter wave coupled line:

To find the open circuit impedance matrix we will consider even and odd mode excitations

separately, then find the result by superposition. I1, I2, I3, I4are the total port currents. i1,i3 are the currents driving the line in even mode, i3, i4are in odd modes. Note that

VA1,VB

1voltages on line A and B due to the current i1,

VA2,VB


VA3,VB


VA4,VB

4voltages on line A and B due to the current i4.

At the evenmode, the impedance seen from ports 1 and 2 is

Zein=-jZoe cot b l. (21)

The voltage on the line can be expressed as

VA1(z)= VB

1(z) =Ve+[e-jb (z-l)+ ejb (z-l)]=2Ve+cosb (l-z), (22)


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The voltages at port 1 and 2 is for z=0;

VA1(0)= VB

1(0)=2Ve+cos b l=i1Zine .(23)

Using (21) in (23),

V(z)=-jZoe . (24)

In the same way voltage equation can be written in terms of i3

VA3(z)= VB

3(z)=-jZoe . (25)

In the odd mode, the line is driven by i2. The impedance seen from 1 and 2

is

Zoin=-jZoe

cot b l. (26)

The voltage on the line A can be expressed as, the voltage on B is just negative of thefollowing,

VA2(z)=- VB

2(z)=Vo+[e-jb (z-l)+ ejb (z-l)]=2Vo+cosb (l-z), (27)The voltages at port 1 and 2 is for z=0;

VA2(0)=- VB

2(0)=2Vo+cos b l=i2Zino .

Using (26) in (28),

VA2(z)=- VB

2(z)=-jZoo . (28)

In the same way voltage equation can be written in terms of i4

VA4(z)=- VB

4(z)=-jZoo . (29)

Using the voltage equations the voltage at port 1 is found as (z=0)


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V1= VA1(0)+VA

2(0) + VA3(0)+VA

4(0) (30)

=-j(Zoe i1 + Zoo i2)cotq - j(Zoe i3 + Zoo i4)cscq .

Also total currents on the line can be expressed in terms of even and odd

mode currents:

I1=i1+i2,

I2=i1-i2,

I3=i3-i4,

I4=i3+i4. (31)

From (31) w can drive,

i1=0.5(I1+I2),

i2=0.5(I1-I2),

i3=0.5(I3+I4),

i4=0.5(I3-I4),

Using the above results in (30):

V1=-j0.5(Zoe(I1+I2)+Zoo(I1-I2))cotq -j0.5(Zoe(I3+I4)+Zoo(I4-I3))cscq (33)

from symetry z-parameters can be found

Z11=Z22=Z33=Z44=-j0.5(Zoe+Zoo)cotq ,

Z12=Z21=Z34=Z43=-j0.5(Zoe-Zoo)cotq ,

Z13=Z31=Z24=Z42=-j0.5(Zoe-Zoo)cscq ,

Z14=Z41=Z32=Z23=-j0.5(Zoe+Zoo)cscq ,

(34)

We can analyze the filter characteristic by calculating the image impedance (Zi) and the

propagation constant.


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for q =pi/2 Zi=0.5(Zoe-Zoo) (35)

When q goes to 0 or pi, Zi goes to infinity; meaning that stop band. Cut off frequency from(35)

cosq1=-cosq2=(Zoe-Zoo)/(Zoe+Zoo) (36)

Also propagation constant is determined by

cosb = . (37)

Design of Coupled Line Band pass filters:

A coupled line can be modelled by two transmission line and an admittance inverter in

between as shown in figure 8.

figure9:Equivalent circuit for coupled line

Firstly, let us compute ABCD matrix for this circuit:

(38)


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Choosing q =p /2, we can find . (39)

The propagation constant is from the matrix

. (40)

Using (35) and (39)

JZo2= (Zoe-Zoo)/2, (41)

And using (40) and (37), also assuming sinq 1

(42)

Solving (41) and (42) we obtain even and odd impedances

Zoe=Zo[1+JZo+(JZo)2],

Zoo=Zo[1-JZo+(JZo)2],

Now, we cascade N+1 coupled line as in figure 9

figure 10:Cascaded coupled lines


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figure 11: Equivalent circuit for a transmission line with length l /2.

When we look at the equivalent circuit of two adjacent pair of N+1 cascaded coupledline, there are a transmission line with l /2 length and a J inverter. Using ABCD matrices wecan show that a transmission line with l /2 length can be modelled by a shunt LC circuit(Fig 11) and an admittance inverter between these LC circuits which transforms shunt LC

to series circuit (Fig 12). In this way, we can use low pass filter prototype element values

to determine Jn.

figure 12:Equivalent circuit for admittance inverter

the ABCD matrix for T network folowed by transformer is given by

If we equate it to the ABCD matrix of transmission line with length l/2 we can find thefollowing

Z12=jZo/sin2q

Z11=-jZocot2q

Z11-Z12=-jZocotq

the transformer in the equivalent circuit (Fig 11) shifts the filter response without changing

the amplitude so we can eliminate it. Then we have left with T network. For w=wo+Dwwhere wo is the center frequency


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2q=bl=( wo+Dw )p/wo (45)

for q~p/2 the series arms impedance of T network become zero, parallel arm impedance is

Z12=jZo/(sinp(1+Dw/wo))~-jZowo/(p(w-wo)) (46)

The impedance of parallel LC circuit near resonance is given by

Z= when we equate these impedances

Z= (46)

Using wo2=1/LC

Using inductance and capacitance values following design equations can be derived

k=2,3N (47)

By using (43), even and odd impedances of each coupled line can be found

Finding geometry ratios by using even and odd impedances:

Cascaded coupled lines can be realised in microstrip or stripline form.


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figure 13: Coupled stripline

Stripline: For the geometry in figure 13, even and odd impedances are given by

(48)

Using (48) we can derive s/d and w/d ratios:

where a2=Zoe/Zoo

figure 14: Coupled microstrip

Microstrip:to find the ratios for microstrip is more laborious to carry out. Here I used

Akhazard method.Firstly, we should determine equivalent single microstrip shape ratios (w/d)s. Then we can

realte coupled line ratios to single line ratios.

Zose=Zoe/2

Zoso=Zoo/2 for single microstrip

Using even and odd impedances for a single line we can find w/d ratio,

For narrow strips Zos {44-2er}


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for wide strips Zos {44-2er}

where d=59p2/(Zoer1/2)

A DESIGN EXAMLE AND COMPARISON:

I run the program for a low pass filter power loss of 15 dB at 2 GHz and cut off frequency

at 1 GHz. The filter impedance is chosen as 50 ohm. The highest and the lowest practicalimpedance values are 150 and 10 ohm respectively.

The output of the program for e=2.2:

Lengths of lines w/d ratios

l1=0.64cm w1/d= 22.65

l2=14cm w2/d=0.33

l3=0.64cm w3/d=22.65

For the substrate thickness of d=0.5 mm, I found the corresponding widths of lines.Then I designed the circuit by using SONNETthe results was satisfactory. The graph is

drawn according to SONNET, as it can be seen from the graph at 1 GHz 3.6 and 2 GHz 17.7dB power loss is obtained.


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Microwave Filters - Theoretical Information

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