RF Filter Part 1

8/9/2019 RF Filter Part 1

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Project Guide:

Dr.(Prof) Dilip Bhattacharya

Project Group:

1. Tuhin Dutta

2. Tanmoy Dey

3. Soumyajit Kundu

4. Arani Ali Khan

5. Sourav Ghosh6. Dipanjan Mondal

Acknowledgement


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We deem it a pleasure to acknowledge our deep sense ofgratitude to our project guide Prof. D. B. who directed and

guided us with his advice and constant inspiration whicheased the task of working with a previously unknowntopic and hence technical report.

We wish to reciprocate in full measure the kindnessshown by our H.O.D. (E.C.E dept.) who encouraged us and

inspired us in successfully attending the project classesand finally completing this technical report.

We also wish to give our heartiest thanks to each other ofour peer group for giving there helping hands throughoutthis entire period of time.

Contents


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1. Introduction

2. Basic RC Filter3. Filter components

4. Constant k-Type Filter

5. Insertion Loss Method

6. Maximally Flat Filter

7. Chebyshev Filter8. Low Pass Filter prototype

9. Impedance and Frequency Scaling

10. Design Specification (Problem)

11. Filter Implementation using Lamped elements

12. Implementation using Micro Strip Line

13. Micro strip Discontinuities

CERTIFICATION


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It is here by certified that the project report on RF filterwhich is being submitted during the period August 2009to April 2010 as a complete fulfillment of the

requirement of ECE -783, is a record of the work done bythe candidates, which they have carried under theguidance ofProf. D. Bhattacharya.

Mr. Santu Sarkar.

Prof. D. Bhattacharya.Head Of The Department ECE, Director

Academy Of Technology, AcademyOf Technology

Aedconagar, Hooghly,Aedconagar, Hooghly,

WB , PIN 712121 WB , PIN 712121

Signature-

Signature-

(Mr. Santu Sarkar)

(Prof. D. Bhattacharya)


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Introduction

A filter is a four terminal network which is designed generally byreactive elements to pass a certain band of frequencies andattenuate fqs outside this band .The band of frequency which istransmitted freely through the filter is called the attenuation orstop band of the filter .The frequency at which the transition froma stop band to pass band and vice versa occurs called cutofffrequency.

Usage of filter :

Filters are used in power supply systems, circuits fq telephonecircuit, and instruments & in other fields. Filter output is used in aload.

Types of filter :-

i) Low pass filter- This type of filter allows transmission fromzero to a cut off frequency fc and attenuates all frequency

above fc.ii) High pass filter-This types of filter allows transmission of fq

above a certain cut off frequency fc , & attenuate allfrequency lying below fc.

iii) Band pass filter-A band pass filter allows transmission ofall fq lying above a fixed minimum frequency f1 and belowa fixed minimum frequency f2.So it allows within the band

(f2-f1) and attenuates frequencies outside this band.iv) Band stop filter- This filter allows transmission of all

frequency from zero to f1 and from f2 to infinity andattenuate all frequencies lying within the range (f2 f1).

Ideal filter characteristic:


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Ideal filters are those that give zeroattenuation and constant impedance in thepass band and offer infinite attenuation forall frequencies outside the pass band.

As it is shown in fig. But all practical filters are non ideal.

Filter component:

Basically resistance(R), capacitance(c) & inductance (L) areused as filter component. In low pass filter resistance,capacitance& inductance & an op amp is used.

Basic RC filters:

Series RC work as low pass and highpass filters. The RC low pass circuit and itsbehavior are shown here.

So at dc and low frequencies the output voltage

is about the same as the input voltage and at

higher frequencies the reactances ofthe capacitance decreases resulting in adecreased output voltage .

So , where

The magnitude of the output voltage is

| | = The ratio of to is thegain Av of the circuit.

The magnitude of the gain is :

|Av|= | / |=


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So for lower fqs ->0 .|Av|=1

So for lower fqs ->infinite .|Av|=0.

So the frq response will be looked as following fig.

This is a frequency response for a low passfilter. If we change the place of resistance&capacitance the circuit will behave as highpass filter. At high frequencies the capacitivereactance is negligible, so the o/p voltage isnearly equal to . At lower frequencies theincreased values of voltage drop across C,causes to decrease.

Here gain Av= V0/Vi=R/(R+jXc) = j CR/1+j CR.

Hence |Av|=|Vo/Vi|= CR/

We know Xc= 1/j CSo if -> 0 so Xc->infinite means c work as open circuit so Vowill decrease,Av also decrease.

If ->infinite so Xc-> 0 means c works asshort circuit path.

This fq response curve is shown in following figure.

Different component used in filters:

For RC filter we intentionally use the resistance, but we can alsouse inductance. But lowpass filter the W=2*pi*f, f is very low.Again we know Xl= jWL ,so for const. Xl the expinductance shouldbe very high. So the inductor should be bulky & very expencive.

Thus resistance is used & power is lossed for the usage ofresistance.But for driving the load power same as input signalpower is required.


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So for amplify the power an op- amp is used. Thus we can drivethe load with a same power as like as input signal power. Thusthe circuit configuration will be as follows for lowpass filter.

Here op amp is called activeelement. Thus low pass active filter isformed. But for high pass filter fq ishigh, so small inductance is needed.So there is no need for design bulkyinductor. Thus we can use smallinductors for good compatibility ofthe filter. Also the power loss factoris not worked as like s RC filter,because inductors so calledresistance is very low. Thus for MHzfq range usage filter , we preferseries LC filter .So there is no needfor op-amp.

LC filter network:

To illustrate about LC network we have to introduce K- constfilter. So we discuss T section of low pass filter as shown bellow-


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From the general T network we get the expression where z-parameters are introduced.

When = 0 ,then

= | = 0, then ( )

= | =0 = .

Now, we know for a symmetrical & reciprocal network thecharacteristic impedence is

Z0 = Zi1 = Zi2= = .

Where Zi1= and Zi2= . These are the expression for asymmetrical & reciprocal network.

Here Z11 = Z12 for symmetrical.

Here Z12 = Z21 for reciprocal.

=

is characteristic impedence for T section =

is characteristic impedence for section =


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Where =j & =1/j .

Since are reactances of opposite sign, the product

is real. The term is imaginary when | | > 1. The cut-off

takes place when the characteristic impedence becomes reactive.Therefore the transmission from the pass-band to the stop band

occurs when =0 or =-1.

The quantity is called nominal characteristic resistance of

the filter & is denoted by RK , Thus

= RK = = =K.

Here K is a const. The filter is termed as constant K filter. For thisfilter two cases occur :-

Case 1 :- Reactances of are opposite in signs, so the term

is real & when is less then 1. Then 1+ is also real. So the

characteristic impedence is real and also resistive in nature.Therefore when the filter is terminated by a load having a valueequal to the characteristic impedence, it will absorb real power

from the source. Since the filter is composed of purely reactiveelements, theoretically all the received power from the source willbe terminated to the load.

Case 2 :- The term is imaginary , when | | > 1. & the

characteristic impedence is reactive. Therefore the transition fromthe pass band to the stop band occurs. In the stop-band as thecharacteristic impedence is reactive, so the filter is terminated by

its characteristic reactance and it does not absorb the real powerfrom the source. Thus the power will reflected by the load.

Frequency dependence of characteristic impedence:-


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So the characteristic impedence of T section is

*

*

*

*

Similarly, characteristic impedence of corresponding - section is

= /

= /

Here normalized characteristic impedence / and / is

plotted against . It is observed that is real in the pass band

and reaches zero value at the cut-off frequency. is reactiveafter the cut-off frequency , where is grater then unity. The

reactive impedances are multiplied by j , so that they can beplotted as real variables.


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Filter design by insertion-loss method:

The power loss ratio of a network is defined as the availableor incident power divided by actual power delivered to the load;thus

= 1/ .

Where is the input reflection co-efficient for a lossless networkterminated in a resistive load impedance, = . The insertionloss, measured in decibels, L = .when the terminatingload impedence equals the internal impedence of the generatorat the input end.

In general the insertion loss is defined as the ratio of powerdelivered to the load when connected directly to the generator tothe power delivered, when the filter is inserted.

For a passive network it is clear that the reflected powercant exceed the incident power and hence one restriction on ()is | ()| 1.If the normalized input impedence of the network is

() = () + j ()

We have, () = -1) / +1) =[ () -1 + j ()] /[ () +1+ j

()]is an even function of and is an odd function of .

Hence (-) = [ () - 1 - j ()] /[ () +1 - j ()] =

And thus = = .


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It is apparent from this relation that = is an evenfunction of and must therefore contain only even function of .

= =

Where M and N are real & non-negative polynomials in .

The power loss ratio can now expressed as

= 1+ = 1+

It shows that, must be an even polynomial in . Since itequals . Hence we replace = . And denote = P(

. So we get= 1+ .

Maximally Flat Filter Characteristic :

The power loss ratio for a maximally flat low-pass filter(Butterworth) is obtained by choosing the polynomial Q equal tounity and choosing equal to .

Hence = 1+ .

The pass band is the region from to the cut-off frequency . The maximum value of in the pass band is 1+ .For , thepower loss ratio increases indefinitely at a rate dependent on theexponent 2N, which is related to the number of filter sectionemployed.

Equal ripple or chebyshev filter characteristic :


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The power loss ratio for equal ripplefilter is chosen as = 1+ .

Where is the chebyshev

polynomial of degree N. In besidesfig, characteristic response of equalripples and maximally flat filter areshown.

Ladder circuit for low-pass filter protype:-

For normalized low-pass design where the source impedenceis 1 ohm and the cut-off frequency is =1.However the elementvalue for the ladder type circuit of the figure below can becalculated.

Classification and notification of used elements:

Here,


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Generator resistance or conductance

Inductance or capacitance for k = 1 to N.

Load resistance for =shunt capacitor.

Load conductance for = series inductor.

Impedence and frequency scaling:

In the prototype design the source resistance can beobtained by multiplying the impedances of the prototype designby . Then, if we assume prime denote the impedence scaled

quantities, we have the new filter component value given by

For frequency scaling we have to change the cut-off frequency ofa low-pass filter prototype from unity to .Thus we scale thefrequency dependence of the filter by the factor 1/ , which isaccomplished by replacing by .

Thus .

The new element values are determined by applying thesubstitution of with , to the series reactances andsuscptances j of the prototype.

Thus,

.


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.

Design specification:

Design a lowpass (Butterworth) filter which has a cut off

fq 4 GHz. At stop band at frequency 5 GHz ,where

minimum attenuation will be 30 db. Assume 50 ohm

environment.

Solution: -

Formula for number of component used is :-

= 5GHz ; c = 4 GHz ;A = 30 dB.

So by calculating, we get n=16. So number of component used inthe filter is 16.

Using constant prototype method, where prototype began with ashunt element. So resulting primary circuit look like following:-

The g- parameter values we can calculate. Where, It is seenthat if :-

k is odd , then gk=capacitor

k is even then gk = Inductor.


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Formula for g- parameter calculation is :-

gk = 2*sin [(2k-1)*pi/2n]

Where for n=16 and k = 1 ,2,3.16 . g values can be calculated

.The inductor(L) & capacitor(C) values can also calculated byimpedance scaling (because design is in 50 ohm environment) asfollows:-

C1=0.155pf

L2=1.1548nH

C3=0.7501pf

L4=2.524nH

C5=1.2302pf

L6=3.5090nH

C7=1.5320pf

L8=3.9597nH

C9=1.5838pfL10=3.8075432nH

C11=1.4036pf

L12=3.07570nH

C13=1.00967pf

L14=1.87544nH

C15=0.4621947pf

L16=0.389929nH

So by using this values PSpice program , we can observe theresponse. PSpice program code is given as follows:-


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PROGRAMME:-

R1 1 2 50ohm

C1 2 0 0.155pf

L2 2 3 1.1548nH

C3 3 0 0.7501pf

L4 3 4 2.524nH

C5 4 0 1.2302pf

L6 4 5 3.5090nH

C7 5 0 1.5320pf

L8 5 6 3.9597nH

C9 6 0 1.5838pf

L10 6 7 3.8075432nH

C11 7 0 1.4036pf

L12 7 8 3.075470nHC13 8 0 1.00967pf

L14 8 9 1.87544nH

C15 9 0 0.461947pf

L16 9 10 0.389929nH

R2 10 0 50ohm

Vin 1 0 ac 1v

. AC LIN 200 1 20G

. PROBE

.END


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We get the output response in the PSpice o/p window. If theresponse does not match with the desired response, then wehave to iterate the inductor and capacitor values by trial and errorcorrection method. After correction we have to use the changedinductor and capacitor values. Thus we get the output responseshown in following figure.

MICROWAVE FILTER DESIGNING

In order to design all type of filters it is very laborites to designindividually different types of filters. So designer designs a low-pass or high-pass prototype. This prototype filter is transformed in

desired form like band-pass or band-stop filter.

There are three steps to complete a microwave filter.

1. Design the prototype low-pass or high-pass filter withdesired pass-band and stop-band specification.

2. Transform it into desired form band-pass or band-reject.

3. Implement the filter elements using transmission line (using

microstrip or stripline etc).

The frequency and impedance scaling must be done beforeimplement it.

There are two techniques to design the filter.


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1. Image parameter method.

2. Insertion loss method.

Image parameter method is not suitable for filter designing it is

some time used to design solid state travelling wave amplifier.

Insertion loss method is widely used in microwave filter designing.

We have used this method to design the filter.

Problem(Chebyshev filter):

Design a low pass filter having cut off frequency 4GHz.Pass band ripple 1dB for Chebyshev filter. Stop band

attenuation 30dB at 50GHz.

Soln.

Chebyshev Filter Design:Ripple Magnitude:

10log10 (1+am2) = 1 dB

Or, log10 (1+am2) = 0.1

Or, 1+am2 = 100.1

Or, am2 = 100.1 -1 = 0.2589

Or, am = 0.5088

Stop band attenuation at x :

30 dB = 10 log10 [1 + am2cosh2(n cosh-1 x)]

Or, 3 = log10 [1 + 0.2589 cosh2(n cosh-1(1.25))]


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Or, 3 = log10 [1 + 0.2589 cosh2(n 0.6931)]

Or, 999/0.2589 = cosh2(n 0.6931)

Or, 62.11789 = cosh(n 0.6931)

Or, n= 6.957 7.

Therefore, the filter order should be 7. So, the number ofelements is 7.

= ln[cot (Am/17.37)] = 2.8558 where Am = rippleheight = 1 dB

p1 = sin (/14) = 0.22252 p2 = sin

(3/14) = 0.62349p3 = sin (5/14) = 0.90097 p4 = sin(7/14) = 1p5 = sin (9/14) = 0.90097 p6 = sin(11/14) = 0.62349p7 = sin (13/14) = 0.22252

q1 = sinh2(2.8558/14) + sin2(/7) = 0.0422 +0.18825

= 0.23045q2 = 0.65346 q3 =0.9927q4 = 0.9927 q5 =0.65346q6 = 0.23045 q7 =0.0422

The g-parameters:

g0 = 1 g1 = 2p1/[sinh(/2n)] =(2*0.22252)/0.2054 = 2.167g2 = (4p1p2)/(q1g1) = 1.111 g3 = (4p2p3)/(q2g2)= 3.095g4 = (4p3p4)(q3g3) = 1.17298 g5 = (4p4p5)/(q4g4)


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= 3.095g6 = (4p5p6)/(q5g5) = 1.111 g7 = (4p6p7)/(q6g6)= 2.167g8 = 1.

The lamped element values for cascaded sections for 50 ohmterminations are obtained as

Ck = gk/ZLc and Lk = gk ZL/c

C1 = 2.167/(50*2*4*109)=1.7244 pFC3 = 3.095/(50*2*4*109)

=2.4629 pFC5 = 3.095/(50*2*4*109)=2.4629 pFC7 = 2.167/(50*2*4*109)=1.7244 pF

L2 =(1.111*50)/(2*4*109)=2.21026 nHL4 =(1.17298*50)/(2*4*109)

=2.333 nHL6 =(1.111*50)/(2*4*109)=2.21026 nH

Circuit Diagram:

Pspice program for Chebyshev Filter (7 elements):


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Vs 1 0 1VRs 2 1 50 ohmC1 2 0 1.7244 pFC3 3 0 2.4629 pF

C5 4 0 2.4629 pFC7 5 0 1.7244 pFL2 2 3 2.21026 nHL4 3 4 2.333 nHL6 4 5 2.21026 nHR0 5 0 50 ohm.ac lin 200 1 6GHz.probe.end

At cut off frequency, the amplitude of output voltage = 445.065mV.

Filter Implementation:

Lumped element filter design works well at low frequencies. Twoproblem arise at microwave frequency:

i. Lumped element such as inductors and capacitorsare generally available only for a limited range ofvalues and are also difficult to implement atmicrowave frequency.

ii. At microwave frequency the distance betweencomponents is not negligible.

Richards transformation is used to convert lumped elements totransmission line sections, while Kurodas identities can be usedto separate filter elements by using transmission line sections.Because such additional transmission line sections do not affectthe filter response, this type of design is called redundant filtersynthesis.


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Richards Transformation:

The transformation,

=tan l =tan (l/vp)

The transformation is introduced to synthesis an LC- networkusing open and short circuited transmission lines.

The reactance of an inductor can be written as

jXL = jL = jL tan l

And the susceptance of a capacitor can be written asjBC = jC = jC tan l

The result indicates that an inductor can be replaced with a shortcircuited stub of length l and characteristic impedance L, while acapacitor can be replaced with an open circuit stub of length land characteristic impedance 1/C.

Cut off occurs at unity frequency for a L.P.F. prototype to obtain

the same cut off for Richards transformed filter shows that

= 1 = tan l

which gives, l=/8, where is the wavelength of a line at the cutoff frequency c. These lines are called commensurate lines.


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Kurodas Identities:

n2

n2:1

n21:

n2

n2

n2 = 1 +

n2

n2

n2

n2

n2

(d)

(c)

(a)

(b)

Z1Z1

Z2Z2

Z1 Z1

Z2/Z1

Z2

Z1

Z1

Z1

Z2Z1

Z

Lumped element circuit:


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g-parameters of the derived circuit are:

g0=1,g1=2.167,g2=1.111,

g3=3.095,g4=1.173,g5=3.095,g6=1.111,g7=2.167, g8=1.

Replacing capacitors by open stubs and inductors by short stubsusing Richards transformation, we get


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Using Kurodas Identities:


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After impedance scaling the characteristic impedances are:

1. Z(g1)=67.3, 2.Z0=194.4,

3. Z(g2)=54.185, 4.

Z0=103.425,5. Z(g3)=20.92, 6.Z0=74.495,

7. Z(g4)=16.125, 8.Z0=71.815,

9. Z(g5)=16.15, 10.Z0=71.335,

11. Z(g6)=20.265, 12. Z0=84.205,

13. Z(g7)= 44.305, 14.Z0=179.95,


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15. Z0=50.

Length of the elements:

l=/8

Now =c/f= (3*1010)/ (4*109)=7.5 cm

Therefore, l=/8=0.9375 cm

w/h ratio calculation:

e=(r+1)/2 + (r-1)/[2(1+12d/w)]

Z0= (60/ e) ln(8d/w + w/4d)

for w/d 1

120/[ e + 1.393 + 0.667 ln (w/d + 1.444)] forw/d 1.

Z0= 74.145 (for w/d=1 & r=4)

51.037 (for w/d=2 & r=4)


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For Z0 > 51.037 we use the formula

For Z0 < 51.037 we use the formula

w/d ratios of the filter circuit:


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1. Z(g1)=67.3, A= (Z0/60) (5/2) +(3/5)(0.23 + .00275)

= 1.9280

w/d = 8eA/(e2A -2)= 1.2149

2. Z0= 194.4, A=5.2774, w/d=0.04085

3. Z(g2)=54.185, A=1.5824,w/d=1.79547

4. Z0=103.425, A=2.8799,w/d=0.45193

5. Z(g3)=20.92, B= 377/(2 Z0r)=14.154,

w/d=6.9405

6. Z0=74.495, A=2.1176, w/d=0.99124

7. Z(g4)=16.125, B=18.3625,w/d=9.51497

8. Z0=71.815, A=2.047, w/d=1.0686

9. Z(g5)=16.15, B=18.334,w/d=9.4974

10. Z0=71.335, A=2.034, w/d= 1.0832

11. Z(g6)=20.265, B=14.611, w/d=7.21866

12. Z0=84.205, A=2.3735, w/d=0.7584

13. Z(g7)=44.305, B=6.6831,w/d=2.4884

14. Z0=179.95, A=4.8966, w/d=0.0598


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15. Z0=50, B=5.9219, w/d=2.0531

Microstrip discontinuities

1. Open end discontinuity:Open end discontinuity frequentlyoccurs in microstrip circuits. The equivalent circuit of anopen end is represented by an excess capacitance COC ,which can be transformed into equivalent length oftransmission line, lOC .


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COC/w = exp {2.3026Ci(r)[log(w/h)](i-1)} pF/m

lOC/h = (COC/w)(cZomw)/hr

lOC/h = 0.412 [(re + .03)/(re 0.258)][(w/h + 0.264)/(w/h + 0.8)]

Calculation of lOC for the open ends:

1. Z(g1)=67.3 & w/h=1.2149

re=(5/2) + (3/2)/[1+12(1/1.2149)]= 2.9548

lOC/h = 0.412 [(re + .03)/(re 0.258)][(w/h + 0.264)/(w/h + 0.8)]

= 0.364967

lOC=0.579 mm

2. Z(g2)=54.185 & w/h=1.79547

re=3.041143

lOC/h=0.3924598

lOC=0.623 mm

3. Z(g3)=20.92 & w/h=6.9405

re=3.4080106

lOC/h=0.451399421


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lOC=0.71658 mm

4. Z(g4)=16.125 & w/h=9.51497

re=3.4975

lOC/h=0.4578694

lOC=0.7267 mm

5. Z(g5)=16.15 & w/h=9.4974

re=3.497012619

lOC/h=0.457837269

lOC=0.7254 mm

6. Z(g6)=20.265 & w/h=7.21866

re=3.4193

lOC/h=0.452321149

lOC=0.7180 mm

7. Z(g7)=44.305 & w/h=2.4884

re=3.121643

lOC/h=0.412040534

lOC=0.65405 mm

Filer implementation

We designed LPF using lumped elements at low frequencies, but

two problms arise at microwave frequencies.

First, lumped elements such as inductors and capacitors aregenerally available only for a limited range of values. Second,they are very difficult to implement at microwave frequencies.

There are three types of filter implementation,


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(i) Filter design using stubs.(ii) Stepped impedance LPF design.(iii) Coupled line filter.

Stepped impedance LPF:

It is relatively an easier way of filter implementation in micro stripor strip line. Here series inductors can be replaced as highimpedance (Z0=Zh) and shunt capacitors can be replaced as lowimpedance (Z0=Zl) line sections.

It is very popular because it takes less space comparatively otherfilter design using stubs.

Disadvantage:

Electrical performance is not good due to approximation involvedinto it.

Approximate equivalent circuits for short T.L sections:

We begin by finding the approximate equiv. circuits for a shortlength of T.L having either a very large or very small char.Impedance.

Now the ABCD parameters of length l, having char. Impedance Z0can be shown as,


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Fig:1

Now Z parameters can be obtained as,

Z11=Z22=A/C=-jZ0cotbl .. (i)

Z12=Z21=1/C=-jZ0cosecbl .. (ii)

The series elements interms of Z paramet

Z11-Z12=jZ0tan(bl/2) ..(iii)

Shunt element is Z12= 1/C

Fig:2

T equivalent circuit for a T.L

Sections having bl


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B=0 ..(viii)

Fig:3

Now for blbl=R0L/Z0(from scaling equans.)

Here Z0=Zh; so bl=R0L/Zh ..(xi)

And the electrical length of the capacitor,

B=Y0bl=>bl=BZ0=CZl/R0 ..(xii)


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Where R0 is the filter impedance & l and C are normalizedelement values of the LPF.

Process of filter implementation for stepped impedance:

1. calculate w/d [where w is the width & d is thediameter]

2. find bl.(electrical length).3. find b [b=2/(g); (g)=(0)/(eff)]4. Then calculate length (l).

Problem:

Design a stepped impedance LPF having a maximally

flat response and cut off freq of 4GHz. The highest

practical band attenuation 30 db at 5GHz. The

highest practical line impedance is 100 ohm and low

impedance is 20 ohm. [Given =4; d=0.158cm]

consider f=3.5GHz.

SOLUTION:

From the pre computation at 4 GHz cut off,it can be shown that

the order of the low pass filter N=16.

Now we know the prototype values i.e value of g(k) for 16elements.

Now Zh=100 ohm, Zl= 20 ohm (REF: LUDWIG)

w/d= 8e^A/(8e^2A-2) when w/d


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When w/d>2.. (ii)

Now A= Zh/60 {(+1)/2}^1/2 + (-1)/ (+1) {0.23+0.11/}=2.7896

(iii)

B= 377 /2Zl()^1/2=14.796 (iv)

Now here for w/d, there are two relations.

But which one will be used for inductor & which one for capacitorcan be done by the following ways [taking w/d=1]

(eff)= (+1/2) + (-1)/2 (1/(1+12 d/w)^1/2)... (v)

=2.916 [where (eff) =2.916]

Now, Z0 = 60/( (eff)^1/2) ln(8d/w +w/4d) where w/d1

(vi)

=51.037 [ where (eff)=3.06]

Here for equn (vi), Z0= 85.44 but we have consideredZh=100>Z0, so for inductor w/d always


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But for equn (vii), Z0= 51.037 but we have considered Zl=202.

Now for equn. (i), w/d for inductor =0.49. so, w= 0.08cm=.8mm

And from equn (ii) w/d for capacitor =7.335. so w= 11.59mm.

Now we have to calculate the length through the following waysfor inductor and capacitor separately.

For capacitor electrical length,

b=[2 *((eff))^1/2 *f]/c . (viii)

For capacitor dielectric, (eff)=3.42

So, b=(6.28*1.85*3.5*10^9)/(3*10^10) = 1.3554 [wherebl=CZl/R0 ]

Now for inductor the electrical length,

b= (6.28*1.672*3.5*10^9)/(3*10^10)=1.225 [where (eff )=2.798] [where bl=R0L/Zh ]

now, g1=.196; so for capacitor bl1= .196*20/50=.0784= 4.49=>l1=.578mm

g2=.58057; so for inductor,bl2= .58057*50/100=.290285=16.64=>l2=2.36mm

Thus the odd number will be for capacitor and even no. will be for

inductor.Similarly,

g3=.9428; so for capacitor bl3=(.9428*20)/50=21.618=>l3=2.78mm


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g4=1.2688; so for inductor,bl4=36.37=>l4=5.179mm

g5=1.546; so for capacitor, bl5=.6184=32.74=>l5=4.56mm

g6=1.7637; so for inductor, bl6=.8818=50.55 ,l6=7.1983mm

Now in this following way we can find the others length.

Sl. No. Zl or

Zh

(in

ohm)

bl(i)

(in

degree)

w(i)

(in

mm)

l(i)

(in mm)


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1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

20

100

20

100

20

100

20

100

20

100

20

100

20

100

20

100

0

4.5

16.6

4

21.6

18

36.3

7

32.7

4

50.55

43.8

9

57.04

45.6

3

54.85

40.4

44.31

29.0

9

27.02

13.31

5.61

11.5

9

0.8

11.5

9

0.8

11.5

9

0.8

11.59

0.8

11.5

9

0.8

11.59

0.8

11.

59

0.8

11.

59

0.8

0.578

2.36

2.78

5.179

4.56

7.1983

5.65

8.12

5.87

7.81183

5.206

6.31

3.75

3.8481

1.713

0.8


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fig4: Steped impedance filter design

Fig5: Microstrip layout


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Documents

RF Filter Part 1