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EE 524 PROJECT 1

MICROSTRIP

LOWPASS AND HIGHPASS

FILTER DESIGN

SUBMITTED BY:

Göksenin BOZDAĞ

SUBMITTED TO:

Asst. Prof. Dr. Sevinç AYDINLIK BECHTELER

December-2011

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CONTENTS

ABSTRACT…………………………………………………………………………………....................... 3

A)MICROSTRIP CHEBSYSEHEV LOWPASS FILTER..………………………………………….. 4

1.CHEBSYSEHEV RESPONSE………………………………………………..……………………… 5

2.CHEBSYSEHEV LOWPASS FILTER DESIGN………………………………………………….. 5

B) MICROSTRIP CHEBSYSEHEV HIGHPASS FILTER ..…………………………………………. 7

1.MICROSTRIP HIGHPASS FILTER DESIGN……………………………………………………… 10

C)REFERENCES…………………………………………………………………………………………………. 27

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ABSTRACT

The ambition of the project is designing Microstrip Chebsysehev Low Pass and High Pass

Filters. Some of the line calculations and design simulations have been done by the

computer program, QUCS. Realization of the filters have not been done but the realization

procedure and realization results will be added to this report as soon as it is possible.

Finally, 5th

degree Microstrip Chebsysehev Low Pass and 3rd

degree Microstrip Chebsysehev

High Pass Filters are designed and simulated.

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A) MICROSTRIP CHEBSYSEHEV LOWPASS FILTER

1)CHEBSYSEHEV RESPONSE

The Chebysehev response that exhibits the equal-ripple passband and

maximally flat stopband. The amplitude-squared transfer function that

describes this type of response is

LAr represents the ripple in dB and ɛ represents the ripple constant.

Tn(Ω) is a Chebysehev function of the first kind of order n.

General Response of Lowpass Chebysehev

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While desining this type of lowpass filters we use some coefficients. The coefficients are got

by calculation according to the formula given below or more generally look-up tables are

used.

Where

For the required passband ripple LAr in dB, the minimum stopband attenuation LAs in dB at

Ω = Ωs, the degree of a Chebyshev lowpass prototype, which will meet this specification, can

be found by

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2)CHEBSYSEHEV LOWPASS FILTER DESİGN

Example: Design a lowpass filter whose input and output are matched to a 50 Ω impedance

and cut-off frequency 2,5 GHz, equi-ripple 0.5 dB, and rejection of at least 40 dB at

approximately twice of the cut-off frequency.

Solution: Firstly, we determine the degree of filter according to the formula.

According to table g1=g5=1.7058, g2=g4=1.2296, g3=2.5408.

Π type lowpass prototype

LAs = 40 dB LAr = 0.5 dB

Then, n ≥4.8

So we can use n=5 coefficients from the table

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Secondly; Lumped elements are changed with their microstrip equivalents. (Richard’s

Transformation). Replacing inductors and capacitors by series and shunt stubs.

Third step is using unit elements converting short circuits to open circuits. (Kudora’s

Identities) N2 =1 + (Z2/Z1)

While Kuroda’s identities are applying, unit elements are put both left and right sight.

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For this figure we should calculate N2

.

N2

Z1=1 and (1/N2Z2)=1.7058 Then; N

2=1+(Z2/Z1) = 1.5862 (second identity is used)

(Z1N2) = 1 Z1= 1/ 1.5862 = 0.6304 (1/ N

2 Z2) = 1.7058 Z2 = 0.3696

When we put the calculated values, we get the circuit below.

Now we use the first identity for short circuits at center.

Z2/N2=0.3696 and Z1/N

2=1.2296 Then; N

2= 1 + (Z2/Z1) =1.3006

Z1 and Z2 are calculated as Z2=0.4807 Z1=1.5992

Lastly, we have to convert outer short circuits to open circuits so we again add unit elements

and we again apply Kuroda’s first identity.

Z2/N2 = 1 and Z1/N

2 = 0.6304 Then; N

2 = 1 + (Z2/Z1) = 2.5863

Z1 and Z2 are calculated as Z2=2.5863 Z1=1.6304

All shorts are converted to opens by applying Kuroda’s identities twice.

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At fourth step, all impedances are de-normalized, microstrip equivalents are calculated.

Impedance (Ω) Width (mm) Eeff Length (mm)

Z1=Z5=2.5859*50=129.299 0.305 3.00596 Qc=c/(fc*8*√����) = 8.65166

L1=L4=1.6304*50=81.5200 1.080 3.16777 Qc=c/(fc*8*√����) = 8.42780

Z2=Z4=0.4793*50=23.9659 8.190 3.76791 Qc=c/(fc*8*√����) = 7.72753

L2=L3=1.5992*50=79.9600 1.130 3.17625 Qc=c/(fc*8*√����) = 8.41655

Z3=0.3936*50 =19.6800 10.48 3.85611 Qc=c/(fc*8*√����) = 7.63865

Figure 1 Chebsyshe

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Figure 1 Chebsyshev Lowpass Filter fc=2.5 GHz Er=4.5 t=1.5 mm

S11 and S12 response of the designed filt

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Figure 2

of the designed filter. According to this figure we can obviously see that both lowpass

cut-off frequency sliding are acceptable.

lowpass characteristics and

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B) MICROSTRIP CHEBSYSEHEV HIGHPASS FILTER

While designing highpass filter, we fallow a different procedure from lowpass filter. For

constructing highpass design procedure, we are helped by optimum distributed highpass

filter circuit scheme and its’ look-up table. The type of filter consists of a cascade of shunt

short-circuited stubs of electrical length Qc at some specified frequency fc (usually the cutoff

frequency of high pass), separated by connecting lines (unit elements) of electrical length

2Qc. Although the filter consists of only n stubs, it has an insertion function of degree 2n – 1

in frequency so that its highpass response has 2n – 1 ripples. Characteristic response of the

filter is calculated by the formula given below.

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For highpass applications, the filter has a primary passband from Qc to π - Qc with a

cutoff at Qc. The harmonic passbands occur periodically, centered at Q = 3π/2, 5π/2.

1)MICROSTRIP HIGHPASS FILTER DESIGN

Example: Design a highpass filter whose input and output are matched to a 50 Ω impedance

and cut-off frequency 2,5 GHz with equi-ripple 0.1 dB.

Solution: Let we assume that our filter is third degree and Qc is 35 degree. Of course we can

choose another filter degree and Qc degree but for third degree necessary impedances can

not be get from the used substrate FR4 ( Er=4.5 t=1.5 mm).

Impedances are de-normalized as Zi=Z0/Yi Zi,i+1=Z0/Yi,i+1.

Generally length of short stubs are Qc and length of lines are 2Qc in design procedure of this

filter type. On the other hand, according to my simulation results, I get the optimum results

when I take length of short stubs are λ/8 and length of lines are 2λ/8. Additionaly, instead

of using square root of Eff, I used Eff.

Impedance (Ω) Width (mm) Eeff Length (mm)

Z1=Z3=50/0.40104=124.6758 0.31969 3.01026 Qc=c/(fc*8*����) = 4.9830

L1,2=L2,3=50/1.05378=47.448 3.07166 3.43108 Qc=2c/(fc*8*����)= 8.7436

Z2=50/0.48294=103.5325 0.57839 3.07357 Qc=c/(fc*8*����) = 4.8803

Figure 3 Chebsyshev

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Figure 3 Chebsyshev Highpass Filter fc=2.5 GHz Er=4.5 t=1.5 mm

S11 and S12 response of the designed highpass

cut

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Figure 4

hpass filter. According to this figure we can obviously see that both

ut-off frequency sliding are acceptable. (fc=2.5 GHz)

both highpass characteristics and

Figure 5

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Figure 5 Chebsyshev Highpass Filter fc=3 GHz Er=4.5 t=1.5 mm

S11 and S12 response of the designed highpass

cut

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Figure 6

hpass filter. According to this figure we can obviously see that both

ut-off frequency sliding are acceptable. (fc=3 GHz)

both highpass characteristics and

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E) REFERENCES

� Books

1. Microstrip Filters for RF/Microwave Applications, Jia-Sheng Hong, M. J. Lancaster

2. RF Circuit Design, R.Ludwig – P. Bretechko

3. Microwave Engineering, David M. Pozar