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**Microwave Filter Design**

By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Eng Engineering Campus USM Nibong Tebal 14300 SPS Penang

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Contents Composite filter LC ladder filter Microwave filter 2

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Composite filter m<0.6 for m-derived section is to place the pole near the cutoff frequency(wc) For 1/2 p matching network , we choose the Z’1 and Z’2 of the circuit so that 3

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Image method Let’s say we have image impedance for the network Zi1 and Zi2 Where Zi1= input impedance at port 1 when port 2 is terminated with Zi2 Zi2= input impedance at port 2 when port 1 is terminated with Zi1 Where Zi2= V2 / I2 Then @ and V1 = - Zi1 I1 4

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ABCD for T and p network 5

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**Image impedance in T and p network**

Substitute ABCD in terms of Z1 and Z2 Substitute ABCD in terms of Z1 and Z2 Image impedance Image impedance Propagation constant Propagation constant 6

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Composite filter 7

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**Constant-k section for Low-pass filter using T-network**

If we define a cutoff frequency And nominal characteristic impedance Then Zi T= Zo when w=0 8

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continue Propagation constant (from page 11), we have Two regions can be considered w<wc : passband of filter --> Zit become real and g is imaginary (g= jb ) since w2/wc2-1<1 w>wc : stopband of filter_--> Zit become imaginary and g is real (g= a ) wc a,b stopband passband wc a Mag p b w 9 w

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**Constant-k section for Low-pass filter using p-network**

Zi p= Zo when w=0 Propagation constant is the same as T-network 10

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**Constant-k section for high-pass filter using T-network**

If we define a cutoff frequency And nominal characteristic impedance Then Zi T= Zo when w = 11

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**Constant-k section for high-pass filter using p-network**

Zi p= Zo when w= Propagation constant is the same for both T and p-network 12

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Composite filter 13

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**m-derived filter T-section**

Constant-k section suffers from very slow attenuation rate and non-constant image impedance . Thus we replace Z1 and Z2 to Z’1 and Z’2 respectively. Let’s Z’1 = m Z1 and Z’2 to obtain the same ZiT as in constant-k section. Solving for Z’2, we have 14

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**Low -pass m-derived T-section**

For constant-k section and Propagation constant where 15

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**continue 16 If we restrict 0 < m < 1 and**

Thus, both equation reduces to Then When w < wc, eg is imaginary. Then the wave is propagated in the network. When wc<w <wop, eg is positive and the wave will be attenuated. When w = wop, eg becomes infinity which implies infinity attenuation. When w>wop, then eg become positif but decreasing.,which meant decreasing in attenuation. 16

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**Comparison between m-derived section and constant-k section**

M-derived section attenuates rapidly but after w>wop , the attenuation reduces back . By combining the m-derived section and the constant-k will form so called composite filter.This is because the image impedances are nonconstant. 17

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**High -pass m-derived T-section**

and Propagation constant where 18

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**continue 19 Thus wop< wc If we restrict 0 < m < 1 and**

Thus, both equation reduces to Then When w < wop , eg is positive. Then the wave is gradually attenuated in the networ as function of frequency. When w = wop, eg becomes infinity which implies infinity attenuation. When wc>w >wop, eg is becoming negative and the wave will be propagted. 19

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continue a wop wc w M-derived section seem to be resonated at w=wop due to serial LC circuit. By combining the m-derived section and the constant-k will form composite filter which will act as proper highpass filter. 20

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**m-derived filter p-section**

Note that The image impedance is 21

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**Low -pass m-derived p-section**

For constant-k section Then and Therefore, the image impedance reduces to The best result for m is 0.6which give a good constant Zip . This type of m-derived section can be used at input and output of the filter to provide constant impedance matching to or from Zo . 22

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Composite filter 23

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**Matching between constant-k and m-derived**

The image impedance ZiT does not match Zip, I.e The matching can be done by using half- p section as shown below and the image impedance should be Zi1= ZiT and Zi2=Zip It can be shown that Note that 24

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Example #1 Design a low-pass composite filter with cutoff frequency of 2GHz and impedance of 75W . Place the infinite attenuation pole at 2.05GHz, and plot the frequency response from 0 to 4GHz. Solution For high f- cutoff constant -k T - section or Rearrange for wc and substituting, we have 25

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continue For m-derived T section sharp cutoff 26

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continue For matching section m=0.6 27

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continue A full circuit of the filter 28

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Simplified circuit

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continue Pole due to m=0.2195 section Pole due to m=0.6 section 30

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**N-section LC ladder circuit (low-pass filter prototypes)**

Prototype beginning with serial element Prototype beginning with shunt element 31

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**Type of responses for n-section prototype filter**

Maximally flat or Butterworth Equal ripple or Chebyshev Elliptic function Linear phase Equal ripple Elliptic Linear phase Maximally flat 32

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**Maximally flat or Butterworth filter**

For low -pass power ratio response Prototype elements Series R=Zo g0 = gn+1 = 1 Shunt G=1/Zo where C=1 for -3dB cutoff point n= order of filter wc= cutoff frequency Series element No of order (or no of elements) Shunt element k= 1,2,3…….n Where A is the attenuation at w1 point and w1>wc 33

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Example #2 Calculate the inductance and capacitance values for a maximally-flat low-pass filter that has a 3dB bandwidth of 400MHz. The filter is to be connected to 50 ohm source and load impedance.The filter must has a high attenuation of 20 dB at 1 GHz. Solution Prototype values g0 = g 3+1 = 1 First , determine the number of elements Thus choose an integer value , I.e n=3 34

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continue 35

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or 36

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**Equi-ripple filter 37 For low -pass power ratio response**

Chebyshev polinomial where Cn(x)=Chebyshev polinomial for n order and argument of x n= order of filter wc= cutoff frequency Fo=constant related to passband ripple Where Lr is the ripple attenuation in pass-band 37

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Continue Prototype elements where Series element Shunt element 38

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Example #3 Design a 3 section Chebyshev low-pass filter that has a ripple of 0.05dB and cutoff frequency of 1 GHz. From the formula given we have F1= F2= a1=1.0 a2=2.0 b1=2.043 g1 = g3 = g2= 39

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**Transformation from low-pass to high-pass**

Series inductor Lk must be replaced by capacitor C’k Shunts capacitor Ck must be replaced by inductor L’k 40

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**Transformation from low-pass to band-pass**

where Now we consider the series inductor normalized Thus , series inductor Lk must be replaced by serial Lsk and Csk Impedance= series 41

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**continue 42 Now we consider the shunt capacitor**

Shunts capacitor Ck must be replaced by parallel Lpk and Cpk Admittance= parallel 42

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**Transformation from low-pass to band-stop**

where Now we consider the series inductor --convert to admittance Thus , series inductor Lk must be replaced by parallel Lpk and Cskp admittance = parallel 43

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Continue Now we consider the shunt capacitor --> convert to impedance Shunts capacitor Ck must be replaced by parallel Lpk and Cpk 44

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Example #4 Design a band-pass filter having a 0.5 dB ripple response, with N=3. The center frequency is 1GHz, the bandwidth is 10%, and the impedance is 50W. Solution From table 8.4 Pozar pg 452. go=1 , g1=1.5963, g2=1.0967, g3= , g4= 1.000 Let’s first and third elements are equivalent to series inductance and g1=g3, thus 45

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continue Second element is equivalent to parallel capacitance, thus 46

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**Implementation in microstripline**

Equivalent circuit A short transmission line can be equated to T and p circuit of lumped circuit. Thus from ABCD parameter( refer to Fooks and Zakareviius ‘Microwave Engineering using microstrip circuits” pg 31-34), we have Model for series inductor with fringing capacitors Model for shunt capacitor with fringing inductors 47

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**48 p-model with C as fringing capacitance**

T-model with L as fringing inductance ZoC should be low impedance ZoL should be high impedance 48

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**Example #5 Note: For more accurate calculate for difference Zo 49**

From example #3, we have the solution for low-pass Chebyshev of ripple 0.5dB at 1GHz, Design a filter using in microstrip on FR4 (er=4.5 h=1.5mm) Let’s choose ZoL=100W and ZoC =20 W. Note: For more accurate calculate for difference Zo 49

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**continue More may be needed to obtain sufficiently stable solutions 50**

The new values for L1=L3= 7nH-0.75nH= 6.25nH and C2=3.543pF-0.369pF=3.174pF Thus the corrected value for d1,d2 and d3 are More may be needed to obtain sufficiently stable solutions 50

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**Now we calculate the microstrip width using this formula (approximation)**

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**Implementation using stub**

Richard’s transformation At cutoff unity frequency,we have x=1. Then The length of the stub will be the same with length equal to l/8. The Zo will be difference with short circuit for L and open circuit for C.These lines are called commensurate lines. 52

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Kuroda identity It is difficult to implement a series stub in microstripline. Using Kuroda identity, we would be able to transform S.C series stub to O.C shunt stub d d=l/8 53

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Example #6 Design a low-pass filter for fabrication using micro strip lines .The specification: cutoff frequency of 4GHz , third order, impedance 50 W, and a 3 dB equal-ripple characteristic. Protype Chebyshev low-pass filter element values are g1=g3= = L1= L3 , g2 = = C2 , g4=1=RL Zo Zo Using Richard’s transform we have ZoL= L=3.3487 and Zoc=1/ C=1/0.7117=1.405 54

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**Using Kuroda identity to convert S.C series stub to O.C shunt stub.**

We have and thus Substitute again, we have and 55

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**Band-pass filter from l/2 parallel coupled lines**

Microstrip layout Equivalent admittance inverter Equivalent LC resonator 57

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**Required admittance inverter parameters**

The normalized admittance inverter is given by where A B C D where E 58

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Example #7 Design a coupled line bandpass filter with n=3 and a 0.5dB equi-ripple response on substrate er=10 and h=1mm. The center frequency is 2 GHz, the bandwidth is 10% and Zo=50W. We have g0=1 , g1=1.5963, g2=1.0967, g3=1.5963, g4= 1 and W=0.1 A C D E 59

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**B B D E 60 The required resonator**

Using the graph Fig 7.30 in Pozar pg388 we would be able to determine the required s/h and w/h of microstripline with er=10. For others use other means. 60

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Thus we have For sections 1 and 4 s/h= > s=0.45mm and w/h=0.7--> w=0.7mm For sections 2 and 3 s/h=1.3 --> s=1.3mm and w/h=0.95--> w=0.95mm 61

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**Band-pass and band-stop filter using quarter-wave stubs**

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Example #8 Design a band-stop filter using three quarter-wave open-circuit stubs . The center frequency is 2GHz , the bandwidth is 15%, and the impedance is 50W. Use an equi-ripple response, with a 0.5dB ripple level. We have g0=1 , g1=1.5963, g2=1.0967, g3=1.5963, g4= 1 and W=0.1 Note that: It is difficult to impliment on microstripline or stripline for characteristic > 150W. 63

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**Capacitive coupled resonator band-pass filter**

where i=1,2,3….n 64

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Example #9 Design a band-pass filter using capacitive coupled resonators , with a 0.5dB equal-ripple pass-band characteristic . The center frequency is 2GHz, the bandwidth is 10%, and the impedance 50W. At least 20dB attenuation is required at 2.2GHz. First , determine the order of filter, thus calculate prototype From Pozar ,Fig 8.27 pg 453 , we have N=3 n gn ZoJn Bn Cn qn x pF o x pF o x pF o x pF - 65

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**Other shapes of microstripline filter**

Rectangular resonator filter Interdigital filter U type filter 66

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**Wiggly coupled line j1= p/2 j2= p/4 67**

The design is similar to conventional edge coupled line but the layout is modified to reduce space. Modified Wiggly coupled line to improve 2nd and 3rd harmonic rejection. l/8 stubs are added. j1= p/2 j2= p/4 67

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