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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 2 1.Composite filter 2.LC ladder filter 3.Microwave filter

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

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Image method Let’s say we have image impedance for the network Z i1 and Z i2 Where Z i1 = input impedance at port 1 when port 2 is terminated with Z i2 Z i2 = input impedance at port 2 when port 1 is terminated with Z i1 Then 4 @ Where Z i2 = V 2 / I 2 and V 1 = - Z i1 I 1

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ABCD for T and network 5

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Image impedance in T and network 6 Image impedance Propagation constant Substitute ABCD in terms of Z 1 and Z 2

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Composite filter 7

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Constant-k section for Low-pass filter using T-network 8 If we define a cutoff frequency And nominal characteristic impedance Then Z i T = Z o when =0

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continue 9 Propagation constant (from page 11), we have Two regions can be considered Z it become real and is imaginary ( = j ) since 2 / c 2 -1<1 > c : stopband of filter_--> Z it become imaginary and is real ( = ) since 2 / c 2 -1<1 cc Mag cc passband stopband

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Constant-k section for Low-pass filter using -network 10 Z i = Z o when =0 Propagation constant is the same as T-network

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Constant-k section for high-pass filter using T-network 11 If we define a cutoff frequency And nominal characteristic impedance Then Z i T = Z o when =

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Constant-k section for high-pass filter using -network 12 Z i = Z o when = Propagation constant is the same for both T and -network

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Composite filter 13

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m-derived filter T-section 14 Constant-k section suffers from very slow attenuation rate and non-constant image impedance. Thus we replace Z 1 and Z 2 to Z’ 1 and Z’ 2 respectively. Let’s Z’ 1 = m Z 1 and Z’ 2 to obtain the same Z iT as in constant-k section. Solving for Z’ 2, we have

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Low -pass m-derived T-section 15 For constant-k section and Propagation constant where

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continue 16 If we restrict 0 < m < 1 and Thus, both equation reduces to Then When op, then e become positif but decreasing.,which meant decreasing in attenuation.

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Comparison between m-derived section and constant-k section 17 M-derived section attenuates rapidly but after > op, 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.

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High -pass m-derived T-section 18 and Propagation constant where

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continue 19 If we restrict 0 < m < 1 and Thus, both equation reduces to Then When op, e is becoming negative and the wave will be propagted. Thus op < c

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continue 20 op cc M-derived section seem to be resonated at = op 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.

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m-derived filter -section 21 Note that The image impedance is

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

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Composite filter 23

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Matching between constant-k and m-derived 24 The image impedance Z iT does not match Z i , I.e The matching can be done by using half- section as shown below and the image impedance should be Z i1 = Z iT and Z i2 =Z i It can be shown that Note that

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Example #1 25 Design a low-pass composite filter with cutoff frequency of 2GHz and impedance of 75 . 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 c and substituting, we have

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continue 26 For m-derived T section sharp cutoff

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continue 27 For matching section m=0.6

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continue 28 A full circuit of the filter

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Simplified circuit

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

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N-section LC ladder circuit (low-pass filter prototypes) 31 Prototype beginning with serial element Prototype beginning with shunt element

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Type of responses for n-section prototype filter 32 Maximally flat or Butterworth Equal ripple or Chebyshev Elliptic function Linear phase Maximally flat Equal rippleEllipticLinear phase

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Maximally flat or Butterworth filter 33 For low -pass power ratio response g 0 = g n+1 = 1 where C=1 for -3dB cutoff point n= order of filter c = cutoff frequency No of order (or no of elements) Where A is the attenuation at point and 1 > c Prototype elements k= 1,2,3…….n Series element Shunt element Series R=Z o Shunt G=1/Z o

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Example #2 34 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. g 0 = g 3+1 = 1 First, determine the number of elements Solution Thus choose an integer value, I.e n=3 Prototype values

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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 where C n (x)=Chebyshev polinomial for n order and argument of x n= order of filter c = cutoff frequency F o =constant related to passband ripple Chebyshev polinomial Where Lr is the ripple attenuation in pass-band

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Continue 38 Prototype elements where Series element Shunt element

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Example #3 39 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 g 2 = 1.1132 g 1 = g 3 = 0.8794 F 1 =1.4626F 2 = 1.1371 a 1 =1.0a 2 =2.0 b 1 =2.043

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Transformation from low-pass to high-pass 40 Series inductor L k must be replaced by capacitor C’ k Shunts capacitor C k must be replaced by inductor L’ k

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Transformation from low-pass to band-pass 41 Thus, series inductor L k must be replaced by serial L sk and C sk where and Now we consider the series inductor Impedance= series normalized

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continue 42 Shunts capacitor C k must be replaced by parallel L pk and C pk Now we consider the shunt capacitor Admittance= parallel

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Transformation from low-pass to band-stop 43 Thus, series inductor L k must be replaced by parallel L pk and C skp where and Now we consider the series inductor --convert to admittance admittance = parallel

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Continue 44 Shunts capacitor C k must be replaced by parallel L pk and C pk Now we consider the shunt capacitor --> convert to impedance

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Example #4 45 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 50 . Solution From table 8.4 Pozar pg 452. g o =1, g 1 =1.5963, g 2 =1.0967, g 3 = 1.5963, g 4 = 1.000 Let’s first and third elements are equivalent to series inductance and g 1 =g 3, thus

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continue 46 Second element is equivalent to parallel capacitance, thus

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Implementation in microstripline 47 Equivalent circuit A short transmission line can be equated to T and 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

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48 -model with C as fringing capacitance -model with L as fringing inductance Z oL should be high impedance Z oC should be low impedance

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Example #5 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 ( r =4.5 h=1.5mm) Let’s choose Z oL =100 and Z oC =20 . Note: For more accurate calculate for difference Z o

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continue 50 The new values for L 1 =L 3 = 7nH-0.75nH= 6.25nH and C 2 =3.543pF-0.369pF=3.174pF Thus the corrected value for d 1,d 2 and d 3 are More may be needed to obtain sufficiently stable solutions

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

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

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Kuroda identity 53 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= /8

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Example #6 54 Design a low-pass filter for fabrication using micro strip lines.The specification: cutoff frequency of 4GHz, third order, impedance 50 , and a 3 dB equal-ripple characteristic. Protype Chebyshev low-pass filter element values are g 1 =g 3 = 3.3487 = L 1 = L 3, g 2 = 0.7117 = C 2, g 4 =1=R L Using Richard’s transform we have Z oL = L=3.3487Z oc =1/ C=1/0.7117=1.405and ZoZo ZoZo

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Using Kuroda identity to convert S.C series stub to O.C shunt stub. thus We have and Substitute again, we have and 55

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Band-pass filter from /2 parallel coupled lines 57 Microstrip layout Equivalent admittance inverter Equivalent LC resonator

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

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Example #7 59 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=50 . We have g 0 =1, g 1 =1.5963, g 2 =1.0967, g 3 =1.5963, g 4 = 1 and =0.1 A C D E

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60 B B D E 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 r =10. For others use other means. The required resonator

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61 Thus we have For sections 1 and 4 s/h=0.45 --> 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

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Band-pass and band-stop filter using quarter-wave stubs 62 Band-pass Band-stop

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Example #8 63 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 g 0 =1, g 1 =1.5963, g 2 =1.0967, g 3 =1.5963, g 4 = 1 and =0.1 Note that: It is difficult to impliment on microstripline or stripline for characteristic > 150

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Capacitive coupled resonator band-pass filter 64 where i=1,2,3….n

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Example #9 65 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 From Pozar,Fig 8.27 pg 453, we have N=3 prototype ng n Z o J n B n C n n 11.59630.31376.96x10 -3 0.554pF155.8 o 21.09670.11872.41x10 -3 0.192pF166.5 o 31.09670.11872.41x10 -3 0.192pF155.8 o 41.00000.31376.96x10 -3 0.554pF-

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Other shapes of microstripline filter 66 Rectangular resonator filter U type filter Interdigital filter

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Wiggly coupled line 67 1 = /2 2 = /4 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. /8 stubs are added.

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