Download presentation

Presentation is loading. Please wait.

Published byDesirae Alloway Modified about 1 year ago

1
Microwave Filter Design By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Eng Engineering Campus USM Nibong Tebal SPS Penang

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

3
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

4
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 Where Z i2 = V 2 / I 2 and V 1 = - Z i1 I 1

5
ABCD for T and network 5

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

7
Composite filter 7

8
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

9
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

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

11
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 =

12
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

13
Composite filter 13

14
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

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

16
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.

17
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.

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

19
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

20
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.

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

22
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.

23
Composite filter 23

24
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

25
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

26
continue 26 For m-derived T section sharp cutoff

27
continue 27 For matching section m=0.6

28
continue 28 A full circuit of the filter

29
Simplified circuit

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

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

32
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

33
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

34
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

35
continue 35

36
or 36

37
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

38
Continue 38 Prototype elements where Series element Shunt element

39
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 = g 1 = g 3 = F 1 =1.4626F 2 = a 1 =1.0a 2 =2.0 b 1 =2.043

40
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

41
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

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

43
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

44
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

45
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 = , g 4 = Let’s first and third elements are equivalent to series inductance and g 1 =g 3, thus

46
continue 46 Second element is equivalent to parallel capacitance, thus

47
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

48
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

49
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

50
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

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

52
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.

53
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

54
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 = = L 1 = L 3, g 2 = = 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

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

56
56

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

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

59
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

60
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

61
61 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

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

63
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

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

65
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 x pF155.8 o x pF166.5 o x pF155.8 o x pF-

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

67
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.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google