1. Microwave Filter DesignBy
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang

2. Contents
Composite filter
LC ladder filter
Microwave filter 2

3. 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

4. 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

5. ABCD for T and p network 5

6. 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

7. Composite filter 7

8. 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

9. 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

10. 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

11. 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

12. 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

13. Composite filter 13

14. 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

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

16. 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

17. 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

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

19. 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

20. 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

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

22. 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

23. Composite filter 23

24. 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

25. 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

26. continue
For m-derived T section sharp cutoff 26

27. continue
For matching section
m=0.6 27

28. continue
A full circuit of the filter 28

29. Simplified circuit

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

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

32. 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

33. 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

34. 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

35. continue 35

36. or 36

37. 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

38. Continue
Prototype elements
where
Series element
Shunt element 38

39. 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

40. 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

41. 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

42. continue 42 Now we consider the shunt capacitor
Shunts capacitor Ck must be replaced by parallel Lpk and Cpk
Admittance= parallel 42

43. 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

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

45. 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

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

47. 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

48. 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

49. 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

50. 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

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

52. 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

53. 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

54. 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

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

56. 56

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

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

59. 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

60. 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

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 61

62. Band-pass and band-stop filter using quarter-wave stubs62

63. 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

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

65. 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

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

67. 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