# Microwave Filter Design

## Presentation on theme: "Microwave Filter Design"— Presentation transcript:

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

Contents Composite filter LC ladder filter Microwave filter 2

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

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

ABCD for T and p network 5

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

Composite filter 7

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

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

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

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

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

Composite filter 13

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

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

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

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

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

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

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

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

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

Composite filter 23

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

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

continue For m-derived T section sharp cutoff 26

continue For matching section m=0.6 27

continue A full circuit of the filter 28

Simplified circuit

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

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

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

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

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

continue 35

or 36

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

Continue Prototype elements where Series element Shunt element 38

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

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

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

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

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

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

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

continue Second element is equivalent to parallel capacitance, thus 46

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 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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