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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

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

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

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

If we define a cutoff frequency

And nominal characteristic impedance

Zi T= Zo when w=0

Then

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 )
- since w2/wc2-1<1

wc

a,b

stopband

passband

a

wc

p

Mag

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

Zi T= Zo when w =

Then

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

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

continue

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

continue

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

w

wop

wc

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

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

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

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

33

Where A is the attenuation at w1 point and w1>wc

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

34

Thus choose an integer value , I.e n=3

continue

35

or

36

Equi-ripple filter

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

37

Where Lr is the ripple attenuation in pass-band

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=1.4626 F2= 1.1371

a1=1.0 a2=2.0

b1=2.043

g1 = g3 = 0.8794

g2= 1.1132

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

and

where

- Now we consider the series inductor

normalized

- Thus , series inductor Lk must be replaced by serial Lsk and Csk

Impedance= series

41

continue

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

and

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= 1.5963, g4= 1.000

Let’s first and third elements are equivalent to series inductance and g1=g3, thus

45

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

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

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

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= 3.3487 = L1= L3 , g2 = 0.7117 = 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

Band-pass filter from l/2 parallel coupled lines

Microstrip layout

Equivalent admittance inverter

Equivalent LC resonator

57

Required admittance inverter parameters

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

D

E

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

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

61

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

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 Cnqn

1 1.5963 0.3137 6.96x10-3 0.554pF 155.8o

2 1.0967 0.1187 2.41x10-3 0.192pF 166.5o

3 1.0967 0.1187 2.41x10-3 0.192pF 155.8o

4 1.0000 0.3137 6.96x10-3 0.554pF -

65

Wiggly coupled line

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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