IIT Bombay Course Code : EE 611 Department: Electrical Engineering - - PowerPoint PPT Presentation

IIT Bombay

Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in

EE 611 Lecture 14 Jayanta Mukherjee Page 1

Lecture 14

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Topics Covered

• Multisection Bandpass Filter
• Higher Order Bandpass Filter
• Direct Coupled Filters

EE 611 Lecture 10 Jayanta Mukherjee Page 2 EE 611 Lecture 14 Jayanta Mukherjee

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An Example

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filter to bandpass a Design

9 r r cr cr L

= = − = × × = = × = = = ≈ = =

−

776 . 2 25 . 5 / 2 tan 10 3 303 . 10 2 2 ) 25 . 5 ( 50 25 . 5

1 8 9

π π β π ω l pF

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An Example

1.85 1.90 1.95 2.00 2.05 2.10 2.15 1.80 2.20 0.2 0.4 0.6 0.8 0.0 1.0

freq, GHz pow(mag(S(2,1)),2)

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filter.

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N for the Q loaded total the is 1/ Q where S : is gain insertion the N

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filters design to need we response filter

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sharpness the improve To

T 21

∆ ε = + =

• N

f T

Q

2 2

1 1

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Higher Order Filters

0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Normalized Frequency (f/fr) Gain |S21|2 N=1 N=2 N=3 N=4

resonators shunt and series g alternatin section multi using d implemente be can Filters Order Higher

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Multisection BandPass Filter

fr1 QE1 fr2 fr3 QE3 QE2 VG Port 1 Port 2 Z0 Z0

N ... 1,2,3 r for f .... f f 2 1 2 sin Q Q : to according resonators select the to need design we h butterwort For the

rN r2 r1 T Er

=      = = =       − =

• π

N r

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Implementation of Multisection Filters with only shunt resonators

• If only shunt resonators are available it is possible to implement

series resonators using an inverter:

• To demonstrate the equivalence we use the ADCD matrices:

It results that: Z’(ω)/Z0 = Y(ω)/Y0 and the shunt resonator in shunt is now a series resonator in series

Y fr QE λ/4 λ/4 Z0 Z0 fr2 QE2 Z’

( )

               

Z' Series Y Shunt

        − =         − =       ×         ×       =         1 ' 1 1 ) ( 1 1 ) ( 1 Z Z Y Y j j Y Y j j D C B A

Inverter Inverter

ω ω ω

below circuit bandpass

• rder

high The

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Implementation of Multisection Filters with only shunt resonators

fr2 QE1 λ/4 Z0 λ/4 Z0 fr1 QE1 Port 1 fr3 QE3

Port 2

Z0 Z0

fr1 QE1 fr2 fr3 QE3 QE2 VG Port 1 Port 2 Z0 Z0

resonators section

• multi

following in the d implemente then is

e r e e e e e e e e e e e e e e e e e e r r r r

l l j j j j j j j j j j 2 2 2 cos 2 sin 2 sin 2 cos sin sin cos 2 sin cos 2 sin cos sin sin cos 1 1 cos sin sin cos cos sin sin cos

2 2

− = ⇒ =

• 

     =         − − =       ×       ×       =       − − λ θ π φ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ φ φ φ φ

ω e r e 2 e 2 l line at thru l line resonance at resonator

2

• :

select we if cos cos

e r e

                              IIT Bombay

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Exact Circuit equivalent of the Shunt Resonator

fr QE Z0 Z0 le le l C C Port 1 Port 2

layout final following in the results inverter /4 the to i resonator each by d contribute l length extra the g ubstractin

ei

λ S

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Final Layout of 3rd Order Band Pass Filter

fr QE Z0 Z0 λ/4 - le3 - le2 fr QE fr3 QE3 Z0 Port 2 le2 le3 le2 le1 λ/4 – le1 - le2 λ/4 λ/4 Z0 VG Resonator 2 Resonator 1 Resonator 3 Port 2 l3~λ/2 l2~λ/2 l1~λ/2 Port 1

1660 . 2 8096 . 2 2 tan , 04 . 290 25 6 5 sin 1199 . 2 , 9109 . 2 2 tan 11 . 409 1822 . 8 * 50 , 50 6 3 sin 1660 . 2 , 8096 . 2 2 tan 04 . 290 8007 . 5 50 25

3 3 3 1 3 3 3 3 2 2 2 1 2 2 2 2 1 1 1 1 1 1 1

= − = =         − = = = = = = − = =         − = = = = = = = − = =         − = = = × = = = = = =

• −

− − r e r cr r r cr T r e r cr r r cr T r e r cr r r cr

l X Z l X Q Q l X Z l X Q Q l X Z l X D φ π β π β φ π φ π β π β φ Ω π φ π β π β φ Ω π Ω , , 6 1 sin Q Q : Soln 50 Z Note 50. Q with 3) (N 3

• rder
• f

filter h Butterwort a esign

T 1 T

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An Example

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Direct Coupled Filters

Z0 Z0 VG Z0 Z0 Z0 λ/4 λ/4 λ/4 λ/4 λ/4 λ/4 λ/4 Port 1 Port 2 Open Open Open Open Z1 Z2 Z3 Z4 Z0 VG Z0 Z0 Z0 λ/4 λ/4 λ/4 λ/4 λ/4 λ/4 λ/4 Port 1 Port 2 Short Z1 Z2 Z3 Z4 Short Short Short Z0

Bandstop BandPass

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Direct Coupled Filters

• The characteristic impedances of the stubs are given by

Bandstop: Z0n=(4Z0QT)/πgn Bandpass: Z0n=(πZ0)/(4QTgn) Where gn are the filter coefficients (Ch 7). For example for A 3rd order Butterworth filter: g1=1, g2=2 and g3=1

• Problem with this design is that at high QT the Zn’s may not

be realizable

• Eg for QT=10 and N=3 Z1=3.93 ohm, Z2=1.06 ohm and

Z3=3.93 ohm which are quite small