Page 1 IIT Bombay Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in Lecture 14 EE 611 Lecture 14 Jayanta Mukherjee
Page 2 IIT Bombay Topics Covered • Multisection Bandpass Filter • Higher Order Bandpass Filter • Direct Coupled Filters EE 611 EE 611 Lecture 10 Lecture 14 Jayanta Mukherjee Jayanta Mukherjee
Page 3 IIT Bombay An Example Design a bandpass filter to operate at 2 GHz which wil l have 3 dB cutoff frequencie s at 1950 and 2050 MHz. Assume 50 ohm, air - filled lines. Find the length of the intermedia te transmiss ion line. = = ≈ Sol : - First, since Q 2000/100 20, us the chart to find X /Z 5 . 25 L cr 0 - Since we have 50 ohm lines and desire 2000 MHz as the center frequency, find C from 1 = = X 50 ( 5 . 25 ) cr ω C r 1 = ) ( ) = C 0 . 303 pF ( π × 9 50 5.25 2 2 10 - The exact line length can now be found from the resonance condition : ( ) π × 9 2 2 10 l ( ) − β = = π − = 1 l tan 2 / 5 . 25 2 . 776 r × 8 3 10 = l 0.0663 m EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 4 IIT Bombay An Example 1.0 0.8 pow(mag(S(2,1)),2) 0.6 0.4 0.2 0.0 1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15 2.20 freq, GHz EE 611 EE 611 Lecture 10 Lecture 14 Jayanta Mukherjee Jayanta Mukherjee
Page 5 IIT Bombay • To improve the sharpness of filter response we need to design filters of higher Higher Order Filters orders. • In Chapter 7 we will see how we can build higher order filters using the " Insertion Loss Synthesis" . In this chapter we only use the results. For a Butterwort h filter of order N the insertion gain is : 1 2 = ( ) S 21 + ε 2 N Q 1 T f = ∆ where Q 1/ is the total loaded Q for the N order filter. T 0.9 N=1 0.8 N=2 N=3 0.7 N=4 Gain |S 21 | 2 0.6 0.5 0.4 0.3 0.2 0.1 0 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 Normalized Frequency (f/f r ) EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 6 IIT Bombay Multisection BandPass Filter • Higher Order Filters can be implemente d using multi section alternatin g series and shunt resonators f r2 Q E2 Z 0 Port 2 f r1 f r3 V G Z 0 Q E1 Q E3 Port 1 • For the butterwort h design we need to select the resonators according to : − 2 r 1 = π Q Q sin = Er T 2 for r 1,2,3 ... N N = = = f f .... f r1 r2 rN EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 7 IIT Bombay Implementation of Multisection Filters with only shunt resonators • If only shunt resonators are available it is possible to implement series resonators using an inverter: Q E2 f r2 λ /4 λ /4 Z’ f r Y Z 0 Z 0 Q E • To demonstrate the equivalence we use the ADCD matrices: ( ) ω ω Y Z ( ) ' 1 0 0 j 0 j A B 1 1 ω = × × = − = − Y ( ) Y Z 1 0 0 j 0 j 0 C D Y 0 1 0 1 0 Inverter Inverter Shunt Y Series Z' It results that: Z’( ω )/Z 0 = Y( ω )/Y 0 and the shunt resonator in shunt is now a series resonator in series EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 8 IIT Bombay Implementation of Multisection Filters with only shunt resonators • The high order bandpass circuit below f r2 Q E2 Z 0 Port 2 f r1 f r3 V G Z 0 Q E1 Q E3 Port 1 • is then implemente d in the following multi - section resonators λ /4 Z 0 λ /4 f r2 f r3 f r1 Port 2 Z 0 Port 1 Q E1 Q E1 Q E3 Z 0 Z 0 EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 9 IIT Bombay Exact Circuit equivalent of the Shunt Resonator l e l e l 0 C C f r Port 1 Z 0 Z 0 Port 2 Q E − φ φ θ θ θ θ j j j cos sin cos sin 1 0 cos sin r r = e e × × e e φ − φ θ θ θ θ j sin cos j sin cos 0 1 j sin cos r r e e e e ω resonator at resonance line l thru at line l e r e θ θ θ − θ θ θ 2 2 j cos 2 sin 2 j cos sin 2 cos sin = = e e e e e e θ θ θ θ θ − θ 2 2 j sin 2 cos 2 2 j cos sin cos sin e e e e e e λ • φ = π θ ⇒ = − r if we select : - 2 l 2 l r e e 2 EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 10 IIT Bombay Final Layout of 3 rd Order Band Pass Filter • S ubstractin g the extra length l contribute d by each resonator i to the ei λ /4 inverter results in the following final layout λ /4 – l e1 - l e2 λ /4 - l e3 - l e2 l e1 l e2 l e2 l e3 0 0 Z 0 Port 2 λ /4 λ /4 f r f r f r3 V G Q E Q E Q E3 Z 0 Z 0 Z 0 l 1 ~ λ /2 Port 1 Port 2 l 2 ~ λ /2 l 3 ~ λ /2 Resonator 1 Resonator 3 Resonator 2 EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 11 IIT Bombay An Example • = = = Ω D esign a Butterwort h filter of order 3 (N 3) with Q 50. Note Z 50 T 0 Soln : π 1 = = = × = Ω Q Q sin 25 , X 50 5 . 8007 290 . 04 1 T cr 1 6 2 Z − φ = β = π − = β = π − φ = 1 0 l tan 2 . 8096 , 2 l 0 . 1660 r r r e r 1 1 1 1 X cr 1 π 3 = = = = Ω Q Q sin 50 , X 50 * 8 . 1822 409 . 11 2 T cr 2 6 2 Z − φ = β = π − = β = π − φ = 1 0 l tan 2 . 9109 , 2 l 0 . 1199 r 2 r 2 r e 2 r 2 X cr 2 π 2 Z 5 − = = = φ = β = π − = 1 0 Q Q X l sin 25 , 290 . 04 , tan 2 . 8096 3 T cr 3 r 3 r 3 6 X cr 3 β = π − φ = 2 l 0 . 1660 r e 3 r 3 EE 611 EE 611 Lecture 10 Lecture 14 Jayanta Mukherjee Jayanta Mukherjee
Page 12 IIT Bombay Direct Coupled Filters λ /4 λ /4 λ /4 Z 0 Z 0 Z 0 Z 0 Port 2 λ /4 λ /4 λ /4 λ /4 Port 1 Z 0 V G Bandstop Z 1 Z 2 Z 3 Z 4 Short Short Short Short λ /4 λ /4 λ /4 Z 0 Z 0 Z 0 Z 0 Port 2 λ /4 λ /4 λ /4 BandPass λ /4 Port 1 V G Z 0 Z 1 Z 2 Z 3 Z 4 Open Open Open Open EE 611 EE 611 Lecture 14 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee
Page 12 IIT Bombay Direct Coupled Filters •The characteristic impedances of the stubs are given by Bandstop: Z 0n =(4Z 0 Q T )/ π g n Bandpass: Z 0n =( π Z 0 )/(4Q T g n ) Where g n are the filter coefficients (Ch 7). For example for A 3 rd order Butterworth filter: g 1 =1, g 2 =2 and g 3 =1 • Problem with this design is that at high Q T the Z n ’s may not be realizable • Eg for Q T =10 and N=3 Z 1 =3.93 ohm, Z 2 =1.06 ohm and Z 3 =3.93 ohm which are quite small EE 611 EE 611 Lecture 10 Lecture 14 Jayanta Mukherjee Jayanta Mukherjee
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