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

Page 1 IIT Bombay Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in Lecture 15 EE 611 Lecture 15 Jayanta Mukherjee

Page 2 IIT Bombay Topics Covered • General Introduction to filter synthesis • Insertion Loss method • Filter Prototypes • Scaling of prototypes EE 611 EE 611 Lecture 10 Lecture 15 Jayanta Mukherjee Jayanta Mukherjee

Page 3 IIT Bombay Filter Synthesis • We have seen how narrowband bandstop or band pass filters can be constructed using resonators • For narrowband cases, lumped element models of the distributed network were good enough to get good performance predictions • For wider bandwidths this will not be the case; we need design methods specifically for distributed circuits • Design methods for lumped element filters exist, and can be adapted in some cases to distributed circuits. EE 611 EE 611 Lecture 15 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee

Page 4 IIT Bombay Filter Synthesis (2) • Let’s start by reviewing the “insertion loss” synthesis method, which allows control of filter bandwidth and performance through a set of “prototype” designs. L I ( ω ) Stopband Pass- band 3 dB ω c ω EE 611 EE 611 Lecture 15 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee

Page 5 IIT Bombay Filter Synthesis (2) • Insertion loss synthesis applies for either lumped element or distribute d circuits, so we will focus on lumped designs first. 1 • Filters are designed to achieve desired values of , which determines the 2 S 21 insertion loss of the filter. • Most common designs produce either maximally flat (Butterwor th) or Chebyshev responses. For maximally flat low - pass filters : 2 N   ω 1   = + ω 1 where is the frequency at which 3 dB loss is obtained.   ω c 2   S c 21 • For Chebyshev low - pass filter :   ω 1   = + 2 2 1 k T where T is a Chebyshev polynomial and k is a constant. For   ω N N 2   S c 21 ω < ω ± 2 , an equal ripple response in the range 1 k is obtained c ω > ω while for higher losses are obtained. c EE 611 EE 611 Lecture 10 Lecture 15 Jayanta Mukherjee Jayanta Mukherjee

Page 6 IIT Bombay Synthesis Example for N=3 Butterworth • = = Ω ω = Let us consider t he case of N 3 with Z 1 and 1 Hz. 0 c 1 = 2 = G S + ω T 21 6 1 • 2 + 2 = The filter is lossless : thus we have S 1 resulting in : S 11 21 ω − 6 6 s 2 = − 2 = = = * S 1 S S ( s ) S ( s ) + ω − 11 21 11 11 6 6 1 1 s • = = ± ± = ± 6 1 - s 0 admits 6 roots : s 0.5 j0.866, and s 1. As the filter is realizable 1,2,3,4 5,6 and passive we select the left hand plane roots to identify S (s) : 11 ± 3 s = S + + + 11 3 2 s 2 s 2 s 1 + + + = = ± 3 2 One can verify that s 2s 2 s 1 0 admits the following 3 roots s -0.5 j0.866, 1,2 = and s -1 5 • We select the positive sign for S and convert S to input impedance Z : 11 11 in + + + + 3 2 1 S 2 s 2 s 2 s 1 = = = 11 Z using Z 1ohm − + + in 0 2 1 S 2 s 2 s 1 11 EE 611 EE 611 Lecture 15 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee

Page 7 IIT Bombay Cauer Expansion • Dividing the higher order polynomial by the Z in,1 Y in,2 Z in,3 lower polynomial we get : 1 + + + 3 2 R 2s 2s 2s 1 = = + 1 Z Z in,1 ext,1 + + 2 g 3 D 2s 2s 1 g 1 + s 1 = + s + + 2 V G 2s 2s 1 g 4 g 2 • = We have extracted an inductor L 1H = = Port 1 Port 2 since Z Ls s . ext,1 D • = Let us divide D by R : Y 1 in,2 R 1 + + 2 2s 2s 1 R 1 = = + = + 2 Y 2 s + ext , 2 + s R s 1 1 1 • = = We have extracted a capacitor since C 2 F since Y 2s ext,2 + R s 1 • = = = + 1 s Let us now divide R by R : Z 1 1 2 in,3 R 1 2 • = = We have now extracted an inductor L 1H and resistor R 1 ohm EE 611 EE 611 Lecture 15 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee

Page 8 IIT Bombay Alternate Implementation • If we select the negative sign solution - s 3 = S + + + 11 3 2 s 2 s 2 s 1 a, different prototype will result : 1 g 2 Port 2 V G Port 1 g 1 g 3 g 4 EE 611 EE 611 Lecture 10 Lecture 15 Jayanta Mukherjee Jayanta Mukherjee

Page 9 IIT Bombay Prototypes 1 g 1 g 3 g 5 V G Port 1 g 4 Port 2 g 2 • F or the maximally flat filters g is always 1, so the source and load impedances are equal. + N 1 However for Chebyshev filters with an even number of sections, the source and load impedances are not equal. • In this case an impedance transform er can be used if needed. Even number section Chebyshev filters are usually avoided. • ω ω − Book provides plots of the attenuatio n versus the frequency deviation / 1 for the c maximally flat (p 450) and Chebyshev (p 453) prototypes versus the number of sections. • = ω = These protypes assume that Z 1 and also the " breakpoint " frequencie s 1. 0 c • We need " scaling" technique s to obtain more general filters. EE 611 EE 611 Lecture 10 Lecture 15 Jayanta Mukherjee Jayanta Mukherjee

Page 10 IIT Bombay Scaling of Prototypes 1 Z 0 g 2 L 2 Port 2 Port 2 V G Port 1 V G Port 1 g 1 g 3 C 1 C 3 g 4 Z 0 • We can make a low - pass prototype with source (and usually load) impedance Z instead of 1 0 just by multiplyin g the original inductors by Z and dividing the original capacitors by Z . 0 0 • ω We can change the breakpoint frequency from 1 to just by dividing the original inductors c ω ω by and dividing the original capacitors by . c c • ' ' Thus, the new inductors L and capacitanc e C values in an impedance k k and frequency scaled low - pass filter are Z g g = = ' ' 0 k k L C where g are the prototype values. ω ω k k k Z c 0 c • It is also possible to generate high pass, band pass and band stop filters from suitable transform ations of the low - pass prototype. EE 611 EE 611 Lecture 15 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee

Page 11 IIT Bombay Transformation from Low Pass to High Pass • ω A high pass filter is obtained by transfo rming in the original low - pass ω ω filter to - / . Note then high frequencie s in the new filter wil l correspond c to low frequencie s in the original. • This transform ation wind s up switching inductors and capacitors in the low pass prototype. • Values for an impedance and frequency scaled high pass filter are then Z Z 1 1 = = = = ' ' 0 0 C and L k ω ω k ω ω Z L Z g C g 0 c k 0 c k c k c k • As in the impedance matching case equal ripple filters provide better cutoff than maximally flat filters at the expense of ripple in the passband. EE 611 EE 611 Lecture 10 Lecture 15 Jayanta Mukherjee Jayanta Mukherjee

Page 12 IIT Bombay Transformation from Low Pass to High Pass 1 g 3 g 5 g 1 Low Pass Prototype V G Port 1 g 4 Port 2 g 2 Z 0 C k ’ C k ’ C k ’ L k ’ V G Port 1 L k ’ High Pass Prototype Z 0 Port 2 EE 611 EE 611 Lecture 15 Jayanta Mukherjee Lecture 10 Jayanta Mukherjee

Page 13 IIT Bombay Transformation from Low Pass to Band Pass • Because band pass filters have a two sided response, the transform ω from a low - pass prototype replaces the original with   ω ω ω ω 1 -   − ∆ = ω ω 0 2 1 , where , and define the upper and lower   ω ∆ ω ω 2 1   0 0 ω = ω ω passband frequencie s, and . 0 1 2 • ω ω ω This transform maps to 0 in the original low pass prototype, and and 0 1 2 ± to 1, the breakpoint s of the original filter. • This transform ation can be achieved only if the original inductors and capacitors are each trans formed into LC circuits. • For a frequency and impedance scaled band - pass filter, inductors in the original prototype are replaced by a series LC circuit, with valu es ∆ ω g Z g = = = = ' k ' 0 0 k L and C or f and Q k ∆ ω k ω k π k ∆ Z g 2 2 0 0 0 k EE 611 EE 611 Lecture 10 Lecture 15 Jayanta Mukherjee Jayanta Mukherjee