EE247 Lecture 4. EECS 247 Lecture 4: Filters 2005 H.K. Page 1. This Lecture

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1 EE247 Lecture 4 Lst lecture Active biquds Sllen Key & TowThoms Integrtor bsed filters Signl flowgrph concept First order integrtor bsed filter Second order integrtor bsed filter & biquds High order & high Q filters Cscded biquds Sensitivity of cscded biqud EECS 247 Lecture 4: Filters 2005 H.K. Pge This Lecture Ldder Type Filters For simplicity, will strt with ll pole ldder type filters Convert to integrtor bsed form Exmple shown Then will ttend to high order ldder type filters incorporting zeros Implement the sme 7 th order elliptic filter in the form of ldder C with zeros Find level of sensitivity to component vritions Compre with cscde of biquds Convert to integrtor bsed form utilizing SFG techniques Exmple shown Effect of Integrtor NonIdelities on Filter Performnce EECS 247 Lecture 4: Filters 2005 H.K. Pge 2

2 LC Ldder Filters LowPss C Filter V in Rs C L2 C3 L4 C5 Mde of resistors, inductors, nd cpcitors Doubly terminted (with R s ) or singly terminted (w/o R s ) Doubly terminted LC ldder filters Lowest sensitivity to component vritions when R S =R L EECS 247 Lecture 4: Filters 2005 H.K. Pge 3 LC Ldder Filters LowPss C Filter V in Rs C L2 C3 L4 C5 Design: CAD tools Mtlb Spice Filter tbles A. Zverev, Hndbook of filter synthesis, Wiley, 967. A. B. Willims nd F. J. Tylor, Electronic filter design, 3 rd edition, McGrwHill, 995. EECS 247 Lecture 4: Filters 2005 H.K. Pge 4

LC Ldder Filter Design Exmple Design LPF with mximlly flt pssbnd: f3db = 0MHz, fstop = 20MHz Rs >27dB Mximlly flt pssbnd Butterworth Determine minimum filter order : Use of Mtlb or Tbles Here tbles

3 LC Ldder Filter Design Exmple Design LPF with mximlly flt pssbnd: f3db = 0MHz, fstop = 20MHz Rs >27dB Mximlly flt pssbnd Butterworth Determine minimum filter order : Use of Mtlb or Tbles Here tbles used fstop / f3db = 2 Rs >27dB 3dB Stopbnd Attenution db Minimum Filter Order 5th order Butterworth 2 Νοrmlized w From: Willims nd Tylor, p. 237 EECS 247 Lecture 4: Filters 2005 H.K. Pge 5 LC Ldder Filter Design Exmple Find vlues for L & C from Tble: Note L &C vlues normlized to w 3dB = Denormliztion: Multiply ll L Norm, C Norm by: L r = R/w 3dB C r = /(RXw 3dB ) R is the vlue of the source nd termintion resistor (choose both Ω for now) Then: L= L r xl Norm C= C r xc Norm From: Willims nd Tylor, p..3 EECS 247 Lecture 4: Filters 2005 H.K. Pge 6

4 LC Ldder Filter Design Exmple Find vlues for L & C from Tble: Normlized vlues: C Norm =C5 Norm =0.68 C3 Norm = 2.0 L2 Norm = L4 Norm =.68 Denormliztion: Since w 3dB =2πx0MHz L r = R/w 3dB = 5.9 nh C r = /(RXw 3dB )= 5.9 nf R = C=C5=9.836nF, C3=3.83nF L2=L4=25.75nH From: Willims nd Tylor, p..3 EECS 247 Lecture 4: Filters 2005 H.K. Pge 7 Vin Rs=Ohm LowPss C Filter Mgnitude Response Simultion L2=25.75nH C 9.836nF L4=25.75nH C3 3.83nF C nF =Ohm SPICE simultion Results 6 db pssbnd ttenution due to double termintion Mgnitude (db) dB Frequency [MHz] EECS 247 Lecture 4: Filters 2005 H.K. Pge 8

5 LowPss C Ldder Filter Conversion to Integrtor Bsed Active Filter V in V V 2 V3 V4 V5 V6 Rs I L2 I3 L4 I 5 C C3 C5 I 2 I4 I 6 I 7 Use KCL & KVL to derive equtions: I V = Vin V 2, V 2 2 =, V 3 = V 2 V sc 4 I I V =, V 5 = V 4 V 6, V 6 = V o = V 6 sc 3 sc 5 V V3 I =, I 2 = I I 3, I Rs 3 = sl2 V5 V6 I 4 = I 3 I 5, I 5 =, I 6 = I 5 I 7, I7 = sl4 EECS 247 Lecture 4: Filters 2005 H.K. Pge 9 Vin V I LowPss C Ldder Filter Signl Flowgrph I V = V 2 in V 2, V 2 =, V 3 = V 2 V sc 4 I 4 I V, V V V, V 6 4 = 5 = = V o = V 6 sc 3 sc 5 V V3 I =, I 2 = I I 3, I3 = Rs sl2 V5 V6 I 4 = I 3 I 5, I 5 =, I 6 = I 5 I 7, I7 = sl4 V 2 Rs sc I 2 V3 V4 V5 V6 sl2 sc3 sl4 sc5 I3 I4 I 5 I 6 I 7 SFG EECS 247 Lecture 4: Filters 2005 H.K. Pge 0 Vo

6 LowPss C Ldder Filter Signl Flowgrph V in V V 2 V3 V4 V5 V6 Rs I L2 I3 L4 I 5 C C3 C5 I 2 I4 I 6 I 7 Vin V I V 2 Rs sc I 2 V3 V4 V5 V6 sl2 sc3 sl4 sc5 Vo I3 I4 I 5 I 6 I 7 SFG EECS 247 Lecture 4: Filters 2005 H.K. Pge Vin V I LowPss C Ldder Filter Normlize V 2 Rs sc I 2 V3 V4 V5 V6 sl2 sc3 sl4 sc5 Vo I 3 I4 I 5 I 6 I 7 Vin V V * R Rs V 2 V3 V4 V5 V6 * * R R sc * R sc * sl 3R sl sc * 2 4 5R V V 2 3 V 4 V 5 V 6 Vo V 7 * R EECS 247 Lecture 4: Filters 2005 H.K. Pge 2

7 Vin V V * R Rs LowPss C Ldder Filter Synthesize V 2 V3 V4 V5 V6 * * R R sc * R sc * sl 3R sl sc * 2 4 5R V V 2 3 V 4 V 5 V 6 Vo V 7 * R V in * R Rs V 2 V4 V6 sτ 2 sτ sτ 3 sτ 4 sτ 5 * R V 3 V 5 EECS 247 Lecture 4: Filters 2005 H.K. Pge 3 LowPss C Ldder Filter Integrtor Bsed Implementtion V in * R Rs V 2 V4 V6 sτ 2 sτ sτ 3 sτ 4 sτ 5 * R V 3 * L2 * * L4 * * = C.R, 2 = = C.R, C.R, C.R, C.R * 2 3 = 3 4 = = * 4 5 = 5 R R τ τ τ τ τ V 5 Building Block: RC Integrtor V 2 = V src EECS 247 Lecture 4: Filters 2005 H.K. Pge 4

8 Negtive Resistors Vo V V2 Vo V2 Vo V V2 V EECS 247 Lecture 4: Filters 2005 H.K. Pge 5 Synthesize V 2 V 3 V 4 V 5 EECS 247 Lecture 4: Filters 2005 H.K. Pge 6

9 Frequency Response V 2 V 3 V 4 V 5 EECS 247 Lecture 4: Filters 2005 H.K. Pge 7 Scle Node Voltges Scle by fctor s V 2 V 3 V 4 V 5 EECS 247 Lecture 4: Filters 2005 H.K. Pge 8

10 Mximizing Signl Hndling by Node Voltge Scling Scle by fctor s EECS 247 Lecture 4: Filters 2005 H.K. Pge 9 Filter Noise Totl the output:.4 µv rms (noiseless opmps) Tht s excellent, but the cpcitors re very lrge (nd the resistors smll high power dissiption). Not possible to integrte. Suppose our ppliction llows higher noise in the order of 40 µv rms EECS 247 Lecture 4: Filters 2005 H.K. Pge 20

11 Scle to Meet Noise Trget Scle cpcitors nd resistors to meet noise objective s = 0 4 Noise: 4 µv rms (noiseless opmps) EECS 247 Lecture 4: Filters 2005 H.K. Pge 2 Completed Design V 2 5 th order ldder filter Finl design utilizing: Node scling Finl R & C scling bsed on noise considertions V 3 V 5 V 4 EECS 247 Lecture 4: Filters 2005 H.K. Pge 22

12 Sensitivity C mde (rbitrrily) 50% (!) lrger thn its nominl vlue 0.5 db error t bnd edge 3.5 db error in stopbnd Looks like very low sensitivity EECS 247 Lecture 4: Filters 2005 H.K. Pge 23 Differentil 5 th Order Lowpss Filter V in Since ech signl nd its inverse redily vilble, elimintes the need for negtive resistors! Differentil design hs the dvntge of even order hrmonic distortion nd common mode spurious pickup utomticlly cncels Disdvntge: Double resistor nd cpcitor re! EECS 247 Lecture 4: Filters 2005 H.K. Pge 24

13 C Ldder Filters Including Trnsmission Zeros All poles Rs L2 L4 V in C C3 C5 Poles & Zeros C2 C4 C6 Rs L2 L4 L6 V in C C3 C5 C7 EECS 247 Lecture 4: Filters 2005 H.K. Pge 25 C Ldder Filter Design Exmple Design bsebnd filter for CDMA IS95 receiver with the following specs. Filter frequency msk shown on the next pge Allow enough mrgin for mnufcturing vritions Assume pssbnd mgnitude vrition of.8db Assume the 3dB frequency cn vry by 8% due to mnufcturing tolernces Assume ny phse impirment cn be compensted in the digitl domin * Note this is the sme exmple s for cscde of biqud while the specifictions re given closer to rel industry cse EECS 247 Lecture 4: Filters 2005 H.K. Pge 26

C Ldder Filter Design Exmple CDMA IS95 Receive Filter Frequency Msk 0 Mgnitude (db) 44 46 600k 700k 900k.2M Frequency [Hz] EECS 247 Lecture 4: Filters 2005 H.K.

14 C Ldder Filter Design Exmple CDMA IS95 Receive Filter Frequency Msk 0 Mgnitude (db) k 700k 900k.2M Frequency [Hz] EECS 247 Lecture 4: Filters 2005 H.K. Pge 27 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Since phse impirment cn be corrected for, use filter type with mx. cutoff slope/pole Elliptic Design filter freq. response to fll well within the freq. msk Allow mrgin for component vritions For the pssbnd ripple, llow enough mrgin for ripple chnge due to component & temperture vritions Pssbnd ripple 0.2dB For stopbnd rejection dd few db mrgin 444=48dB Design to spec.: fpss = 650 khz Rpss = 0.2 db fstop = 750 khz Rstop = 48 db Use Mtlb or filter tbles to decide the min. order for the filter (sme s cscded biqud exmple) 7 th Order Elliptic EECS 247 Lecture 4: Filters 2005 H.K. Pge 28

15 C LowPss Ldder Filter Design Exmple: CDMA IS95 Receive Filter 7 th order Elliptic C2 C4 C6 V in Rs C L2 C3 L4 C5 L6 C7 Use filter tbles to determine LC vlues EECS 247 Lecture 4: Filters 2005 H.K. Pge 29 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Spec. fpss = 650 khz Rpss = 0.2 db fstop = 750 khz Rstop = 49 db Use filter tbles to determine LC vlues Tble from: A. Zverev, Hndbook of filter synthesis, Wiley, 967 Elliptic filters tbulted wrt reflection coeficient ρ Rpss= 0 log( ρ 2 ) Since Rpss=0.2dB ρ=20% Use tble ccordingly EECS 247 Lecture 4: Filters 2005 H.K. Pge 30

Pge 3 Tble from Zverev pge #28 & 282: Normlized component vlues: C=.7677 C2=0.9393 L2=.9467 C3=.

16 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Tble from Zverev book pge #28 & 282: Since our spec. is Amin=44dB dd 5dB mrgin & design for Amin=49dB EECS 247 Lecture 4: Filters 2005 H.K. Pge 3 Tble from Zverev pge #28 & 282: Normlized component vlues: C=.7677 C2= L2=.9467 C3=.534 C4=.0098 L4= C5= C6=0.72 L6= C7= EECS 247 Lecture 4: Filters 2005 H.K. Pge 32

17 C Filter Frequency Response Frequency msk superimposed Frequency response well within spec. Mgnitude (db) Frequency [khz] EECS 247 Lecture 4: Filters 2005 H.K. Pge 33 Pssbnd Detil Pssbnd well within spec Mgnitude (db) Frequency [khz] EECS 247 Lecture 4: Filters 2005 H.K. Pge 34

18 C Ldder Filter Sensitivity The design hs the sme spec.s s the previous exmple implemented with cscded biquds To compre the sensitivity of C ldder versus cscdedbiquds: Chnged ll Ls &Cs one by one by 2% in order to chnge the pole/zeros by % (similr test s for cscded biqud) Found frequency response most sensitive to L4 vritions Note tht by vrying L4 both poles & zeros re vried EECS 247 Lecture 4: Filters 2005 H.K. Pge 35 RCL Ldder Filter Sensitivity Component vrition in C filter: Increse L4 by 2% Decrese L4 by 2% Mgnitude (db) L4 nom L4 low L4 high Frequency [khz] EECS 247 Lecture 4: Filters 2005 H.K. Pge 36

19 RCL Ldder Filter Sensitivity dB Mgnitude (db) dB Frequency [khz] EECS 247 Lecture 4: Filters 2005 H.K. Pge 37 Cscde of Biquds Sensitivity Component vrition in Biqud 4 (highest Q pole): Increse w p4 by % Decrese w z4 by % 2.2dB 0 Mgnitude (db) dB kHz 600kHz Frequency [Hz] MHz High Q poles High sensitivity in Biqud reliztions EECS 247 Lecture 4: Filters 2005 H.K. Pge 38

20 Sensitivity Comprison for CscdedBiquds versus C Ldder 7 th Order elliptic filter % chnge in pole & zero pir Pssbnd devition Stopbnd devition Cscded Biqud 2.2dB (29%) 3dB (40%) C Ldder 0.2dB (2%).7dB (2%) Doubly terminted LC ldder filters Significntly lower sensitivity compred to cscdedbiquds prticulrly within the pssbnd EECS 247 Lecture 4: Filters 2005 H.K. Pge 39 7 th order Elliptic C Ldder Filter Design Exmple: CDMA IS95 Receive Filter C2 C4 C6 V in Rs C L2 C3 L4 C5 L6 C7 Previously designed integrtor bsed ldder filters without trnsmission zeros Question: How do we implement the trnsmission zeros in the integrtorbsed version? EECS 247 Lecture 4: Filters 2005 H.K. Pge 40

21 Integrtor Bsed Ldder Filters How Do We Implement Trnsmission zeros? C V in V I V V 2 3 Rs L2 I 3 C I 2 I 4 V 4 C3 I 5 Use KCL & KVL to derive : I I C V 3 2 = V s 4 ( C C ) C C I3 I5 C V V 4 = s 2 ( C 3 C ) C 3 C Voltge Controlled Voltge Source! EECS 247 Lecture 4: Filters 2005 H.K. Pge 4 Integrtor Bsed Ldder Filters Trnsmission zeros C V in V I V V 2 3 Rs L2 I 3 C C I 2 V 4 I 5 ( ) ( C 3 C ) C V 4 C C C I V 4 2 C 3 C Replce shunt cpcitor with voltge controlled voltge sources: I I3 C V V 2 = s 4 ( C C ) C C I3 I5 C V V 4 = s 2 ( C 3 C ) C 3 C EECS 247 Lecture 4: Filters 2005 H.K. Pge 42

22 V in LC Ldder Filters Trnsmission zeros V I V V 2 3 V 4 I 5 Rs L2 I 3 ( C C ) ( C 3 C ) C V C 4 C C I 2 I V 2 4 C 3 C C C C C C 3 C Vin V V 2 V3 V4 s C C Rs ( ) sl2 s( C C ) 3 I I 2 I 3 I 4 Vo EECS 247 Lecture 4: Filters 2005 H.K. Pge 43 Integrtor Bsed Ldder Filters Higher Order Trnsmission zeros V 2 C V 4 C b V 6 Convert zero generting Cs in C loops to voltgecontrolled voltge sources C C 3 V2 V4 ( ) ( ) C C C 3 C C b C C V V 4 C C 2 C C 3 C V b 6 C 3 C b C 5 V 6 ( C 5 C b) C V b 4 C 3 C b EECS 247 Lecture 4: Filters 2005 H.K. Pge 44

23 V in Vin V V 2 * R Rs s( C C ) R * Higher Order Trnsmission zeros V 4 I V 5 5 I L4 4 ( C 3 C C b) C V 2 C 3 C C V b 6 C 3 C b C C C b C C C C 3 C C 3 b V V 2 V3 Rs I L2 I 3 ( C C ) C V 4 C I C 2 V V 2 V3 V4 * R sl2 sr * ( C C C ) 3 b V6 I 7 V 3 V 4 V 5 V 6 I 6 ( C 5 C b) C b V 4 C 5 C b C b C 5 C b V5 V6 * R * sl4 sr ( C5 Cb) Vo V 7 * R EECS 247 Lecture 4: Filters 2005 H.K. Pge 45 Exmple: 5 th Order Chebyshev II Filter 5 th order Chebyshev II Tble from: Willims & Tylor book, p..2 50dB stopbnd ttenution f 3dB =0MHz EECS 247 Lecture 4: Filters 2005 H.K. Pge 46

24 Reliztion with Integrtor V V V C V = sc C R i 2 3 ( ) * Rs C C V EECS 247 Lecture 4: Filters 2005 H.K. Pge 47 5 th Order Butterworth Filter V 2 From: Lecture 4 pge 22 V 3 V 4 V 5 EECS 247 Lecture 4: Filters 2005 H.K. Pge 48

25 OpmpRC Simultion V 2 V 3 V 4 V 5 EECS 247 Lecture 4: Filters 2005 H.K. Pge 49 V in Seventh Order Differentil LowPss Filter Including Trnsmission Zeros Trnsmission zeros implemented with coupling cpcitors EECS 247 Lecture 4: Filters 2005 H.K. Pge 50

26 Effect of Integrtor NonIdelities on Filter Performnce Idel Intg. C V in R opmpdc gin= ω H(s) = o s ωo = /RC EECS 247 Lecture 4: Filters 2005 H.K. Pge 5 Idel Integrtor Qulity Fctor Idel Intg. ω H(s) = o= s ωo jω Since Q is defined s: H Q ( jω) = R( ω) jx( ω) X ( ω) = R ( ω) Then: intg. Q idel = EECS 247 Lecture 4: Filters 2005 H.K. Pge 52

27 Effect of Integrtor NonIdelities on Filter Performnce Idel Intg. Rel Intg. ωo H(s) = H(s) s s s ( s o)( p2)( ω p3)... EECS 247 Lecture 4: Filters 2005 H.K. Pge 53 Effect of Integrtor Finite DC Gin on Q log H ( s) ω P = 0 ω 0 ω o ω π ArctnP 2 ωo ψ 90 o P ωo Phseled@ ωo (in rdin) Exmple: P/ ω 0 =/00 EECS 247 Lecture 4: Filters 2005 H.K. Pge 54

28 Effect of Integrtor Finite DC Gin on Q Phse ω 0 Mgnitude (db) Droop in the pssbnd Droop in the pssbnd Normlized Frequency EECS 247 Lecture 4: Filters 2005 H.K. Pge 55 Effect of Integrtor NonDominnt Poles log H ( s) ω 0 P2P3 ω o π 2 ω Arctn ωo p i= 2 i ψ 90 o ωo p i= 2 i Phse ωo (in rdin) Exmple: ω 0 /P2 =/00 EECS 247 Lecture 4: Filters 2005 H.K. Pge 56

29 Effect of Integrtor NonDominnt Poles Phse ω 0 Mgnitude (db) Peking in the pssbnd Peking in the pssbnd In extreme cses could result in oscilltion! Normlized Frequency EECS 247 Lecture 4: Filters 2005 H.K. Pge 57 Effect of Integrtor NonDominnt Poles & Finite DC Gin on Q log H ( s ) ω P = 0 ψ ω 0 P2P3 ω o ω π Arctn 2 P ωo ω Arctn o p 90 i= 2 i 90 o Note tht the two terms hve different signs Cn cncel ech other s effect EECS 247 Lecture 4: Filters 2005 H.K. Pge 58

30 Integrtor Qulity Fctor Rel Intg. Bsed on the definition of Q nd ssuming tht: H(s) s... ( s s )( p2)( ω p3) o ωo << & >> p 2,3,... It cn be shown tht in the vicinity of unityginfrequency: Q intg. rel ω o i= 2 p i Phse ω 0 Phse ω 0 EECS 247 Lecture 4: Filters 2005 H.K. Pge 59 Exmple: Effect of Integrtor Finite Q on Bndpss Filter Behvior Idel Idel Integrtor DC gin=00 Integrtor 00.ω o EECS 247 Lecture 4: Filters 2005 H.K. Pge 60

31 Exmple: Effect of Integrtor Finite Q on Filter Behvior Idel Integrtor DC gin=00 & 00. ω ο EECS 247 Lecture 4: Filters 2005 H.K. Pge 6 Summry Effect of Integrtor NonIdelities on Q Q intg. idel = Q intg. rel ω o p i= 2 i Amplifier DC gin reduces the overll Q in the sme mnner s series/prllel resistnce ssocited with pssive elements Amplifier poles locted bove integrtor unitygin frequency enhnce the Q! If nondominnt poles close to unitygin freq. Oscilltion Depending on the loction of unityginfrequency, the two terms cn cncel ech other out! EECS 247 Lecture 4: Filters 2005 H.K. Pge 62

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