Overview Of LectureGeneralized Transfer FunctionMaximally Flat Magnitude FilterButterworth Frequency ResponseButterworth AttenuationPoles Of Butterworth FilterExample Butterworth DesignDesign Calculations 1Design Calculations 2Design Calculations 3Comments On Critical FrequenciesDesign Calculations 4Design Calculations 5Bandpass StructureHSPICE Simulated PerformanceChebyshev FilterChebyshev PassbandChebyshev ResponseStopband AttenuationChebyshev VS ButterworthChebyshev Pole LocationsSome Helpful IdentitiesPole InvestigationPole SolutionExample Chebyshev DesignPole LocationsChebyshev Transfer FunctionChebyshev ResponsesSecond Order LP ButterworthSecond Order LP ChebyshevButterworth VS ChebyshevMagnitude Response ComparisonPhase Response ComparisonEnvelope Delay2nd Order Envelope DelaysDelay Response ComparisonGeneralized 2nd Order Delay2nd Order Delay ComparisonsComments On Envelope DelayBessel FilterBessel MathematicsBessel Transfer FunctionBessel Magnitude ResponsesBessel Phase Angle ResponsesBessel VS ButterworthBessel/Butterworth MagnitudeBessel/Butterworth Phase AngleEE 541Class LectureWeeks 7 - 10(additional slides to be supplied)Prof. John Choma, ProfessorDepartment of Electrical Engineering-ElectrophysicsUniversity of Southern CaliforniaUniversity Park; MC: 0271; PHE #604Los Angeles, California 90089-0271213-740-4692 [USC Office]213-740-7581 [USC Fax][email protected] Filter ApproximationsFall 2006 SemesterUniversity of Southern California EE 541: Choma182Overview Of LectureOverview Of Lecturez Classical Filter Forms Maximally Flat Magnitude Butterworth Maximally Flat Magnitude Chebyshev Equal Ripple Bessel Maximally Flat Delayz Second Order Confirmation Of Resultsz Example Filter RealizationsUniversity of Southern California EE 541: Choma183n22kkk1H( jω)1cω==+∑()k1kkk kjjj1cab bc−−==−−∑2222nnP(ω )H( jω)P(ω )bω=+z Magnitude Squared Function nth Order Network Containing m Zeros Normalized To Zero Frequency Gain, H(0) Squared Magnitude Function Is Even Function Of Frequency ω Maclaurin Series ck= 0, k = 1, 2, 3, ... m ... n-1Eliminates Frequency Dependence In PassbandEquivalent To Setting First (2n-1) Derivatives w/r ω To ZeroImplies ak=bk, k = 1, 2, 3 … mImplies bk= 0, k = m+1, m+2, … n-1z Resultant Maximally FlatMagnitude Expression246 2m2123 m246 2m 2n123 m n1aω a ω a ω a ωH( jω)1bω b ω b ω b ω b ω+++++=+++++ ++Generalized Transfer FunctionGeneralized Transfer FunctionUniversity of Southern California EE 541: Choma184()222n22bbP(ω )H( jω)ωP(ω )Pωω=+()22nb bb ω P ω=z Transfer Function 3-dB Bandwidth Is ωb Implication Alternate Formz Butterworth P(ω2) = 1 Bandwidth Remains ωb Normalized Frequency: y = ω/ωb nth Order Butterworth Squared Magnitude Transfer Functionz Complex Frequency p = jy = jω/ωb= s/ωb Resultant Transfer Function In Normalized s-Plane2222nnP(ω )H( jω)P(ω )bω=+()22n1H( jy)1y=+() ()22n n2n11H(p)1jp 11p==+− +−Maximally Flat Magnitude FilterMaximally Flat Magnitude FilterUniversity of Southern California EE 541: Choma185Butterworth Frequency ResponseButterworth Frequency Response00.20.40.60.811.20.10 0.16 0.25 0.40 0.63 1.00 1.58 2.51 3.98 6.31 10.00Normalized Frequency, yNormalized Squared Transfer MagnitudeOrder (n) = 2n = 4n = 7Ideal Lowpass Brick Wall00.20.40.60.811.20.10 0.16 0.25 0.40 0.63 1.00 1.58 2.51 3.98 6.31 10.00Normalized Frequency, yNormalized Squared Transfer MagnitudeOrder (n) = 2n = 4n = 7Ideal Lowpass Brick WallUniversity of Southern California EE 541: Choma186()()a2naa2aaaa11H( jy )A1yA1An2y ylogloglog log=≤+−≥≈z Determines Requisite Order Of Filterz Strategy Let Transfer Function Be Attenuated By Factor Of AaAt A Normalized Frequency Of ya yaAnd AaAre Both LargerThan Onez Example 45 dB Down, One OctaveAbove 3-dB Bandwidth 45 dB Means Aa= 177.8 1 Octave Above ωbMeans ya= 2 Result Is n ≥7.474 Practical Realization Must Therefore EntailsAn 8thOrder Butterworth, Maximally FlatMagnitude Filter()161H( jy)1y=+Butterworth AttenuationButterworth AttenuationUniversity of Southern California EE 541: Choma187z Poles Are At p = pkz Observations Left Half Plane Poles Ascribed To H(p) Right Half Plane Poles Ascribed To H(-p) Poles Lie On Unit Circle In Normalized Complex Frequency Plane Poles Lie On Circle Of Radius Equal To 3-dB Bandwidth In Complex Frequency Plane PolesSymmetric About Real And Imaginary Axes In Complex PlaneReal Poles Exist Only For Odd Filter Order, n() ()()()()()nn12n 2nkkj2k n 1πn12nk2k n 1 πj2nk11p 0p 1p1p,k0,1,2,2n1ee+++++++− = ⇒ = −=− ===−()2n2n21H(p)11pH(p) H(p)H( p)=+−=−Poles Of Butterworth FilterPoles Of Butterworth FilterUniversity of Southern California EE 541: Choma188α pcα Lsω BB2π(400 MHz)ZRR50== ====Ωz Specifications Lossless Bandpass StructureCenter Frequency: ωc= 2π(1.8 GHz)3-dB Bandwidth: Bc= 2π(400 MHz)Load Termination: RL= 50 ΩSource Resistance: Rs= 50 ΩAttenuation: 40 dB (Aa= 100) At 3-Times Bc(ya= 3) Design StrategyRealize Lowpass Prototype¾For Convenience, Choose Lowpass BW Equal To Bandpass BW¾Frequency Transform Prototype To Bandpass StructureTasks¾Find Requisite Butterworth Poles¾Determine Lowpass Prototype Transfer Function¾Use Lossless Criteria To Determine Input Impedance¾Realize Filter Topologyz Normalizationααα α ααLZω 19.89 nH C 1 Z ω 7.958 pF====Example Butterworth DesignExample Butterworth DesignUniversity of Southern California EE 541: Choma189()()()()()o41321H(p)pp pp pp pp pp=−−−−−()()2k n 1 π 2k 6 πjj2n 10kp,k0,1,2,9ee++ +== =()()22aaA 1 100 1n4.1922n35y2log loglog log−−≥=⇒==z RequiredFilterOrderz Pole Locationsp0= ej108°= -0.3090 + j0.9511 p5= ej288°= 0.3090 - j0.9511p1= ej144°= -0.8090 + j0.5878 p6= ej324°= 0.8090 - j0.5878p2= ej180°= -1.0000 + j0 p7= ej360°= 1.0000 + j0p3= ej216°= -0.8090 - j0.5878 p8= ej396°= 0.8090 + j0.5878p4= ej252°= -0.3090 - j0.9511 p9= ej432°= 0.3090 + j0.9511 Retain Only Left Half Plane Poles To Compute H(p) Transfer FunctionLHPRHP()()()221H(p)p1 p 0.618 p 1 p 1.618 p 1=+++++Design Calculations 1Design Calculations 1University of Southern California EE 541: