DIGITAL SIGNAL PROCESSING LECTURE 11 Fall 2010 2K8-5th Semester Tahir Muhammad [email protected] Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer, and Buck, ©1999-2000 Prentice Hall Inc. FILTER DESIGN TECHNIQUES | Filter: y Any discrete-time system that modifies certain frequencies | Freque ncy -sel ective filters pass ooynly ceceartain freqeqeuen cy components and totally reject all others | Filter Design Steps: y SifitiSpecification | Problem or application specific y Approximation of specification with a discrete-time system | Our focus is to go from spec to discrete-time system y Implementation | Realization of discrete-time systems depends on target technology | Desired filter is generally implemented with digital computation y It is used to filter a samppgled and digitized continuous-time signal Chapter 7: Filter Design Techniques 1 FILTER DESIGN TECHNIQUES | We have alread y sttdidudied the use of discre te-time systems to implement a continuous-time system y If our specifications are given in continuous time we can use: jω x[n] y[n] HHjTH e = HjTc ( ω / ) jω ( ) xc(t) C/D H(e ) D/C yr(t) ωπ< y It can be noted that the specifications for filters are typically given in frequency domain Chapter 7: Filter Design Techniques 2 FILTER SPECIFICATIONS | SifitiSpecifications y Passband 0.99≤Ω=≤Ω≤Hjeff () 1.01 0 2π ( 2000) y Stopband Hjeff ()Ω≤0.001 2π ( 3000) ≤Ω | Parameters δ1 = 0.01 δ2 = 0.001 Ω=p 2π () 2000 Ω = 2π 3000 | Specs in dB s ( ) y Ideal passband gain =20log(1) = 0 dB y Max passband gain = 20log(1.01) = 0.086dB y Max stopband gain = 20log(0.001) = -60 dB Chapter 7: Filter Design Techniques 6 Butterworth Lowpass Filters | Passband is designed to be maximally flat | The magnitude-squared function is of the form 2 1 2 1 H jΩ = c () 2N Hc ()s = 2N 1 + ()jΩ / jΩc 1 + ()s / jΩc 1 / 2N (jπ / 2N)(2k +N−1) sk = ()− 1 (jΩc )= Ωce for k = 0,1,...,2N - 1 Chebyshev Filters | Equiripple in the passband and monotonic in the stopband | Or equiripple in the stopband and monotonic in the passband 2 1 −1 Hc ()jΩ = 2 2 VN ()x = cos()Ncos x 1 + ε VN ()Ω / Ωc ELLEPTICAL FILTER DISCRETE SIGNAL PROCESSING 7.1 Discrete-Time IIR Filter Design from Continuous-Time Filters Chapter 7: Filter Design Techniques 13 IIR FILTER DESIGN 7.1.1 FILTER DESIGN BY IMPULSE INVARIANCE | RbRemember ilimpulse iiinvariance y Mapping a continuous-time impulse response to discrete-time y Mapping a continuous-time frequency response to discrete-time ∞ jω ⎛⎞ωπ2 hn[]==+ Thdc() nT d H() e∑ H c⎜⎟ j j k k =−∞ ⎝⎠TTdd | If the continuous-time filter is bbdliidandlimited to Hjcd( Ω=) 0 Ω≥π /T jω ⎛⎞ω He( ) = Hc ⎜⎟ j ω ≤ π ⎝⎠Td | In this design procedure: y Start with discrete-time spec in terms of ω (Td has no role in the process) y Go to continuous-time Ω= ω /T and design continuous-time filter y Use impulse invariance to map it back to discrete-time ω= ΩT | Works best for practical filters due to possible aliasing Chapter 7: Filter Design Techniques 16 IMPULSE INVARIANCE OF SYSTEM FUNCTIONS DSP | DlDevelop impul se invari ance relltiation bbtetween system fftiunctions | Partial fraction expansion of transfer function of continuous-time filter: N Ak Hsc ( ) = ∑ k =1 ss− k | Corresponding impulse response: N ⎧ stk ⎪∑ Ak et≥ 0 htc ()= ⎨ k =1 ⎪ ⎩ 00t < | Impulse response of discrete-time filter: NNn snTkd sT kd hn[]== Thdc() nT d∑∑ TAe d k un[] = TA d k() e un[] | System function: kk==11 N TA Hz= dk () ∑ sTkd −1 k =1 1− ez sT | Pole s=sk in s-plane transforms into a pole at e kd in the z-plane Chapter 7: Filter Design Techniques 17 EXAMPLE DSP | IlImpulse iiinvariance applilided to BttButterwor th jω 0.89125≤≤He( ) 1 0 ≤≤ω 0.2π jω H (e ) ≤ 0.17783 0.3π ≤≤ωπ | Since sampling rate Td cancels out we can assume Td=1 | Map spec to continuous time 0.89125≤Ω≤Hj( ) 1 0 ≤Ω≤ 0.2π Hj()Ω≤0.17783 0.3ππ ≤Ω≤ | Butterworth filter is monotonic so spec will be satisfied if Hc ()j0.2π ≥ 0.89125 and Hc (j0.3π) ≤ 0.17783 2 1 Hc ()jΩ = 2N 1 + ()jΩ / jΩc | DiDetermine N and Ωc to satifisfy these condidiitions Chapter 7: Filter Design Techniques 18 EXAMPLE CONT’D DSP | Satisfy both constrains 22NN22 ⎛⎞0.2ππ⎛⎞ 1 ⎛⎞ 0.3 ⎛⎞ 1 1+=⎜⎟⎜⎟ and 1 += ⎜⎟ ⎜⎟ ⎝⎠ΩΩcc⎝⎠0.89125 ⎝⎠ ⎝⎠ 0.17783 | Solve these equations to get N = 5.8858≅Ω= 6 and c 0.70474 | N must be an integer so we round it up to meet the spec | Poles of transfer function 1/12 ()()jkπ /12 2+ 11 sjekcc=−()1( Ω) =Ω for k = 0,1,...,11 | The transfer function 0.12093 Hs= ( ) 22 2 (ss++0.364 0.4945)( s + 0.9945 s + 0.4945)( s + 1.3585 s + 0.4945) | Mapping to z-domain 0.2871−−+ 0.4466zz−−11 2.1428 1.1455 Hz()=+ 1−+ 1.2971zz−12 0.6949−−− 1 −+ 1.0691 zz 12 0.3699 1.8557− 0.6303z−1 + 1− 0.9972zz−−12+ 0.257 Chapter 7: Filter Design Techniques 19 DSP EXAMPLE CONT’D Chapter 7: Filter Design Techniques 20 FILTER DESIGN BY BILINEAR TRANSFORMATION DSP | AidAvoids the aliiliasing problem of impul se invari ance | Maps the entire s-plane onto the unit-circle in the z-plane y Nonlinear transformation y Frequency response subject to warping | Bilinear transformation 21⎛⎞− z−1 s = ⎜⎟−1 Tzd ⎝⎠1+ | Transformed system function ⎡ 21⎛⎞− z−1 ⎤ Hz( ) = Hc ⎢⎥⎜⎟−1 ⎣⎦Td ⎝⎠1+ z | Again Td cancels out so we can ignore it | We can solve the transformation for z as 1/2+ ()Ts1/2/2++ΩσTjT z ==d dds = σ +Ωj 1/21−−−Ω()Tsdddσ T /2/2 jT | Maps the left-half s-plane into the inside of the unit-circle in z y Stable in one domain would stay in the other Chapter 7: Filter Design Techniques 21 BILINEAR TRANSFORMATION DSP | On the unit circle the transform becomes 1/2+ jTΩ ze==d jω 1/2− jTΩ d | To derive the relation between ω and Ω − jω /2 21⎛⎞− ej− jω 2⎡⎤2sin/2ej(ω ) 2 ⎛⎞ω sj==+Ω==⎜⎟−−jjωωσ ⎢⎥/2 tan ⎜⎟ Tedd⎝⎠12cos/22+ Te⎣⎦⎢⎥(ω ) T d⎝⎠ | Which yields 2 ⎛⎞ω ⎛⎞ΩTd Ω=tan⎜⎟ or ω =2arctan ⎜⎟ Td ⎝⎠22⎝⎠ Chapter 7: Filter Design Techniques 22 DSP BILINEAR TRANSFORMATION Chapter 7: Filter Design Techniques EXAMPLE DSP | Bilinear transform applied to Butterworth jω 0.89125≤≤He( ) 1 0 ≤≤ω 0.2π jω H (e ) ≤≤≤0. 17783 0 .3 π ωπ | Apply bilinear transformation to specifications 20.2⎛⎞π 0.89125≤Ω≤Hj( ) 1 00tan ≤Ω≤ tan ⎜⎟ Td ⎝⎠2 20.3⎛⎞π Hj()Ω≤0.17783 tan ⎜⎟ ≤Ω<∞ Td ⎝⎠2 | We can assume Td=1 and apply the specifications to 2 1 Hjc ()Ω= 2N 1/+ΩΩ( c ) | To ge2 22NN22 ⎛⎞2tan0.1ππ⎛⎞ 1 ⎛ 2tan0.15 ⎞ ⎛⎞ 1 1and11+=⎜⎟⎜⎟ and 1 + ⎜ ⎟ = ⎜⎟ ⎝⎠ΩΩcc⎝⎠0.89125 ⎝ ⎠ ⎝⎠ 0.17783 Chapter 7: Filter Design Techniques 24 EXAMPLE CONT’D DSP | SlSolve N and Ωc 22 ⎡⎤⎛⎞⎛⎞⎛⎞⎛⎞11 log−− 1 1 ⎢⎥⎜⎟⎜⎟⎜⎟⎜⎟ ⎣⎦⎢⎥⎝⎠⎝⎠⎝⎠⎝⎠0.17783 0.89125 N ==≅5. 305 6 Ωc = 0.766 2log⎣⎦⎡⎤ tan()() 0.15ππ tan 0.1 | The resulting transfer function has the following poles 1/12 ( jπ /12)( 2k + 11) sjekcc=−()1( Ω) =Ω for k = 0,1,...,11 | Resulting in 0.20238 H s = c ( ) 222 (ss++0.3996 0.5871)( ss ++ 1.0836 0.5871)( ss ++ 1.4802 0.5871) | Applying the bilinear transform yields 6 −1 0.0007378( 1+ z ) Hz= ( ) −−12 −− 12 ()()1−+ 1.2686zz 0.7051 1 −+ 1.0106 zz 0.3583 1 × −12− (1− 0.9044zz+ 0.2155 ) Chapter 7: Filter Design Techniques 25 DSP EXAMPLE CONT’D Chapter 7: Filter Design Techniques 26.