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IIR filter design, Digital signal processing

This chapter discusses the design of IIR digital filters. It begins with preliminaries on the three steps of filter design: specifications, approximations, and implementation. It then covers characteristics of prototype analog filters, including Butterworth filters. Butterworth filters have a maximally flat magnitude response. The chapter describes designing IIR filters by first designing an analog prototype filter, then transforming it to the digital domain.

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Chapter 7 IIR Filter Design Content Preliminaries Characteristics of Prototype Analog Filters Analog-to-Digital Filter Transformations Frequency Transformations
Preliminaries  How to design a digital filter  First: Specifications The design of a digital filter is carried out in three steps: Before we can design a filter, we must have some specifications. These specifications are determined by the applications.  Second: Approximations Once the specifications are defined, we use various concepts and mathematics to come up with a filter description that approximates the given set of specifications. This step is the topic of filter design.
Preliminaries  Third: Implementation The product of the above step is a filter description in the form of either a difference equation, or a system function, or an impulse response. From this description we implement the filter in hardware or software on a computer. In this and the next chapter we will discuss in detail only the second step, which is the conversion of specification into a filter description.
Preliminaries  In many applications, digital filters are used to implement frequency-selective operations;  Therefore, specifications are required in the frequency- domain in terms of the desired magnitude and phase response of the filter;  Generally a linear phase response in the passband is desirable;  An FIR filter is possible to have an exact linear phase;  An IIR filter is impossible to have linear phase in passband. Hence we will consider magnitude-only specifications.  The specifications
Preliminaries There are two ways to give the magnitude specifications  Absolute specifications Provide a set of requirements on the magnitude response function and generally used for FIR filters.)( ωj eH πωaeH eHa s j p j ≤≤≤ ≤≤≤− ||)( ||1)(1 2 1 ω ωω ω ω ],0[ pω Passband ],[ πωs Stopband ],[ sp ωω Transition band The ending frequency of the passband. Bandwidthpω The beginning frequency of the stopband.sω The tolerance (or ripple) in passband1a The tolerance (or ripple) in stopband2a
Preliminaries  Relative specifications (dB) Provide requirements in decibels (dB). This approach is the most popular one in practice and used for both FIR and IIR filters 2 0 2 1 0 1 lg20)(lg20 )( )( lg20 )1lg(20)(lg20 )( )( lg20 aeH eH eH aeH eH eH s s p p j j j j j j −=−== −−=−== ω ω ω ω δ δ The maximum tolerable passband ripple1δ pR The minimum tolerable stopband attenuation2δ sA
Preliminaries  Examples  In a certain filter’s specifications the passband ripple is 0.25dB, and the stopband attenuation is 50dB. Determine the a1 and a2. 0.003210,lg2050 0.0284101),1lg(2025.0 ) 20 50 ( 222 ) 20 25.0 ( 111 ==−== =−=−−== − − aa aa δ δ
Preliminaries dB60001.0lg20lg20 0.1755dB)02.01lg(20)1lg(20 22 11 =−=−= =−−=−−= a a δ δ  Given the passband tolerance a1=0.02 and the stopband tolerance a2=0.001, determine the passband ripple and the stopband attenuation1δ 2δ
Preliminaries  The basic technique of IIR filter design  IIR filters have infinite-length impulse responses, hence they can be matched to analog filters.  Analog filter design is a mature and well developed field.  We can begin the design of a digital filter in the analog domain and then convert the design into the digital domain
Preliminaries There are two approaches to this basic technique Approach 1 Design analog lowpass filter Apply freq. band transformation s → s Apply filter transformation s → z Designed IIR filter Approach 2 Design analog lowpass filter Apply filter transformation s → z Apply freq. band transformation z → z Designed IIR filter return
Characteristics of Prototype Analog Filters  Magnitude-squared function sa pa A jH jH Ω≥Ω≤Ω≤ Ω≤Ω≤Ω≤ + ||, 1 )(0 ||,1)( 1 1 2 2 2 2 ε  Let be the frequency response of an analog filter )( ΩjHa is a passband ripple parameterε is the passband cutoff frequency in rad/secpΩ is the stopband cutoff frequency in rad/secsΩ is a stopband attenuation parameterA sa pa A jH jH Ω=Ω=Ω Ω=Ω + =Ω at 1 )( at 1 1 )( 2 2 2 2 ε
Characteristics of Prototype Analog Filters Ω= =Ω jsaa sHjH )()(  The properties of 2 )( ΩjHa Ω= ∗ −=Ω−Ω=ΩΩ=Ω jsaaaaaaa sHsHjHjHjHjHjH )()()()()()()( 2 )(tha is a real function The poles and zeros of are distributed in a mirror-image symmetry with respect to the axis. For real filters, poles and zeros occur in complex conjugate pairs. Ωj )()( sHsH aa − 22 2 )()()( s aaa jHsHsH −=Ω Ω=−
Characteristics of Prototype Analog Filters Ωj)()( sHsH aa − σ 0 2 2 s-plane
Characteristics of Prototype Analog Filters  How to construct )(sHa )(sHa is the system function of the analog filter. It must be causal and stable. Then all poles of must lie within the left half-plane. )(sHa )()( sHsH aa −All left-half poles of should be assigned to )(sHa )(sHa )( sHa − Zeros are not uniquely determined. They can be halved between and . (Zeros in each half must occur in complex conjugate pairs) If a minimum-phase filter is required, the left-half zeros should be assigned to )(sHa
 Examples )36)(49( )25(16 )( 22 22 2 Ω+Ω+ Ω− =ΩjHa )36)(49( )25(16 )()()( 22 22 2 22 ss s jHsHsH s aaa −− + =Ω=− −=Ω poles 6,7 ±=±= ss 2th order zeros 5js ±= We can assign left-half poles and a pair of conjugate zeros to 6,7 −=−= ss 5js ±= )(sHa )6)(7( )25( )( 2 0 ++ + = ss sK sHa 4 )()( 0 00 =∴ Ω= =Ω= K jHsH asa 4213 1004 )6)(7( )25(4 )( 2 22 ++ + = ++ + = ss s ss s sHa
Characteristics of Prototype Analog Filters  Butterworth lowpass filters  This filter is characterized by the property that its magnitude response is flat in both passband and stopband. The magnitude-squared function of an Nth -order lowpass filter is given by N c a jH 2 2 1 1 )(       Ω Ω + =Ω
Characteristics of Prototype Analog Filters  The properties of Butterworth lowpass filters  At , for all N1)( =ΩjHa0=Ω 707.0 2 1 )( ==ΩjHa At , for all N, which implies a 3dB attenuation at cΩ=Ω cΩ=Ω )( ΩjHa is a monotonically decreasing function of Ω )( ΩjHa approaches an ideal lowpass filter as ∞→N )( ΩjHa 0=Ω is maximally flat at since derivatives of all orders exist and are equal to zero
Characteristics of Prototype Analog Filters  The poles and zeros of )()( sHsH aa − N c N N c N c js aaa js j j s jHsHsH 22 2 2 / 2 )( )( 1 1)()()( Ω+ Ω =       Ω + =Ω=− =Ω Nkejs N kj cc N k 2,,2,1,)()1( ) 2 12 2 1( 2 1 =Ω=Ω−= −+π
Characteristics of Prototype Analog Filters )()( sHsH aa − There are 2N poles of , which are equally distributed on a circle of radius with angular spacing of radians. cΩ N π  If the N is odd, there are poles on real axis.  If the N is even, there are not poles on real axis.  The poles are symmetrically located with respect to the imaginary axis.  A pole never falls on the imaginary axis, and falls on the real axis only if N is odd.
Characteristics of Prototype Analog Filters ∏= − Ω = N k k N c a ss sH 1 )( )( Nkes N kj ck ,,2,1, ) 2 12 2 1( =Ω= −+π In general, we consider and this results in a normalized Butterworth analog prototype filter rad/s1=Ωc )(sHan )()( c ana sHsH Ω = When designing an actual filter with , we can simply do a replacement for s, that is )(sHa rad/s1≠Ωc
 Designing equations Given , two parameters are required to determine a Butterworth lowpass filters : 21 ,,, δδsp ΩΩ cN Ω, 2 2 1 2 1lg20 1lg20 δ δ =               Ω Ω +Ω=Ω =               Ω Ω +Ω=Ω N c s s N c p p at at Solving these two equations for cN Ω,
 , )lg(2 )110/()110(lg 1010 21 NNN s p ′= Ω Ω       −− =′ δδ pΩ Since the actual N chosen is larger than required, specifications can be either met or exceeded at or sΩ pΩTo satisfy the specifications exactly at sΩTo satisfy the specifications exactly at N s c 2 10 110 2 − Ω =Ω δN p c 2 10 110 1 − Ω =Ω δ
 Example  Determine the system function of 3th -order Butterworth analog lowpass filter. Suppose rad/s2=Ωc 62 2 2 1 1 1 1)(       Ω+ =       Ω Ω+ =Ω N c a jH 6,,2,1,2 ) 6 12 2 1( == −+ kes kj k π 64 1 1)()( 6 s sHsH aa − =− 884 8 ))()(( )( 23 321 3 +++ = −−− Ω = sssssssss sH c a Solution:
Design the above filter with normalized Butterworth analog prototype filter. See table 6-4 on page 261 1,)( 02 210 0 == ++++ = NN N an aa sasasaa d sH  in case of 1)0( =jHa00 ad = 3=NFor 2,2 21 == aaWe can find 32 32 2 488 8 ) 2 () 2 (2) 2 (21 1 )()( ssssss sHsH ssana +++ = +++ = = = 2 sss c = Ω →
 Design a lowpass Butterworth filter to satisfy: rad/s1020fordB1 4 1 ×≤Ω≤≤ πδPassband Stopband rad/s105.12fordB15 4 2 ××≥Ω≥ πδ Solution: 15 105.12 1lg20 1 102 1lg20 2 4 2 4 =               Ω ×× + =               Ω × + N c N c π π rad/s105.12dB,15 rad/s102dB,1 4 s2 4 p1 ××=Ω= ×=Ω= πδ πδ
rad/s1013.12 4 ××=Ω πc 5.8858 105.12 102 lg2 )110/()110(lg )lg(2 )110/()110(lg 4 4 10 15 10 1 1010 21 =       ×× ×       −− = Ω Ω       −− =′ π π δδ s p N 6=N 4 12 10 15 4 2 10 101.12792 110 105.12 110 2 ××= − ×× = − Ω =Ω π π δN s c
Look for table 6-4 on page 261 864.3,464.7,142.9,464.7,864.3 54321 ===== aaaaa 65432 65 5 4 4 3 3 2 21 864.3464.7142.9464.7864.31 1 1 1 )( ssssss ssasasasasa sHan ++++++ = ++++++ = === ×× = Ω = 4 1013.12 )()()( π s sans sana sHsHsH c 6554103152202429 29 102.74103.76103.27101.90106.97101.28 101.28 ssssss +×+×+×+×+×+× ×
Look for table 6-6 on page 263 259.0966.0 707.0707.0 966.0259.0 4,3 5,2 6,1 js js js ±−= ±−= ±−=To construct a cascade structure )1.001.93)(1.001.41)(1.000.52( 1 ))()()()()(( 1 )( 222 654321 ++++++ = −−−−−− = ssssss ssssssssssss sHan 4 1013.12 ×× = Ω → π ss s c )105.04101.37)(105.0410)(105.04103.69( 1028.1 )( 952952942 29 ×+×+×++×+×+ × = ssssss sHa
Characteristics of Prototype Analog Filters  Chebyshev lowpass filters  There are two types of Chebyshev filters Chebyshev-I: equiripple in the passband and monotonic in the stopband. Chebyshev-II: monotonic in the passband and equiripple in the stopband. Chebyshev filters can provide lower order than Butterworth filters for the same specifications.       Ω Ω+ =Ω c N a C jH 22 2 1 1)( ε Chebyshev-I
N is the order of the filter is the Nth -order Chebyshev polynomial given by 2 NC    Ω Ω= > ≤ = − c N x xxN xxN xC where 1||),ch(ch 1||),coscos( )( -1 1 xxCxC xCxxCxC NNN == −= −+ )(,1)( )()(2)( 10 11 is the passband ripple factor.ε 10 << ε       Ω Ω+ =Ω c N a C jH 22 2 1 1)( ε
      Ω Ω+ =Ω c N a C jH 22 1 1)( ε  The properties of Chebyshev lowpass filters  At : 0=Ω evenisfor 1 1)0( oddisfor1)0( 2 NjH NjH a a ε+ = =  At : cΩ=Ω NjH ca allfor 2 1 1)( ε+ =Ω  For : cΩ≤Ω≤0 2 1 1~1betweenoscillates ε+ Ω)( jHa cΩ>Ω For : 0tollymonotonicadecreases)( ΩjHa sΩ=Ω For : A jH sa 1)( =Ω
 Designing equations Given , two parameters are required to determine a Chebyshev-I filter: pAAssc ,,,ΩΩ N,ε 110 1.02 −= pA ε       Ω Ω         − ≥ − − c s As ch ch N 1 1.0 1 110 ε                 − Ω=Ω − ε 1101 1.0 1 sA cs ch N ch    Ω=Ω − ε 11 1 3 ch N chcdB Note: this is only for dBc 3Ω<Ω
 Determine system function To determine a causal and stable , we must find the poles of and select the left half-plane poles for . The poles are obtained by finding the roots of )(sHa )()( sHsH aa − )(sHa 01 22 =      Ω + c N j sCε It can be shown that if are the (left half-plane) roots of the above polynomial, then Nkjs kkk ,,2,1 =Ω+= σ     − Ω=Ω     − Ω−= N k b N k a ck ck 2 )12( cos)( 2 )12( sin)( π π σ Nk ,,2,1 =
111 2 ++= εε γ )( 2 1),( 2 1 1111 NNNN ba −− +=−= γγγγ ∏= − = N k k a ss KsH 1 )( )(     + = evenisN, 1 1 oddisN,1 )0( 2 ε jHa Where K is a normalizing factor chosen to make
 Determine poles by geometric method The poles of fall on an ellipse with major axis and minor axis . )()( sHsH aa − cbΩ caΩ σ Ωj cbΩcaΩ N π
 Determine the system function of 2th -order Chebyshev-I lowpass filter. Suppose andrad/s1=Ωc dB1=pA 2589.0110110 1.01.02 =−=−= Ap ε 2 0 2 10 0 0977.11025.1 )( ss d ssaa d sHa ++ = ++ = 0977.1 1025.1 1 0 = = a a 0.8913 2589.1 1 1 1)0( 2 == + = ε jHa 9827.0,8913.0 1025.1 )( 0 0 0 =∴=== d d sH sa  Examples Solution:
 Design a lowpass Chebyshev-I filter to satisfy: rad/s102 4 ×=Ω πc Passband cutoff: dB1=pAPassband ripple: Stopsband cutoff: rad/s105.12 4 ××=Ω πs Stopband attenuation: dB15=sA Solution: 5088.0110110 1.01.0 =−=−= pA ε 3.1978 )5.1( (10.8761) 110 1 1 1 1.0 1 ==       Ω Ω         − ≥ − − − − ch ch ch ch N c s As ε
4=N 5088.0=εdB1=pA 9528.0,4539.1,7426.02756.0 3210 ==== aaaa ， 432 0 43 3 2 210 0 9528.04539.17426.02756.0 )( ssss d ssasasaa d sHan ++++ = ++++ = 0.8913 2589.1 1 1 1)0( 2 == + = ε jHa 0.2456,0.8913 2756.0 )( 0 0 0 =∴=== d d sH san
432 9528.04539.17426.02756.0 2456.0)( ssss sHan ++++ = 4 102 × = Ω → π sss c 434291418 18 105.9866105.7398101.8420104.2954 103.8278 )( ssss sHa +×+×+×+× × =
Analog-to-Digital Filter Transformations  Impulse invariance transformation  Definition To design an IIR filter having a unit sample response h(n) that is the sampled version of the impulse response of the analog filter. That is )()( nThnh a= T : Sampling interval Tjj eeT Ω =Ω= ω ω or, Since this is a sampling operation, the analog and digital frequencies are related by
The system function and are related by)(sHa)(zH ∑ ∞ −∞= = −= k aez k T jsH T zH sT ) 2 ( 1 )( π This implies a mapping from the s-plane to the z-plane T π T π− σ Ωj 0 T π3 T π3− ]Re[z ]Im[ zj
Analog-to-Digital Filter Transformations  Properties  Using ]Re[s=σ UC)theof(outside1|z|intomaps0 UC)the(on1|z|intomaps0 UC)theof(inside1|z|intomaps0 >> == << σ σ σ  Since the entire left half of the s-plane maps into the unit circle, a causal and stable analog filter maps into a causal and stable digital filter.  All semi-infinite left strips of width map into . Thus this mapping is not unique but a many-to-one mapping T/2π 1|| <z
Analog-to-Digital Filter Transformations πω ωω <= ||),( 1 )( T jH T eH a j then There will be no aliasing. Frequency response ∑ ∞ −∞= − = k a j T k jH T eH ) 2 ( 1 )( πωω TT jHjH aa πω ≥Ω==Ω ||for0)()(If  To minimize the effects of aliasing, the T should be selected sufficiently small.  If the filter specifications are given in digital frequency domain, we cannot reduce aliasing by selecting T.  Aliasing occurs if the filter is not exactly band-limited
Analog-to-Digital Filter Transformations  Digitalizing of analog filters ∑= − = N k k k a ss A sH 1 )( Using partial fraction expansion, expand into)(sHa The corresponding impulse response is ∑= − == N k ts kaa tueAsHLth k 1 1 )()]([)( ∑∑ == === N k nTs k N k nTs ka nueAnueAnThnh kk 11 )()()()()( To sample the )(tha
The z-transform of is)(nh ∑∑∑∑ = − ∞ = = − ∞ −∞= − − === N k Ts k n N k nTs k n n ze A zeAznhzH k k 1 1 0 1 1 1 )()()( Conclusions: ∑= − = N k k k a ss A sH 1 )(Compared with  The pole in s-plane is mapped to the pole in z-planeks Tsk e  The partial fraction expansion coefficient of is the same as that of )(sHa )(zH  The zeros in the two domains do not satisfy the same relationship
Analog-to-Digital Filter Transformations  Advantages and disadvantages  The digital filter impulse response is similar to that of a analog filter. This means we can get a good approximations in time domain.  Due to the presence of aliasing, this method is useful only when the analog filter is essentially band-limited to a lowpass or bandpass filter in which there are no oscillations in the stopband.  It is a stable design and that the frequencies and are linearly related. So a linear phase analog filter can be mapped to a linear phase digital filter. Ω ω
 Design procedure  Choose T and determine the analog frequencies  Transform analog poles into digital poles to obtain the digital filter TT s s p p ωω =Ω=Ω , Given the digital lowpass filter specifications 21 ,,, δδωω sp  Design an analog filter using the specifications)(sHa 21 ,,, δδsp ΩΩ ∑= − = N k k k a ss A sH 1 )(  Using partial fraction expansion, expand into)(sHa ∑= − − = N k Ts k ze A zH k 1 1 1 )(
Analog-to-Digital Filter Transformations  Examples  Transform 3 1 1 1 34 2 )( 2 + − + = ++ = ssss sHa into a digital filter using the impulse invariance method in which T=1 )(zH 21 1 4231 31 311 0183.04177.01 3181.0 )(1 )( 11 )( −− − −−−−− −−− −−−− +− = ++− − = − − − = zz z ezeez eeTz ez T ez T zH TTT TT TT
Analog-to-Digital Filter Transformations  Bilinear transformation  Definition This is a conformal mapping that transforms the -axis into the unit circle in the z-plane only once, thus avoiding aliasing of frequency components. This mapping is the best transformation method. Ωj       Ω ⋅=Ω 2 tan 1T c Tj Tj T j T j T j T j e e c ee ee cj 1 1 11 11 1 1 22 22 Ω− Ω− Ω − Ω Ω − Ω + − ⋅= + − ⋅=Ω 1 1 1 1 1 1 1 1 − − − − + − ⋅= + − ⋅= z z c e e cs Ts Ts sc sc z − + =
Ts Ts e e cs 1 1 1 1 − − + − ⋅= Ts ez 1 = 1 1 1 1 − − + − ⋅= z z cs T π T π− 1Ωj 1σ0 s1-plane ]Re[z ]Im[ zj 0 z-plane 0 Ωj σ s-plane
Analog-to-Digital Filter Transformations  Parameter c T c T c T c 2 then, 22 tan 1 11 =Ω≈Ω Ω ⋅≈      Ω ⋅=Ω  Keeping a good corresponding relationship between the analog filter and the digital filter in low frequencies. i.e. in low frequencies1Ω≈Ω       Ω=      ⋅=      Ω ⋅=Ω 2 cotthen 2 tanc 2 tan 1 c c cc c c T c ωω  Keeping a good corresponding relationship between the analog filter and the digital filter in a specific frequency (for example, in the cutoff frequency, )Tcc 1Ω≈ω
 Properties  Using , we obtainΩ+= js σ 22 22 )( )( ||, )( )( Ω+− Ω++ = Ω−− Ω++ = − + = σ σ σ σ c c z jc jc sc sc z So 1||0,1||0,1||0 >→>=→=<→< zzz σσσ  Using , we obtainωj ez = Ω=⋅= + − ⋅= + − ⋅= − − − − jjc e e c z z cs j j ) 2 tan( 1 1 1 1 1 1 ω ω ω The imaginary axis maps onto the unit circle in a one-to-one fashion. Hence there is no aliasing in the frequency domain.  The entire left half-plane maps into the inside of the unit circle. Hence this is a stable transformation.
Analog-to-Digital Filter Transformations  Advantages and disadvantages  It is a stable design;  There is no aliasing;  There is no restriction on the type of filter that can be transformed;. Ω ω The frequencies and are not linearly related. So a linear phase analog filter cannot be mapped to a linear phase digital filter.
 Design procedure  Choose a value for T. We may set T=1 ) 2 tan( 2 ), 2 tan( 2 s s p p TT ωω =Ω=Ω Given the digital lowpass filter specifications 21 ,,, δδωω sp  Prewarp the cutoff frequencies and ; that ispω sω  Design an analog filter to meet the specifications 21 ,,, δδsp ΩΩ )(sHa  Finally, set ) 1 12 ()( 1 1 − − + − = z z T HzH a and simplify to obtain as a rational function in)(zH 1− z
Analog-to-Digital Filter Transformations  Examples  Transform into a digital filter using the bilinear transformation. Choose T=1 34 2 )( 2 ++ = ss sHa 21 21 1 1 2 1 1 1 1 1 1 1 0.070.131 13.00.2713.0 3 1 1 24 1 1 2 2 ) 1 1 2() 1 12 ()( −− −− − − − − − − = − − −− ++ = +      + − +      + − = + − = + − = zz zz z z z z z z H z z T HzH a T a
 Design the digital Chebyshev-I filter using bilinear transformation. The specifications are: dB15,3.0 dB1,2.0 2 1 == == δπω δπω s p Solution  Let T=1 1.0191)15.0tan(2) 2 tan( 2 0.6498)1.0tan(2) 2 tan( 2 ===Ω ===Ω π ω π ω s s p p T T  Prewarp the cutoff frequencies
 Design an analog Chebyshev-I filter to meet the specifications 21 ,,, δδsp ΩΩ )(sHa 0.04920.20380.61406192.0 0438.0 )( 234 ++++ = ssss sHa )0.64931.55481)(0.84821.49961( )1(0018.0 0.55072.29253.82903.05431 0.00180.00730.01100.00730018.0 )( 2121 41 4321 4321 −−−− − −−−− −−−− +−+− + = +−+− ++++ = zzzz z zzzz zzzz zH
Analog-to-Digital Filter Transformations  Comparison of three filters Using different prototype analog filters will give out different N and the minimum stopband attenuations. dB15,3.0 dB1,2.0 2 1 == == δπω δπω s p Given the digital filter specifications: prototype Order N Stopband Att. Butterworth 6 15 dB Chebyshev-I 4 25 dB Elliptic 3 27 dB return
Frequency Transformations  Introduction The treatment in the preceding section is focused primarily on the design of digital lowpass IIR filters. If we wish to design a highpass or a bandpass or a bandstop filter, it is a simple matter to take a lowpass prototype filter and perform a frequency transformation.  Frequency transformations in the analog domain  Frequency transformations in the digital domain There are two approaches to perform the frequency transformation
Frequency Transformations Approach 1 Analog lowpass filter Frequency transformation s → s Filter transformation s → z Designed IIR filter Approach 2 Analog lowpass filter Filter transformation s → z Frequency transformation z → z Designed IIR filter
Frequency Transformations  Specifications of frequency-selective filters  Lowpass filter 21 ,,, δδωω sp  highpass filter 21 ,,, δδωω ps  bandpass filter 212211 ,,,,, δδωωωω spps  bandstop filter 212211 ,,,,, δδωωωω pssp
Frequency Transformations  Frequency transformations in the digital domain )(zHL the given prototype lowpass digital filter )(ZHd the desired frequency-selective digital filter )( 11 −− = ZGzDefine a mapping of the form Such that )( 11)()( −− = = ZGzLd zHZH To do this, we simply replace everywhere in by the function 1− z )(zHL )( 1− ZG
Frequency Transformations Given that is a stable and causal filter, we also want to be stable and causal. This imposes the following requirements: )(zHL )(ZHd  The unit circle of the z-plane must map onto the unit circle of the Z-plane  The inside of the unit circle of the z-plane must also map onto the inside of the unit circle of the Z-plane. 1− Z)( 1− ZG must be a rational function in so that )(ZHd is implementable.
Frequency Transformations Let and be the frequency variables of and , respectively. That is . Then θ ω z Z ωθ jj eZez == , )](arg[ )()( ω ωωθ j eGjjjj eeGeGe − −−− == )](arg[,1)( ωω jj eGθeG −− −== Hence the is an all-pass function)( 1− ZG 1||, 1 )( 1 1 1 11 < − − == ∏= − ∗− −− k N k k k a Za aZ ZGz By choosing an appropriate order N and the coefficients , we can obtain a variety of mappingska
Frequency Transformations  Frequency transformation formulae  Lowpass - Lowpass 1 1 1 1 − − − − − = Z Z z α α ]2/)sin[( ]2/)sin[( cc cc ωθ ωθ α + − = : Cutoff frequency of new digital filtercω cθ The cutoff frequency of prototype lowpass digital filter
Frequency Transformations  Lowpass - Highpass 1 1 1 1 − − − + + −= Z Z z α α ]2/)cos[( ]2/)cos[( cc cc ωθ ωθ α − + −= : Cutoff frequency of new digital filtercω
Frequency Transformations  Lowpass - Bandpass 11 1 2 2 2 1 1 2 1 +− +− −= −− −− − ZZ ZZ z αα αα 1 1 , 1 2 2 tan) 2 cot(,cos ]2/)cos[( ]2/)cos[( 21 12 0 12 12 + − = + = − == − + = k k k k k c α β α θωω ω ωω ωω β : lower cutoff frequency of bandpass digital filter1ω : upper cutoff frequency of bandpass digital filter2ω : center frequency of the passband0ω
Frequency Transformations  Lowpass - Bandstop : lower cutoff frequency of bandstop digital filter1ω : upper cutoff frequency of bandstop digital filter2ω : center frequency of the stopband0ω k k k k c + − = + = − == − + = 1 1 , 1 2 2 tan) 2 tan(,cos ]2/)cos[( ]2/)cos[( 21 12 0 12 12 α β α θωω ω ωω ωω β 11 1 2 2 2 1 1 2 1 +− +− = −− −− − ZZ ZZ z αα αα
Frequency Transformations  Design procedure  Determine the specifications of the digital prototype lowpass filter;  Determine the specifications of the analog prototype lowpass filter;  Design the analog prototype lowpass filter;  Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation;  Perform the frequency transformation in digital domain to obtain the desired frequency-selective filters.
Frequency Transformations  Examples  Given the specifications of Chebyshev-I lowpass filter dB15,3.0 dB1,2.0 2 1 == == δπθ δπθ s p Design a highpass filter with the above tolerances but with passband beginning at πω 6.0=p and its system function )6493.05548.11)(8482.04996.11( )1(01836.0 )( 2121 41 −−−− − +−+− + = zzzz z zHL
Frequency Transformations Solution 3820.0 ]2/)6.02.0cos[( ]2/)6.02.0cos[( −= − + −= ππ ππ α )4019.00416.11)(7647.05561.01( )1(0243.0 )()( 2121 41 3820.01 3820.0 1 1 1 −−−− − − − −= ++++ − = = − − − ZZZZ Z zHZH Z Z zLd
Frequency Transformations  Using the Chebyshev-I prototype to design a highpass digital filter to satisfy dB15,46.0 dB1,6.0 == == ss pp A R πω πω  Determine the specifications of the digital prototype lowpass filter Solution πθ 2.0=p Choose the passband frequency with a reasonable value: Determine the stopband frequency by 1 1 1 1 − − − + + −= Z Z z α α ) 1 arg( 1 ω ω ω ω θ α α θ α α j j j j j e e e e e − − − − − + + −−=→ + + −=
3820.0 ]2/)6.02.0cos[( ]2/)6.02.0cos[( ]2/)cos[( ]2/)cos[( −= − + −= − + −= ππ ππ ωθ ωθ α pp pp π α α θ ω ω 3.0) 3820.01 3820.0 arg() 1 arg( 46.0 46.0 = − − −= + + −= − − − − j j j j s e e e e s s  Determine the specifications of the analog prototype lowpass filter Set T = 1 and prewarp the cutoff frequencies 1.0191)15.0tan(2) 2 tan( 2 0.6498)1.0tan(2) 2 tan( 2 ===Ω ===Ω π θ π θ s s p p T T
 Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,,, ΩΩ 0.04920.20380.61406192.0 0438.0 )( 234 ++++ = ssss sHa )0.64931.55481)(0.84821.49961( )1(0018.0 0.55072.29253.82903.05431 0.00180.00730.01100.00730018.0 )( 2121 41 4321 4321 −−−− − −−−− −−−− +−+− + = +−+− ++++ = zzzz z zzzz zzzz zHL  Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation
 Perform the frequency transformation in digital domain to obtain the desired digital highpass filter )4019.00416.11)(7647.05561.01( )1(0243.0 )()( 2121 41 1 1 1 1 −−−− − + + −= ++++ − = = − − − ZZZZ Z zHZH Z Z zLh α α
Frequency Transformations  Using the Chebyshev-I prototype to design a bandpass digital filter to satisfy dB15,7.0,2.0 dB1,5.0,4.0 21 21 === === sss ppp A R πωπω πωπω  Determine the specifications of the digital prototype lowpass filter Solution πθ 2.0=p Choose the passband frequency with a reasonable value:
Determine the stopband frequency by π αα αα θ ωω ωω 69.0) 1 arg( 22 22 1 2 2 21 2 = +− +− −−= −− −− ss ss jj jj s ee ee 1584.0 ]2/)4.05.0cos[( ]2/)4.05.0cos[( ]2/)cos[( ]2/)cos[( 12 12 = − + = − + = ππ ππ ωω ωω β pp pp 0515.2 2 2.0 tan) 2 4.05.0 cot( = − = πππ k 11 1 2 2 2 1 1 2 1 +− +− −= −− −− − ZZ ZZ z αα αα 3446.0 1 1 ,2130.0 1 2 21 = + − == + = k k k k α β α
 Determine the specifications of the analog prototype lowpass filter Set T = 1 and prewarp the cutoff frequencies 3.7842)0.3450tan(2) 2 tan( 2 0.6498)1.0tan(2) 2 tan( 2 ===Ω ===Ω π θ π θ s s p p T T  Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,,, ΩΩ 4656.07134.0 4149.0 )( 2 ++ = ss sHa
21 21 5157.01997.11 )1(0704.0 )( −− − +− +× = zz z zHL  Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation  Perform the frequency transformation in digital domain to obtain the desired digital bandpass filter 4321 42 1 0.71060.48147020.15731.01 0205.00410.00205.0 )()( 1 1 2 2 2 1 1 2 1 −−−− −− +− +− −= +−+− +− = = −− −− − ZZZZ ZZ zHZH ZZ ZZ zLbp αα αα
Frequency Transformations  Using the Chebyshev-I prototype to design a bandstop digital filter to satisfy dB20,65.0,35.0 dB1,75.0,25.0 21 21 === === sss ppp A R πωπω πωπω  Determine the specifications of the digital prototype lowpass filter Solution πθ 2.0=p Choose the passband frequency with a reasonable value:
Determine the stopband frequency by π αα αα θ ωω ωω 0.1919) 1 arg( 11 11 1 2 2 21 2 = +− +− −= −− −− ss ss jj jj s ee ee 0 ]2/)25.075.0cos[( ]2/)25.075.0cos[( ]2/)cos[( ]2/)cos[( 12 12 = − + = − + = ππ ππ ωω ωω β pp pp 0.1584 2 2.0 tan) 2 25.075.0 tan( = − = πππ k 11 1 2 2 2 1 1 2 1 +− +− = −− −− − ZZ ZZ z αα αα 0.7265 1 1 ,0 1 2 21 = + − == + = k k k α β α
 Determine the specifications of the analog prototype lowpass filter Set T = 1 and prewarp the cutoff frequencies 6217.0)0.0959tan(2) 2 tan( 2 0.3168)1.0tan(2) 2 tan( 2 ===Ω ===Ω π θ π θ s s p p T T  Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,,, ΩΩ 0.01561243.03131.0 0156.0 )( 23 +++ = sss sHa
321 321 0.73352.36922.62251 0.00160.00490.00490.0016 )( −−− −−− ++− +++ = zzz zzz zHL  Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation  Perform the frequency transformation in digital domain to obtain the desired digital bandstop filter )0.3391)(0.7761.2481)(0.7761.2481( )1(0.132 )()( 22121 32 11 1 2 2 2 1 1 2 1 −−−−− − +− +− = −+−++ +× = = −− −− − ZZZZZ Z zHZH ZZ ZZ zLbs αα αα return
0 30 40 50 60 0.707 1 Magnitude Response Analog frequency in rad/s N=2 N=4 N=8 N=16 return )( ΩjHa
0 2 4 6 0.707 1 Magnitude Response Analog frequency in rad/s Amplitude 0 2 4 6 -3 -1 0 1 3 Phase Response Analog frequency in rad/s Phaseinrad )( ΩjHa return
0 2 3 5 x 10 4 0 0.1778 0.8913 1 Magnitude Response Analog frequency in pi units H 0 2 3 5 x 10 4 -30 -15 -10 Magnitude in dB Analog frequency in pi units decibels 0 2 3 5 x 10 4 -1 -0.5 0 0.5 1 Phase Response Analog frequency in pi units P 0 0.5 1 1.5 x 10 -4 0 10000 20000 Impulse Response time in seconds ha(t) return
0 2 3 5 x 10 4 0 0.1778 0.8913 1 Magnitude Response Analog frequency in pi units H 0 2 3 5 x 10 4 -30 -15 -10 Magnitude in dB Analog frequency in pi units decibels 0 2 3 5 x 10 4 -1 -0.5 0 0.5 1 Phase Response Analog frequency in pi units P 0 0.5 1 1.5 x 10 -4 -5000 0 5000 10000 15000 20000 Impulse Response time in seconds ha(t) return
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(0 xC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(2 xC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(3 xC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(4 xC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(5 xC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(6 xC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 return
0 2 3 5 0.8913 1 Magnitude Response Analog frequency in rad/s Amplitude 4=N
0 2 3 5 0.8913 1 Magnitude Response Analog frequency in rad/s Amplitude return 5=N
0 1 2 3 5 0.8913 1 Magnitude Response Analog frequency in rad/s Amplitude 0 1 2 3 5 -3 -2 -1 0 1 Phase Response Analog frequency in rad/s Phaseinrad return )( ΩjHa
0 2 3 5 x 10 4 0 0.1778 0.8913 1 Magnitude Response Analog frequency in pi units H 0 2 3 5 x 10 4 -30 -15 -10 Magnitude in dB Analog frequency in pi units decibels 0 2 3 5 x 10 4 -1 -0.5 0 0.5 1 Phase Response Analog frequency in pi units Phaseinpiunits 0 1 2 3 4 x 10 -4 -5000 0 5000 10000 15000 20000 Impulse Response time in seconds ha(t) return
0 pi/T 2*pi/T 0.2 0.4 0.6 0 pi 2*pi 0.2 0.4 0.6 )( ΩjHa )( ωj eH 1=T Ω ω
0 pi/T 2*pi/T 0.2 0.4 0.6 0 pi 2*pi 0.2 0.4 0.6 )( ΩjHa )( ωj eH 1.0=T Ω ω return
0 0.2 0.3 1 0.1778 0.8913 1 Magnitude Response Frequency in pi 0 0.2 0.3 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi 0 0.2 0.3 1 -15 -10 Magnitude Response in dB Frequency in pi 0 0.2 0.3 1 2 4 6 8 10 Group Delay Frequency in pi return
0 pi/T 2*pi/T 0.2 0.4 0.6 Magnitude Response frequency in rad/s 0 pi 2*pi 0.2 0.4 0.6 frequency in rad/sample )( ΩjHa )( ωj eH 1=T Ω ω
0 pi/T 2*pi/T 0.2 0.4 0.6 Magnitude Response frequency in rad/s 0 pi 2*pi 0.2 0.4 0.6 frequency in rad/sample )( ΩjHa )( ωj eH 1.0=T Ω ω return
0 0.2 0.3 1 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.2 0.3 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.2 0.3 1 -15 -10 Magnitude Response in dB Frequency in pi units 0 0.2 0.3 1 3 6 9 12 15 Group Delay Frequency in pi units samples return
0 0.2 0.3 1 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.6 1 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.2 0.3 1 -15 -10 Magnitude Response in dB Frequency in pi units 0 0.6 1 -15 -10 Magnitude Response in dB Frequency in pi units return
0 0.46 0.6 1 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.46 0.6 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.46 0.6 1 -15 -10 Magnitude Response in dB Frequency in pi units 0 0.46 0.6 1 2 4 6 8 10 Group Delay Frequency in pi units samples return
0 0.2 0.4 0.5 0.7 1 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.2 0.4 0.5 0.7 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.2 0.4 0.5 0.7 1 -100 -80 -60 -40 -20 0 Magnitude Response in dB Frequency in pi units 0 0.2 0.4 0.5 0.7 1 3 6 9 12 15 Group Delay Frequency in pi units samples return
0 0.250.35 0.650.75 1 0.1 0.8913 1 Magnitude Response Frequency in pi units 0 0.250.35 0.650.75 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.250.35 0.650.75 1 -30 -20 -10 0 Magnitude Response in dB Frequency in pi units 0 0.250.35 0.650.75 1 2 4 6 8 10 Group Delay Frequency in pi units samples return

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IIR filter design, Digital signal processing

  • 1.
    Chapter 7 IIR FilterDesign Content Preliminaries Characteristics of Prototype Analog Filters Analog-to-Digital Filter Transformations Frequency Transformations
  • 2.
    Preliminaries  How todesign a digital filter  First: Specifications The design of a digital filter is carried out in three steps: Before we can design a filter, we must have some specifications. These specifications are determined by the applications.  Second: Approximations Once the specifications are defined, we use various concepts and mathematics to come up with a filter description that approximates the given set of specifications. This step is the topic of filter design.
  • 3.
    Preliminaries  Third: Implementation Theproduct of the above step is a filter description in the form of either a difference equation, or a system function, or an impulse response. From this description we implement the filter in hardware or software on a computer. In this and the next chapter we will discuss in detail only the second step, which is the conversion of specification into a filter description.
  • 4.
    Preliminaries  In manyapplications, digital filters are used to implement frequency-selective operations;  Therefore, specifications are required in the frequency- domain in terms of the desired magnitude and phase response of the filter;  Generally a linear phase response in the passband is desirable;  An FIR filter is possible to have an exact linear phase;  An IIR filter is impossible to have linear phase in passband. Hence we will consider magnitude-only specifications.  The specifications
  • 5.
    Preliminaries There are twoways to give the magnitude specifications  Absolute specifications Provide a set of requirements on the magnitude response function and generally used for FIR filters.)( ωj eH πωaeH eHa s j p j ≤≤≤ ≤≤≤− ||)( ||1)(1 2 1 ω ωω ω ω ],0[ pω Passband ],[ πωs Stopband ],[ sp ωω Transition band The ending frequency of the passband. Bandwidthpω The beginning frequency of the stopband.sω The tolerance (or ripple) in passband1a The tolerance (or ripple) in stopband2a
  • 6.
    Preliminaries  Relative specifications(dB) Provide requirements in decibels (dB). This approach is the most popular one in practice and used for both FIR and IIR filters 2 0 2 1 0 1 lg20)(lg20 )( )( lg20 )1lg(20)(lg20 )( )( lg20 aeH eH eH aeH eH eH s s p p j j j j j j −=−== −−=−== ω ω ω ω δ δ The maximum tolerable passband ripple1δ pR The minimum tolerable stopband attenuation2δ sA
  • 7.
    Preliminaries  Examples  Ina certain filter’s specifications the passband ripple is 0.25dB, and the stopband attenuation is 50dB. Determine the a1 and a2. 0.003210,lg2050 0.0284101),1lg(2025.0 ) 20 50 ( 222 ) 20 25.0 ( 111 ==−== =−=−−== − − aa aa δ δ
  • 8.
    Preliminaries dB60001.0lg20lg20 0.1755dB)02.01lg(20)1lg(20 22 11 =−=−= =−−=−−= a a δ δ  Given thepassband tolerance a1=0.02 and the stopband tolerance a2=0.001, determine the passband ripple and the stopband attenuation1δ 2δ
  • 9.
    Preliminaries  The basictechnique of IIR filter design  IIR filters have infinite-length impulse responses, hence they can be matched to analog filters.  Analog filter design is a mature and well developed field.  We can begin the design of a digital filter in the analog domain and then convert the design into the digital domain
  • 10.
    Preliminaries There are twoapproaches to this basic technique Approach 1 Design analog lowpass filter Apply freq. band transformation s → s Apply filter transformation s → z Designed IIR filter Approach 2 Design analog lowpass filter Apply filter transformation s → z Apply freq. band transformation z → z Designed IIR filter return
  • 11.
    Characteristics of PrototypeAnalog Filters  Magnitude-squared function sa pa A jH jH Ω≥Ω≤Ω≤ Ω≤Ω≤Ω≤ + ||, 1 )(0 ||,1)( 1 1 2 2 2 2 ε  Let be the frequency response of an analog filter )( ΩjHa is a passband ripple parameterε is the passband cutoff frequency in rad/secpΩ is the stopband cutoff frequency in rad/secsΩ is a stopband attenuation parameterA sa pa A jH jH Ω=Ω=Ω Ω=Ω + =Ω at 1 )( at 1 1 )( 2 2 2 2 ε
  • 12.
    Characteristics of PrototypeAnalog Filters Ω= =Ω jsaa sHjH )()(  The properties of 2 )( ΩjHa Ω= ∗ −=Ω−Ω=ΩΩ=Ω jsaaaaaaa sHsHjHjHjHjHjH )()()()()()()( 2 )(tha is a real function The poles and zeros of are distributed in a mirror-image symmetry with respect to the axis. For real filters, poles and zeros occur in complex conjugate pairs. Ωj )()( sHsH aa − 22 2 )()()( s aaa jHsHsH −=Ω Ω=−
  • 13.
    Characteristics of PrototypeAnalog Filters Ωj)()( sHsH aa − σ 0 2 2 s-plane
  • 14.
    Characteristics of PrototypeAnalog Filters  How to construct )(sHa )(sHa is the system function of the analog filter. It must be causal and stable. Then all poles of must lie within the left half-plane. )(sHa )()( sHsH aa −All left-half poles of should be assigned to )(sHa )(sHa )( sHa − Zeros are not uniquely determined. They can be halved between and . (Zeros in each half must occur in complex conjugate pairs) If a minimum-phase filter is required, the left-half zeros should be assigned to )(sHa
  • 15.
     Examples )36)(49( )25(16 )( 22 22 2 Ω+Ω+ Ω− =ΩjHa )36)(49( )25(16 )()()(22 22 2 22 ss s jHsHsH s aaa −− + =Ω=− −=Ω poles 6,7 ±=±= ss 2th order zeros 5js ±= We can assign left-half poles and a pair of conjugate zeros to 6,7 −=−= ss 5js ±= )(sHa )6)(7( )25( )( 2 0 ++ + = ss sK sHa 4 )()( 0 00 =∴ Ω= =Ω= K jHsH asa 4213 1004 )6)(7( )25(4 )( 2 22 ++ + = ++ + = ss s ss s sHa
  • 16.
    Characteristics of PrototypeAnalog Filters  Butterworth lowpass filters  This filter is characterized by the property that its magnitude response is flat in both passband and stopband. The magnitude-squared function of an Nth -order lowpass filter is given by N c a jH 2 2 1 1 )(       Ω Ω + =Ω
  • 17.
    Characteristics of PrototypeAnalog Filters  The properties of Butterworth lowpass filters  At , for all N1)( =ΩjHa0=Ω 707.0 2 1 )( ==ΩjHa At , for all N, which implies a 3dB attenuation at cΩ=Ω cΩ=Ω )( ΩjHa is a monotonically decreasing function of Ω )( ΩjHa approaches an ideal lowpass filter as ∞→N )( ΩjHa 0=Ω is maximally flat at since derivatives of all orders exist and are equal to zero
  • 18.
    Characteristics of PrototypeAnalog Filters  The poles and zeros of )()( sHsH aa − N c N N c N c js aaa js j j s jHsHsH 22 2 2 / 2 )( )( 1 1)()()( Ω+ Ω =       Ω + =Ω=− =Ω Nkejs N kj cc N k 2,,2,1,)()1( ) 2 12 2 1( 2 1 =Ω=Ω−= −+π
  • 19.
    Characteristics of PrototypeAnalog Filters )()( sHsH aa − There are 2N poles of , which are equally distributed on a circle of radius with angular spacing of radians. cΩ N π  If the N is odd, there are poles on real axis.  If the N is even, there are not poles on real axis.  The poles are symmetrically located with respect to the imaginary axis.  A pole never falls on the imaginary axis, and falls on the real axis only if N is odd.
  • 20.
    Characteristics of PrototypeAnalog Filters ∏= − Ω = N k k N c a ss sH 1 )( )( Nkes N kj ck ,,2,1, ) 2 12 2 1( =Ω= −+π In general, we consider and this results in a normalized Butterworth analog prototype filter rad/s1=Ωc )(sHan )()( c ana sHsH Ω = When designing an actual filter with , we can simply do a replacement for s, that is )(sHa rad/s1≠Ωc
  • 21.
     Designing equations Given, two parameters are required to determine a Butterworth lowpass filters : 21 ,,, δδsp ΩΩ cN Ω, 2 2 1 2 1lg20 1lg20 δ δ =               Ω Ω +Ω=Ω =               Ω Ω +Ω=Ω N c s s N c p p at at Solving these two equations for cN Ω,
  • 22.
     , )lg(2 )110/()110(lg 1010 21 NNN s p ′= Ω Ω       −− =′ δδ pΩ Sincethe actual N chosen is larger than required, specifications can be either met or exceeded at or sΩ pΩTo satisfy the specifications exactly at sΩTo satisfy the specifications exactly at N s c 2 10 110 2 − Ω =Ω δN p c 2 10 110 1 − Ω =Ω δ
  • 23.
     Example  Determinethe system function of 3th -order Butterworth analog lowpass filter. Suppose rad/s2=Ωc 62 2 2 1 1 1 1)(       Ω+ =       Ω Ω+ =Ω N c a jH 6,,2,1,2 ) 6 12 2 1( == −+ kes kj k π 64 1 1)()( 6 s sHsH aa − =− 884 8 ))()(( )( 23 321 3 +++ = −−− Ω = sssssssss sH c a Solution:
  • 24.
    Design the abovefilter with normalized Butterworth analog prototype filter. See table 6-4 on page 261 1,)( 02 210 0 == ++++ = NN N an aa sasasaa d sH  in case of 1)0( =jHa00 ad = 3=NFor 2,2 21 == aaWe can find 32 32 2 488 8 ) 2 () 2 (2) 2 (21 1 )()( ssssss sHsH ssana +++ = +++ = = = 2 sss c = Ω →
  • 25.
     Design alowpass Butterworth filter to satisfy: rad/s1020fordB1 4 1 ×≤Ω≤≤ πδPassband Stopband rad/s105.12fordB15 4 2 ××≥Ω≥ πδ Solution: 15 105.12 1lg20 1 102 1lg20 2 4 2 4 =               Ω ×× + =               Ω × + N c N c π π rad/s105.12dB,15 rad/s102dB,1 4 s2 4 p1 ××=Ω= ×=Ω= πδ πδ
  • 26.
  • 27.
    Look for table6-4 on page 261 864.3,464.7,142.9,464.7,864.3 54321 ===== aaaaa 65432 65 5 4 4 3 3 2 21 864.3464.7142.9464.7864.31 1 1 1 )( ssssss ssasasasasa sHan ++++++ = ++++++ = === ×× = Ω = 4 1013.12 )()()( π s sans sana sHsHsH c 6554103152202429 29 102.74103.76103.27101.90106.97101.28 101.28 ssssss +×+×+×+×+×+× ×
  • 28.
    Look for table6-6 on page 263 259.0966.0 707.0707.0 966.0259.0 4,3 5,2 6,1 js js js ±−= ±−= ±−=To construct a cascade structure )1.001.93)(1.001.41)(1.000.52( 1 ))()()()()(( 1 )( 222 654321 ++++++ = −−−−−− = ssssss ssssssssssss sHan 4 1013.12 ×× = Ω → π ss s c )105.04101.37)(105.0410)(105.04103.69( 1028.1 )( 952952942 29 ×+×+×++×+×+ × = ssssss sHa
  • 29.
    Characteristics of PrototypeAnalog Filters  Chebyshev lowpass filters  There are two types of Chebyshev filters Chebyshev-I: equiripple in the passband and monotonic in the stopband. Chebyshev-II: monotonic in the passband and equiripple in the stopband. Chebyshev filters can provide lower order than Butterworth filters for the same specifications.       Ω Ω+ =Ω c N a C jH 22 2 1 1)( ε Chebyshev-I
  • 30.
    N is theorder of the filter is the Nth -order Chebyshev polynomial given by 2 NC    Ω Ω= > ≤ = − c N x xxN xxN xC where 1||),ch(ch 1||),coscos( )( -1 1 xxCxC xCxxCxC NNN == −= −+ )(,1)( )()(2)( 10 11 is the passband ripple factor.ε 10 << ε       Ω Ω+ =Ω c N a C jH 22 2 1 1)( ε
  • 31.
          Ω Ω+ =Ω c N a C jH 22 1 1)( ε  The propertiesof Chebyshev lowpass filters  At : 0=Ω evenisfor 1 1)0( oddisfor1)0( 2 NjH NjH a a ε+ = =  At : cΩ=Ω NjH ca allfor 2 1 1)( ε+ =Ω  For : cΩ≤Ω≤0 2 1 1~1betweenoscillates ε+ Ω)( jHa cΩ>Ω For : 0tollymonotonicadecreases)( ΩjHa sΩ=Ω For : A jH sa 1)( =Ω
  • 32.
     Designing equations Given, two parameters are required to determine a Chebyshev-I filter: pAAssc ,,,ΩΩ N,ε 110 1.02 −= pA ε       Ω Ω         − ≥ − − c s As ch ch N 1 1.0 1 110 ε                 − Ω=Ω − ε 1101 1.0 1 sA cs ch N ch    Ω=Ω − ε 11 1 3 ch N chcdB Note: this is only for dBc 3Ω<Ω
  • 33.
     Determine systemfunction To determine a causal and stable , we must find the poles of and select the left half-plane poles for . The poles are obtained by finding the roots of )(sHa )()( sHsH aa − )(sHa 01 22 =      Ω + c N j sCε It can be shown that if are the (left half-plane) roots of the above polynomial, then Nkjs kkk ,,2,1 =Ω+= σ     − Ω=Ω     − Ω−= N k b N k a ck ck 2 )12( cos)( 2 )12( sin)( π π σ Nk ,,2,1 =
  • 34.
  • 35.
     Determine polesby geometric method The poles of fall on an ellipse with major axis and minor axis . )()( sHsH aa − cbΩ caΩ σ Ωj cbΩcaΩ N π
  • 36.
     Determine thesystem function of 2th -order Chebyshev-I lowpass filter. Suppose andrad/s1=Ωc dB1=pA 2589.0110110 1.01.02 =−=−= Ap ε 2 0 2 10 0 0977.11025.1 )( ss d ssaa d sHa ++ = ++ = 0977.1 1025.1 1 0 = = a a 0.8913 2589.1 1 1 1)0( 2 == + = ε jHa 9827.0,8913.0 1025.1 )( 0 0 0 =∴=== d d sH sa  Examples Solution:
  • 37.
     Design alowpass Chebyshev-I filter to satisfy: rad/s102 4 ×=Ω πc Passband cutoff: dB1=pAPassband ripple: Stopsband cutoff: rad/s105.12 4 ××=Ω πs Stopband attenuation: dB15=sA Solution: 5088.0110110 1.01.0 =−=−= pA ε 3.1978 )5.1( (10.8761) 110 1 1 1 1.0 1 ==       Ω Ω         − ≥ − − − − ch ch ch ch N c s As ε
  • 38.
    4=N 5088.0=εdB1=pA 9528.0,4539.1,7426.02756.0 3210==== aaaa ， 432 0 43 3 2 210 0 9528.04539.17426.02756.0 )( ssss d ssasasaa d sHan ++++ = ++++ = 0.8913 2589.1 1 1 1)0( 2 == + = ε jHa 0.2456,0.8913 2756.0 )( 0 0 0 =∴=== d d sH san
  • 39.
  • 40.
    Analog-to-Digital Filter Transformations Impulse invariance transformation  Definition To design an IIR filter having a unit sample response h(n) that is the sampled version of the impulse response of the analog filter. That is )()( nThnh a= T : Sampling interval Tjj eeT Ω =Ω= ω ω or, Since this is a sampling operation, the analog and digital frequencies are related by
  • 41.
    The system functionand are related by)(sHa)(zH ∑ ∞ −∞= = −= k aez k T jsH T zH sT ) 2 ( 1 )( π This implies a mapping from the s-plane to the z-plane T π T π− σ Ωj 0 T π3 T π3− ]Re[z ]Im[ zj
  • 42.
    Analog-to-Digital Filter Transformations Properties  Using ]Re[s=σ UC)theof(outside1|z|intomaps0 UC)the(on1|z|intomaps0 UC)theof(inside1|z|intomaps0 >> == << σ σ σ  Since the entire left half of the s-plane maps into the unit circle, a causal and stable analog filter maps into a causal and stable digital filter.  All semi-infinite left strips of width map into . Thus this mapping is not unique but a many-to-one mapping T/2π 1|| <z
  • 43.
    Analog-to-Digital Filter Transformations πω ωω <=||),( 1 )( T jH T eH a j then There will be no aliasing. Frequency response ∑ ∞ −∞= − = k a j T k jH T eH ) 2 ( 1 )( πωω TT jHjH aa πω ≥Ω==Ω ||for0)()(If  To minimize the effects of aliasing, the T should be selected sufficiently small.  If the filter specifications are given in digital frequency domain, we cannot reduce aliasing by selecting T.  Aliasing occurs if the filter is not exactly band-limited
  • 44.
    Analog-to-Digital Filter Transformations Digitalizing of analog filters ∑= − = N k k k a ss A sH 1 )( Using partial fraction expansion, expand into)(sHa The corresponding impulse response is ∑= − == N k ts kaa tueAsHLth k 1 1 )()]([)( ∑∑ == === N k nTs k N k nTs ka nueAnueAnThnh kk 11 )()()()()( To sample the )(tha
  • 45.
    The z-transform ofis)(nh ∑∑∑∑ = − ∞ = = − ∞ −∞= − − === N k Ts k n N k nTs k n n ze A zeAznhzH k k 1 1 0 1 1 1 )()()( Conclusions: ∑= − = N k k k a ss A sH 1 )(Compared with  The pole in s-plane is mapped to the pole in z-planeks Tsk e  The partial fraction expansion coefficient of is the same as that of )(sHa )(zH  The zeros in the two domains do not satisfy the same relationship
  • 46.
    Analog-to-Digital Filter Transformations Advantages and disadvantages  The digital filter impulse response is similar to that of a analog filter. This means we can get a good approximations in time domain.  Due to the presence of aliasing, this method is useful only when the analog filter is essentially band-limited to a lowpass or bandpass filter in which there are no oscillations in the stopband.  It is a stable design and that the frequencies and are linearly related. So a linear phase analog filter can be mapped to a linear phase digital filter. Ω ω
  • 47.
     Design procedure Choose T and determine the analog frequencies  Transform analog poles into digital poles to obtain the digital filter TT s s p p ωω =Ω=Ω , Given the digital lowpass filter specifications 21 ,,, δδωω sp  Design an analog filter using the specifications)(sHa 21 ,,, δδsp ΩΩ ∑= − = N k k k a ss A sH 1 )(  Using partial fraction expansion, expand into)(sHa ∑= − − = N k Ts k ze A zH k 1 1 1 )(
  • 48.
    Analog-to-Digital Filter Transformations Examples  Transform 3 1 1 1 34 2 )( 2 + − + = ++ = ssss sHa into a digital filter using the impulse invariance method in which T=1 )(zH 21 1 4231 31 311 0183.04177.01 3181.0 )(1 )( 11 )( −− − −−−−− −−− −−−− +− = ++− − = − − − = zz z ezeez eeTz ez T ez T zH TTT TT TT
  • 49.
    Analog-to-Digital Filter Transformations Bilinear transformation  Definition This is a conformal mapping that transforms the -axis into the unit circle in the z-plane only once, thus avoiding aliasing of frequency components. This mapping is the best transformation method. Ωj       Ω ⋅=Ω 2 tan 1T c Tj Tj T j T j T j T j e e c ee ee cj 1 1 11 11 1 1 22 22 Ω− Ω− Ω − Ω Ω − Ω + − ⋅= + − ⋅=Ω 1 1 1 1 1 1 1 1 − − − − + − ⋅= + − ⋅= z z c e e cs Ts Ts sc sc z − + =
  • 50.
    Ts Ts e e cs 1 1 1 1 − − + − ⋅= Ts ez1 = 1 1 1 1 − − + − ⋅= z z cs T π T π− 1Ωj 1σ0 s1-plane ]Re[z ]Im[ zj 0 z-plane 0 Ωj σ s-plane
  • 51.
    Analog-to-Digital Filter Transformations Parameter c T c T c T c 2 then, 22 tan 1 11 =Ω≈Ω Ω ⋅≈      Ω ⋅=Ω  Keeping a good corresponding relationship between the analog filter and the digital filter in low frequencies. i.e. in low frequencies1Ω≈Ω       Ω=      ⋅=      Ω ⋅=Ω 2 cotthen 2 tanc 2 tan 1 c c cc c c T c ωω  Keeping a good corresponding relationship between the analog filter and the digital filter in a specific frequency (for example, in the cutoff frequency, )Tcc 1Ω≈ω
  • 52.
     Properties  Using, we obtainΩ+= js σ 22 22 )( )( ||, )( )( Ω+− Ω++ = Ω−− Ω++ = − + = σ σ σ σ c c z jc jc sc sc z So 1||0,1||0,1||0 >→>=→=<→< zzz σσσ  Using , we obtainωj ez = Ω=⋅= + − ⋅= + − ⋅= − − − − jjc e e c z z cs j j ) 2 tan( 1 1 1 1 1 1 ω ω ω The imaginary axis maps onto the unit circle in a one-to-one fashion. Hence there is no aliasing in the frequency domain.  The entire left half-plane maps into the inside of the unit circle. Hence this is a stable transformation.
  • 53.
    Analog-to-Digital Filter Transformations Advantages and disadvantages  It is a stable design;  There is no aliasing;  There is no restriction on the type of filter that can be transformed;. Ω ω The frequencies and are not linearly related. So a linear phase analog filter cannot be mapped to a linear phase digital filter.
  • 54.
     Design procedure Choose a value for T. We may set T=1 ) 2 tan( 2 ), 2 tan( 2 s s p p TT ωω =Ω=Ω Given the digital lowpass filter specifications 21 ,,, δδωω sp  Prewarp the cutoff frequencies and ; that ispω sω  Design an analog filter to meet the specifications 21 ,,, δδsp ΩΩ )(sHa  Finally, set ) 1 12 ()( 1 1 − − + − = z z T HzH a and simplify to obtain as a rational function in)(zH 1− z
  • 55.
    Analog-to-Digital Filter Transformations Examples  Transform into a digital filter using the bilinear transformation. Choose T=1 34 2 )( 2 ++ = ss sHa 21 21 1 1 2 1 1 1 1 1 1 1 0.070.131 13.00.2713.0 3 1 1 24 1 1 2 2 ) 1 1 2() 1 12 ()( −− −− − − − − − − = − − −− ++ = +      + − +      + − = + − = + − = zz zz z z z z z z H z z T HzH a T a
  • 56.
     Design thedigital Chebyshev-I filter using bilinear transformation. The specifications are: dB15,3.0 dB1,2.0 2 1 == == δπω δπω s p Solution  Let T=1 1.0191)15.0tan(2) 2 tan( 2 0.6498)1.0tan(2) 2 tan( 2 ===Ω ===Ω π ω π ω s s p p T T  Prewarp the cutoff frequencies
  • 57.
     Design ananalog Chebyshev-I filter to meet the specifications 21 ,,, δδsp ΩΩ )(sHa 0.04920.20380.61406192.0 0438.0 )( 234 ++++ = ssss sHa )0.64931.55481)(0.84821.49961( )1(0018.0 0.55072.29253.82903.05431 0.00180.00730.01100.00730018.0 )( 2121 41 4321 4321 −−−− − −−−− −−−− +−+− + = +−+− ++++ = zzzz z zzzz zzzz zH
  • 58.
    Analog-to-Digital Filter Transformations Comparison of three filters Using different prototype analog filters will give out different N and the minimum stopband attenuations. dB15,3.0 dB1,2.0 2 1 == == δπω δπω s p Given the digital filter specifications: prototype Order N Stopband Att. Butterworth 6 15 dB Chebyshev-I 4 25 dB Elliptic 3 27 dB return
  • 59.
    Frequency Transformations  Introduction Thetreatment in the preceding section is focused primarily on the design of digital lowpass IIR filters. If we wish to design a highpass or a bandpass or a bandstop filter, it is a simple matter to take a lowpass prototype filter and perform a frequency transformation.  Frequency transformations in the analog domain  Frequency transformations in the digital domain There are two approaches to perform the frequency transformation
  • 60.
    Frequency Transformations Approach 1 Analog lowpassfilter Frequency transformation s → s Filter transformation s → z Designed IIR filter Approach 2 Analog lowpass filter Filter transformation s → z Frequency transformation z → z Designed IIR filter
  • 61.
    Frequency Transformations  Specificationsof frequency-selective filters  Lowpass filter 21 ,,, δδωω sp  highpass filter 21 ,,, δδωω ps  bandpass filter 212211 ,,,,, δδωωωω spps  bandstop filter 212211 ,,,,, δδωωωω pssp
  • 62.
    Frequency Transformations  Frequencytransformations in the digital domain )(zHL the given prototype lowpass digital filter )(ZHd the desired frequency-selective digital filter )( 11 −− = ZGzDefine a mapping of the form Such that )( 11)()( −− = = ZGzLd zHZH To do this, we simply replace everywhere in by the function 1− z )(zHL )( 1− ZG
  • 63.
    Frequency Transformations Given thatis a stable and causal filter, we also want to be stable and causal. This imposes the following requirements: )(zHL )(ZHd  The unit circle of the z-plane must map onto the unit circle of the Z-plane  The inside of the unit circle of the z-plane must also map onto the inside of the unit circle of the Z-plane. 1− Z)( 1− ZG must be a rational function in so that )(ZHd is implementable.
  • 64.
    Frequency Transformations Let andbe the frequency variables of and , respectively. That is . Then θ ω z Z ωθ jj eZez == , )](arg[ )()( ω ωωθ j eGjjjj eeGeGe − −−− == )](arg[,1)( ωω jj eGθeG −− −== Hence the is an all-pass function)( 1− ZG 1||, 1 )( 1 1 1 11 < − − == ∏= − ∗− −− k N k k k a Za aZ ZGz By choosing an appropriate order N and the coefficients , we can obtain a variety of mappingska
  • 65.
    Frequency Transformations  Frequencytransformation formulae  Lowpass - Lowpass 1 1 1 1 − − − − − = Z Z z α α ]2/)sin[( ]2/)sin[( cc cc ωθ ωθ α + − = : Cutoff frequency of new digital filtercω cθ The cutoff frequency of prototype lowpass digital filter
  • 66.
    Frequency Transformations  Lowpass- Highpass 1 1 1 1 − − − + + −= Z Z z α α ]2/)cos[( ]2/)cos[( cc cc ωθ ωθ α − + −= : Cutoff frequency of new digital filtercω
  • 67.
    Frequency Transformations  Lowpass- Bandpass 11 1 2 2 2 1 1 2 1 +− +− −= −− −− − ZZ ZZ z αα αα 1 1 , 1 2 2 tan) 2 cot(,cos ]2/)cos[( ]2/)cos[( 21 12 0 12 12 + − = + = − == − + = k k k k k c α β α θωω ω ωω ωω β : lower cutoff frequency of bandpass digital filter1ω : upper cutoff frequency of bandpass digital filter2ω : center frequency of the passband0ω
  • 68.
    Frequency Transformations  Lowpass- Bandstop : lower cutoff frequency of bandstop digital filter1ω : upper cutoff frequency of bandstop digital filter2ω : center frequency of the stopband0ω k k k k c + − = + = − == − + = 1 1 , 1 2 2 tan) 2 tan(,cos ]2/)cos[( ]2/)cos[( 21 12 0 12 12 α β α θωω ω ωω ωω β 11 1 2 2 2 1 1 2 1 +− +− = −− −− − ZZ ZZ z αα αα
  • 69.
    Frequency Transformations  Designprocedure  Determine the specifications of the digital prototype lowpass filter;  Determine the specifications of the analog prototype lowpass filter;  Design the analog prototype lowpass filter;  Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation;  Perform the frequency transformation in digital domain to obtain the desired frequency-selective filters.
  • 70.
    Frequency Transformations  Examples Given the specifications of Chebyshev-I lowpass filter dB15,3.0 dB1,2.0 2 1 == == δπθ δπθ s p Design a highpass filter with the above tolerances but with passband beginning at πω 6.0=p and its system function )6493.05548.11)(8482.04996.11( )1(01836.0 )( 2121 41 −−−− − +−+− + = zzzz z zHL
  • 71.
  • 72.
    Frequency Transformations  Usingthe Chebyshev-I prototype to design a highpass digital filter to satisfy dB15,46.0 dB1,6.0 == == ss pp A R πω πω  Determine the specifications of the digital prototype lowpass filter Solution πθ 2.0=p Choose the passband frequency with a reasonable value: Determine the stopband frequency by 1 1 1 1 − − − + + −= Z Z z α α ) 1 arg( 1 ω ω ω ω θ α α θ α α j j j j j e e e e e − − − − − + + −−=→ + + −=
  • 73.
    3820.0 ]2/)6.02.0cos[( ]2/)6.02.0cos[( ]2/)cos[( ]2/)cos[( −= − + −= − + −= ππ ππ ωθ ωθ α pp pp π α α θ ω ω 3.0) 3820.01 3820.0 arg() 1 arg( 46.0 46.0 = − − −= + + −=− − − − j j j j s e e e e s s  Determine the specifications of the analog prototype lowpass filter Set T = 1 and prewarp the cutoff frequencies 1.0191)15.0tan(2) 2 tan( 2 0.6498)1.0tan(2) 2 tan( 2 ===Ω ===Ω π θ π θ s s p p T T
  • 74.
     Design ananalog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,,, ΩΩ 0.04920.20380.61406192.0 0438.0 )( 234 ++++ = ssss sHa )0.64931.55481)(0.84821.49961( )1(0018.0 0.55072.29253.82903.05431 0.00180.00730.01100.00730018.0 )( 2121 41 4321 4321 −−−− − −−−− −−−− +−+− + = +−+− ++++ = zzzz z zzzz zzzz zHL  Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation
  • 75.
     Perform thefrequency transformation in digital domain to obtain the desired digital highpass filter )4019.00416.11)(7647.05561.01( )1(0243.0 )()( 2121 41 1 1 1 1 −−−− − + + −= ++++ − = = − − − ZZZZ Z zHZH Z Z zLh α α
  • 76.
    Frequency Transformations  Usingthe Chebyshev-I prototype to design a bandpass digital filter to satisfy dB15,7.0,2.0 dB1,5.0,4.0 21 21 === === sss ppp A R πωπω πωπω  Determine the specifications of the digital prototype lowpass filter Solution πθ 2.0=p Choose the passband frequency with a reasonable value:
  • 77.
    Determine the stopbandfrequency by π αα αα θ ωω ωω 69.0) 1 arg( 22 22 1 2 2 21 2 = +− +− −−= −− −− ss ss jj jj s ee ee 1584.0 ]2/)4.05.0cos[( ]2/)4.05.0cos[( ]2/)cos[( ]2/)cos[( 12 12 = − + = − + = ππ ππ ωω ωω β pp pp 0515.2 2 2.0 tan) 2 4.05.0 cot( = − = πππ k 11 1 2 2 2 1 1 2 1 +− +− −= −− −− − ZZ ZZ z αα αα 3446.0 1 1 ,2130.0 1 2 21 = + − == + = k k k k α β α
  • 78.
     Determine thespecifications of the analog prototype lowpass filter Set T = 1 and prewarp the cutoff frequencies 3.7842)0.3450tan(2) 2 tan( 2 0.6498)1.0tan(2) 2 tan( 2 ===Ω ===Ω π θ π θ s s p p T T  Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,,, ΩΩ 4656.07134.0 4149.0 )( 2 ++ = ss sHa
  • 79.
    21 21 5157.01997.11 )1(0704.0 )( −− − +− +× = zz z zHL  Transformthe analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation  Perform the frequency transformation in digital domain to obtain the desired digital bandpass filter 4321 42 1 0.71060.48147020.15731.01 0205.00410.00205.0 )()( 1 1 2 2 2 1 1 2 1 −−−− −− +− +− −= +−+− +− = = −− −− − ZZZZ ZZ zHZH ZZ ZZ zLbp αα αα
  • 80.
    Frequency Transformations  Usingthe Chebyshev-I prototype to design a bandstop digital filter to satisfy dB20,65.0,35.0 dB1,75.0,25.0 21 21 === === sss ppp A R πωπω πωπω  Determine the specifications of the digital prototype lowpass filter Solution πθ 2.0=p Choose the passband frequency with a reasonable value:
  • 81.
    Determine the stopbandfrequency by π αα αα θ ωω ωω 0.1919) 1 arg( 11 11 1 2 2 21 2 = +− +− −= −− −− ss ss jj jj s ee ee 0 ]2/)25.075.0cos[( ]2/)25.075.0cos[( ]2/)cos[( ]2/)cos[( 12 12 = − + = − + = ππ ππ ωω ωω β pp pp 0.1584 2 2.0 tan) 2 25.075.0 tan( = − = πππ k 11 1 2 2 2 1 1 2 1 +− +− = −− −− − ZZ ZZ z αα αα 0.7265 1 1 ,0 1 2 21 = + − == + = k k k α β α
  • 82.
     Determine thespecifications of the analog prototype lowpass filter Set T = 1 and prewarp the cutoff frequencies 6217.0)0.0959tan(2) 2 tan( 2 0.3168)1.0tan(2) 2 tan( 2 ===Ω ===Ω π θ π θ s s p p T T  Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,,, ΩΩ 0.01561243.03131.0 0156.0 )( 23 +++ = sss sHa
  • 83.
    321 321 0.73352.36922.62251 0.00160.00490.00490.0016 )( −−− −−− ++− +++ = zzz zzz zHL  Transformthe analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation  Perform the frequency transformation in digital domain to obtain the desired digital bandstop filter )0.3391)(0.7761.2481)(0.7761.2481( )1(0.132 )()( 22121 32 11 1 2 2 2 1 1 2 1 −−−−− − +− +− = −+−++ +× = = −− −− − ZZZZZ Z zHZH ZZ ZZ zLbs αα αα return
  • 84.
    0 30 4050 60 0.707 1 Magnitude Response Analog frequency in rad/s N=2 N=4 N=8 N=16 return )( ΩjHa
  • 85.
    0 2 46 0.707 1 Magnitude Response Analog frequency in rad/s Amplitude 0 2 4 6 -3 -1 0 1 3 Phase Response Analog frequency in rad/s Phaseinrad )( ΩjHa return
  • 86.
    0 2 35 x 10 4 0 0.1778 0.8913 1 Magnitude Response Analog frequency in pi units H 0 2 3 5 x 10 4 -30 -15 -10 Magnitude in dB Analog frequency in pi units decibels 0 2 3 5 x 10 4 -1 -0.5 0 0.5 1 Phase Response Analog frequency in pi units P 0 0.5 1 1.5 x 10 -4 0 10000 20000 Impulse Response time in seconds ha(t) return
  • 87.
    0 2 35 x 10 4 0 0.1778 0.8913 1 Magnitude Response Analog frequency in pi units H 0 2 3 5 x 10 4 -30 -15 -10 Magnitude in dB Analog frequency in pi units decibels 0 2 3 5 x 10 4 -1 -0.5 0 0.5 1 Phase Response Analog frequency in pi units P 0 0.5 1 1.5 x 10 -4 -5000 0 5000 10000 15000 20000 Impulse Response time in seconds ha(t) return
  • 88.
    -1 -0.8 -0.6-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(0 xC
  • 89.
    -1 -0.8 -0.6-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(2 xC
  • 90.
    -1 -0.8 -0.6-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(3 xC
  • 91.
    -1 -0.8 -0.6-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(4 xC
  • 92.
    -1 -0.8 -0.6-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(5 xC
  • 93.
    -1 -0.8 -0.6-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 )(6 xC
  • 94.
    -1 -0.8 -0.6-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -2 -1 0 1 2 3 return
  • 95.
    0 2 35 0.8913 1 Magnitude Response Analog frequency in rad/s Amplitude 4=N
  • 96.
    0 2 35 0.8913 1 Magnitude Response Analog frequency in rad/s Amplitude return 5=N
  • 97.
    0 1 23 5 0.8913 1 Magnitude Response Analog frequency in rad/s Amplitude 0 1 2 3 5 -3 -2 -1 0 1 Phase Response Analog frequency in rad/s Phaseinrad return )( ΩjHa
  • 98.
    0 2 35 x 10 4 0 0.1778 0.8913 1 Magnitude Response Analog frequency in pi units H 0 2 3 5 x 10 4 -30 -15 -10 Magnitude in dB Analog frequency in pi units decibels 0 2 3 5 x 10 4 -1 -0.5 0 0.5 1 Phase Response Analog frequency in pi units Phaseinpiunits 0 1 2 3 4 x 10 -4 -5000 0 5000 10000 15000 20000 Impulse Response time in seconds ha(t) return
  • 99.
    0 pi/T 2*pi/T 0.2 0.4 0.6 0pi 2*pi 0.2 0.4 0.6 )( ΩjHa )( ωj eH 1=T Ω ω
  • 100.
    0 pi/T 2*pi/T 0.2 0.4 0.6 0pi 2*pi 0.2 0.4 0.6 )( ΩjHa )( ωj eH 1.0=T Ω ω return
  • 101.
    0 0.2 0.31 0.1778 0.8913 1 Magnitude Response Frequency in pi 0 0.2 0.3 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi 0 0.2 0.3 1 -15 -10 Magnitude Response in dB Frequency in pi 0 0.2 0.3 1 2 4 6 8 10 Group Delay Frequency in pi return
  • 102.
    0 pi/T 2*pi/T 0.2 0.4 0.6 MagnitudeResponse frequency in rad/s 0 pi 2*pi 0.2 0.4 0.6 frequency in rad/sample )( ΩjHa )( ωj eH 1=T Ω ω
  • 103.
    0 pi/T 2*pi/T 0.2 0.4 0.6 MagnitudeResponse frequency in rad/s 0 pi 2*pi 0.2 0.4 0.6 frequency in rad/sample )( ΩjHa )( ωj eH 1.0=T Ω ω return
  • 104.
    0 0.2 0.31 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.2 0.3 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.2 0.3 1 -15 -10 Magnitude Response in dB Frequency in pi units 0 0.2 0.3 1 3 6 9 12 15 Group Delay Frequency in pi units samples return
  • 105.
    0 0.2 0.31 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.6 1 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.2 0.3 1 -15 -10 Magnitude Response in dB Frequency in pi units 0 0.6 1 -15 -10 Magnitude Response in dB Frequency in pi units return
  • 106.
    0 0.46 0.61 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.46 0.6 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.46 0.6 1 -15 -10 Magnitude Response in dB Frequency in pi units 0 0.46 0.6 1 2 4 6 8 10 Group Delay Frequency in pi units samples return
  • 107.
    0 0.2 0.40.5 0.7 1 0.1778 0.8913 1 Magnitude Response Frequency in pi units 0 0.2 0.4 0.5 0.7 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.2 0.4 0.5 0.7 1 -100 -80 -60 -40 -20 0 Magnitude Response in dB Frequency in pi units 0 0.2 0.4 0.5 0.7 1 3 6 9 12 15 Group Delay Frequency in pi units samples return
  • 108.
    0 0.250.35 0.650.751 0.1 0.8913 1 Magnitude Response Frequency in pi units 0 0.250.35 0.650.75 1 -1 -0.5 0 0.5 1 Phase Response Frequency in pi piunits 0 0.250.35 0.650.75 1 -30 -20 -10 0 Magnitude Response in dB Frequency in pi units 0 0.250.35 0.650.75 1 2 4 6 8 10 Group Delay Frequency in pi units samples return

Editor's Notes

  • #2 We now turn our attention to the inverse problem of designing systems from the given specifications. It is an important as well as a difficult problem. In digital signal processing there are two important types of systems. The first type of systems perform signal filtering in the time domain and hence are called digital filters. The second type of systems provide signal representation in the frequency domain and are called spectrum analyzers.
  • #7 W=0时的幅度被归一化为1。
  • #23 .
  • #48 Given the digital lowpass filter specifications Wp, Ws, Rp, As, we want to determine H(z) by first designing an equivalent analog filter and then mapping it into the desired digital filter.
  • #59 Clearly, the elliptic prototype gives the best design. However, if we compare their phase responses, then the elliptic design has the most nonlinear phase response in the passband. Bilinear transformation is used in this comparison.
  • #60 The treatment in the preceding section is focused primarily on the design of digital lowpass IIR filters. If we wish to design a highpass or a bandpass or a bandstop filter, it is a simple matter to take a lowpass prototype filter and perform a frequency transformation. There are two methods to perform the frequency transformation Frequency transformations in the analog domain To perform the frequency transformation in the analog domain and then to convert the analog filter into a corresponding digital filter by mapping of the s-plane into the z-plane. Frequency transformations in the digital domain To convert the analog lowpass filter into a digital lowpass filter and then to transform the digital lowpass filter into the desired digital filter by a digital transformation.
  • #85 (digital signal processing\chap6\butterworth_filter_design\butterworth_demo.m)
  • #86 (digital signal processing\chap6\butterworth_filter_design\butterworth_filter1.m)
  • #88 该图是用通带特性计算的OmegaC，因此通带精确满足指标，而阻带有富裕。 (digital signal processing\chap6\butterworth_filter_design\butterworth_filter3.m)
  • #89 (digital signal processing\chap6\chebyshev_filter_design\chebyshev_polynomial\CNx.m), N=0
  • #95 (digital signal processing\chap6\chebyshev_filter_design\chebyshev_polynomial\CNx.m) 特点： 1，不论N为多少，曲线都通过（1，1）点 2，N为奇数时，曲线奇对称，曲线都通过（0，0）点 3，N为偶数时，曲线偶对称，
  • #96 (digital signal processing\chap6\chebyshev_filter_design\chebyshev1_demo.m) N=4
  • #97 (digital signal processing\chap6\chebyshev_filter_design\chebyshev1_demo.m) N=5
  • #98 (digital signal processing\chap6\chebyshev_filter_design\chebyshev_filter1.m)
  • #99 (digital signal processing\chap6\chebyshev_filter_design\chebyshev_filter3.m)
  • #100 (digital signal processing\chap6\impulse invariance\example1.m) T=1
  • #101 (digital signal processing\chap6\impulse invariance\example1.m) T=0.1
  • #102 (digital signal processing\chap6\impulse invariance\example3.m)
  • #103 (digital signal processing\chap6\bilinear\example1.m) T=1
  • #104 (digital signal processing\chap6\bilinear\example1.m) T=0.1
  • #105 (digital signal processing\chap6\bilinear\example3.m)
  • #106 (digital signal processing\chap6\frequency_transformation\example1.m)
  • #107 (digital signal processing\chap6\frequency_transformation\highpass_example.m)
  • #108 (digital signal processing\chap6\frequency_transformation\bandpass_example.m)
  • #109 (digital signal processing\chap6\frequency_transformation\bandstop_example.m)