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Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters IIR Filter Design Procedure Analog Filters DESIGN OF ANALOG IIR FILTERS By Arun A. Balakrishnan Asst. Professor Dept. of AE&I, RSET Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Infinite Impulse Response(IIR) IIR filters are computationally more efficient It has feedback or poles and require fewer coefficients Feedback can result in unstable filters, if coefficients deviate from their true value due to coefficient scaling or quantization In general, IIR filter can be expressed as −1 −N b0+b1z +:::+bN z H(z) = −1 −M 1+a1z +:::+aM z N P −k bk z k=0 = M P −k 1+ ak z k=1 where ak and bk are filter coefficient Transfer function can be factorized as H(z) = k (z−z1)(z−z2):::(z−zN ) (z−p1)(z−p2):::(z−pM ) N M P P y(n) = bk x(n − k) − ak y(n − k) k=0 k=1 Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Design Procedure Filter specification Coefficients calculation Appropriate structure selection Simulation(optional) Filter implementation Analog low pass filters Butterworth Filter Chebyshev Filter Type I Chebyshev Filter Type II Chebyshev Filter Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Filter specifications Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Filter specification !p;!s;!c = digital passband, stopband, 3-dB cut off frequency in radians Ωp; Ωs; Ωc = analog passband, stopband, 3-dB cut off frequency in radians = parameter specifying allowable passband λ = parameter specifying allowable stopband δp = tolerance of magnitude response in passband δs = tolerance of magnitude response in stopband p δp = 2 1 − δp q 2 2 (1 + δp) − δs λ = δs Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Butterworth Filters Analog Low Pass Butterworth Filter Magnitude function jH(Ω)j = 1 = 1 N = 1; 2; 3; ::: 2N 1=2 2 2N 1=2 [1+(Ω=Ωc ) ] [1+ (Ω=Ωp) ] where N = order of filter and Ωc = cutoff frequency Function is monotonically decreasing magnitude response approaches ideal LPF as N increases Ω < Ωc; jH(jΩ)j ≈ 1 and Ω > Ωc; jH(jΩ)j decreases rapidly At Ω = Ωc; jH(jΩ)j = 0:707 = −3dB Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Butterworth Filters Transfer function of a stable filter magnitude square function of a normalized Butterworth filter jH(Ω)j2 = 1 N = 1; 2; 3; ::: 1+(Ω)2N s To obtain transfer function, put Ω = j since H(s)H(−s) evaluated at s = jΩ is equal to jH(Ω2)j H(s)H(−s) = 1 = 1 = 1 s 2N 1+(−1)N s2N 1+(−s2)N 1+( j ) function has poles in left half and right half of s-plane due to H(s) and H(−s) If H(s) has root in left half, H(−s) has root in right half of s-plane N roots are given by 1 + −s2 = 0 2N j2kπ jπk=N for N odd, s = 1 = e =) sk = e ; k = 1; 2;:::; 2N for N even, 2N j(2k−1)π jπ(2k−1)=2N s = −1 = e =) sk = e ; k = 1; 2;:::; 2N Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Butterworth Filters Transfer function of a stable filter all poles lie on a unit circle 360◦ poles are located at an angular separation 2N To ensure stability, only left half poles are considered jφk π (2k−1)π sk = e where φk = 2 + 2N ; k = 1; 2;:::; N Table: List of Butterworth Polynomials N Denominator of H(s) 1 sp+ 1 2 s2 + 2s + 1 3 (s + 1)(s2 + s + 1) 4 (s2 + 0:76537s + 1)(s2 + 1:8477s + 1) 0 unnormalized poles are sk = sk Ωc Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Butterworth Filters Order N of a Butterworth Filter 2 1 Magnitude squared function jH(Ω)j = 2 2N N = 1; 2; 3; ::: 1+ (Ω=Ωp) Taking logarithm on both sides 2 h 2 2N i 10 log jH(Ω)j = 10 log 1 − 10 log 1 + (Ω=Ωp) At Ω = Ωp attenuation is αp 2 2 20 log jH(Ωp)j = −αp = −10 log(1 + ) =) αp = 10 log(1 + ) = (100:1αp − 1)1=2 At Ω = Ωs attenuation is αs 2 2N 20 log jH(Ωs)j = −αs = −10 log(1 + (Ωs=Ωp) ) r 0:1α log 10 s −1 2N 0:1αp 2 Ωs 0:1α 10 −1 =) = 10 s − 1=) N ≥ Ωp log Ωs Ωp log A λ q 100:1αs −1 Ωp N ≥ where A = = 0:1α and k = 1 p − Ωs log( k ) 10 1 Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Butterworth Filters Ω Prove that Ω = p = Ωs c (100:1αp −1)1=2N (100:1αs −1)1=2N 2 1 1 jH(Ω)j = 2N = 2 2N 1+(Ω=Ωc ) 1+ (Ω=Ωp) 2N 2N 2N Ω = 2 Ω =) Ωp = 2 Ωc Ωp Ωc 2N Ωp 0:1αp Ωp = 10 − 1 =) Ωc = 0:1α Ωc (10 p −1)1=2N 2N 1=2N 0:1αs 0:1αs Ωs 10 −1 10 −1 We know = 0:1α =) Ωs = Ωp 0:1α Ωp 10 p −1 10 p −1 1=2N 1=2N 0:1αs Ω = Ω 100:1αp − 1 10 −1 s c 100:1αp −1 Ω = Ωs c (100:1αs −1)1=2N Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Butterworth Filters Steps to design an analog Butterworth lowpass filter From the given specification find the order of filter N Round of it to the next higher integer Find the transfer function H(s) for Ωc = 1 rad/sec for value of N Calculate the value of cut off frequency Ωc Find the required transfer function for above value of Ωc by substituting s ! s in H(s) Ωc Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Butterworth Filters Problem Design an analog Butterworth filter that has - 2dB passband attenuation at a frequency of 20 rad/sec and atleast -10dB stopband attenuation at 30 rad/sec. Answer N ≥ 3:37 =) N = 4 1 H(s) = (s2+0:76537s+1)(s2+1:8477s+1) Ωc = 21:3868 Transfer function for Ωc = 21:3868 rad/sec is obtained by s s ! 21:3868 0:20921×106 H(s) = (s2+16:3686s+457:394)(s2+39:5176s+457:394) Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Chebyshev Filters Analog Low Pass Chebyshev Filter Type I - all pole filter that exhibits equiripple behavior in pass band and monotonic characteristics in stopband Type II - contains both poles and zeros and exhibits monotonic behavior in pass band and equiripple behavior in stopband Magnitude squared function Design of Analog IIR Filters, Sep 4, 5 2 1 jH(Ω)j = N = 1; 2; 3; ::: 1+2C2 Ω N Ωp th where CN (x) = N order Chebyshev polynomial IIR Filter Design Procedure Analog Filters Chebyshev Filters Chebyshev Polynomial Nth order Chebyshev polynomial is cos(N cos−1(x)); jxj < 1 (Passband) C (x) = N cosh(N cosh−1(x)); jxj > 1 (Stopband) CN (x) = 2xCN−1(x) − CN−2(x); N > 1 C0(x) = 1, C1(x) = x and CN (1) = 1 Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Chebyshev Filters Order N of a Chebyshev Filter Magnitude squared function 2 1 jH(Ω)j = 2 2 N = 1; 2; 3; ::: 1+ CN (Ω=Ωp) Taking logarithm on both sides 2 2 2 10 log jH(Ω)j = 10 log 1 − 10 log 1 + CN (Ω=Ωp) At Ω = Ωp attenuation is αp 2 20 log jH(Ωp)j = −αp = −10 log(1 + )[* CN (1) = 1] 2 0:1αp 1=2 =) αp = 10 log(1 + ) =) = (10 − 1) At Ω = Ωs attenuation is αs 2 2 20 log jH(Ωs)j = −αs = −10 log(1 + CN (Ωs=Ωp)) r 0:1α cosh−1 10 s −1 0:1αp 2 2 Ωs 0:1α 10 −1 =) C = 10 s − 1=) N ≥ N Ωp cosh−1 Ωs Ωp −1 cosh A λ q 100:1αs −1 Ωp N ≥ where A = = 0:1α and k = −1 1 p − Ωs cosh ( k ) 10 1 Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Chebyshev Filters Pole Locations of a Chebyshev Filter roots are 1 + 2C2 −js = 0 N Ωp h i C −js = ±j/ = cos N cos−1 −js N Ωp Ωp Defining cos−1 −js = φ − jθ Ωp ±j/ = cos [Nφ − jNθ] = cos(Nφ) cos(jNθ) + sin(Nφ) sin(jNθ) = cos(Nφ) cosh(Nθ) + j sin(Nφ) sinh(Nθ) Equating real and imaginary parts cos(Nφ) cosh(Nθ) = 0 (1) sin(Nφ) sinh(Nθ) = ±1/ (2) Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Chebyshev Filters Pole Locations of a Chebyshev Filter (2k−1)π since cosh(Nθ) > 0 for θ real =) φ = 2N ; k = 1; 2;:::; N 1 −1 1 substituting in (2) sin(Nφ) = ±1 ) θ = ± N sinh Left half pole locations sk = jΩp cos(φ − jθ) sk = jΩp [cos(φ) cosh(θ) + j sin(φ) sinh(θ)] sk = Ωp [− sin(φ) sinh(θ) + j cos(φ) cosh(θ)] Identity p cosh−1(x) = ln(x + x 2 − 1) p p sinh−1(x) = ln(x+ x 2 + 1) =) sinh−1(−1) = ln(−1+ −2 + 1) −1 −1 p or µ = esinh ( ) = −1 + −2 + 1 Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Chebyshev Filters Pole Locations of a Chebyshev Filter 1 sinh(θ) = sinh sinh−1(−1) N 1 sinh−1(−1) − 1 sinh−1(−1) e[ N ] − e [ N ] eθ − e−θ = sinh θ = 2 * 2 1 1 h sinh−1(−1)i N h − sinh−1(−1)i N − 1 − 1 e e µ N − µ N = = 2 2 1 − 1 µ N +µ N similarly cosh(θ) = 2 " 1 − 1 ! 1 − 1 !# µ N − µ N µ N + µ N s = Ω − sin(φ) + j cos(φ) k p 2 2 = −a sin(φ) + jb cos(φ) Design of Analog IIR Filters, Sep 4, 5 IIR Filter Design Procedure Analog Filters Chebyshev Filters contd..
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