Analog Lowpass Filter Specifications

Analog Lowpass Filter Specifications Analog Lowpass Filter Specifications • Typical magnitude response H a ( jΩ) of an analog lowpass filter may be given as indicated below 1 • In the passband, defined by 0 ≤ Ω ≤ Ω p , we require 1 − δ p ≤ H a ( jΩ ) ≤ 1 + δ p , Ω ≤ Ω p i.e., H a ( jΩ) approximates unity within an error of ± δ p • In the stopband, defined by Ω s ≤ Ω < ∞ , we require H a ( jΩ ) ≤ δ s , Ω s ≤ Ω < ∞ 2 i.e., H a ( jΩ) approximates zero within an error of δ s Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Lowpass Filter Specifications • • • • • Analog Lowpass Filter Specifications Ω p - passband edge frequency Ω s - stopband edge frequency δ p - peak ripple value in the passband δ s - peak ripple value in the stopband Peak passband ripple α p = − 20 log10 (1 − δ p ) dB • Magnitude specifications may alternately be given in a normalized form as indicated below • Minimum stopband attenuation α s = − 20 log10 (δ s ) dB 3 4 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Lowpass Filter Specifications Analog Lowpass Filter Design • Two additional parameters are defined - • Here, the maximum value of the magnitude in the passband assumed to be unity • (1) Transition ratio k = 1 / 1 + ε 2 - Maximum passband deviation, given by the minimum value of the magnitude in the passband For a lowpass filter k < 1 (2) Discrimination parameter k1 = Usually k1 << 1 • 1 - Maximum stopband magnitude A 5 Ωp Ωs ε A2 − 1 6 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 1 Butterworth Approximation 7 • The magnitude-square response of an N-th order analog lowpass Butterworth filter is given by 1 2 H a ( jΩ ) = 1 + (Ω / Ω c ) 2 N 2 • First 2 N − 1 derivatives of H a ( jΩ) at Ω = 0 are equal to zero • The Butterworth lowpass filter thus is said to have a maximally-flat magnitude at Ω = 0 Butterworth Approximation • Gain in dB is G (Ω) = 10 log10 H a ( jΩ) 2 • As G (0) = 0 and G (Ωc ) = 10 log10 (0.5) = −3.0103 ≅ −3 dB Ω c is called the 3-dB cutoff frequency 8 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Butterworth Approximation Butterworth Approximation • Typical magnitude responses with Ωc = 1 • Two parameters completely characterizing a Butterworth lowpass filter are Ωc and N • These are determined from the specified bandedges Ω p and Ω s , and minimum passband magnitude 1 / 1 + ε 2, and maximum stopband ripple 1 / A Butterworth Filter N=2 N=4 N = 10 Magnitude 1 0.8 0.6 0.4 0.2 0 0 1 2 3 Ω 9 10 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Butterworth Approximation Butterworth Approximation • Ωc and N are thus determined from 2 1 = 1 H a ( jΩ p ) = 1 + (Ω p / Ω c ) 2 N 1 + ε 2 2 1 = 1 H a ( jΩ s ) = 1 + ( Ω s / Ω c ) 2 N A2 • Solving the above we get log [( A2 − 1) / ε 2 ] log10 (1 / k1 ) N = 1 ⋅ 10 = 2 log10 (Ω s / Ω p ) log10 (1 / k ) 11 12 Copyright © 2005, S. K. Mitra • Since order N must be an integer, value obtained is rounded up to the next highest integer • This value of N is used next to determine Ωc by satisfying either the stopband edge or the passband edge specification exactly • If the stopband edge specification is satisfied, then the passband edge specification is exceeded providing a safety margin Copyright © 2005, S. K. Mitra 2 Butterworth Approximation • Transfer function of an analog Butterworth lowpass filter is given by ΩcN ΩcN H a ( s) = C = N = DN ( s ) s + ∑lN=−01d s l ∏lN=1 ( s − pl ) l Butterworth Approximation • Example - Determine the lowest order of a Butterworth lowpass filter with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHz • Now 10 log10 ⎛⎜ 1 2 ⎞⎟ = −1 ⎝1 + ε ⎠ which yields ε 2 = 0.25895 and 10 log10 ⎛⎜ 12 ⎞⎟ = − 40 ⎝A ⎠ which yields A2 = 10,000 where pl = Ωc e j[π ( N + 2l−1) / 2 N ] , 1 ≤ l ≤ N • Denominator DN (s ) is known as the Butterworth polynomial of order N 13 14 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Chebyshev Approximation Butterworth Approximation A2 − 1 = 196.51334 1= k1 • Therefore ε 1 = Ωs = 5 k Ωp and • The magnitude-square response of an N-th order analog lowpass Type 1 Chebyshev filter is given by 1 2 H a ( s) = 2 2 1 + ε TN (Ω / Ω p ) • Hence where TN (Ω) is the Chebyshev polynomial of order N: log (1 / k1 ) N = 10 = 3.2811 log10 (1 / k ) ⎧ cos( N cos −1 Ω), TN (Ω) = ⎨ −1 ⎩cosh( N cosh Ω), • We choose N = 4 15 Ω ≤1 Ω >1 16 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Chebyshev Approximation Chebyshev Approximation • If at Ω = Ω s the magnitude is equal to 1/A, then 2 1 = 12 H a ( jΩ s ) = 2 2 1 + ε TN (Ω s / Ω p ) A • Typical magnitude response plots of the analog lowpass Type 1 Chebyshev filter are shown below Type 1 Chebyshev Filter Magnitude • Solving the above we get cosh −1 ( A2 − 1 / ε ) cosh −1 (1 / k1 ) = N= cosh −1 (Ω s / Ω p ) cosh −1 (1 / k ) • Order N is chosen as the nearest integer greater than or equal to the above value N=2 N=3 N=8 1 0.8 0.6 0.4 0.2 0 0 1 2 3 Ω 17 18 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 3 Chebyshev Approximation Chebyshev Approximation • The magnitude-square response of an N-th order analog lowpass Type 2 Chebyshev (also called inverse Chebyshev) filter is given by 1 2 H a ( jΩ ) = 2 ⎡T (Ω / Ω p ) ⎤ 1+ ε 2 ⎢ N s ⎥ ⎣ TN (Ω s / Ω) ⎦ • Typical magnitude response plots of the analog lowpass Type 2 Chebyshev filter are shown below Type 2 Chebyshev Filter Magnitude 0.6 0.4 0.2 where TN (Ω) is the Chebyshev polynomial of order N 19 N=3 N=5 N=7 1 0.8 0 0 1 2 20 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Elliptic Approximation Chebyshev Approximation • The square-magnitude response of an elliptic lowpass filter is given by 1 2 H a ( jΩ ) = 2 2 1 + ε RN ( Ω / Ω p ) where RN (Ω) is a rational function of order N satisfying RN (1 / Ω) = 1 / RN (Ω) , with the roots of its numerator lying in the interval 0 < Ω < 1 and the roots of its denominator lying in the interval 1 < Ω < ∞ • The order N of the Type 2 Chebyshev filter is determined from given ε , Ω s , and A using cosh −1 ( A2 − 1 / ε ) cosh −1 (1 / k1 ) = N= cosh −1 (Ω s / Ω p ) cosh −1 (1 / k ) • Example - Determine the lowest order of a Chebyshev lowpass filter with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHz - N= 21 3 Ω cosh −1 (1 / k1 ) = 2.6059 cosh −1 (1 / k ) 22 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Elliptic Approximation Elliptic Approximation • For given Ω p , Ω s , ε , and A, the filter order can be estimated using 2 log10 (4 / k1 ) N≅ log10 (1 / ρ ) • Example - Determine the lowest order of a elliptic lowpass filter with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHz Note: k = 0.2 and 1 / k1 = 196.5134 • Substituting these values we get ρ0 = 0.00255135, k '= 0.979796, where k ' = 1 − k 2 ρ0 = 1 − k ' 2(1 + k ') ρ = ρ0 + 2( ρ0 )5 + 15( ρ0 )9 + 150( ρ0 )13 23 ρ = 0.0025513525 • and hence N = 2.23308 • Choose N = 3 24 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 4 Analog Lowpass Filter Design Elliptic Approximation • Example - Design an elliptic lowpass filter of lowest order with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHz • Code fragments used • Typical magnitude response plots with Ω p = 1 are shown below Elliptic Filter N=3 N=4 Magnitude 1 [N, Wn] = ellipord(Wp, Ws, Rp, Rs, ‘s’); 0.8 0.6 0.2 0 [b, a] = ellip(N, Rp, Rs, Wn, ‘s’); Wp = 2*pi*1000; Ws = 2*pi*5000; Rp = 1; Rs = 40; with 0.4 0 1 2 3 Ω 25 26 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Design of Analog Highpass, Highpass, Bandpass and Bandstop Filters Analog Lowpass Filter Design • Gain plot Lowpass Elliptic Filter Gain, dB 0 -20 -40 -60 0 2000 4000 Frequency, Hz 6000 27 28 • Steps involved in the design process: Step 1 - Develop of specifications of a prototype analog lowpass filter H LP (s) from specifications of desired analog filter HD (s ) using a frequency transformation Step 2 - Design the prototype analog lowpass filter Step 3 - Determine the transfer function HD (s ) of desired analog filter by applying the inverse frequency transformation to H LP (s) Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Design of Analog Highpass, Highpass, Bandpass and Bandstop Filters Analog Highpass Filter Design • Spectral Transformation: • Let s denote the Laplace transform variable of prototype analog lowpass filter H LP (s) and ŝ denote the Laplace transform variable of desired analog filter HD (sˆ) • The mapping from s-domain to ŝ -domain is given by the invertible transformation s = F (sˆ) H ( s ) = H LP ( s ) s = F ( sˆ) ˆ • Then D H LP ( s ) = HD ( sˆ) sˆ = F −1 ( s ) 29 Ω p Ω̂ p sˆ where Ω p is the passband edge frequency of H LP ( s) and Ω̂ p is the passband edge frequency of H HP ( sˆ) • On the imaginary axis the transformation is ˆp Ω pΩ Ω=− ˆ Ω s= 30 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 5 Analog Highpass Filter Design Analog Highpass Filter Design Ω=− ˆp Ω pΩ ˆ Ω Ω Ω p Ωs Stopband − Ω s − Ω p Stopband Ω 0 Passband Stopband Passband ˆ p −Ω ˆs −Ω 0 Passband Ω̂ s Lowpass Ω̂ Highpass Ω̂ p 31 • Example - Design an analog Butterworth highpass filter with the specifications: Fˆ p = 4 kHz, Fˆs = 1 kHz, α p = 0.1 dB, α s = 40 dB • Choose Ω p = 1 • Then 2πFˆ p Fˆ p 4000 = = =4 Ωs = 2πFˆs Fˆs 1000 • Analog lowpass filter specifications: Ω p = 1, Ω s = 4 , α p = 0.1 dB, α s = 40 dB 32 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Bandpass Filter Design Analog Highpass Filter Design • Code fragments used [N, Wn] = buttord(1, 4, 0.1, 40, ‘s’); [B, A] = butter(N, Wn, ‘s’); [num, den] = lp2hp(B, A, 2*pi*4000); • Spectral Transformation ˆ o2 sˆ 2 + Ω s = Ωp ˆ p2 − Ω ˆ p1 ) sˆ(Ω • Gain plots -20 -20 -40 -60 33 -80 0 where Ω p is the passband edge frequency ˆ p1 and Ω ˆ p 2 are the lower of H LP (s ), and Ω and upper passband edge frequencies of desired bandpass filter H BP (sˆ) Highpass Filter 0 Gain, dB Gain, dB Prototype Lowpass Filter 0 -40 -60 2 4 6 Ω 8 10 -80 0 2 4 6 8 10 Frequency, kHz Copyright © 2005, S. K. Mitra 34 Copyright © 2005, S. K. Mitra Analog Bandpass Filter Design • On the imaginary axis the transformation is ˆ 2 −Ω ˆ2 Ω Ω = −Ω p o ˆ Bw Ω ˆ p2 − Ω ˆ p1 is the width of where Bw = Ω passband and Ω̂ o is the passband center frequency of the bandpass filter • Passband edge frequency ± Ω p is mapped ˆ p1 and ± Ω ˆ p 2, lower and upper into m Ω passband edge frequencies 35 Analog Bandpass Filter Design Ω = −Ω p ˆ o2 − Ω ˆ2 Ω ˆ Bw Ω Ω p Ωs Stopband − Ω s − Ω p Stopband Ω 0 Passband Stopband Passband Stopband Passband Stopband ˆ ↓ −Ω ˆ s1 0 Ω ˆ s1 ↓ Ω̂ o ↓ Ω ˆ s2 ˆ s2 ↓ − Ω −Ω ˆ p2 o− Ω ˆ p1 −Ω ˆ p1 Ω ˆ p2 Ω Lowpass Ω̂ Bandpass 36 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 6 Analog Bandpass Filter Design Analog Bandpass Filter Design • Stopband edge frequency ± Ω s is mapped ˆ s 2 , lower and upper ˆ s1 and ± Ω into m Ω stopband edge frequencies • Also, ˆ o2 = Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 Ω ˆ p1Ω ˆ p2 > Ω ˆ s1Ω ˆ s2 • Case 1: Ω ˆ ˆ ˆ ˆ s 2 we can either To make Ω p1Ω p 2 = Ω s1Ω increase any one of the stopband edges or decrease any one of the passband edges as shown below • If bandedge frequencies do not satisfy the above condition, then one of the frequencies needs to be changed to a new value so that the condition is satisfied 37 ← Passband ← Stopband → ˆ s1 Ω → Stopband ˆ p1 Ω ˆ p2 Ω ˆ s2 Ω Ω̂ 38 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Bandpass Filter Design Analog Bandpass Filter Design ˆ s1Ω ˆ s2 / Ω ˆ p2 ˆ p1 to Ω (1) Decrease Ω larger passband and shorter leftmost transition band ˆ p1Ω ˆ p2 / Ω ˆ s2 ˆ s1 to Ω (2) Increase Ω ˆ o2 = Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 • Note: The condition Ω ˆ p2 can also be satisfied by decreasing Ω which is not acceptable as the passband is reduced from the desired value • Alternately, the condition can be satisfied ˆ s 2 which is not acceptable by increasing Ω as the upper stop band is reduced from the desired value No change in passband and shorter leftmost transition band 39 40 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Bandpass Filter Design Analog Bandpass Filter Design ˆ p1Ω ˆ p2 < Ω ˆ s1Ω ˆ s2 • Case 2: Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s 2 we can either To make Ω decrease any one of the stopband edges or increase any one of the passband edges as shown below → Passband → ← ← Stopband ˆ s1 Ω ˆ p1 Ω ˆ p 2 to Ω ˆ s1Ω ˆ s2 / Ω ˆ p1 (1) Increase Ω larger passband and shorter rightmost transition band ˆ p1Ω ˆ p2 / Ω ˆ s1 ˆ s 2 to Ω (2) Decrease Ω No change in passband and shorter rightmost transition band ˆ p2 Ω ˆ s2 Ω Stopband Ω̂ 41 42 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 7 Analog Bandpass Filter Design Analog Bandpass Filter Design • Example - Design an analog elliptic bandpass filter with the specifications: Fˆ p1 = 4 kHz, Fˆ p 2 = 7 kHz, Fˆs1 = 3 kHz ˆ Fs 2 = 8 kHz, α p = 1 dB, α s = 22 dB ˆ o2 = Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 • Note: The condition Ω ˆ p1 can also be satisfied by increasing Ω which is not acceptable as the passband is reduced from the desired value • Alternately, the condition can be satisfied ˆ s1 which is not acceptable by decreasing Ω as the lower stopband is reduced from the desired value 43 • Now Fˆ p1Fˆ p 2 = 28 × 106 and Fˆs1Fˆs 2 = 24 × 106 • Since Fˆ p1Fˆ p 2 > Fˆs1Fˆs 2 we choose Fˆ p1 = Fˆs1Fˆs 2 / Fˆ p 2 = 3.571428 kHz 44 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Bandpass Filter Design Analog Bandpass Filter Design • Code fragments used [N, Wn] = ellipord(1, 1.4, 1, 22, ‘s’); [B, A] = ellip(N, 1, 22, Wn, ‘s’); [num, den] = lp2bp(B, A, 2*pi*4.8989795, 2*pi*25/7); • We choose Ω p = 1 • Hence 24 − 9 Ωs = = 1.4 (25 / 7) × 3 • Gain plot Prototype Lowpass Filter • Analog lowpass filter specifications: Ω p = 1, Ω s = 1.4 , α p = 1 dB, α s = 22 dB 45 0 Gain, dB Gain, dB 46 Copyright © 2005, S. K. Mitra -20 -40 -60 0 -20 -40 2 4 Ω 6 8 -60 0 5 10 15 Frequency, kHz © 2005, S. K. Mitra Copyright Analog Bandstop Filter Design Analog Bandstop Filter Design • On the imaginary axis the transformation is ˆB Ω Ω = Ωs 2 w 2 ˆ o −Ω ˆ Ω ˆ ˆ where Bw = Ω s 2 − Ω s1 is the width of stopband and Ω̂o is the stopband center frequency of the bandstop filter • Stopband edge frequency ± Ω s is mapped ˆ s1and ± Ω ˆ s 2 , lower and upper into m Ω stopband edge frequencies • Spectral Transformation ˆ s2 − Ω ˆ s1 ) sˆ(Ω s = Ωs 2 ˆ 2o sˆ + Ω where Ω s is the stopband edge frequency ˆ s1 and Ω ˆ s 2 are the lower of H LP ( s ) , and Ω and upper stopband edge frequencies of the desired bandstop filter H BS ( sˆ) 47 Bandpass Filter 0 48 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 8 Analog Bandstop Filter Design Analog Bandstop Filter Design • Passband edge frequency ± Ω p is mapped ˆ p1 and ± Ω ˆ p 2 , lower and upper into m Ω passband edge frequencies • Also, Ω p Ωs Stopband − Ω s − Ω p Stopband Ω 0 Passband Passband Stopband Passband Stopband Passband ˆ o ↓ −Ω ˆ p1 ↓ Ω̂ o ↓ Ω ˆ p10 Ω ˆ p2 ˆ p2 ↓ − Ω −Ω ˆ s2 ˆ s1 ˆ s1 ˆ s2 −Ω Ω −Ω Ω ˆ o2 = Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 Ω • If bandedge frequencies do not satisfy the above condition, then one of the frequencies needs to be changed to a new value so that the condition is satisfied Lowpass Ω̂ Bandpass 49 50 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Bandstop Filter Design Analog Bandstop Filter Design ˆ p1Ω ˆ p2 > Ω ˆ s1Ω ˆ s2 • Case 1: Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s 2 we can either • To make Ω increase any one of the stopband edges or decrease any one of the passband edges as shown below Passband Passband Stopband 51 ˆ p 2 to Ω ˆ s1Ω ˆ s2 / Ω ˆ p2 (1) Decrease Ω larger high-frequency passband and shorter rightmost transition band ˆ p1Ω ˆ p2 / Ω ˆ s2 ˆ s 2 to Ω (2) Increase Ω No change in passbands and shorter rightmost transition band ˆ p1 Ω ˆ s1 Ω ˆ s2 Ω ˆ p2 Ω Ω̂ 52 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Analog Bandstop Filter Design Analog Bandstop Filter Design ˆ o2 = Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 • Note: The condition Ω can also be satisfied by decreasing ˆ p1 which is not acceptable as the low- Ω frequency passband is reduced from the desired value • Alternately, the condition can be satisfied ˆ s1 which is not acceptable by increasing Ω as the stopband is reduced from the desired value ˆ p1Ω ˆ p2 < Ω ˆ s1Ω ˆ s2 • Case 1: Ω ˆ ˆ ˆ ˆ s 2 we can either • To make Ω p1Ω p 2 = Ω s1Ω decrease any one of the stopband edges or increase any one of the passband edges as shown below 53 Passband 54 Copyright © 2005, S. K. Mitra Passband Stopband ˆ p1 Ω ˆ s1 Ω ˆ s2 Ω ˆ p2 Ω Ω̂ Copyright © 2005, S. K. Mitra 9 Analog Bandstop Filter Design Analog Bandstop Filter Design ˆ o2 = Ω ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 • Note: The condition Ω ˆ p2 can also be satisfied by increasing Ω which is not acceptable as the highfrequency passband is decreased from the desired value • Alternately, the condition can be satisfied ˆ s 2 which is not acceptable by decreasing Ω as the stopband is decreased ˆ p1 to Ω ˆ s1Ω ˆ s2 / Ω ˆ p1 (1) Increase Ω larger passband and shorter leftmost transition band ˆ s1 to Ω ˆ p1Ω ˆ p2 / Ω ˆ s1 (2) Decrease Ω No change in passbands and shorter leftmost transition band 55 56 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 10 Digital Filter Design Digital Filter Specifications • Usually, either the magnitude and/or the phase (delay) response is specified for the design of digital filter for most applications • In some situations, the unit sample response or the step response may be specified • In most practical applications, the problem of interest is the development of a realizable approximation to a given magnitude response specification • Objective - Determination of a realizable transfer function G(z) approximating a given frequency response specification is an important step in the development of a digital filter • If an IIR filter is desired, G(z) should be a stable real rational function • Digital filter design is the process of deriving the transfer function G(z) 1 Copyright © 2005, S. K. Mitra 2 Digital Filter Specifications Digital Filter Specifications • We discuss in this course only the magnitude approximation problem • There are four basic types of ideal filters with magnitude responses as shown below HLP(e j ) HHP (e j ) 1 1 – c 0 – c c – c2 – c1 c 1 –1 3 0 HBS (e j ) HBP (e j ) c1 c2 – c2 – c1 c1 Copyright © 2005, S. K. Mitra c2 Copyright © 2005, S. K. Mitra 4 • As the impulse response corresponding to each of these ideal filters is noncausal and of infinite length, these filters are not realizable • In practice, the magnitude response specifications of a digital filter in the passband and in the stopband are given with some acceptable tolerances • In addition, a transition band is specified between the passband and stopband Copyright © 2005, S. K. Mitra Digital Filter Specifications Digital Filter Specifications jω • As indicated in the figure, in the passband, defined by 0 ≤ ω ≤ ω p , we require that G (e jω ) ≅ 1 with an error ± δ p, i.e., • For example, the magnitude response G (e ) of a digital lowpass filter may be given as indicated below 1 − δ p ≤ G ( e jω ) ≤ 1 + δ p , ω ≤ ωp • In the stopband, defined by ωs ≤ ω ≤ π, we require that G (e jω ) ≅ 0 with an error δ s , i.e., G ( e jω ) ≤ δ s , ω s ≤ ω ≤ π 5 Copyright © 2005, S. K. Mitra 6 Copyright © 2005, S. K. Mitra 1 Digital Filter Specifications 7 • ω p - passband edge frequency • ωs - stopband edge frequency • δ p - peak ripple value in the passband • δ s - peak ripple value in the stopband • Since G (e jω ) is a periodic function of ω, and G (e jω ) of a real-coefficient digital filter is an even function of ω • As a result, filter specifications are given only for the frequency range 0 ≤ ω ≤ π Copyright © 2005, S. K. Mitra Digital Filter Specifications • Specifications are often given in terms of jω loss function A (ω) = − 20 log10 G (e ) in dB • Peak passband ripple α p = − 20 log10 (1 − δ p ) dB • Minimum stopband attenuation α s = − 20 log10 (δ s ) dB 8 Copyright © 2005, S. K. Mitra Digital Filter Specifications Digital Filter Specifications • Magnitude specifications may alternately be given in a normalized form as indicated below • Here, the maximum value of the magnitude in the passband is assumed to be unity • 1 / 1 + ε 2 - Maximum passband deviation, given by the minimum value of the magnitude in the passband • 1 - Maximum stopband magnitude A 9 Copyright © 2005, S. K. Mitra Digital Filter Specifications • For the normalized specification, maximum value of the gain function or the minimum value of the loss function is 0 dB • Maximum passband attenuation α max = 20 log10 1 + ε 2 dB • For δ p << 1, it can be shown that α max ≅ − 20 log10 (1 − 2 δ p ) dB ( 11 ) Copyright © 2005, S. K. Mitra 10 Copyright © 2005, S. K. Mitra Digital Filter Specifications • In practice, passband edge frequency Fp and stopband edge frequency Fs are specified in Hz • For digital filter design, normalized bandedge frequencies need to be computed from specifications in Hz using Ω p 2π Fp ωp = = = 2π FpT FT FT Ω 2π Fs ωs = s = = 2π Fs T FT FT 12 Copyright © 2005, S. K. Mitra 2 Digital Filter Specifications • Example - Let Fp = 7 kHz, Fs = 3 kHz, and FT = 25 kHz • Then 2π(7 × 103 ) ωp = = 0.56π 25 × 103 ωs = 2π(3 × 103 ) = 0.24π 25 × 103 13 Copyright © 2005, S. K. Mitra Selection of Filter Type • The transfer function H(z) meeting the frequency response specifications should be a causal transfer function • For IIR digital filter design, the IIR transfer function is a real rational function of z −1: p0 + p1z −1 + p2 z − 2 + L + pM z − M , M≤N d 0 + d1z −1 + d 2 z − 2 + L + d N z − N • H(z) must be a stable transfer function and must be of lowest order N for reduced 14 computational complexity H ( z) = Copyright © 2005, S. K. Mitra Selection of Filter Type Selection of Filter Type • For FIR digital filter design, the FIR transfer function is a polynomial in z −1 with real coefficients: • Advantages in using an FIR filter (1) Can be designed with exact linear phase, (2) Filter structure always stable with quantized coefficients • Disadvantages in using an FIR filter - Order of an FIR filter, in most cases, is considerably higher than the order of an equivalent IIR filter meeting the same specifications, and FIR filter has thus higher computational complexity N H ( z ) = ∑ h[n] z −n n =0 • For reduced computational complexity, degree N of H(z) must be as small as possible • If a linear phase is desired, the filter coefficients must satisfy the constraint: h[n] = ± h[ N − n] 15 Copyright © 2005, S. K. Mitra 16 Digital Filter Design: Basic Approaches Digital Filter Design: Basic Approaches • Most common approach to IIR filter design (1) Convert the digital filter specifications into an analog prototype lowpass filter specifications • (2) Determine the analog lowpass filter transfer function H a (s ) • (3) Transform H a (s ) into the desired digital transfer function G ( z ) 17 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 18 • This approach has been widely used for the following reasons: (1) Analog approximation techniques are highly advanced (2) They usually yield closed-form solutions (3) Extensive tables are available for analog filter design (4) Many applications require digital simulation of analog systems Copyright © 2005, S. K. Mitra 3 Digital Filter Design: Basic Approaches Digital Filter Design: Basic Approaches • An analog transfer function to be denoted as P ( s) H a ( s) = a Da ( s ) • Basic idea behind the conversion of H a (s ) into G ( z ) is to apply a mapping from the s-domain to the z-domain so that essential properties of the analog frequency response are preserved • Thus mapping function should be such that – Imaginary ( jΩ ) axis in the s-plane be where the subscript “a” specifically indicates the analog domain • A digital transfer function derived from H a ( s ) shall be denoted as P( z ) G( z) = D( z ) 19 Copyright © 2005, S. K. Mitra 20 Digital Filter Design: Basic Approaches Copyright © 2005, S. K. Mitra • Three commonly used approaches to FIR filter design (1) Windowed Fourier series approach (2) Frequency sampling approach (3) Computer-based optimization methods 22 IIR Digital Filter Design: Bilinear Transformation Method • Digital filter design consists of 3 steps: (1) Develop the specifications of Ha ( s ) by applying the inverse bilinear transformation to specifications of G(z) (2) Design Ha (s ) (3) Determine G(z) by applying bilinear transformation to Ha ( s ) • As a result, the parameter T has no effect on G(z) and T = 2 is chosen for convenience • Above transformation maps a single point in the s-plane to a unique point in the z-plane and vice-versa • Relation between G(z) and Ha ( s ) is then given by G ( z ) = Ha ( s ) 2 ⎛⎜ 1− z −1 ⎞⎟ s= T ⎜ 1+ z −1 ⎟ ⎝ ⎠ Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Bilinear Transformation • Bilinear transformation −1 ⎞ ⎛ s = 2 ⎜⎜ 1 − z −1 ⎟⎟ T ⎝1 + z ⎠ 23 Copyright © 2005, S. K. Mitra Digital Filter Design: Basic Approaches • FIR filter design is based on a direct approximation of the specified magnitude response, with the often added requirement that the phase be linear • The design of an FIR filter of order N may be accomplished by finding either the length-(N+1) impulse response samples {h[n]} or the (N+1) samples of its frequency response H (e jω ) 21 mapped onto the unit circle of the z-plane – A stable analog transfer function be mapped into a stable digital transfer function 24 Copyright © 2005, S. K. Mitra 4 Bilinear Transformation Bilinear Transformation • Inverse bilinear transformation for T = 2 is z = 1+ s 1− s • Mapping of s-plane into the z-plane • For s = σo + jΩo 2 2 (1 + σo ) + jΩo 2 (1 + σo ) + Ω o z= ⇒z = (1 − σo ) − jΩo (1 − σo ) 2 + Ωo2 σo = 0 → z = 1 • Thus, σo < 0 → z < 1 σo > 0 → z > 1 25 Copyright © 2005, S. K. Mitra 26 Bilinear Transformation Bilinear Transformation jω • For z = e with T = 2 we have − jω e − jω / 2 (e jω / 2 − e − jω / 2 ) jΩ = 1 − e − jω = − jω / 2 jω / 2 − jω / 2 1+ e e (e +e ) j 2 sin(ω / 2) = = j tan(ω / 2) 2 cos(ω / 2) or Ω = tan(ω / 2) 27 Copyright © 2005, S. K. Mitra • Mapping is highly nonlinear • Complete negative imaginary axis in the splane from Ω = −∞ to Ω = 0 is mapped into the lower half of the unit circle in the z-plane from z = −1 to z = 1 • Complete positive imaginary axis in the splane from Ω = 0 to Ω = ∞ is mapped into the upper half of the unit circle in the z-plane from z = 1 to z = −1 28 Ω 29 Ω = tan(ω/2) Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Bilinear Transformation Bilinear Transformation • Nonlinear mapping introduces a distortion in the frequency axis called frequency warping • Effect of warping shown below Copyright © 2005, S. K. Mitra • Steps in the design of a digital filter (1) Prewarp (ω p , ωs ) to find their analog equivalents (Ω p , Ω s ) (2) Design the analog filter H a (s ) (3) Design the digital filter G(z) by applying bilinear transformation to H a (s ) • Transformation can be used only to design digital filters with prescribed magnitude response with piecewise constant values • Transformation does not preserve phase response of analog filter 30 Copyright © 2005, S. K. Mitra 5 IIR Digital Filter Design Using Bilinear Transformation IIR Digital Filter Design Using Bilinear Transformation • Example - Consider Ωc s + Ωc • Applying bilinear transformation to the above we get the transfer function of a first-order digital lowpass Butterworth filter • Rearranging terms we get H a ( s) = 1+ z −1 where α= Ωc (1 + z −1 ) = (1 − z −1 ) + Ωc (1 + z −1 ) 31 Copyright © 2005, S. K. Mitra 32 Copyright © 2005, S. K. Mitra IIR Digital Filter Design Using Bilinear Transformation IIR Digital Filter Design Using Bilinear Transformation • Example - Consider the second-order analog notch transfer function s 2 + Ωo2 H a ( s) = 2 s + B s + Ωo2 for which H a ( jΩo ) = 0 33 H a ( j 0) = H a ( j∞) = 1 • Ωo is called the notch frequency • If H a ( jΩ 2 ) = H a ( jΩ1 ) = 1 / 2 then B = Ω 2 − Ω1 is the 3-dB notch bandwidth Copyright © 2005, S. K. Mitra • Then = 35 1+α 1 − 2β z + z ⋅ 2 1 − 2 β (1 + α ) z −1 + α z −2 1 + Ωo2 − B 1 − tan( Bw / 2) = 1 + Ωo2 + B 1 + tan( Bw / 2) 1 − Ωo2 β= = cos ωo Copyright © 2005, S. K. Mitra 1 + Ωo2 34 IIR Digital Filter Design Using Bilinear Transformation • Thus 0.954965 − 1.1226287 z −1 + 0.954965 z − 2 1 − 1.1226287 z −1 + 0.90993 z − 2 • The gain and phase responses are shown below G( z) = 2 0 -10 -20 -30 β = 0.587785 Copyright © 2005, S. K. Mitra −1 where α = Gain, dB α = 0.90993 G ( z ) = H a ( s ) s =1− z 1+ z −1 2 2 (1 + Ωo ) − 2(1 − Ωo ) z −1 + (1 + Ωo2 ) z −2 (1 + Ωo2 + B) − 2(1 − Ωo2 ) z −1 + (1 + Ωo2 − B ) z −2 −1 −2 = IIR Digital Filter Design Using Bilinear Transformation • Example - Design a 2nd-order digital notch filter operating at a sampling rate of 400 Hz with a notch frequency at 60 Hz, 3-dB notch bandwidth of 6 Hz • Thus ωo = 2π(60 / 400) = 0.3π Bw = 2π(6 / 400) = 0.03π • From the above values we get 1 − Ωc 1 − tan(ωc / 2) = 1 + Ωc 1 + tan(ωc / 2) Phase, radians G ( z ) = H a ( s ) s =1− z −1 −1 G ( z ) = 1 − α ⋅ 1 + z −1 2 1−α z 36 -40 0 50 100 150 Frequency, Hz 200 1 0 -1 -2 0 50 100 150 Frequency, Hz 200 Copyright © 2005, S. K. Mitra 6 IIR Lowpass Digital Filter Design Using Bilinear Transformation IIR Lowpass Digital Filter Design Using Bilinear Transformation • Example - Design a lowpass Butterworth digital filter with ω p = 0.25π, ωs = 0.55π , α p ≤ 0.5 dB, and α s ≥ 15 dB • Thus A2 = 31.622777 ε 2 = 0.1220185 • Prewarping we get Ω p = tan(ω p / 2) = tan(0.25π / 2) = 0.4142136 Ω s = tan(ωs / 2) = tan(0.55π / 2) = 1.1708496 • The inverse transition ratio is 1 Ωs = = 2.8266809 k Ωp • The inverse discrimination ratio is • If G (e j 0 ) = 1 this implies 20 log10 G (e j 0.25π ) ≥ − 0.5 20 log10 G (e j 0.55π ) ≤ −15 37 Copyright © 2005, S. K. Mitra 38 N= • 3rd-order lowpass Butterworth transfer function for Ωc = 1 is 1 Han ( s ) = ( s + 1)( s 2 + s + 1) • We choose N = 3 • To determine Ωc we use 2 1 1 = 2N 1 + (Ω p / Ω c ) 1 + ε2 39 Copyright © 2005, S. K. Mitra • Denormalizing to get Ω c = 0.588148 we arrive at s ⎞ Ha ( s ) = Han ⎛⎜ ⎟ ⎝ 0.588148 ⎠ 40 Copyright © 2005, S. K. Mitra IIR Lowpass Digital Filter Design Using Bilinear Transformation Design of IIR Highpass, Highpass, Bandpass, Bandpass, and Bandstop Digital Filters • Applying bilinear transformation to H a (s ) we get the desired digital transfer function G ( z ) = H a ( s ) s =1− z • First Approach (1) Prewarp digital frequency specifications of desired digital filter GD (z ) to arrive at frequency specifications of analog filter H D (s ) of same type (2) Convert frequency specifications of H D (s ) into that of prototype analog lowpass filter H LP (s ) −1 1+ z −1 1 0 -10 Gain, dB Magnitude • Magnitude and gain responses of G(z) shown below: 0.8 0.6 0.4 41 0 -20 -30 0.2 0 0.2 0.4 0.6 ω/π 0.8 1 -40 0 0.2 0.4 0.6 Copyright © 2005, S. K. Mitra • We then get Ω c = 1.419915 (Ω p ) = 0.588148 log10 (1 / k1 ) = 2.6586997 log10 (1 / k ) H a ( jΩ p ) = A2 − 1 = 15.841979 ε IIR Lowpass Digital Filter Design Using Bilinear Transformation IIR Lowpass Digital Filter Design Using Bilinear Transformation • Thus 1 = k1 0.8 1 Copyright © 2005, S. K. Mitra ω/π 42 (3) Design analog lowpass filter H LP (s ) Copyright © 2005, S. K. Mitra 7 Design of IIR Highpass, Highpass, Bandpass, Bandpass, and Bandstop Digital Filters Design of IIR Highpass, Highpass, Bandpass, Bandpass, and Bandstop Digital Filters (4) Convert H LP (s ) into H D (s ) using inverse frequency transformation used in Step 2 (5) Design desired digital filter GD (z ) by applying bilinear transformation to H D (s ) 43 Copyright © 2005, S. K. Mitra • Second Approach (1) Prewarp digital frequency specifications of desired digital filter GD (z ) to arrive at frequency specifications of analog filter H D (s ) of same type (2) Convert frequency specifications of H D (s ) into that of prototype analog lowpass filter H LP (s ) 44 IIR Highpass Digital Filter Design Design of IIR Highpass, Highpass, Bandpass, Bandpass, and Bandstop Digital Filters • Design of a Type 1 Chebyshev IIR digital highpass filter • Specifications: Fp = 700 Hz, Fs = 500 Hz, α p = 1 dB, α s = 32 dB, FT = 2 kHz • Normalized angular bandedge frequencies (3) Design analog lowpass filter H LP (s ) (4) Convert H LP (s ) into an IIR digital transfer function GLP (z ) using bilinear transformation (5) Transform GLP (z ) into the desired digital transfer function GD (z ) • We illustrate the first approach 45 Copyright © 2005, S. K. Mitra 46 Copyright © 2005, S. K. Mitra IIR Highpass Digital Filter Design • Prewarping these frequencies we get ˆ p = tan(ω p / 2) = 1.9626105 Ω ˆ s = tan(ωs / 2) = 1.0 Ω • MATLAB code fragments used for the design [N, Wn] = cheb1ord(1, 1.9626105, 1, 32, ’s’) [B, A] = cheby1(N, 1, Wn, ’s’); [BT, AT] = lp2hp(B, A, 1.9626105); [num, den] = bilinear(BT, AT, 0.5); • For the prototype analog lowpass filter choose Ωp =1 ˆ Ω Ω • Using Ω = − p p we get Ω s = 1.962105 ˆ Ω • Analog lowpass filter specifications: Ω p = 1, 0 Gain, dB -10 Ω s = 1.926105 , α p = 1dB, α s = 32 dB Copyright © 2005, S. K. Mitra 2π Fp 2π × 700 = = 0.7 π FT 2000 2π Fs 2π × 500 ωs = = = 0.5π FT 2000 ωp = IIR Highpass Digital Filter Design 47 Copyright © 2005, S. K. Mitra -20 -30 -40 48 -50 0 0.2 0.4 0.6 ω/π 0.8 1 Copyright © 2005, S. K. Mitra 8 IIR Bandpass Digital Filter Design IIR Bandpass Digital Filter Design • Design of a Butterworth IIR digital bandpass filter • Specifications: ω p1 = 0.45π , ω p 2 = 0.65π , ωs1 = 0.3π, ωs 2 = 0.75π, α p = 1 dB, α s = 40 dB • Prewarping we get ˆ p1 = tan(ω p1 / 2) = 0.8540807 Ω ˆ p 2 = tan(ω p 2 / 2) = 1.6318517 Ω ˆ p2 − Ω ˆ p1 = 0.777771 • Width of passband Bw = Ω ˆ o2 = Ω ˆ p1Ω ˆ p 2 = 1.393733 Ω ˆ s1Ω ˆ s 2 = 1.23010325 ≠ Ω ˆ o2 Ω ˆ s1and ˆ • We therefore modify Ω s1 so that Ω ˆ exhibit geometric symmetry with Ωs2 ˆ o2 respect to Ω ˆ s1 = 0.5773031 • We set Ω • For the prototype analog lowpass filter we choose Ω p = 1 ˆ s1 = tan(ωs1 / 2) = 0.5095254 Ω ˆ Ω s 2 = tan(ωs 2 / 2) = 2.41421356 49 Copyright © 2005, S. K. Mitra 50 IIR Bandpass Digital Filter Design IIR Bandpass Digital Filter Design • MATLAB code fragments used for the design ˆ2 ˆ2 • Using Ω = −Ω p Ω o − Ω we get ˆ Bw Ω Ωs = [N, Wn] = buttord(1, 2.3617627, 1, 40, ’s’) [B, A] = butter(N, Wn, ’s’); [BT, AT] = lp2bp(B, A, 1.1805647, 0.777771); [num, den] = bilinear(BT, AT, 0.5); 1.393733 − 0.3332788 = 2.3617627 0.5773031× 0.777771 • Specifications of prototype analog Butterworth lowpass filter: Ω p = 1 , Ω s = 2.3617627 , α p = 1 dB, α s = 40 dB 51 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 0 Gain, dB -10 -20 -30 -40 -50 52 0 0.2 0.4 0.6 0.8 1 ω/π Copyright © 2005, S. K. Mitra IIR Bandstop Digital Filter Design IIR Bandstop Digital Filter Design • Design of an elliptic IIR digital bandstop filter • Specifications: ωs1 = 0.45π , ωs 2 = 0.65π , ω p1 = 0.3π, ω p 2 = 0.75π , α p = 1 dB, α s = 40 dB • Prewarping we get ˆ s 2 = 1.6318517, ˆ s1 = 0.8540806, Ω Ω ˆ p 2 = 2.4142136 ˆ p1 = 0.5095254, Ω Ω ˆ s2 − Ω ˆ s1 = 0.777771 • Width of stopband Bw = Ω ˆ p1and Ω ˆ p2 ˆ p1so that Ω • We therefore modify Ω exhibit geometric symmetry with respect to ˆ o2 Ω 53 ˆ o2 = Ω ˆ s 2Ω ˆ s1 = 1.393733 Ω ˆ ˆ ˆ o2 Ω p 2Ω p1 = 1.230103 ≠ Ω Copyright © 2005, S. K. Mitra ˆ p1 = 0.577303 • We set Ω • For the prototype analog lowpass filter we choose Ω s = 1 ˆB Ω • Using Ω = Ω s 2 w 2 we get ˆ o −Ω ˆ Ω 0.5095254 × 0.777771 Ωp = = 0.4234126 1.393733 − 0.3332787 54 Copyright © 2005, S. K. Mitra 9 IIR Bandstop Digital Filter Design • MATLAB code fragments used for the design [N, Wn] = ellipord(0.4234126, 1, 1, 40, ’s’); [B, A] = ellip(N, 1, 40, Wn, ’s’); [BT, AT] = lp2bs(B, A, 1.1805647, 0.777771); [num, den] = bilinear(BT, AT, 0.5); 0 Gain, dB -10 -20 -30 -40 -50 0 55 0.2 0.4 0.6 0.8 1 ω/π Copyright © 2005, S. K. Mitra 10 Least Integral-Squared Error Design of FIR Filters Least Integral-Squared Error Design of FIR Filters • Let H d (e jω ) denote the desired frequency response • Since H d (e jω ) is a periodic function of ω with a period 2π, it can be expressed as a Fourier series • In general, H d (e jω ) is piecewise constant with sharp transitions between bands • In which case, {hd [n]} is of infinite length and noncausal • Objective - Find a finite-duration {ht [n]} of length 2M+1 whose DTFT H t (e jω ) approximates the desired DTFT H d (e jω ) in some sense ∞ H d (e jω ) = ∑ hd [n]e − jωn n = −∞ where hd [n] = 1 π 1 jω jωn ∫ H d (e )e dω, − ∞ ≤ n ≤ ∞ 2π − π 2 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Least Integral-Squared Error Design of FIR Filters Least Integral-Squared Error Design of FIR Filters • Using Parseval’s relation we can write • Commonly used approximation criterion Minimize the integral-squared error 1 π jω jω 2 Φ= ∫ H t (e ) − H d (e ) dω 2π − π where 2 n = −∞ M 2 − M −1 ∞ n = −∞ n = M +1 = ∑ ht [n] − hd [ n] + ∑ hd2 [n] + ∑ hd2 [n] n=− M M H t (e jω ) = ∑ ht [n]e − jωn n=− M 3 ∞ Φ = ∑ ht [n] − hd [n] 4 • It follows from the above that Φ is minimum when ht [n] = hd [n] for − M ≤ n ≤ M • ⇒ Best finite-length approximation to ideal infinite-length impulse response in the mean-square sense is obtained by truncation Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Impulse Responses of Ideal Filters Least Integral-Squared Error Design of FIR Filters • Ideal lowpass filter HLP(e j ) • A causal FIR filter with an impulse response h[n] can be derived from ht [n] by delaying: h[n] = ht [n − M ] • The causal FIR filter h[n] has the same magnitude response as ht [n] and its phase response has a linear phase shift of ωM radians with respect to that of ht [n] hLP [n] = 1 – c 0 HHP (e j ) 1 6 Copyright © 2005, S. K. Mitra c • Ideal highpass filter - – c 5 sin ωc n πn , − ∞ ≤ n ≤ ∞ 0 c ⎧ 1 − ωc , n=0 π ⎪ hHP [n] = ⎨ sin(ω n) ⎪− πnc , n ≠ 0 ⎩ Copyright © 2005, S. K. Mitra 1 Impulse Responses of Ideal Filters Impulse Responses of Ideal Filters • Ideal bandstop filter - • Ideal bandpass filter - HBS (e j ) HBP (e j ) 1 –1 – c2 – c1 – c2 – c1 c1 c2 c2 ⎧ 1 − (ωc 2 − ωc1 ) , n=0 π ⎪ hBS [n] = ⎨ sin(ω n) sin(ωc 2n) ⎪ πnc1 − πn , n ≠ 0 ⎩ ⎧sin(ωc 2 n) − sin(ωc1n) , n ≠ 0 πn ⎪ πn hBP [n] = ⎨ ωc 2 ωc1 ⎪ n=0 π − π , ⎩ 7 c1 8 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Impulse Responses of Ideal Filters Impulse Responses of Ideal Filters • Ideal multiband filter HML (e j A5 A1 A4 A2 A3 0 • Ideal discrete-time Hilbert transformer - H ML (e jω ) = Ak , ) 1 2 3 4 ⎧ j, − π < ω < 0 H HT (e jω ) = ⎨ ⎩− j , 0 < ω < π ωk −1 ≤ ω ≤ ωk , k = 1, 2,K, L for n even ⎧ 0, hHT [n] = ⎨ ⎩2/π n, for n odd sin(ω n) hML [n] = ∑ ( Al − Al +1) ⋅ πnL L l =1 9 10 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Impulse Responses of Ideal Filters Gibbs Phenomenon • Gibbs phenomenon - Oscillatory behavior in the magnitude responses of causal FIR filters obtained by truncating the impulse response coefficients of ideal filters • Ideal discrete-time differentiator H DIF (e jω ) = jω, 0≤ ω ≤π 1.5 11 Magnitude n=0 ⎧⎪ 0, hDIF [ n] = ⎨cos π n ⎪⎩ n , n ≠ 0 12 Copyright © 2005, S. K. Mitra N = 20 N = 60 1 0.5 0 0 0.2 0.4 0.6 0.8 1 ω/π Copyright © 2005, S. K. Mitra 2 Gibbs Phenomenon 13 • As can be seen, as the length of the lowpass filter is increased, the number of ripples in both passband and stopband increases, with a corresponding decrease in the ripple widths • Height of the largest ripples remain the same independent of length • Similar oscillatory behavior observed in the magnitude responses of the truncated versions of other types of ideal filters Gibbs Phenomenon • Gibbs phenomenon can be explained by treating the truncation operation as an windowing operation: ht [n] = hd [n] ⋅ w[ n] • In the frequency domain H t ( e jω ) = 14 1 2π π jϕ j ( ω− ϕ) ) dϕ ∫ H d (e ) Ψ ( e −π • where H t (e jω ) and Ψ (e jω ) are the DTFTs of ht [n] and w[n] , respectively Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Gibbs Phenomenon Gibbs Phenomenon • Thus H t (e jω ) is obtained by a periodic continuous convolution of H d (e jω ) with Ψ (e jω ) 15 • If Ψ (e jω ) is a very narrow pulse centered at ω = 0 (ideally a delta function) compared to variations in H d (e jω ), then H t (e jω ) will approximate H d (e jω ) very closely • Length 2M+1 of w[n] should be very large • On the other hand, length 2M+1 of ht [n] should be as small as possible to reduce computational complexity 16 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra Gibbs Phenomenon Gibbs Phenomenon • A rectangular window is used to achieve simple truncation: ⎧1, 0 ≤ n ≤ M wR [n] = ⎨ ⎩0, otherwise • Presence of oscillatory behavior in H t (e jω ) is basically due to: – 1) hd [n] is infinitely long and not absolutely • Oscillatory behavior can be explained by jω examining the DTFT ΨR (e ) of wR [n] : Rectangular window 30 Amplitude 20 jω 18 Copyright © 2005, S. K. Mitra side lobe M=4 0 -10 -1 summable, and hence filter is unstable – 2) Rectangular window has an abrupt transition to zero 17 main lobe M = 10 10 -0.5 0 ω/π 0.5 1 • ΨR (e ) has a main lobe centered at ω = 0 • Other ripples are called sidelobes Copyright © 2005, S. K. Mitra 3 Gibbs Phenomenon 19 • Main lobe of ΨR (e jω ) characterized by its width 4π /( 2 M + 1) defined by first zero crossings on both sides of ω = 0 • As M increases, width of main lobe decreases as desired • Area under each lobe remains constant while width of each lobe decreases with an increase in M • Ripples in H t (e jω ) around the point of discontinuity occur more closely but with no decrease in amplitude as M increases Copyright © 2005, S. K. Mitra Gibbs Phenomenon • Rectangular window has an abrupt transition to zero outside the range − M ≤ n ≤ M , which results in Gibbs phenomenon in H t (e jω ) • Gibbs phenomenon can be reduced either: (1) Using a window that tapers smoothly to zero at each end, or (2) Providing a smooth transition from passband to stopband in the magnitude specifications 20 Copyright © 2005, S. K. Mitra 4 Fixed Window Functions Fixed Window Functions 1 Copyright © 2005, S. K. Mitra • Plots of magnitudes of the DTFTs of these windows for M = 25 are shown below: -20 -40 -40 -60 -100 0.2 0.4 0.6 0.8 -100 1 0 0.2 0.4 0.6 ω/π ω/π Hamming window Blackman window 0 -20 -40 -40 Gain, dB 0 -20 -60 -100 2 -60 -80 0 -80 0.8 1 -60 -80 0 0.2 0.4 0.6 0.8 1 -100 0 0.2 ω/π 0.4 0.6 0.8 1 ω/π Copyright © 2005, S. K. Mitra Fixed Window Functions • Magnitude spectrum of each window characterized by a main lobe centered at ω = 0 followed by a series of sidelobes with decreasing amplitudes • Parameters predicting the performance of a window in filter design are: • Main lobe width • Relative sidelobe level Copyright © 2005, S. K. Mitra 0 -20 -80 Fixed Window Functions 3 Hann window 0 Gain, dB Gain, dB Rectangular window Gain, dB • Using a tapered window causes the height of the sidelobes to diminish, with a corresponding increase in the main lobe width resulting in a wider transition at the discontinuity • Hann: w[n] = 0.5 + 0.5 cos( 2π n ), − M ≤ n ≤ M 2M + 1 • Hamming: w[n] = 0.54 + 0.46 cos( 2π n ), − M ≤ n ≤ M 2M + 1 • Blackman: w[n] = 0.42 + 0.5 cos( 2π n ) + 0.08 cos( 4π n ) 2M + 1 2M + 1 • Main lobe width ∆ ML - given by the distance between zero crossings on both sides of main lobe • Relative sidelobe level Asl - given by the difference in dB between amplitudes of largest sidelobe and main lobe 4 Fixed Window Functions Copyright © 2005, S. K. Mitra Fixed Window Functions • Distance between the locations of the maximum passband deviation and minimum stopband value ≅ ∆ ML • Observe H t (e j (ωc + ∆ω) ) + H t (e j (ωc − ∆ω) ) ≅ 1 • Thus, H t (e jωc ) ≅ 0.5 • Passband and stopband ripples are the same 5 Copyright © 2005, S. K. Mitra • Width of transition band ∆ω = ω s − ω p < ∆ ML 6 Copyright © 2005, S. K. Mitra 1 Fixed Window Functions Fixed Window Functions • To ensure a fast transition from passband to stopband, window should have a very small main lobe width • To reduce the passband and stopband ripple δ, the area under the sidelobes should be very small • Unfortunately, these two requirements are contradictory • In the case of rectangular, Hann, Hamming, and Blackman windows, the value of ripple does not depend on filter length or cutoff frequency ωc , and is essentially constant • In addition, ∆ω ≈ c M where c is a constant for most practical purposes 7 Copyright © 2005, S. K. Mitra 8 Fixed Window Functions Fixed Window Functions • Rectangular window - ∆ ML = 4π /(2M + 1) Asl = 13.3 dB, α s = 20.9 dB, ∆ω = 0.92π / M • Hann window - ∆ ML = 8π /(2M + 1) Asl = 31.5 dB, α s = 43.9 dB, ∆ω = 3.11π / M • Hamming window - ∆ ML = 8π /(2M + 1) Asl = 42.7 dB, α s = 54.5 dB, ∆ω = 3.32π / M • Blackman window - ∆ ML = 12π /( 2M + 1) Asl = 58.1 dB, α s = 75.3 dB, ∆ω = 5.56π / M 9 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra • Filter Design Steps (1) Set ωc = (ω p + ωs ) / 2 (2) Choose window based on specified α s (3) Estimate M using ∆ω ≈ c M 10 Copyright © 2005, S. K. Mitra FIR Filter Design Example FIR Filter Design Example • Lowpass filter of length 51 and ωc = π / 2 Lowpass Filter Designed Using Hamming window 0 Gain, dB Gain, dB Lowpass Filter Designed Using Hann window 0 -50 • An increase in the main lobe width is associated with an increase in the width of the transition band • A decrease in the sidelobe amplitude results in an increase in the stopband attenuation -50 -100 -100 0 0.2 0.4 0.6 0.8 0 1 0.2 0.4 0.6 0.8 1 ω/π ω/π Gain, dB Lowpass Filter Designed Using Blackman window 0 -50 -100 11 0 0.2 0.4 0.6 ω/π 0.8 1 Copyright © 2005, S. K. Mitra 12 Copyright © 2005, S. K. Mitra 2 Adjustable Window Functions Adjustable Window Functions • Dolph-Chebyshev window can be designed with any specified relative sidelobe level while the main lobe width adjusted by choosing length appropriately • Filter order is estimated using 2.056α s − 16.4 N= 2.85( ∆ω) • Dolph-Chebyshev Window M w[n] = 1 [1γ + 2 ∑ Tk (β cos k ) cos 2nkπ ], 2M + 1 2M + 1 2M + 1 k =1 −M ≤n≤M amplitude of sidelobe where γ= main lobe amplitude β = cosh( 1 cosh −1 1γ ) 2M and ⎧ cos(l cos −1 x ), x ≤1 Tl ( x) = ⎨ −1 ⎩cosh(l cosh x), x > 1 13 Copyright © 2005, S. K. Mitra 14 Properties of Dolph-Chebyshev window: • All sidelobes are of equal height • Stopband approximation error of filters designed have essentially equiripple behavior • For a given window length, it has the smallest main lobe width compared to other windows resulting in filters with the smallest transition band • Gain response of a Dolph-Chebyshev window of length 51 and relative sidelobe level of 50 dB is shown below Dolph-Chebyshev Window Gain, dB 0 -20 -40 -60 0.2 0.4 0.6 0.8 Copyright © 2005, S. K. Mitra Adjustable Window Functions Adjustable Window Functions -80 0 where ∆ω is the normalized transition bandwidth, e.g, for a lowpass filter ∆ω = ωs − ω p 1 ω/π 15 Copyright © 2005, S. K. Mitra 16 Adjustable Window Functions Adjustable Window Functions • β controls the minimum stopband attenuation of the windowed filter response • β is estimated using • Kaiser Window - I 0{β 1 − (n / M ) } , −M ≤n≤M I 0 (β) where β is an adjustable parameter and I 0 (u ) is the modified zeroth-order Bessel function of the first kind: ∞ (u / 2) r I 0 (u ) = 1 + ∑ [ ]2 r! r =1 • Note I 0 (u ) > 0 for u > 0 20 (u / 2) r • In practice I 0 (u ) ≅ 1 + ∑ [ ]2 r! r =1 17 w[n] = 2 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra 0.1102( α s −8.7 ), ⎧⎪ β= ⎨0.5842( α s − 21)0.4 + 0.07886( α s − 21), 0, ⎪⎩ • Filter order is estimated using N= for α s > 50 for 21 ≤ α s ≤ 50 for α s < 21 αs − 8 2.285(∆ω) where ∆ω is the normalized transition bandwidth 18 Copyright © 2005, S. K. Mitra 3 FIR Filter Design Example FIR Filter Design Example sin(0.4π n) • Hence ht [ n] = ⋅ w[n], − 12 ≤ n ≤ 12 πn where w[n] is the n-th coefficient of a length-25 Kaiser window with β = 3.3953 -20 Gain, dB 0 -20 -40 -80 0 20 Impulse Responses of FIR Filters with a Smooth Transition 0.2 0.4 0.6 0.8 1 -80 0 0.2 0.4 0.6 0.8 1 ω/π Copyright © 2005, S. K. Mitra Impulse Responses of FIR Filters with a Smooth Transition • First-order spline passband-to-stopband transition • Pth-order spline passband-to-stopband transition ωc = (ω p + ωs ) / 2 ωc / π, ⎧ ⎪ hLP [n] = ⎨⎛ 2 sin( ∆ω n / 2 P) ⎞ P sin(ωc n) ⎪⎩⎜⎝ ∆ω n / 2 P ⎟⎠ ⋅ π n ∆ω = ωs − ω p ωc / π, ⎧⎪ hLP [n] = ⎨ 2 sin(∆ω n / 2) sin(ωc n) ⋅ πn ⎪⎩ ∆ω n -40 -60 ω/π Copyright © 2005, S. K. Mitra 21 0 -60 • Choose N = 24 implying M =12 19 Lowpass filter designed with Kaiser window Kaiser Window Gain, dB • Specifications: ω p = 0.3π , ωs = 0.5π , α s = 40 dB • Thus ωc = (ω p + ωs ) / 2 = 0.4π δ s = 10−α s / 20 = 0.01 β = 0.5842(19)0.4 + 0.07886 ×19 = 3.3953 32 N= = 22.2886 2.285(0.2π) n=0 n=0 n >0 n >0 Copyright © 2005, S. K. Mitra 22 Copyright © 2005, S. K. Mitra Lowpass FIR Filter Design Example • Example Magnitude 1 P = 1, N = 40 P = 2, N = 60 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 ω/π 23 Copyright © 2005, S. K. Mitra 4 [dsp TIPS&TRICKS] Greg Berchin Precise Filter Design Y ou have just been assigned to a new project at work, in which the objective is to replace an existing analog system with a functionally equivalent digital system. Your job is to design a digital filter that matches the magnitude and phase response of the existing system’s analog filter over a broad frequency range. You are running out of ideas. The bilinear transform and impulse invariance methods provide poor matches to the analog filter response, particularly at high frequencies. Fast convolution requires more computational resources than you have and creates more input/output latency than you can tolerate. What will you do? In this article, we describe an obscure but simple and powerful method for designing a digital filter that approximates an arbitrary magnitude and phase response. If applied to the problem above, it can create a filter roughly comparable in computational burden and latency to that created by the bilinear transform method, with fidelity approaching that of fast convolution. In addition, the method described here can also be applied to a wide variety of other system identification tasks. The filter design method we present is called frequency-domain least-squares (FDLS) [1]–[3]. The FDLS algorithm produces a transfer function that approximates an arbitrary frequency response. The input to the algorithm is a set of magnitude and phase values at a large number (typically thousands) of arbitrary frequencies between 0 Hz and half the sampling rate. The algorithm’s output is a set of transfer function coefficients. The FDLS algorithm is quite flexible in that it can create transfer functions con- taining poles and zeros (infinite response filters), only zeros (finite response filters), or only poles (autoregressive networks). The algorithm uses nothing more esoteric than basic linear algebra. Before we can see how the technique works, we need to review some basic linear algebra and matrix concepts. BACKGROUND First let us recall that, in order to uniquely solve a system of equations, we need as many equations as unknowns. For example, the single equation with one unknown 5x = 7 has the unique solution x = 7/5. But the single equation with two unknowns 5x + 2y = 7 has multiple solutions x = (7 − 2y )/5 that depend on the unspecified y-value. If another equation, −6x + 4y = 9, is added to the equation above, there are unique solutions for both x and y that can be found algebraically or by matrix inversion (denoted in the following by a “−1” superscript): 5 2 −6 4 x 7 = y 9 x 5 = y −6 1 = 38 16 7 8 = −1 7 9 1 − 16 7 5 9 32 9 − 16 . 21 45 + 16 32 2 4 Let us consider what happens if we add another equation, x + y = 5, to the pair that we already have above (we will see later why we might want to do this). There are no values of x and y that satisfy all three equations simultaneously. To address this case, matrix algebra provides the “pseudoinverse,” which determines the values of x and y that come, in the IEEE SIGNAL PROCESSING MAGAZINE [137] JANUARY 2007 least-squares sense, as close as possible to satisfying all three equations. The solution is then given by     5 2 7 x  −6 4  =  9  or y 1 1 5    −1 5 2 T 5 2 x   ≈  −6 4   −6 4  y 1 1 1 1  T   5 2 7 for THIS case: ×  −6 4   9  p_inv(X) = 1 1 5 [inv(X'X)][X'] 0.3716 ≈ , 2.8491 where T denotes the matrix transpose. Of course, the mathematical derivation of the matrix inverse and pseudoinverse, and the definition of least-squares, can be found in any basic linear algebra text [1]. And while our more mathematically inclined readers will point out that there are better ways than this to compute the pseudoinverse, this method is adequate for our example. You may also remember that filter specifications are commonly expressed in terms of passband width and flatness, transition band width, and stopband attenuation. There may also be some general specifications about phase response “DSP Tips and Tricks” introduces practical design and implementation signal processing algorithms that you may wish to incorporate into your designs. We encourage readers to submit their contributions to Associate Editors Rick Lyons (r.lyons@ieee.org) or Britt Rorabaugh (dspboss@aol.com). 1053-5888/07/$25.00©2007IEEE [dsp TIPS&TRICKS] continued or time-domain performance, but the exact magnitude and phase responses are usually left to the designer’s discretion. However, an important exception occurs when a digital filter is to be used to emulate an analog filter. This is traditionally a very difficult problem, because analog systems are described by Laplace transforms using integration and differentiation, whereas digital systems are described by z-transforms using delay. Since the conversion between them is nonlinear, the response of an analog system can only be approximated by a digital system and vice-versa. Let us assume that the transfer function of our digital filter (i.e., the mathematical description of the relationship between the filter’s input and output) is in a standard textbook form [2] given by Y(z ) b0 + b1 z −1 + · · · + bN z −N = , U(z ) 1 + a1 z −1 + · · · + aD z −D where U(z ) is the z-transform of the input signal, Y(z ) is the z-transform of the output signal, and the a and b factors are realvalued coefficients. Furthermore, we assume that the filter is causal, meaning that its response to an input does not begin until after the input is applied. Under these assumptions, the time-domain difference equation that implements our filter is y(k) = − a1 y(k − 1) − · · · − aD y(k − D) ues of which are not yet known. We also know that the relationship between input u and output y at any sample time can be inferred from the frequency response value A φ at frequency ω. Combining these two ideas, we obtain one equation in D + N + 1 unknowns Let us conclude our background section with a comment on what a frequency response value means. In a simple example, if the frequency response of a system at a frequency ω1 is given in magnitude/phase form as A1 φ1 , the output amplitude will be A1 times the input amplitude and the output phase will be shifted an angle φ1 relative to the input phase when a steady-state sine wave of frequency ω1 is applied to the system. For instance, if the input to the system described above at time k is u1 (k) = cos(kω1 ts ), where ts is the sampling period (equal to one over the sampling frequency), then the output will be y1 (k) = A1 cos(kω1 ts + φ1 ) . The input and output values at any sample time can be determined in a similar manner. For example, the input sample value N samples in the past was u1 (k − N) = cos ((k − N )ω1 ts ) and the output sample value D samples in the past was y1 (k − D) = A1 cos ((k − D) ω1 ts + φ1 ). For our purposes, since k represents the current sample time, its value can conveniently be set to zero. y1 (0) = [−y1 (−1) . . . − y1 (−D) a  1  ..   .     aD   u1 (0) . . . u1 (−N )]   b0  .    .   ..  bN (Note that the current-sample index k has been set to zero.) If we repeat using A2 φ2 at a different frequency ω2 , we obtain a second equation in D + N + 1 unknowns as shown in (a) at the bottom of the page. And if we repeat at many more different frequencies M than we have unknowns D + N + 1, we know from our review of linear algebra that the pseudoinverse will compute values for the set of coefficients a1 . . . aD and b0 . . . bN that come as close as possible to solving all of the equations, which is exactly what we need to design our filter. So now we can write (b), shown at the bottom of the page. We can denote the y1 (0) . . . yM (0) column vector above as Y, the matrix as X, and the a1 . . . bN col- FDLS FILTER APPROXIMATION Based on our review of the pseudoinverse, transfer function, and frequency response, we know that the output is a combination of present and past input and output values, each scaled by a set of b or a coefficients (respectively), the val- + b0 u(k) + · · · + bN u(k − N ), where the a and b coefficients are exactly the same as in the transfer function above, k is the time index, u(k) and y(k) are the current values of the input and output (respectively), u(k − N ) was the input value N samples in the past, and y(k − D) was the output value D samples in the past. We can write the equation above in matrix form as y(k) = [−y(k − 1) . . . − y(k − D) a  1  ..   .     aD   u(k) . . . u(k − N )]   b0  .    .   ..  −y1 (−1) . . . y1 (0) = y2 (0) −y2 (−1) . . . −y1 (−D) −y2 (−D) u1 (0) . . . u2 (0) . . . a  1  ..   .     u1 (−N)   aD  .  u2 (−N)  b 0    .   ..  (a) bN    −y1 (−1) . . . y1 (0)  y2 (0)   −y2 (−1) . . .     ..  =  ..  .   . yM (0) −yM (−1) . . . −y1 (−D) −y2 (−D) .. . u1 (0) u2 (0) .. . −yM (−D) uM (0) . . . bN IEEE SIGNAL PROCESSING MAGAZINE [138] JANUARY 2007 ... ... a  1   ..  u1 (−N)  .    u2 (−N)  aD     ..  b0  .    .  .  uM (−N)  ..  bN (b) We can now summarize our filter design trick as follows: 1) Select the numerator order N and the denominator order D, where N and D do not have to be equal and either one (but not both) may be zero. (We have found no “rule of thumb” for defining N and D; they are best determined experimentally.) 2) Define the M separate input um cosine sequences, each of length (N + 1). 3) Compute the M separate output ym cosine sequences, each of length D (based on Am φ m ). 4) Fill the X matrix with the input um and output ym cosine sequences. 5) Fill the Y vector with the M output cosine values, ym (0) = Am cos(φ m ). 6) Compute the pseudoinverse; the resulting vector contains the filter coefficients. A numerical example is shown in Figures 1 and 2, which illustrate the magnitude and phase, respectively, of a real-world example analog system (black) and of the associated bilinear transform (blue), impulse invariance (green), and FDLS (red) approximations. The sampling rate is equal to 240 Hz and D = N = 12. The red FDLS graphs are almost completely obscured by the black analog system graphs. In this example, the FDLS errors are often three to four orders of magnitude smaller than those of the other methods. (In Figure 2, the bilinear transform curve is obscured by the FDLS and analog curves at low frequencies and by the impulse invariance curve at high frequencies.) In terms of the computational complexity of an FDIS-designed filter, the number of feedback and feed-forward coefficients is determined by the variables D and N, respectively. As such, an FDIS-designed filter requires (N + D + 1) multiplies and (N + D) additions per filter output sample. Magnitude (dB) (X T X )−1 X T Y ≈ . 0 −10 −20 −30 −40 Black: Analog (Desired) Blue: Bilinear Transform Green: Impulse Invariance Red: FDLS −50 10−2 10−1 100 Frequency (Hz) 101 102 [FIG1] Magnitude responses of the filter designed using the bilinear transform, impulse invariance, and FDLS methods. 150 Phase (degrees) umn vector as . With these notations, Y = X, and the pseudoinverse solves for the vector that contains the desired filter coefficients 100 50 0 −50 −100 −150 10−2 Black: Analog (Desired) Blue: Bilinear Transform Green: Impulse Invariance Red: FDLS 10−1 100 Frequency (Hz) 101 102 [FIG2] Phase responses of the filter designed using the bilinear transform, impulse invariance, and FDLS methods. CONCLUSION FDLS is a powerful method for designing digital filters. As is the case with all approximation techniques, there are circumstances in which the FDLS method works well and others in which it does not. The FDLS method does not replace other filter design methods; it provides one more method from which to choose. FDLS is most useful in cases where a specified frequency response must be duplicated to within tight tolerances over a wide frequency range or when the frequency response of an existing system is known but the coefficients of the system’s transfer function are unknown. It is up to the designer to determine whether to use it in any given situation. Detailed examples and a MATLAB code implementation of the FDLS algorithm are available at http://apollo.ee.columbia. edu/spm/?i=external/tipsandtricks. IEEE SIGNAL PROCESSING MAGAZINE [139] JANUARY 2007 ACKNOWLEDGMENTS My thanks to Jaime Andrés Aranguren Cardona for providing the example shown in Figures 1 and 2, originally posted on the comp.dsp Newsgroup on 9 January 2005, which ultimately led to this article. AUTHOR Greg Berchin (berchin@ieee.org) is a signal processing algorithm engineer who provides contract engineering services from Naperville, Illinois. REFERENCES [1] G. Strang, Linear Algebra and Its Applications, 2nd ed., Orlando, FL: Academic, pp. 103–152, 1980. [2] R. Lyons, Understanding Digital Signal Processing, 2nd ed., Upper Saddle River, NJ: Prentice-Hall, pp. 232–240, 2004. [3] G. Berchin, “A new algorithm for system identification from frequency response information,” master’s thesis, University of California-Davis, 1988 . [4] G. Berchin and M.A. Soderstrand, “A transformdomain least-squares beamforming technique,” in Proc. IEEE Oceans ‘90 Conf., Arlington, VA, Sept. 1990. [SP] FDLS Examples by Greg Berchin and Richard Lyons [January 2007] The following material provides additional examples of the FDLS algorithm described in the January 2007 IEEE Signal Processing magazine DSP Tips & Tricks column article "Precise Filter Design" by Greg Berchin. Algebraic Example Recall the FDLS matrix expression ⎡ a1 ⎤ ⎢ ⎥ ⎡ y1 (0) ⎤ ⎡ − y1 ( −1) K − y1 ( − D ) u1 (0) K u1 ( − N ) ⎤ ⎢ M ⎥ ⎢ y (0) ⎥ ⎢ − y ( −1) K − y ( − D ) u (0) K u ( − N ) ⎥ ⎢a ⎥ 2 2 2 2 ⎥⎢ D ⎥ , ⎢ 2 ⎥=⎢ ⎥ ⎢ b0 ⎥ M M M M ⎢ M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ y M (0)⎦ ⎣ − y M ( −1) K − y M ( − D ) u M (0) K u M ( − N )⎦ ⎢ M ⎥ ⎢ ⎥ ⎣ bN ⎦ which we wrote as Y = XΘ. Each individual element in the Y column vector is of the form Amcos(φm), and each element in the X matrix is of the form A1cos(kω1ts + φ1) or cos(kω1ts). Because all of these elements are of the form of a product (Amplitude)[cos(angle)], each element in Y and X is equal to a constant. Now if, say, D = 10 and N = 9, then: y1(0) = A1cos(φ1) is the first element of the Y column vector and [–y1(–1) ... –y1(–10) u1(0) ... u1(–9)] is the top row of the X matrix expression where –y1(–1) = –A1cos[(–1)ω1ts + φ1] = –A1cos(–ω1ts + φ1) –y1(–2) = –A1cos[(–2)ω1ts + φ1] = –A1cos(–2ω1ts + φ1) ... –y1(–10) = –A1cos[(–10)ω1ts + φ1] = –A1cos(–10ω1ts + φ1) and u1(0) = cos[(0)ω1ts] = 1 u1(–1) = cos[(–1)ω1ts] = cos(–ω1ts) u1(–2) = cos[(–2)ω1ts] = cos(–2ω1ts) ... u1(–9) = cos[(–9)ω1ts] = cos(–9ω1ts). IEEE Copyright © All Rights Reserved Page 1 of 5 So the top row of the X matrix looks like: [–A1cos(–ω1ts + φ1) –A1cos(–2ω1ts + φ1) ... –A1cos(–10ω1ts + φ1) 1 cos(–ω1ts) cos(–2ω1ts) ... cos(–9ω1ts)]. The second row of the X matrix looks like: [–A2cos(–ω2ts + φ2) –A2cos(–2ω2ts + φ2) ... –A2cos(–10ω2ts + φ2) 1 cos(–ω2ts) cos(–2ω2ts) ... cos(–9ω2ts)]. And so on. Numerical Example Here’s an example of the above expressions using actual numbers. Suppose we need to approximate the transfer function coefficients for the system whose frequency magnitude and phase response is that shown in Figure E1. Assume that our discretesystem sample rate is 1000 Hz, thus ts = 10–3 seconds, and N = D = 2. Also assume M = 8 and we have the eight A1 -to- A8 magnitude sample values and the eight φ1 -to- φ8 phase samples, shown as dots in Figure E1, available to us as input values to the FDLS algorithm. M=8 ωm = 2π . fm : 1.5 A5 Linear 1 Desired magnitude response 0.5 A1 (a) A8 0 0 100 f1 = 0 Hz 200 300 Frequency f7 = 106.3 Hz 400 500 ω1 = 2π(0.0) ω2 = 2π(19.69) ω3 = 2π(35.43) ω4 = 2π(51.18) ω5 = 2π(59.05) ω6 = 2π(66.93) ω7 = 2π(106.30) ω8 = 2π(389.76) f8 = 389.76 Hz f4 = 51.18 Hz (b) Phase (radians) 4 φ5 2 φ1 Desired phase response 0 –2 φ8 0 100 200 300 Frequency 400 500 Fig E1 IEEE Copyright © All Rights Reserved Page 2 of 5 In matrix form, the target analog system parameters are ┌ │0.0 │19.6850 │35.4331 fm = │51.1811 │59.0551 │66.9291 │106.299 │389.764 └ ┐ ┌ ┐ │ │0.0 │ │ │0.1237 │ │ │0.2226 │ │ ωmts = │0.3216 │ │ │0.3711 │ │ │0.4205 │ │ │0.6679 │ │ │2.449 │ ┘ └ ┘ ┌ ┐ │0.2172 │ │0.2065 │ │0.1696 │ Am = │0.0164 │ │1.3959 │ │0.6734 │ │0.3490 │ │0.3095 │ └ ┘ ┌ │ 0.0 │–0.0156 │–0.0383 φm = │ 3.0125 │ 2.3087 │ 0.955 │ 0.0343 │ 0.0031 └ ┐ │ │ │ │ │ │ │ │ ┘ where the fm vector is in Hz, the ωmts vector is in radians, and 1 ≤ m ≤ 8. The first two elements of the Y vector are: y1(0) = A1cos(φ1) = 0.2172cos(0) = 0.2172. y2(0) = A2cos(φ2) = 0.2065cos(–0.0156) = 0.2065. The complete Y vector is: Y = ┌ │A1cos(φ1) │A2cos(φ2) │A3cos(φ3) │A4cos(φ4) │A5cos(φ5) │A6cos(φ6) │A7cos(φ7) │A8cos(φ8) └ ┐ ┌ ┐ │ │ 0.2172 │ │ │ 0.2065 │ │ │ 0.1695 │ │ = │–0.0162 │ │ │–0.9390 │ │ │ 0.6605 │ │ │ 0.3488 │ │ │ 0.3095 │ ┘ └ ┘ The two elements of the "y1" part of the first row of the X vector are: –y1(–1) = –A1cos(–ω1ts + φ1) = –0.2172cos(–0 + 0) = –0.2172. –y1(–2) = –A1cos(–2ω1ts + φ1) = –0.2172cos(–0 + 0) = –0.2172. The two elements of the "y8" part of the eighth row of the X vector are: –y8(–1) = –A8cos(–ω8ts + φ8) = –0.3095cos(–2.449 + 0.0031) = 0.2376. –y8(–2) = –A8cos(–2ω8ts + φ8) = –0.3095cos(–4.898 + 0.0031) = –0.562. IEEE Copyright © All Rights Reserved Page 3 of 5 The three elements of the "u1" part of the first row of the X matrix are: u1(0) = cos(0) = 1 u1(–1) = cos(–ω1ts) = cos(–0) = 1 u1(–2) = cos(–2ω1ts) = cos(–0) = 1. The three elements of the "u8" part of the eighth row of the X matrix are: u8(0) = cos(0) = 1 u8(–1) = cos(–ω8ts) = cos(–2.449) = –0.7696 u8(–2) = cos(–2ω8ts) = cos(–4.898) = 0.1845. The complete X matrix is: ┌ │–A1cos(–1ω1ts+φ1) │–A2cos(–1ω2ts+φ2) │–A3cos(–1ω3ts+φ3) X = │–A4cos(–1ω4ts+φ4) │–A5cos(–1ω5ts+φ5) │–A6cos(–1ω6ts+φ6) │–A7cos(–1ω7ts+φ7) │–A8cos(–1ω8ts+φ8) └ ┌ │–0.2172 │–0.2045 │–0.1639 = │ 0.0147 │ 0.5007 │–0.6564 │–0.2812 │ 0.2376 └ –0.2172 –0.994 –0.1502 0.0117 –0.0059 –0.5378 –0.0928 –0.0562 –A1cos(–2ω1ts+φ1) –A2cos(–2ω2ts+φ2) –A3cos(–2ω3ts+φ3) –A4cos(–2ω4ts+φ4) –A5cos(–2ω5ts+φ5) –A6cos(–2ω6ts+φ6) –A7cos(–2ω7ts+φ7) –A8cos(–2ω8ts+φ8) 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.0000 0.9924 0.9753 0.9487 0.939 0.9129 0.7851 –0.7696 cos(0) cos(0) cos(0) cos(0) cos(0) cos(0) cos(0) cos(0) cos(–ω1ts) cos(–ω2ts) cos(–ω3ts) cos(–ω4ts) cos(–ω5ts) cos(–ω6ts) cos(–ω7ts) cos(–ω8ts) cos(–2ω1ts) cos(–2ω2ts) cos(–2ω3ts) cos(–2ω4ts) cos(–2ω5ts) cos(–2ω6ts) cos(–2ω7ts) cos(–2ω8ts) ┐ │ │ │ │ │ │ │ │ ┘ ┐ 1.0000 │ 0.9696 │ 0.9025 │ 0.8002 │. 0.7370 │ 0.6667 │ 0.2328 │ 0.1845 │ ┘ Given the above Y vector and the X matrix, the FDLS algorithm computes the 2nd-order (N = D = 2) transfer function coefficients vector θM=8 as θM=8 = ┌ │–1.8439 │ 0.9842 │ 0.3033 │–0.5762 │ 0.3034 └ ┐ │ │ │. │ │ ┘ IEEE Copyright © All Rights Reserved Page 4 of 5 Treated as filter coefficients, we can write vector θM=8 as: a0 = 1 a1 = –1.8439 a2 = 0.9842 b0 = 0.3033 b1 = –0.5762 b2 = 0.3034 implemented as the recursive filter network shown in Figure E2. u(k) y(k) z –1 1.8439 z –1 0.3033 –0.5762 0.3034 –0.9842 Fig E2 (a) Magnitude (dB) The frequency-domain performance of the filter are the solid red curves shown in Figure E3. There we see that the θM=8 coefficients provide an accurate approximation to the desired frequency response in Figure E1. 0 –20 Desired magnitude response Actual magnitude response –40 0 100 200 300 Frequency 400 500 400 500 (b) Phase (radians) 4 2 Desired phase response Actual phase response 0 –2 0 100 200 300 Frequency Fig E3 IEEE Copyright © All Rights Reserved Page 5 of 5

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