Analog Lowpass Filter Specifications

Analog Lowpass Filter Analog Lowpass Filter Specifications Specifications • Typical magnitude response H a ( j Ω ) of an •In the passband, defined by 0 ≤ Ω ≤ Ω p , we analog lowpass filter may be given as require indicated below 1− δ p ≤ Ha ( jΩ) ≤1+ δ p , Ω ≤ Ω p

i.e., H a ( j Ω ) approximates unity within an error of ± δ p

•In the stopband, defined by Ω s ≤ Ω ≤ ∞ , we require Ha ( jΩ) ≤ δs , Ωs ≤ Ω ≤ ∞

Analog Lowpass Filter Analog Lowpass Filter Specifications Specifications

• Ω p - passband edge frequency • Magnitude specifications may alternately be

•-Ωs stopband edge frequency given in a normalized form as indicated •-δ p peak ripple value in the passband below

•-δs peak ripple value in the stopband • Peak passband ripple

α p = −20log10 (1− δ p ) dB • Minimum stopband attenuation

Analog Lowpass Filter Analog Lowpass Filter Design Specifications • Two additional parameters are defined - • Here, the maximum value of the magnitude in the passband assumed to be unity Ω (1) Transition ratio k = p Ωs •-1/ 1+ ε 2 Maximum passband deviation, given by the minimum value of the For a lowpass filter k <1 magnitude in the passband ε (2) Discrimination parameter k = 1 A2 −1 •-1 Maximum stopband magnitude Usually k <<1 A 1 5 6 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

1 Butterworth Approximation Butterworth Approximation • The magnitude-square response of an N-th order analog lowpass Butterworth filter 2 is given by • Gain in dB is G(Ω) =10log10 Ha ( jΩ) 2 1 Ha ( jΩ) = 2N 1+ (Ω/ Ωc ) • As G ( 0 ) = 0 and 2 • First 2 N − 1 derivatives of H a ( j Ω ) at Ω = 0 G(Ωc ) =10log10 (0.5) = −3.0103 ≅ −3dB

are equal to zero Ωc is called the 3-dB cutoff frequency • The Butterworth lowpass filter thus is said to have a maximally-flat magnitude at Ω = 0 7 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 Filter Butterworth lowpass filter are Ω c and N N = 2 • These are determined from the specified 1 N = 4 0.8 N = 10 bandedgesΩ p and Ω s , and minimum 2

nitude 0.6

Butterworth Approximation Butterworth Approximation • Since order N must be an integer, value • Ω and N are thus determined from c obtained is rounded up to the next highest 2 1 1 Ha ( jΩ p ) = = integer 1+ (Ω / Ω )2N 1+ ε 2 p c • This value of N is used next to determine Ω 2 1 1 c H ( jΩ ) = = by satisfying either the stopband edge or the a s 1+ (Ω / Ω )2N A2 s c passband edge specification exactly • Solving the above we get • If the stopband edge specification is log [(A2 −1)/ε 2 ] log (1/ k ) satisfied, then the passband edge N = 1⋅ 10 = 10 1 specification is exceeded providing a safety 2 log (Ω / Ω ) log (1/ k) 10 s p 10 margin 11 12 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

2 Butterworth Approximation Butterworth Approximation •Example- Determine the lowest order of a • Transfer function of an analog Butterworth Butterworth lowpass filter with a 1-dB cutoff lowpass filter is given by frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHz ΩN ΩN H (s) = C = c = c •Now a D (s) N N −1 l N ⎛ 1 ⎞ N s + ∑ =0 d s ∏ =1(s − p ) 10log ⎜ ⎟ = −1 l l l l 10⎝1+ ε 2 ⎠ where which yields ε 2 = 0.25895 j[π (N +2l−1) / 2N ] p = Ωce , 1≤ l ≤ N l and ⎛ 1 ⎞ • Denominator D ( s ) is known as the 10log ⎜ ⎟ = −40 N 10 A2 Butterworth polynomial of order N ⎝ ⎠ which yields A2 =10,000 13 14 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Butterworth Approximation Chebyshev Approximation 2 • The magnitude-square response of an N-th • Therefore 1 = A −1 =196.51334 k ε order analog lowpass Type 1 Chebyshev filter 1 is given by 1 Ωs and = = 5 2 1 k H (s) = Ω p a 2 2 1+ ε TN (Ω / Ω p )

• Hence where T N ( Ω ) is the Chebyshev polynomial log (1/ k ) of order N: N = 10 1 = 3.2811 log10 (1/ k) ⎧ cos(N cos−1 Ω), Ω ≤1 TN (Ω) = ⎨ −1 • We choose N = 4 ⎩cosh(N cosh Ω), Ω >1 15 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, • Typical magnitude response plots of the then analog lowpass Type 1 Chebyshev filter are 2 1 1 shown below Ha ( jΩs ) = 2 2 = 2 Type 1 Chebyshev Filter 1+ ε TN (Ωs / Ω p ) A

N = 2 1 N = 3 • Solving the above we get 0.8 N = 8 −1 2 −1 cosh ( A −1/ε) cosh (1/ k1)

nitude 0.6 g N = −1 = −1

3 Chebyshev Approximation Chebyshev Approximation • The magnitude-square response of an N-th • Typical magnitude response plots of the order analog lowpass Type 2 Chebyshev analog lowpass Type 2 Chebyshev filter are (also called inverse Chebyshev) filter is shown below given by Type 2 Chebyshev Filter 2 1 N = 3 H ( jΩ) = 1 N = 5 a 2 0.8 N = 7 2 ⎡TN (Ωs / Ω p )⎤

Chebyshev Approximation Elliptic Approximation • The order N of the Type 2 Chebyshev filter • The square-magnitude response of an is determined from given ε , , and A elliptic lowpass filter is given by Ωs using 2 1 −1 2 −1 cosh ( A −1/ε) cosh (1/ k ) Ha ( jΩ) = 2 2 1 1+ ε R (Ω/ Ω ) N = −1 = −1 N p cosh (Ωs / Ω p ) cosh (1/ k) where R N ( Ω ) is a rational function of order •Example- Determine the lowest order of a N satisfying R ( 1 / Ω ) = 1 / R ( Ω ) , with the Chebyshev lowpass filter with a 1-dB cutoff N N frequency at 1 kHz and a minimum attenuation of roots of its numerator lying in the interval 40 dB at 5 kHz - 0 < Ω <1 and the roots of its denominator −1 lying in the interval 1< Ω < ∞ cosh (1/ k1) N = −1 = 2.6059 21 cosh (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 • Example - Determine the lowest order of a elliptic can be estimated using lowpass filter with a 1-dB cutoff frequency at 1 2 log (4/ k ) kHz and a minimum attenuation of 40 dB at 5 kHz N 10 1 ≅ 1/ k =196.5134 log10 (1/ ρ) Note: k = 0.2 and 1 where k'= 1− k 2 • Substituting these values we get k'= 0.979796, ρ0 = 0.00255135, ρ = 1− k' 0 2(1+ k') ρ = 0.0025513525 5 9 13 ρ = ρ0 + 2(ρ0 ) +15(ρ0 ) +150(ρ0 ) • and hence N = 2.23308 • Choose N = 3 23 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 • Typical magnitude response plots with Ω p = 1 frequency at 1 kHz and a minimum are shown below attenuation of 40 dB at 5 kHz Elliptic Filter • Code fragments used N = 3 1 N = 4 [N, Wn] = ellipord(Wp, Ws, Rp, Rs, ‘s’); 0.8 [b, a] = ellip(N, Rp, Rs, Wn, ‘s’);

nitude 0.6 g

Design of Analog Highpass, Analog Lowpass Filter Design Bandpass and Bandstop Filters • Steps involved in the design process: • Gain plot Step 1 - Develop of specifications of a Lowpass Elliptic Filter prototype analog lowpass filter H L P ( s ) 0 from specifications of desired analog filter

-20 HD (s) using a frequency transformation Step 2 - Design the prototype analog Gain, dB -40 lowpass filter

-60 Step 3 - Determine the transfer function H (s) 0 2000 4000 6000 D Frequency, Hz of desired analog filter by applying the inverse frequency transformation to H (s) 27 28 LP Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Design of Analog Highpass, Analog Highpass Filter Design Bandpass and Bandstop Filters •Let s denote the Laplace transform variable • Spectral Transformation: of prototype analog lowpass filter H (s) Ω Ωˆ LP s = p p and sˆ denote the Laplace transform sˆ variable of desired analog filter H (sˆ) D where Ω is the passband edge frequency of • The mapping from s-domain to -domain is p sˆ H (s) and Ωˆ is the passband edge given by the invertible transformation LP p frequency of H (sˆ) s = F(sˆ) HP • On the imaginary axis the transformation is •Then H (sˆ) = H (s) D LP s=F (sˆ) ˆ Ω pΩ p H (s) = H (sˆ) −1 Ω = − LP D sˆ=F (s) Ωˆ 29 30 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

5 Analog Highpass Filter Design Analog Highpass Filter Design •Example- Design an analog Butterworth ˆ highpass filter with the specifications: Ω pΩ p Ω = − ˆ kHz, ˆ kHz, α = 0 . 1 dB, ˆ Fp = 4 Fs =1 p Ω α = 40dB Ω s Stopband Stopband − Ωs − Ω p Ω p Ωs • Choose Ω p =1 Ω Lowpass 0 Passband •Then 2πFˆ Fˆ Ω = p = p = 4000 = 4 s ˆ ˆ 1000 Passband Stopband Passband 2πFs Fs Ωˆ Highpass ˆ ˆ 0 ˆ ˆ • Analog lowpass filter specifications: Ω = 1 , − Ω p − Ωs Ωs Ω p p Ω = 4, α = 0 . 1 dB, α = 40 dB 31 32 s p s Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Analog Highpass Filter Design Analog Bandpass Filter • Code fragments used Design [N, Wn] = buttord(1, 4, 0.1, 40, ‘s’); • Spectral Transformation [B, A] = butter(N, Wn, ‘s’); sˆ2 + Ωˆ 2 s = Ω o [num, den] = lp2hp(B, A, 2*pi*4000); p ˆ ˆ sˆ(Ω p2 − Ω p1) • Gain plots Prototype Lowpass Filter Highpass Filter where Ω p is the passband edge frequency 0 0 ˆ ˆ of HLP (s) , and Ω p1 and Ω p 2 are the lower -20 -20 and upper passband edge frequencies of -40 -40

Gain, dB Gain, dB desired bandpass filter HBP (sˆ) -60 -60

Analog Bandpass Filter Analog Bandpass Filter Design Design • On the imaginary axis the transformation is Ωˆ 2 − Ωˆ 2 Ωˆ 2 − Ωˆ 2 Ω = −Ω o Ω = −Ω o p ˆ p ˆ Ω Bw Ω Bw Stopband Ω Stopband ˆ ˆ − Ωs − Ω p p Ωs where B w = Ω p 2 − Ω p 1 is the width of Ω Lowpass 0 ˆ Passband passband and Ω o is the passband center frequency of the bandpass filter Stopband Passband Stopband Passband Stopband ˆ Bandpass • Passband edge frequency ± Ω is mapped ˆ 0 Ω p −Ωˆ ↓ − Ωo ↓ −Ωˆ Ωˆ ↓ Ωˆ ↓ Ωˆ s2 ˆ s1 s1 o s2 ˆ ˆ −Ω p2 −Ωˆ Ωˆ ˆ into m Ω p1 and± Ω p2 , lower and upper p1 p1 Ω p2 passband edge frequencies 35 36 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

6 Analog Bandpass Filter Analog Bandpass Filter Design Design • Stopband edge frequency ± Ω is mapped ˆ ˆ ˆ ˆ s • Case 1: Ω p1Ω p2 > Ωs1Ωs2 into Ωˆ and± Ωˆ , lower and upper m s1 s2 To make Ω ˆ Ω ˆ = Ω ˆ Ω ˆ we can either stopband edge frequencies p1 p2 s1 s2 increase any one of the stopband edges or •Also, decrease any one of the passband edges as ˆ 2 ˆ ˆ ˆ ˆ Ωo = Ω p1Ω p2 = Ωs1Ωs2 shown below • If bandedge frequencies do not satisfy the ← Passband ← above condition, then one of the frequencies

Analog Bandpass Filter Analog Bandpass Filter Design Design ˆ ˆ ˆ ˆ ˆ 2 ˆ ˆ ˆ ˆ (1) Decrease Ω p 1 to Ωs1Ωs2 / Ω p2 •Note: The condition Ω o = Ω p 1 Ω p 2 = Ω s 1 Ω s2 ˆ larger passband and shorter can also be satisfied by decreasing Ω p 2 leftmost transition band which is not acceptable as the passband is ˆ ˆ ˆ ˆ reduced from the desired value (2) Increase Ω s 1 to Ω p1Ω p2 / Ωs2 No change in passband and shorter • Alternately, the condition can be satisfied ˆ leftmost transition band by increasing Ω s 2 which is not acceptable as the upper stop band is reduced from the desired value 39 40 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Analog Bandpass Filter Analog Bandpass Filter Design Design ˆ ˆ ˆ ˆ Design • Case 2: Ω p1Ω p2 < Ωs1Ωs2 To make Ω ˆ Ω ˆ = Ω ˆ Ω ˆ we can either p1 p2 s1 s2 ˆ ˆ ˆ ˆ decrease any one of the stopband edges or (1) Increase Ω p 2 to Ωs1Ωs2 / Ω p1 increase any one of the passband edges as larger passband and shorter shown below rightmost transition band ˆ ˆ ˆ ˆ → Passband → (2) Decrease Ω s 2 to Ω p1Ω p2 / Ωs1 No change in passband and shorter ← Stopband ← Stopband rightmost transition band Ωˆ ˆ ˆ ˆ ˆ Ωs1 Ω p1 Ω p2 Ωs2 41 42 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

7 Analog Bandpass Filter Analog Bandpass Filter Design Design Design •Example- Design an analog elliptic ˆ 2 ˆ ˆ ˆ ˆ •Note: The condition Ω o = Ω p 1 Ω p 2 = Ω s 1 Ω s2 bandpass filter with the specifications: ˆ ˆ ˆ ˆ can also be satisfied by increasing Ω p 1 Fp1 = 4 kHz, F p 2 = 7 kHz, F s 1 = 3 kHz ˆ which is not acceptable as the passband is Fs2 = 8 kHz, α p = 1 dB, α s = 22 dB reduced from the desired value • Alternately, the condition can be satisfied ˆ ˆ 6 ˆ ˆ 6 • Now F p 1 F p 2 = 28 × 10 and Fs1Fs2 = 24×10 by decreasing ˆ which is not acceptable Ωs1 • Since F ˆ F ˆ > F ˆ F ˆ we choose as the lower stopband is reduced from the p1 p2 s1 s2 ˆ ˆ ˆ ˆ desired value Fp1 = Fs1Fs2 / Fp2 = 3.571428 kHz 43 44 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Analog Bandpass Filter Analog Bandpass Filter Design Design • Code fragments used [N, Wn] = ellipord(1, 1.4, 1, 22, ‘s’); • We choose Ω p =1 [B, A] = ellip(N, 1, 22, Wn, ‘s’); • Hence [num, den] 24 − 9 = lp2bp(B, A, 2*pi*4.8989795, 2*pi*25/7); Ω = =1.4 s (25/ 7)×3 • Gain plot Prototype Lowpass Filter Bandpass Filter 0 0 • Analog lowpass filter specifications: Ω p = 1 , -20 -20 Ωs =1.4, α p = 1 dB, α s = 22 dB Gain, dB -40 Gain, dB -40

Analog Bandstop Filter Design Analog Bandstop Filter Design Analog Bandstop Filter Design • On the imaginary axis the transformation is • Spectral Transformation ˆ ΩBw ˆ ˆ Ω = Ωs sˆ(Ωs2 − Ωs1) Ωˆ 2 − Ωˆ 2 s = Ωs o 2 ˆ 2 ˆ ˆ sˆ + Ω o where B w = Ω s 2 − Ω s 1 is the width of ˆ where Ω s is the stopband edge frequency stopband and Ω o is the stopband center ˆ ˆ of HLP (s) , and Ωs1 and Ω s 2 are the lower frequency of the bandstop filter and upper stopband edge frequencies of the • Stopband edge frequency ± Ω s is mapped ˆ ˆ desired bandstop filter HBS (sˆ) into m Ωs1 and± Ωs2 , lower and upper stopband edge frequencies 47 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 •Also, ˆ ˆ ˆ 2 ˆ ˆ ˆ ˆ into m Ω p1 and± Ω p2 , lower and upper Ωo = Ω p1Ω p2 = Ωs1Ωs2 passband edge frequencies • If bandedge frequencies do not satisfy the Stopband Stopband − Ωs − Ω p Ω p Ωs above condition, then one of the frequencies Ω Lowpass 0 Passband needs to be changed to a new value so that the condition is satisfied Passband StopbandPassband Stopband Passband Ωˆ Bandpass ˆ 0 ˆ ˆ −Ωˆ ↓ − Ωo ↓ −Ωˆ Ω p1↓ Ωo ↓ Ωˆ p2 ˆ ˆ p1 ˆ ˆ p2 −Ωs2 −Ωs1 Ωs1 Ωs2

Analog Bandstop Filter Design Analog Bandstop Filter Design ˆ ˆ ˆ ˆ • Case 1: Ω p1Ω p2 > Ωs1Ωs2 • To make Ω ˆ Ω ˆ = Ω ˆ Ω ˆ we can either p1 p2 s1 s2 ˆ ˆ ˆ ˆ increase any one of the stopband edges or (1) Decrease Ω p 2 to Ωs1Ωs2 / Ω p2 decrease any one of the passband edges as larger high-frequency passband shown below and shorter rightmost transition band ˆ ˆ ˆ ˆ (2) Increase Ω s 2 to Ω p1Ω p2 / Ωs2 No change in passbands and Passband Passband shorter rightmost transition band Stopband Ωˆ ˆ ˆ ˆ ˆ 51 Ω p1 Ωs1 Ωs2 Ω p2 52 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

Analog Bandstop Filter Design Analog Bandstop Filter Design •Note: The condition ˆ 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Ωo = Ω p1Ω p2 = Ωs1Ωs2 • Case 1: Ω p1Ω p2 < Ωs1Ωs2 can also be satisfied by decreasing • To make Ω ˆ Ω ˆ = Ω ˆ Ω ˆ we can either ˆ p1 p2 s1 s2 which is not acceptable as the low- Ω p1 decrease any one of the stopband edges or frequency passband is reduced from the increase any one of the passband edges as desired value shown below • Alternately, the condition can be satisfied ˆ by increasing Ω s 1 which is not acceptable as the stopband is reduced from the desired Passband Passband Stopband value Ωˆ ˆ ˆ ˆ ˆ 53 54 Ω p1 Ωs1 Ωs2 Ω p2 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra

9 Analog Bandstop Filter Design Analog Bandstop Filter Design

ˆ 2 ˆ ˆ ˆ ˆ •Note: The condition Ω o = Ω p 1 Ω p 2 = Ω s 1 Ω s2 ˆ ˆ ˆ ˆ can also be satisfied by increasing Ω ˆ (1) Increase Ω p 1 to Ωs1Ωs2 / Ω p1 p2 larger passband and shorter which is not acceptable as the high- leftmost transition band frequency passband is decreased from the desired value (2) Decrease Ω ˆ to Ωˆ Ωˆ / Ωˆ s1 p1 p2 s1 • Alternately, the condition can be satisfied No change in passbands and ˆ by decreasing Ω s 2 which is not acceptable shorter leftmost transition band as the stopband is decreased 55 56 Copyright © 2005, S. K. Mitra Copyright © 2005, S. K. Mitra