Lecture 20

Digital Signal Processing

Analog Filter Design

Analog Lowpass Filter Specifications • Typical magnitude response H a ( jΩ) of an analog lowpass filter may be given as indicated below

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Copyright © 2010, S. K. Mitra

Analog Lowpass Filter Specifications • 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 2

i.e., H a ( jΩ) approximates zero within an error of δ s Copyright © 2010, S. K. Mitra

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

• Minimum stopband attenuation α s = − 20 log10 (δ s ) dB 3 3

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Analog Lowpass Filter Specifications • Magnitude specifications may alternately be given in a normalized form as indicated below

4 4

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Analog Lowpass Filter Specifications • Here, the maximum value of the magnitude in the passband assumed to be unity •

1 / 1 + ε - Maximum passband deviation, given by the minimum value of the magnitude in the passband 2

• 1 - Maximum stopband magnitude A 5 5

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Analog Lowpass Filter Design • Two additional parameters are defined -

Ωp (1) Transition ratio k = Ωs For a lowpass filter k < 1 (2) Discrimination parameter k1 = Usually k1 << 1

ε A −1 2

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Butterworth Approximation

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• 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 Copyright © 2010, S. K. Mitra

Butterworth Approximation • Gain in dB is G (Ω) = 10 log10 H a ( jΩ)

2

• As G (0) = 0 and G (Ω c ) = 10 log10 (0.5) = −3.0103 ≅ −3 dB Ωc is called the 3-dB cutoff frequency 8 8

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Butterworth Approximation • Typical magnitude responses with Ω c = 1 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 9

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Butterworth Approximation • 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

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Butterworth Approximation • Ωc and N are thus determined from 2 1 1 = H a ( jΩ p ) = 2N 1 + (Ω p / Ω c ) 1+ ε 2 2 1 1 H a ( jΩ s ) = = 2 2N A 1 + (Ω s / Ω c ) • Solving the above we get log10 [( A − 1) / ε ] log10 (1 / k1 ) 1 N= ⋅ = 2 log10 (Ω s / Ω p ) log10 (1 / k ) 2

2

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Butterworth Approximation

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• 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 © 2010, S. K. Mitra

Butterworth Approximation • Transfer function of an analog Butterworth lowpass filter is given by N N Ωc Ωc C = N = H a ( s) = l DN ( s ) s + ∑lN=−01d s ∏lN=1 ( s − pl ) l where j[π ( N + 2 l −1) / 2 N ]

pl = Ωc e , 1≤ l ≤ N • Denominator DN ( s ) is known as the Butterworth polynomial of order N 13 13

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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

⎛ ⎞ 1 10 log10 ⎜ 2 ⎟ = −1 ⎝1 + ε ⎠ which yields ε 2 = 0.25895 and ⎛ ⎞ 1 10 log10 ⎜ 2 ⎟ = − 40 ⎝A ⎠ 2 A = 10,000 which yields

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Butterworth Approximation • Therefore and

1= k1

A − 1 = 196.51334 ε 2

1 = Ωs = 5 k Ωp

• Hence

log10 (1 / k1 ) N= = 3.2811 log10 (1 / k ) • We choose N = 4 15 15

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Chebyshev Approximation • The magnitude-square response of an N-th order analog lowpass Type 1 Chebyshev filter is given by 1 2 H a (s) = 1 + ε 2TN2 (Ω / Ω p ) where TN (Ω) is the Chebyshev polynomial of order N: ⎧ cos( N cos −1 Ω), TN (Ω) = ⎨ −1 ⎩cosh( N cosh Ω),

Ω ≤1 Ω >1

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Chebyshev Approximation • Typical magnitude response plots of the analog lowpass Type 1 Chebyshev filter are shown below Type 1 Chebyshev Filter N=2 N=3 N=8

Magnitude

1 0.8 0.6 0.4 0.2 0

0

1

2

3

Ω

17 17

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Chebyshev Approximation • If at Ω = Ω s the magnitude is equal to 1/A, then 2 1 1 = 2 H a ( jΩ s ) = 2 2 1 + ε TN (Ω s / Ω p ) A • Solving the above we get cosh −1 ( A2 − 1 / ε ) cosh −1 (1 / k1 ) = N= −1 cosh (Ω s / Ω p ) cosh −1 (1 / k ) • Order N is chosen as the nearest integer greater than or equal to the above value 18 18

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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 2 ⎡TN (Ω s / Ω p ) ⎤ 1+ ε ⎢ ⎥ T ( / ) Ω Ω ⎦ ⎣ N s where TN (Ω) is the Chebyshev polynomial of order N 19 19

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Chebyshev Approximation • Typical magnitude response plots of the analog lowpass Type 2 Chebyshev filter are shown below Type 2 Chebyshev Filter N=3 N=5 N=7

Magnitude

1 0.8 0.6 0.4 0.2 0

0

1

2

3

Ω

20 20

Copyright © 2010, S. K. Mitra

Chebyshev Approximation • 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= −1 cosh (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 −1

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cosh (1 / k1 ) = 2.6059 N= −1 cosh (1 / k ) Copyright © 2010, S. K. Mitra

Elliptic Approximation • The square-magnitude response of an elliptic lowpass filter is given by 1 2 H a ( jΩ ) = 1 + ε 2 RN2 (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 < ∞ 22 22

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Elliptic Approximation • Typical magnitude response plots with Ω p = 1 are shown below Elliptic Filter N=3 N=4

Magnitude

1 0.8 0.6 0.4 0.2 0

0

1

2

3

Ω

25 23

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Analog Lowpass Filter Design • 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 [N, Wn] = ellipord\(Wp, Ws, Rp, Rs, ‘s’\);

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[b, a] = ellip\(N, Rp, Rs, Wn, ‘s’\); with Wp = 2*pi*1000; Ws = 2*pi*5000; Rp = 1; Rs = 40; Copyright © 2010, S. K. Mitra

Analog Lowpass Filter Design • Gain plot Lowpass Elliptic Filter

Gain, dB

0 -20 -40 -60

0

2000 4000 Frequency, Hz

6000

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Design of Analog Highpass, Bandpass and Bandstop Filters • 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 ) H

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LP

(

s

)

Copyright © 2010, S. K. Mitra

Design of Analog Highpass, Bandpass and Bandstop Filters • Let s denote the Laplace transform variable of prototype analog lowpass filter H LP (s ) and sˆ denote the Laplace transform variable of desired analog filter HD (sˆ) • The mapping from s-domain to sˆ-domain is given by the invertible transformation s = F (sˆ) • Then HD ( sˆ) = H LP ( s ) s = F ( sˆ) H LP ( s ) = HD ( sˆ) sˆ = F −1 ( s )

29 27

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Analog Highpass Filter Design • Spectral Transformation: ˆp Ω pΩ s= sˆ where Ω p is the passband edge frequency of ˆ p is the passband edge H LP ( s ) and Ω frequency of H HP ( sˆ) • On the imaginary axis the transformation is ˆp Ω pΩ Ω=− ˆ Ω 30 28

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Analog Highpass Filter Design ˆp Ω pΩ Ω=− ˆ Ω Ω Ω p Ωs

Stopband − Ω s − Ω p

Stopband

0 Passband Stopband

Passband

ˆs ˆ p −Ω −Ω

0

Passband

ˆs Ω

ˆp Ω

Ω

Lowpass

ˆ Highpass Ω

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Lowpass to Highpass spectral Transformation:

30

Analog Highpass Filter Design

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• 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 Ωs = = = =4 2πFˆs Fˆs 1000 • Analog lowpass filter specifications: Ω p = 1, Ω s = 4 , α p = 0.1 dB, α s = 40 dB Copyright © 2010, S. K. Mitra

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);

• Gain plots Highpass Filter

0

0

-20

-20

Gain, dB

Gain, dB

Prototype Lowpass Filter

-40 -60

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-80 0

-40 -60

2

4

6 Ω

8

10

-80 0

2

4 6 8 10 Frequency, kHz Copyright © 2010, S. K. Mitra

Analog Bandpass Filter Design • Spectral Transformation ˆ o2 sˆ 2 + Ω s = Ωp ˆ p2 − Ω ˆ p1 ) sˆ(Ω 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) 34 33

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Analog Bandpass Filter Design • On the imaginary axis the transformation is ˆ o2 − Ω ˆ2 Ω Ω = −Ω p ˆ Bw Ω ˆ p2 − Ω ˆ p1 is the width of where Bw = Ω ˆ o is the passband center passband and Ω 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 34

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Analog Bandpass Filter Design 2 2 ˆ ˆ Ωo − Ω Ω = −Ω p ˆ Bw Ω Ω p Ωs

Stopband − Ω s − Ω p

Stopband

Ω

0 Passband

Stopband

Passband

Stopband

Passband

Stopband

ˆ o ↓ −Ω ˆ s1 0 Ω ˆo ↓ Ω ˆ s1 ↓ Ω ˆ s2 ˆ s2 ↓ − Ω −Ω ˆ p2 − Ω ˆ p1 −Ω ˆ p1 Ω ˆ p2 Ω

Lowpass

ˆ Bandpass Ω

36 35

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Lowpass to Bandpass spectral Transformation:

36

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 • 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 37

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Analog Bandpass 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

←

Stopband →

ˆ s1 Ω

→ Stopband ˆ p1 Ω

ˆ p2 Ω

ˆ s2 Ω

ˆ Ω

38 38

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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 No change in passband and shorter leftmost transition band

39 39

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Analog Bandpass 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 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 40 40

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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 p2 Ω

ˆ s2 Ω

Stopband

ˆ Ω

41 41

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Analog Bandpass Filter Design 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 42 42

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Analog Bandpass Filter Design 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 43

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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 Fˆs 2 = 8 kHz, α p = 1 dB, α s = 22 dB

44 44

6 6 ˆ ˆ ˆ ˆ • Now Fp1Fp 2 = 28 × 10 and Fs1Fs 2 = 24 × 10 • Since Fˆ p1Fˆ p 2 > Fˆs1Fˆs 2 we choose ˆ s1Fˆs 2 24 F Fˆp1 = = = 3.42857 kHz Fˆp 2 7 Copyright © 2010, S. K. Mitra

Analog Bandpass Filter Design 24 25 • Bandwidth Bw = 7 − = kHz 7 7 • We choose Ω p = 1 ˆFs1Fˆs 2 − ( Fˆs1 ) 2 24 − 9 Ωs = = = 1.4 Fˆs1 ⋅ Bw ( 25 / 7 ) × 3 • Analog lowpass filter specifications: Ω p = 1 , Ω s = 1.4, α s = 22 dB, α p = 1 dB 45 45

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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);

• Gain plot Prototype Lowpass Filter

Bandpass Filter

46 46

0 Gain, dB

Gain, dB

0 -20 -40 -60 0

-20 -40

2

4 Ω

6

8

-60 0

5 10 15 Frequency, kHz © 2010, S. K. Mitra Copyright

Analog Bandstop Filter Design • Spectral Transformation ˆ s2 − Ω ˆ s1 ) sˆ(Ω s = Ωs ˆ 2o sˆ 2 + 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 47

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Analog Bandstop Filter Design • On the imaginary axis the transformation is ˆ Bw Ω Ω = Ωs 2 ˆ o −Ω ˆ2 Ω ˆ s2 − Ω ˆ s1 is the width of where Bw = Ω ˆ o is the stopband center stopband and Ω 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 48 48

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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 Ω p Ωs

Stopband − Ω s − Ω p

Stopband

Ω

0 Passband Passband

Stopband

Passband

Stopband

Passband

ˆ o ↓ −Ω ˆ p1 ↓ Ω ˆo ↓ Ω ˆ p10 Ω ˆ p2 ˆ p2 ↓ − Ω −Ω ˆ s2 ˆ s1 ˆ s1 ˆ s2 −Ω Ω −Ω Ω

Lowpass

ˆ Bandpass Ω

49 49

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Lowpass to Bandstop spectral Transformation:

50

Analog Bandstop Filter Design • Also,

2 ˆ ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 Ωo = 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

50 51

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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

←

51 52

ˆ p1 Ω

ˆ s1 Ω

Stopband

← ˆ s2 Ω

ˆ p2 Ω

ˆ Ω Copyright © 2010, S. K. Mitra

Analog Bandstop Filter Design 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 52 53

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Analog Bandstop Filter Design 2 ˆ ˆ p1Ω ˆ p2 = Ω ˆ s1Ω ˆ s2 • Note: The condition Ωo = 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 53 54

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Analog Bandstop 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

Passband

→

54 55

ˆ p1 Ω

Stopband →

ˆ s1 Ω

ˆ s2 Ω

ˆ p2 Ω

ˆ Ω Copyright © 2010, S. K. Mitra

Analog Bandstop Filter Design 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

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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 56 57

Copyright © 2010, S. K. Mitra