PDF Analog Lowpass Filter Specifications

(1)

1

Analog

Analog Lowpass Lowpass Filter Filter Specifications Specifications

• Typical magnitude response of an analog lowpass filter may be given as indicated below

) (jΩ H_a

2

Analog

• In the passband, defined by , we require

i.e., approximates unitywithin an error of

) (jΩ H_a

Ωp

≤ Ω

≤ 0

p p

a

p≤H jΩ ≤ +δ Ω≤Ω δ

− ( ) 1 ,

1

) (jΩ H_a

∞

<

Ω

≤ Ω_s

∞

<

Ω

≤ Ω δ

≤

Ω _s _s

a j

H ( ) ,

• In the stopband, defined by , we require

i.e., approximates zerowithin an error of

δp

±

δs

3

Analog

• - passband edge frequency

• - stopband edge frequency

• - peak ripple valuein the passband

• - peak ripple valuein the stopband

• Peak passband ripple

dB

• Minimum stopband attenuation dB Ωp

Ωs

δs

δp

) 1 ( log

20 ₁₀ _p

p=− −δ

α

) ( log 20 ₁₀ _s

s=− δ

α

4

Analog

Analog Lowpass Lowpass Filter Filter Specifications Specifications

• Magnitude specifications may alternately be given in a normalized form as indicated below

5

Analog

• Here, the maximum value of the magnitude in the passband assumed to be unity

• -Maximum passband deviation, given by the minimum value of the magnitude in the passband

• -Maximum stopband magnitude 1 2

/

1 +ε

A 1

6

Analog

Analog Lowpass Lowpass Filter Design Filter Design

• Two additional parameters are defined- (1) Transition ratio

For a lowpass filter

(2) Discrimination parameter Usually

s

k p

Ω

=Ω

2 1

1= −

k Aε

<1 k

1<<1 k

(2)

7

Butterworth Approximation Butterworth Approximation

• The magnitude-square response of an N-th order analog lowpassButterworth filter is given by

• First derivatives of at are equal to zero

• The Butterworth lowpass filter thus is said to have a maximally-flat magnitudeat

1

2N− H_a(jΩ)²

N c a j

H ² ₂

) / ( 1 ) 1

( Ω = + Ω Ω

=0 Ω

=0

Ω 8

Butterworth Approximation Butterworth Approximation

• Gain in dB is

• As and

dB is called the 3-dB cutoff frequency

2

10 ( )

log 10 )

(Ω = H_a jΩ

G 0 ) 0 ( = G

3 0103 . 3 ) 5 . 0 ( log 10 )

(Ω_c = ₁₀ =− ≅−

G Ωc

9

• Typical magnitude responses withΩ_c=1

0 1 2 3

0 0.2 0.4 0.6 0.8 1

Ω

Magnitude

Butterworth Filter N = 2 N = 4 N = 10

10

• Two parameters completely characterizing a Butterworth lowpass filter are and N

• These are determined from the specified bandedges and , and minimum passband magnitude , and maximum stopband ripple

Ωc

Ωp Ω_s 1 2

/

1 +ε

A / 1

11

Butterworth Approximation Butterworth Approximation

• and N are thus determined from

• Solving the above we get Ωc

2 2

2 1

) / ( 1 ) 1

(j A

H _N

c s s

a =

Ω Ω

= + Ω

2 2

2

1 1 )

/ ( 1 ) 1

( = +ε

Ω Ω

= +

Ω _N

c p p

a j H

) / 1 ( log

) / 1 ( log ) / ( log

] / ) 1 [(

log 2 1

10 1 10 10

2 2 10

k k N A

p s

Ω = Ω

⋅ −

= ε

12

• Since order Nmust be an integer, value obtained is rounded up to the next highest integer

• This value of Nis used next to determine 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

Ωc

(3)

13

Butterworth Approximation Butterworth Approximation

• Transfer function of an analog Butterworth lowpass filter is given by

where

• Denominator is known as the Butterworth polynomialof orderN

∏ −

= Ω +∑

= Ω

=

− =

= N

N c N N

N c a N

p s s

d s s

D s C H

1 1

0 ( )

) ) (

(

l l

l

N e

p_l=Ω_c ^j^[π⁽^N⁺²^l⁻¹⁾^/²^N^], 1≤l≤ )

(s D_N

14

• Example- Determine the lowest order of a Butterworth lowpass filter with a 1-dB cutoff frequency at 1kHz and a minimum attenuation of 40 dB at 5kHz

• Now which yields and

which yields 1 1 log 1

10 ₁₀ ₂⎟=−

⎠

⎜ ⎞

⎝

⎛ +ε

25895 .

2=0 ε

1 40 log

10 ₁₀ ₂⎟=−

⎠

⎜ ⎞

⎝

⎛ A

000 ,

2=10 A

15

Butterworth Approximation Butterworth Approximation

• Therefore and

• Hence

• We chooseN= 4

51334 . 1 196

1 ²

1

− =

= ^Aε k

1 =5

Ω

=Ω

p s

k

2811 . ) 3 / 1 ( log

) / 1 ( log

10 1

10 =

= k

N k

16

Chebyshev

Chebyshev Approximation Approximation

• The magnitude-square response of an N-th order analog lowpassType 1 Chebyshev filter is given by

where is the Chebyshev polynomial of orderN:

) / ( 1 ) 1

( ² ₂ ₂

p N

a s T

H = + Ω Ω

ε

⎩⎨

⎧

>

Ω Ω

≤ Ω

= Ω

Ω ⁻₋

1 ), cosh cosh(

1 ),

cos ) cos(

( ₁

1

N T_N N

(Ω) TN

17

Chebyshev

Chebyshev Approximation Approximation

• Typical magnitude response plots of the analog lowpassType 1 Chebyshev filterare shown below

0 1 2 3

0 0.2 0.4 0.6 0.8 1

Ω

Magnitude

Type 1 Chebyshev Filter N = 2 N = 3 N = 8

18

Chebyshev

• If at the magnitude is equal to 1/A, then

• Solving the above we get

• Order Nis chosen as the nearest integer greater than or equal to the above value

Ωs

= Ω

2 2

2

2 1

) / ( 1 ) 1

(j T A

H

p s N s

a =

Ω Ω

= +

Ω ε

) / 1 ( cosh

) / 1 ( cosh ) / ( cosh

) / 1 ( cosh

1 1 1 1

2 1

k k N A

p s

−

− =

Ω Ω

= − ε

(4)

19

Chebyshev

• The magnitude-square response of an N-th order analog lowpassType 2 Chebyshev (also calledinverse Chebyshev)filteris given by

where is the Chebyshev polynomial of orderN

2 2

2

) / (

) / 1 (

) 1 (

⎥⎦⎤

⎢⎣⎡

Ω Ω

Ω + Ω

= Ω

s N

p s N a

T T j

H

ε (Ω) TN

20

Chebyshev

• Typical magnitude response plots of the analog lowpassType 2 Chebyshev filterare shown below

0 1 2 3

0 0.2 0.4 0.6 0.8 1

Ω

Magnitude

Type 2 Chebyshev Filter N = 3 N = 5 N = 7

21

Chebyshev

• The order Nof the Type 2 Chebyshev filter is determined from given , , and A using

• Example- Determine the lowest order of a Chebyshev lowpass filter with a 1-dB cutoff frequency at 1kHz and a minimum attenuation of 40dB at 5kHz-

) / 1 ( cosh

) / 1 ( cosh )

/ ( cosh

) / 1 ( cosh

1 1 1 1

2 1

k k N A

p s

−

− =

Ω Ω

= − ε

ε Ωs

6059 . 2 ) / 1 ( cosh

) / 1 ( cosh

1 1

1 =

= ⁻₋ k N k

22

Elliptic Approximation Elliptic Approximation

• The square-magnitude response of an elliptic lowpass filter is given by

where is a rational function of order Nsatisfying , with the roots of its numerator lying in the interval

and the roots of its denominator lying in the interval

) / ( 1 ) 1

( ² ₂ ₂

p N

a j R

H Ω = + Ω Ω

ε (Ω)

RN

) ( / 1 ) / 1

( Ω = _N Ω

N R

R

1 0<Ω<

∞

<

Ω

<

1

23

Elliptic Approximation Elliptic Approximation

• For given , , , and A, the filter order can be estimated using

where

) / 1 ( log

) / 4 ( log 2

10 1

10 ρ

N≅ k

p ε Ω Ω_s

1 2

' k

k= − ) ' 1 ( 2

' 1

0 k

k

−+ ρ =

13 0 9

0 5 0

0 2(ρ ) 15(ρ ) 150(ρ ) ρ

ρ= + + +

24

Elliptic Approximation Elliptic Approximation

• Example- Determine the lowest order of a elliptic lowpass filter with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40dB at 5kHz Note: k= 0.2and

• Substituting these values we get

• and hence N= 2.23308

• Choose N= 3

5134 . 196 / 1 k₁= , 979796 . '=0

k ρ₀=0.00255135,

0025513525 .

=0 ρ

(5)

25

Elliptic Approximation Elliptic Approximation

• Typical magnitude response plots with are shown below

=1 Ω_p

0 1 2 3

0 0.2 0.4 0.6 0.8 1

Ω

Magnitude

Elliptic Filter N = 3 N = 4

26

Analog

• Example- Design an elliptic lowpass filter of lowest order with a 1-dBcutoff

frequency at 1kHzand a minimum attenuation of 40dBat 5kHz

• Code fragments used

[N, Wn] = ellipord(Wp, Ws, Rp, Rs, ‘s’);

[b, a] = ellip(N, Rp, Rs, Wn, ‘s’);

with Wp = 2*pi*1000;

Ws = 2*pi*5000;

Rp = 1;

Rs = 40;

27

Analog

• Gain plot

0 2000 4000 6000

-60 -40 -20 0

Frequency, Hz

Gain, dB

Lowpass Elliptic Filter

28

Design of Analog

Design of Analog Highpass Highpass, , Bandpass

Bandpass and Bandstop and Bandstop Filters Filters

• Steps involved in the design process:

Step 1- Develop of specifications of a prototype analog lowpass filter from specifications of desired analog filter

using a frequency transformation Step 2- Design the prototype analog lowpass filter

Step 3 -Determine the transfer function of desired analog filter by applying the inverse frequency transformation to

) (s H_LP

) (s H_D

) (s H_LP

29

Design of Analog

Design of Analog Highpass Highpass, , Bandpass

Bandpass and Bandstop and Bandstop Filters Filters

• Let sdenote the Laplace transform variable of prototype analog lowpass filter and denote the Laplace transform variable of desired analog filter

• The mapping from s-domain to -domain is given by the invertible transformation

• Then sˆ

sˆ ) ˆ (s H_D

ˆ)

) (

( ˆ)

( _LP _s _F _s

D s H s

H = ₌

) ( ˆ ¹

) ˆ ( )

( _D _s _F _s

LP s H s

H = = −

ˆ) (s F s=

) (s H_LP

30

Analog

Analog Highpass Highpass Filter Design Filter Design

• Spectral Transformation:

where is the passband edge frequency of and is the passband edge frequency of

• On the imaginary axis the transformation is s ^ps ^p

ˆ Ωˆ

=Ω

Ωp

Ωˆp

Ω Ω

−Ω

=

Ω ˆ

ˆp p

) (s H_LP

ˆ) (s H_HP

(6)

31

Analog

Ω Ω

−Ω

=

Ω ˆ

ˆp p

Ω

Ωs

−ˆ

Ωˆp Ωˆs Ωp

−ˆ ^Highpass

Passband

Passband Stopband

Ωs

− −Ωp Ωp Ω_s

Lowpass Passband

Stopband Stopband

Ω

Ωˆ 0

0

32

Analog

• Example-Design an analog Butterworth highpass filterwith the specifications:

kHz, kHz, dB, dB

• Choose

• Then

• Analog lowpass filter specifications: , , dB, dB

ˆ_p=4

F Fˆ_s=1

=40

α_s α_p=0.1

=1 Ω_p

=1 Ω_p 1000 4

4000 ˆ

ˆ 2 ˆ 2 ˆ

=

= π =

= π Ω

s p s p

s F

F F F

1 .

=0

α_p α_s=40

=4 Ω_s

33

Analog

[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

0 2 4 6 8 10

-80 -60 -40 -20 0

Ω

Gain, dB

Prototype Lowpass Filter

0 2 4 6 8 10

-80 -60 -40 -20 0

Frequency, kHz

Gain, dB

Highpass Filter

34

Analog

Analog Bandpass Bandpass Filter Filter Design

Design

• Spectral Transformation

where is the passband edge frequency of , and and are the lower and upper passband edge frequencies of desired bandpass filter

ˆ ) (ˆ ˆ

ˆ ˆ

1 2

2 2

p p p o

s s s

Ω

− Ω

Ω Ω +

=

Ωp

ˆ_p1

Ω Ωˆ_p₂ ) ˆ (s H_BP )

(s H_LP

35

Analog

Design

• On the imaginary axis the transformation is

where is the width of passband and is the passband center frequencyof the bandpass filter

• Passband edge frequency is mapped into and , lower and upper passband edge frequencies

w o

p ΩB

Ω

− Ω Ω

−

=

Ω ˆ

ˆ ˆ² ²

1

2 ˆ

ˆ_p _p

Bw=Ω −Ω Ωˆo

ˆ_p1

mΩ ±Ωˆ_p2 ±Ω^p

36

Analog

Design

w o

p ΩB

Ω

− Ω Ω

−

=

Ω ˆ

ˆ ˆ² ²

ˆ1 Ωs

− Ωˆo

ˆ_s1 Ω ˆ 1 Ωp

−

Bandpass Passband

Passband Stopband Ωs

Lowpass Passband

Stopband Stopband

Ω

Ωˆ

Stopband Stopband

0

0 Ωˆ_s2

ˆ_p2 Ω

−↓−Ωˆ^o ↓ Ω↓ˆ_p₁ Ω↓ˆ_p₂ ˆ_s2

Ω

−

(7)

37

Analog

Design

• Stopband edge frequency is mapped into and , lower and upper stopband edge frequencies

• Also,

• 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

Ωs

± ˆ 1

Ωs

m ±Ωˆ_s2

2 1 2 1

2 ˆ ˆ ˆ ˆ

ˆo=ΩpΩp =ΩsΩs

Ω

38

Analog

Design

• Case 1:

To make we can either increase any one of the stopband edges or decrease any one of the passband edges as shown below

2 1 2

1ˆ ˆ ˆ

ˆ_pΩ_p >Ω_sΩ_s Ω

ˆ_p1

Ω Ωˆ_p2 ˆ1

Ωs ˆ₂

Ωs

→ →

←

Ωˆ

2 1 2

1ˆ ˆ ˆ

ˆpΩp =ΩsΩs

Ω

Passband

Stopband Stopband

39

Analog

Design

(1) Decrease to

larger passband and shorter leftmost transition band

(2)Increase to

No change in passband and shorter leftmost transition band

ˆ_p1

Ω

ˆ_s1

Ω

2 2 1ˆ /ˆ ˆsΩs Ωp

Ω

2 2 1ˆ /ˆ ˆpΩp Ωs

Ω

40

Analog

Design

• Note: The condition can also be satisfied by decreasing which is not acceptable as the passband is reduced from the desired value

• Alternately, the condition can be satisfied by increasing which is not acceptable as the upper stop band is reduced from the desired value

2 1 2 1

2 ˆ ˆ ˆ ˆ

Ω

ˆ 2

Ωp

ˆ 2

Ωs

41

Analog

Design

• Case 2:

To make we can either decrease any one of the stopband edges or increase any one of the passband edges as shown below

2 1 2

1ˆ ˆ ˆ

ˆpΩp <ΩsΩs

Ω

ˆ_p1

Ω Ωˆ_p2 ˆ1

Ωs ˆ ₂

Ωs

→ →

← ←

Ωˆ

2 1 2

1ˆ ˆ ˆ

ˆpΩp =ΩsΩs

Ω

Passband

Stopband Stopband

42

Analog

Design

(1) Increase to

larger passband and shorter rightmost transition band (2)Decrease to

No change in passband and shorter rightmost transition band

ˆ 2

Ωp

ˆ 2

Ωs

Ω

1 2 1ˆ /ˆ ˆ pΩp Ωs

Ω

(8)

43

Analog

Design

• Note: The condition can also be satisfied by increasing which is not acceptable as the passband is reduced from the desired value

• Alternately, the condition can be satisfied by decreasing which is not acceptable as the lower stopband is reduced from the desired value

2 1 2 1

2 ˆ ˆ ˆ ˆ

Ω

ˆ 1

Ωp

ˆ 1

Ωs

44

Analog

Design

• Example-Design an analog elliptic

bandpass filter with the specifications:

kHz, kHz, kHz kHz, dB, dB

• Now and

• Since we choose

kHz ˆ 4

1=

Fp ˆ 7

2= Fp

ˆ_s₂=8 F

ˆ 3

1= Fs

=1

αp α_s=22

6 2

1ˆ 28 10

ˆ_pF_p = ×

F Fˆ_s₁Fˆ_s₂=24×10⁶

2 1 2

1ˆ ˆ ˆ

ˆpFp FsFs

F >

571428 . ˆ 3 ˆ / ˆ

ˆ_p1=F_s1F_s2 F_p2= F

45

Analog

Design

• We choose

• Hence

• Analog lowpass filter specifications: , , dB, dB

=1 Ω_p

4 . 3 1 ) 7 / 25 (

9

24 =

×

= − Ω_s

=1 Ω_p 4

.

=1

Ω_s α_p=1 α_s=22

46

Analog

Design

[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

0 2 4 6 8

-60 -40 -20 0

Ω

Gain, dB

Prototype Lowpass Filter

0 5 10 15

-60 -40 -20 0

Frequency, kHz

Gain, dB

Bandpass Filter

47

Analog

Analog Bandstop Bandstop Filter Design Filter Design

• Spectral Transformation

where is the stopband edge frequency of , and and are the lower and upper stopband edge frequencies of the desired bandstop filter

Ωs

) (s H_LP

2 2

1 2

ˆ ˆ ˆ ) (ˆ ˆ

s o

s _ss ^s ^s

Ω +

Ω

− Ω Ω

=

ˆ) (s H_BS ˆ_s1

Ω Ωˆ_s₂

48

Analog

• On the imaginary axis the transformation is

where is the width of stopband and is the stopband center frequencyof the bandstop filter

• Stopband edge frequency is mapped into and , lower and upper stopband edge frequencies

2 2 ˆ ˆ

ˆ Ω

− Ω Ω Ω

= Ω

o s w

B

1 2 ˆ

ˆs s

Bw=Ω −Ω Ωˆo

ˆ 1

Ωs

m ˆ 2

Ωs

± ±Ω^s

(9)

49

Analog

• Passband edge frequency is mapped into and , lower and upper passband edge frequencies

Ωp

± ˆ 1

Ωp

m ˆ 2

Ωp

±

ˆ1 Ωs

− Ωˆ^o

ˆ1 Ωs ˆ 1 Ωp

− ^Bandpass

Passband Passband

Ωs

Lowpass Passband

Stopband Stopband

Ω

Ωˆ

Stopband Stopband

0

ˆ 2 Ωs ˆ_p2

Ω

− Ωˆ_p1 ˆ ₂

Ωp Ωo

−ˆ

↓ ↓ ↓ ↓

ˆ 2 Ωs

−

Passband

50

Analog

• Also,

• 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

2 1 2 1

2 ˆ ˆ ˆ ˆ

ˆ_o=Ω_pΩ_p =Ω_sΩ_s Ω

51

Analog

• Case 1:

• To make we can either increase any one of the stopband edges or decrease any one of the passband edges as shown below

2 1 2

1ˆ ˆ ˆ

ˆpΩp >ΩsΩs

Ω

ˆ_p1

Ω ˆ1 Ωˆ_p₂

Ωs ˆ₂

Ωs Ωˆ

2 1 2

1ˆ ˆ ˆ

ˆpΩp =ΩsΩs

Ω

Stopband

Passband Passband

52

Analog

(1) Decrease to

larger high-frequency passband and shorter rightmost transition band (2)Increase to

No change in passbands and shorter rightmost transition band

ˆ 2

Ωp

ˆ 2

Ωs

2 2 1ˆ /ˆ ˆ_sΩ_s Ω_p Ω

2 2 1ˆ / ˆ ˆpΩp Ωs

Ω

53

Analog

• Note: The condition can also be satisfied by decreasing which is not acceptable as the low- frequency passband is reduced from the desired value

• Alternately, the condition can be satisfied by increasing which is not acceptable as the stopband is reduced from the desired value

2 1 2 1

2 ˆ ˆ ˆ ˆ

ˆ_o=Ω_pΩ_p =Ω_sΩ_s Ω

ˆ_p1

Ω

ˆ 1

Ωs

54

Analog

• Case 1:

• To make we can either decrease any one of the stopband edges or increase any one of the passband edges as shown below

2 1 2

1ˆ ˆ ˆ

ˆ pΩp <ΩsΩs

Ω

2 1 2

1ˆ ˆ ˆ

ˆpΩp =ΩsΩs

Ω

ˆ_p1

Ω ˆ ₂

Ωp ˆ_s1

Ω Stopband Ωˆ_s2 Ωˆ

Passband Passband

(10)

55

Analog

(1) Increase to

larger passband and shorter leftmost transition band

(2)Decrease to

No change in passbands and shorter leftmost transition band

ˆ 1

Ωp

ˆ 1

Ωs

Ω

1 2 1ˆ / ˆ ˆpΩp Ωs

Ω

56

Analog

• Note: The condition can also be satisfied by increasing which is not acceptable as the high- frequency passband is decreased from the desired value

• Alternately, the condition can be satisfied by decreasing which is not acceptable as the stopband is decreased

2 1 2 1

2 ˆ ˆ ˆ ˆ

Ω

ˆ 2

Ωp

ˆ 2

Ωs