Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Digital Signal Processing

Electrical Engineering (EE) : Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

The document Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Digital Signal Processing.
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Butter worth Filter Design

The butterworth LP filter of order N is defined as HB(s) HB(-s) = Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Where s = jΩc

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRevOR Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

It has 2N poles

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Ex: for N=3

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev = 1200, 1800, 2400, 3000, 3600, 600

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRevButter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Poles that are let half plane are belongs to desired system function.

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

For a large Ω, magnitude response decreases as Ω -N, indicating the LP nature of this filter.

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

As Ω → ∞

= -20 N log10Ω

= -20 N dB/ Decade = -6 N dB/Octane

As N increases, the magnitude response approaches that of ideal LP filter.
The value of N is determined by Pass & stop band specifications.

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Ex: Design Butterworth LPF for the following specifications.

Pass band:

-1< H ( jΩ) 2 dB ≤ 0 for 0≤ Ω ≤ 1404π ( W π = 1404π )   

Stop band:

H ( jW)2 dB < -60  for W ≥ 8268π   ( Ωs = 8268π )

If the Ωc is given

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Since Ωc is not given, a guess must be made.

The specifications call for a drop of -59dB, In the frequency range from the edge of the pass band (1404π ) to the edge of stop band (8268π ). The frequency difference is equal to 

log2(8268/1404)=2.56 octaves.

1 oct ---- - 6N dB

2.56 ------ ?

=> 2.56 X - 6N dB = -59 dB’s

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Ω2N > 106 Ωc 2 N

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Ω c <1470.3π

Let Ω c =1470.3π

At this Ω c it should satisfy pass band specifications.

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

= 0.59

This result is below the pass band specifications. Hence N=4 is not sufficient.

Let N=5

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

In the pass band  Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Since N=5

Ω c = 2076π

S1 = -2076π

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

1. Magnitude response is smooth, and decreases monotonically as Ω increases from 0 to ∞

2. the magnitude response is maximally flat about Ω =0, in that all its derivatives up to order N are equal to zero at Ω =0

Ex: Ωc=1, N=1

HB( jΩ)2= (1+ Ω 2)-1

The first derivative

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

The second derivative

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

3. The phase response curve approaches  Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev for large Ω , where N is the no. of poles of butterworth circle in the left side of s-plane.

Advantages:

1. easiest to design

2. used because of smoothness of magnitude response .

Disadvantage: 

Relatively large transition range between the pass band and stop band.

Other procedure

When Ω c = 1      Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

If n is even S2N = 1 = e j ( 2k -1)π

The 2N roots will be Sk= Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev k=1,2,….2N

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Therefore: 

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRevButter worth Filter Design Electrical Engineering (EE) Notes | EduRev

If N is odd

S2n =1 = e j 2kπ

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

where θ k = Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRevButter worth Filter Design Electrical Engineering (EE) Notes | EduRev

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev

choosing this value for n, results in two different selections for Ω. If we wish to satisfy our requirement at Ω1 exactly and do better than our req. at Ω2 , we use 

Butter worth Filter Design Electrical Engineering (EE) Notes | EduRev  orButter worth Filter Design Electrical Engineering (EE) Notes | EduRevfor better req at Ω2

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