Butter worth Filter Design

# Butter worth Filter Design Notes | Study Signals and Systems - Electronics and Communication Engineering (ECE)

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Butter worth Filter Design

The butterworth LP filter of order N is defined as HB(s) HB(-s) =

Where s = jΩc

OR

It has 2N poles

Ex: for N=3

= 1200, 1800, 2400, 3000, 3600, 600

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

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

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.

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

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

Ω2N > 106 Ωc 2 N

Ω c <1470.3π

Let Ω c =1470.3π

At this Ω c it should satisfy pass band specifications.

= 0.59

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

Let N=5

In the pass band

Since N=5

Ω c = 2076π

S1 = -2076π

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

The second derivative

3. The phase response curve approaches   for large Ω , where N is the no. of poles of butterworth circle in the left side of s-plane.

1. easiest to design

2. used because of smoothness of magnitude response .

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

Other procedure

When Ω c = 1

If n is even S2N = 1 = e j ( 2k -1)π

The 2N roots will be Sk=  k=1,2,….2N

Therefore:

If N is odd

S2n =1 = e j 2kπ

where θ k =

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

orfor better req at Ω2

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