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**Butter worth Filter Design**

The butterworth LP filter of order N is defined as H_{B}(s) H_{B}(-s) =

Where s = jÎ©_{c}

OR

It has 2N poles

Ex: for N=3

= 120^{0}, 180^{0}, 240^{0}, 300^{0}, 360^{0}, 60^{0}

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

For a large Î©, magnitude response decreases as Î© ^{-N}, indicating the LP nature of this filter.

As Î© â†’ âˆž

= -20 N log_{10}Î©

= -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 ( jÎ©)^{ 2} dB â‰¤ 0 for 0â‰¤ Î© â‰¤ 1404Ï€ ( W Ï€ = 1404Ï€ )

Stop band:

H ( jW)^{2} dB < -60 for W â‰¥ 8268Ï€ ( Î©_{s} = 8268Ï€ )

If the Î©_{c} is given

Since Î©_{c} 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 (1404Ï€ ) to the edge of stop band (8268Ï€ ). The frequency difference is equal to

log_{2}(8268/1404)=2.56 octaves.

1 oct ---- - 6N dB

2.56 ------ ?

=> 2.56 X - 6N dB = -59 dBâ€™s

Î©_{s }^{2N} > 10^{6} Î©_{c} ^{2 N}

Î© c <1470.3Ï€

Let Î© c =1470.3Ï€

At this Î© 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

Î© c = 2076Ï€

S1 = -2076Ï€

1. Magnitude response is smooth, and decreases monotonically as Î© increases from 0 to âˆž

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

Ex: Î©c=1, N=1

H_{B}( jÎ©)^{2}= (1+ Î©^{ 2})-1

The first derivative

The second derivative

3. The phase response curve approaches for large Î© , 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 Î© c = 1

If n is even S^{2N} = 1 = e ^{j ( 2k -1)}Ï€

The 2N roots will be Sk= k=1,2,â€¦.2N

**Therefore: **

**If N is odd**

S^{2}n =1 = e^{ j 2kÏ€}

**where Î¸**_{ k = }

choosing this value for n, results in two different selections for Î©_{c }. If we wish to satisfy our requirement at Î©1 exactly and do better than our req. at Î©2 , we use

or**for better req at **Î©**2**

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