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# Four Types of Linear-Phase FIR Filters

Module by: Ivan Selesnick. E-mail the author

Summary: (Blank Abstract)

## FOUR TYPES OF LINEAR-PHASE FIR FILTERS

Linear-phase FIR filter can be divided into four basic types.

Table 1
Type impulse response
I symmetric length is odd
II symmetric length is even
III anti-symmetric length is odd
IV anti-symmetric length is even

When hn h n is nonzero for 0nN1 0 n N 1 (the length of the impulse response hn h n is NN), then the symmetry of the impulse response can be written as

hn=hN1n h n h N 1 n
(1)
and the anti-symmetry can be written as
hn=hN1n h n h N 1 n
(2)

## TYPE I: ODD-LENGTH SYMMETRIC

The frequency response of a length N=5 N 5 FIR Type I filter can be written as follows.

H f ω= h 0 + h 1 e(i)ω+ h 2 e-2iω+ h 1 e-3iω+ h 0 e-4ω H f ω h 0 h 1 ω h 2 -2 ω h 1 -3 ω h 0 -4 ω
(3)
H f ω=e-2iω( h 0 e2iω+ h 1 eiω+ h 2 + h 1 e(i)ω+ h 0 e-2iω) H f ω -2 ω h 0 2 ω h 1 ω h 2 h 1 ω h 0 -2 ω
(4)
H f ω=e-2iω( h 0 (e2iω+e-2iω)+ h 1 (eiω+e(i)ω)+ h 2 ) H f ω -2 ω h 0 2 ω -2 ω h 1 ω ω h 2
(5)
H f ω=e-2iω(2 h 0 cos2ω+2 h 1 cosω+ h 2 ) H f ω -2 ω 2 h 0 2 ω 2 h 1 ω h 2
(6)
H f ω=Aωeiθω H f ω A ω θ ω
(7)
where θω=-2ω θ ω -2 ω Aω=2 h 0 cos2ω+2 h 1 cosω+ h 2 A ω 2 h 0 2 ω 2 h 1 ω h 2 Note that Aω A ω is real-valued and can be both positive and negative. In general, for a Type I FIR filters of length NN: H f ω=Aωeiθω H f ω A ω θ ω
Aω=hM+2n=0M1hncos(Mn)ω A ω h M 2 n 0 M 1 h n M n ω
(8)
θω=(M)ω θ ω M ω M=N12 M N 1 2

## TYPE II: EVEN-LENGTH SYMMETRIC

The frequency response of a length N=4 N 4 FIR Type II filter can be written as follows.

H f ω= h 0 + h 1 e(i)ω+ h 1 e-2iω+ h 0 e-3iω H f ω h 0 h 1 ω h 1 -2 ω h 0 -3 ω
(9)
H f ω=e-32iω( h 0 e32iω+ h 1 e12iω+ h 1 e-12iω+ h 0 e-32iω) H f ω -3 2 ω h 0 3 2 ω h 1 1 2 ω h 1 -1 2 ω h 0 -3 2 ω
(10)
H f ω=e-32iω( h 0 (e32iω+e-32iω)+ h 1 (e12iω+e-12iω)) H f ω -3 2 ω h 0 3 2 ω -3 2 ω h 1 1 2 ω -1 2 ω
(11)
H f ω=e-32iω(2 h 0 cos32ω+2 h 1 cos12ω) H f ω -3 2 ω 2 h 0 3 2 ω 2 h 1 1 2 ω
(12)
H f ω=Aωeiθω H f ω A ω θ ω
(13)
where θω=-32ω θ ω -3 2 ω Aω=2 h 0 cos32ω+2 h 1 cos12ω A ω 2 h 0 3 2 ω 2 h 1 1 2 ω In general, for a Type II FIR filters of length NN: H f ω=Aωeiθω H f ω A ω θ ω
Aω=2n=0N21hncos(Mn)ω A ω 2 n 0 N 2 1 h n M n ω
(14)
θω=(M)ω θ ω M ω M=N12 M N 1 2

## TYPE III: ODD-LENGTH ANTI-SYMMETRIC

The frequency response of a length N=5 N 5 FIR Type III filter can be written as follows.

H f ω= h 0 + h 1 e(i)ω( h 1 e-3iω h 0 e-4ω) H f ω h 0 h 1 ω h 1 -3 ω h 0 -4 ω
(15)
H f ω=e-2iω( h 0 e2iω+ h 1 eiω( h 1 e(i)ω h 0 e-2iω)) H f ω -2 ω h 0 2 ω h 1 ω h 1 ω h 0 -2 ω
(16)
H f ω=e-2iω( h 0 (e2iωe-2iω)+ h 1 (eiωe(i)ω)) H f ω -2 ω h 0 2 ω -2 ω h 1 ω ω
(17)
H f ω=e-2iω(2i h 0 sin2ω+2i h 1 sinω) H f ω -2 ω 2 h 0 2 ω 2 h 1 ω
(18)
H f ω=e-2iωi(2 h 0 sin2ω+2 h 1 sinω) H f ω -2 ω 2 h 0 2 ω 2 h 1 ω
(19)
H f ω=e-2iωeiπ2(2 h 0 sin2ω+2 h 1 sinω) H f ω -2 ω 2 2 h 0 2 ω 2 h 1 ω
(20)
H f ω=Aωeiθω H f ω A ω θ ω
(21)
where θω=2ω+π2 θ ω -2 ω 2 Aω=2 h 0 sin2ω+2 h 1 sinω A ω 2 h 0 2 ω 2 h 1 ω In general, for a Type III FIR filters of length NN: H f ω=Aωeiθω H f ω A ω θ ω
Aω=2n=0M1hnsin(Mn)ω A ω 2 n 0 M 1 h n M n ω
(22)
θω=Mω+π2 θ ω M ω 2 M=N12 M N 1 2

## TYPE IV: EVEN-LENGTH ANTI-SYMMETRIC

The frequency response of a length N=4 N 4 FIR Type IV filter can be written as follows.

H f ω= h 0 + h 1 e(i)ω( h 1 e-2iω h 0 e-3iω) H f ω h 0 h 1 ω h 1 -2 ω h 0 -3 ω
(23)
H f ω=e-32iω( h 0 e32iω+ h 1 e12iω( h 1 e-12iω h 0 e-32iω)) H f ω -3 2 ω h 0 3 2 ω h 1 1 2 ω h 1 -1 2 ω h 0 -3 2 ω
(24)
H f ω=e-32iω( h 0 (e32iωe-32iω)+ h 1 (e12iωe-12iω)) H f ω -3 2 ω h 0 3 2 ω -3 2 ω h 1 1 2 ω -1 2 ω
(25)
H f ω=e-32iω(2i h 0 sin32ω+2i h 1 sin12ω) H f ω -3 2 ω 2 h 0 3 2 ω 2 h 1 1 2 ω
(26)
H f ω=e-32iωi(2 h 0 sin32ω+2 h 1 sin12ω) H f ω -3 2 ω 2 h 0 3 2 ω 2 h 1 1 2 ω
(27)
H f ω=e-32iωeiπ2(2 h 0 sin32ω+2 h 1 sin12ω) H f ω -3 2 ω 2 2 h 0 3 2 ω 2 h 1 1 2 ω
(28)
H f ω=Aωeiθω H f ω A ω θ ω
(29)
where θω=-32ω+π2 θ ω -3 2 ω 2 Aω=2 h 0 sin32ω+2 h 1 sin12ω A ω 2 h 0 3 2 ω 2 h 1 1 2 ω In general, for a Type IV FIR filters of length NN: H f ω=Aωeiθω H f ω A ω θ ω
Aω=2n=0N21hnsin(Mn)ω A ω 2 n 0 N 2 1 h n M n ω
(30)
θω=Mω+π2 θ ω M ω 2 M=N12 M N 1 2

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