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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
Figure 1
Figure 1 (fourstem.png)

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

h⁢n=h⁢N−1−n h n h N 1 n
(1)
and the anti-symmetry can be written as
h⁢n=−h⁢N−1−n 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(−j)⁢ω+ h 2 ⁢e-2⁢j⁢ω+ h 1 ⁢e-3⁢j⁢ω+ h 0 ⁢e-4⁢ω 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-2⁢j⁢ω⁢( h 0 ⁢e2⁢j⁢ω+ h 1 ⁢ej⁢ω+ h 2 + h 1 ⁢e(−j)⁢ω+ h 0 ⁢e-2⁢j⁢ω) 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-2⁢j⁢ω⁢( h 0 ⁢(e2⁢j⁢ω+e-2⁢j⁢ω)+ h 1 ⁢(ej⁢ω+e(−j)⁢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-2⁢j⁢ω⁢(2⁢ h 0 ⁢cos2⁢ω+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=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
(7)
where θ⁢ω=-2⁢ω θ ω -2 ω A⁢ω=2⁢ h 0 ⁢cos2⁢ω+2⁢ h 1 ⁢cosω+ h 2 A ω 2 h 0 2 ω 2 h 1 ω h 2 Note that A⁢ω 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=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
A⁢ω=h⁢M+2⁢∑n=0M−1h⁢n⁢cos(M−n)⁢ω A ω h M 2 n 0 M 1 h n M n ω
(8)
θ⁢ω=(−M)⁢ω θ ω M ω M=N−12 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(−j)⁢ω+ h 1 ⁢e-2⁢j⁢ω+ h 0 ⁢e-3⁢j⁢ω H f ω h 0 h 1 ω h 1 -2 ω h 0 -3 ω
(9)
H f ⁢ω=e-32⁢j⁢ω⁢( h 0 ⁢e32⁢j⁢ω+ h 1 ⁢e12⁢j⁢ω+ h 1 ⁢e-12⁢j⁢ω+ h 0 ⁢e-32⁢j⁢ω) 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-32⁢j⁢ω⁢( h 0 ⁢(e32⁢j⁢ω+e-32⁢j⁢ω)+ h 1 ⁢(e12⁢j⁢ω+e-12⁢j⁢ω)) 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-32⁢j⁢ω⁢(2⁢ h 0 ⁢cos32⁢ω+2⁢ h 1 ⁢cos12⁢ω) H f ω -3 2 ω 2 h 0 3 2 ω 2 h 1 1 2 ω
(12)
H f ⁢ω=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
(13)
where θ⁢ω=-32⁢ω θ ω -3 2 ω A⁢ω=2⁢ h 0 ⁢cos32⁢ω+2⁢ h 1 ⁢cos12⁢ω 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=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
A⁢ω=2⁢∑n=0N2−1h⁢n⁢cos(M−n)⁢ω A ω 2 n 0 N 2 1 h n M n ω
(14)
θ⁢ω=(−M)⁢ω θ ω M ω M=N−12 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(−j)⁢ω−( h 1 ⁢e-3⁢j⁢ω− h 0 ⁢e-4⁢ω) H f ω h 0 h 1 ω h 1 -3 ω h 0 -4 ω
(15)
H f ⁢ω=e-2⁢j⁢ω⁢( h 0 ⁢e2⁢j⁢ω+ h 1 ⁢ej⁢ω−( h 1 ⁢e(−j)⁢ω− h 0 ⁢e-2⁢j⁢ω)) H f ω -2 ω h 0 2 ω h 1 ω h 1 ω h 0 -2 ω
(16)
H f ⁢ω=e-2⁢j⁢ω⁢( h 0 ⁢(e2⁢j⁢ω−e-2⁢j⁢ω)+ h 1 ⁢(ej⁢ω−e(−j)⁢ω)) H f ω -2 ω h 0 2 ω -2 ω h 1 ω ω
(17)
H f ⁢ω=e-2⁢j⁢ω⁢(2⁢j⁢ h 0 ⁢sin2⁢ω+2⁢j⁢ h 1 ⁢sinω) H f ω -2 ω 2 h 0 2 ω 2 h 1 ω
(18)
H f ⁢ω=e-2⁢j⁢ω⁢j⁢(2⁢ h 0 ⁢sin2⁢ω+2⁢ h 1 ⁢sinω) H f ω -2 ω 2 h 0 2 ω 2 h 1 ω
(19)
H f ⁢ω=e-2⁢j⁢ω⁢ej⁢π2⁢(2⁢ h 0 ⁢sin2⁢ω+2⁢ h 1 ⁢sinω) H f ω -2 ω 2 2 h 0 2 ω 2 h 1 ω
(20)
H f ⁢ω=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
(21)
where θ⁢ω=2⁢⁢ω+π2 θ ω -2 ω 2 A⁢ω=2⁢ h 0 ⁢sin2⁢ω+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=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
A⁢ω=2⁢∑n=0M−1h⁢n⁢sin(M−n)⁢ω A ω 2 n 0 M 1 h n M n ω
(22)
θ⁢ω=M⁢ω+π2 θ ω M ω 2 M=N−12 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(−j)⁢ω−( h 1 ⁢e-2⁢j⁢ω− h 0 ⁢e-3⁢j⁢ω) H f ω h 0 h 1 ω h 1 -2 ω h 0 -3 ω
(23)
H f ⁢ω=e-32⁢j⁢ω⁢( h 0 ⁢e32⁢j⁢ω+ h 1 ⁢e12⁢j⁢ω−( h 1 ⁢e-12⁢j⁢ω− h 0 ⁢e-32⁢j⁢ω)) 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-32⁢j⁢ω⁢( h 0 ⁢(e32⁢j⁢ω−e-32⁢j⁢ω)+ h 1 ⁢(e12⁢j⁢ω−e-12⁢j⁢ω)) 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-32⁢j⁢ω⁢(2⁢j⁢ h 0 ⁢sin32⁢ω+2⁢j⁢ h 1 ⁢sin12⁢ω) H f ω -3 2 ω 2 h 0 3 2 ω 2 h 1 1 2 ω
(26)
H f ⁢ω=e-32⁢j⁢ω⁢j⁢(2⁢ h 0 ⁢sin32⁢ω+2⁢ h 1 ⁢sin12⁢ω) H f ω -3 2 ω 2 h 0 3 2 ω 2 h 1 1 2 ω
(27)
H f ⁢ω=e-32⁢j⁢ω⁢ej⁢π2⁢(2⁢ h 0 ⁢sin32⁢ω+2⁢ h 1 ⁢sin12⁢ω) H f ω -3 2 ω 2 2 h 0 3 2 ω 2 h 1 1 2 ω
(28)
H f ⁢ω=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
(29)
where θ⁢ω=-32⁢ω+π2 θ ω -3 2 ω 2 A⁢ω=2⁢ h 0 ⁢sin32⁢ω+2⁢ h 1 ⁢sin12⁢ω 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=A⁢ω⁢ej⁢θ⁢ω H f ω A ω θ ω
A⁢ω=2⁢∑n=0N2−1h⁢n⁢sin(M−n)⁢ω A ω 2 n 0 N 2 1 h n M n ω
(30)
θ⁢ω=M⁢ω+π2 θ ω M ω 2 M=N−12 M N 1 2

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