FIR Design

Digital Signal Processing

Chapter 3: FIR filter design

Vijaya Krishna A

Vijaya Krishna A EC-305: DSP

Introduction

Filters extract information from signal.

Filters are everywhere: both in natural systems and designedones.

We consider linear, time-invariant filters.

Definition

A filter is a (LTI) system that modifies the spectral content of theinput signal. It passes some frequencies almost unaltered, andblocks/attenuates other frequencies.

Classification:

Analog/DigitalFIR/IIR (among digital filters)Linear phase/nonlinear phase.


Introduction

Analog filters:

Continuous-time systems that act upon continuous-timesignals.Still very much in use.Needed for designing IIR filters.Specified in terms of Laplace and Fourier transforms.Types: Lowpass, highpass, bandpass, bandstop, notch, allpassetc.

Digital filters:

Discrete-time systems that act upon discrete-time signals.Specified in terms z-transform and DTFT.The spectrum is periodic.


Introduction

FIR filters: The output of a causal FIR filter depends on thepresent input and finitely many past inputs.

y(n) =M−1∑k=0

h(k)x(n − k)

H(z) =M−1∑n=0

h(n)z−n

IIR filters: The output of a causal IIR filter depends on thepresent input and finitely many past inputs and outputs.

y(n) = −N∑

k=1

aky(n − k) +M∑

k=0

bkx(n − k)

H(z) =

∑Mk=0 bkz−k

1 +∑N

k=1 akz−k


FIR filter design

The filter design problem: Given H(e jω), find h(n).Consider the ideal lowpass filter

H(e jω) =

{1 |ω| ≤ ωc

0 ωc ≤ |ω| ≤ π

The corresponding h(n) is

h(n) =

{ωcπ n = 0ωcπ

sinωcnωcn

n 6= 0

This filter cannot be realized because it is infinitely long andnon-causal.Paley-Wiener theorem: If h(n) is causal and has finite energy,then ∫ π

−π| log |H(e jω)||dω <∞


Linear Phase

An M-length FIR filter h(n) has linear phase if and only if it iseither

Symmetric

h(n) = h(M − 1− n) n = 0, 1, . . . ,M − 1

or antisymmetric

h(n) = −h(M − 1− n) n = 0, 1, . . . ,M − 1

If h(n) has linear phase, it follows that

z−(M−1)H(z−1) = ±H(z)

H(z) and H(z−1) have the same roots.

If z0 is a root of H(z), then 1/z0 is also a root of H(z).

If z0 is a complex root of H(z), then z∗0 and 1/z∗0 are alsoroots of H(z).


Design of linear phase FIR filters using window functions

Let Hd(e jω) be the desired (ideal) frequency response. Let hd(n)be the corresponding impulse response.

hd(n) =1

2π

∫ π

−πHd(e jω)e jωndω

Let’s consider only the samples of hd(n) between n = 0 andn = M − 1, for some chosen M. This implies

h(n) = hd(n)w(n),

where

w(n) =

{1 n = 0, 1, . . . ,M − 1

0 otherwise

w(n) is the rectangular window function.


Linear phase FIR filters using window functions

Thus,H(e jω) = Hd(e jω) ∗W (e jω)

We have

W (e jω) = e−jωM−12

sin ωM2

sin ω2

Convolution with W (e jω) introduces

Transition band.Passband ripple and stopband ripple.

Transition band has a width of 4πM rad/sample.

Same as the width of the mainlobe of W (e jω).

=⇒ Smaller the required transition band, higher the filterorder M.

Peak sidelobe magnitude is independent of M.


Rectangular window magnitude response: Different M

0 0.5 1 1.5 2 2.5 3 3.5−70

−60

−50

−40

−30

−20

−10

0

ω (rad/sample)

Mag

nitu

de (

dB)

N=5

0 0.5 1 1.5 2 2.5 3 3.5−70

−60

−50

−40

−30

−20

−10

0

ω (rad/sample)

Mag

nitu

de (

dB)

N=11

0 0.5 1 1.5 2 2.5 3 3.5−70

−60

−50

−40

−30

−20

−10

0

ω (rad/sample)

Mag

nitu

de (

dB)

N=31

0 0.5 1 1.5 2 2.5 3 3.5−70

−60

−50

−40

−30

−20

−10

0

ω (rad/sample)

Mag

nitu

de (

dB)

N=61


LPF using rectangular window with M = 31

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

ω (rad/sample)

Mag

nitu

deLinear Scale

0 0.5 1 1.5 2 2.5 3 3.5−100

−80

−60

−40

−20

0

20

ω (rad/sample)

Mag

nitu

de (

dB)

Log Scale


Bartlett window

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

n

Bartlett window (N=31)

0 0.5 1 1.5 2 2.5 3 3.5−100

−80

−60

−40

−20

0

20

ω (rad/sample)

Mag

nitu

de (

dB)

Frequency Response

w(n) = 1−2|n − M−1

2 |M − 1

n = 0, 1 . . . ,M − 1


Hamming window

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

n

Hamming window (N=31)

0 0.5 1 1.5 2 2.5 3 3.5−100

−80

−60

−40

−20

0

20

ω (rad/sample)

Mag

nitu

de (

dB)

Frequency Response

w(n) = 0.54− 0.46 cos2πn

M − 1n = 0, 1 . . . ,M − 1


Magnitude response comparison (M = 31)

0 0.5 1 1.5 2 2.5 3 3.5−160

−140

−120

−100

−80

−60

−40

−20

0

20

ω (rad/sample)

Mag

nitu

de (

dB)

N=31

Rectangular

Bartlett

Hamming


LPF comparison (M = 31)

0 0.5 1 1.5 2 2.5 3 3.5−120

−100

−80

−60

−40

−20

0

20

ω (rad/sample)

Mag

nitu

de (

dB)

LPF with ωc=π/4. (N=31)

RectangularBartlettHamming


Comparison

Window Mainlobe width Peak sidelobe Stopband attenuation

Rectangular 4πM -13 dB -21 dB

Bartlett 8πM -27 dB -25 dB

Hamming 8πM -43 dB -53 dB

Exercises:

1 Find the linear phase FIR filter with the best transition bandfor approximating the following response:

Hd(e jω) =

{e−j2ω |ω| < π

4

0 π4 < |ω| < π

2 Design a linear phase FIR filter for approximating the followingresponse. The stopband ripple must be less than 0.003.

Hd(e jω) =

{0 |ω| < π

4

e−j3ω π4 < |ω| < π


Ideal filter responses

Filtertype

Hd(e jω) hd(n)

Lowpass

{e−jn0ω |ω| < ωc

0 ωc < |ω| < π

{ωcπ n = n0sin(ωc (n−n0))π(n−n0) n 6= n0

Highpass

{0 |ω| < ωc

e−jn0ω ωc < |ω| < π

{1− ωc

π n = n0

− sin(ωc (n−n0))π(n−n0) n 6= n0

Bandpass

0 |ω| < ωc1

e−jn0ω ωc1 < |ω| < ωc2

0 ωc2 < |ω| < π

{ωc2−ωc1π n = n0

sinωc2 (n−n0)−sinωc1 (n−n0)

π(n−n0) n 6= n0

Bandstop

e−jn0ω |ω| < ωc1

0 ωc1 < |ω| < ωc2

e−jn0ω ωc2 < |ω| < π

{1− ωc2−ωc1

π n = n0sinωc1 (n−n0)−sinωc2 (n−n0)

π(n−n0) n 6= n0


Ideal filter responses

Filter type Hd(e jω) hd(n)

Differentiator jω − π < ω < π

{0 n = 0(−1)n

n n 6= 0

Hilberttransformer

{−j 0 < ω < π

j −π < ω < 0

{0 n = 01−(−1)n

πn n 6= 0


Kaiser Window

Kaiser window allows for a trade-off between the mainlobe widthand the sidelobe attenuation.

w(n) =

I0

(β√

1−( n−αα )

2)

I0(β) , 0≤n≤N − 1

0 otherwise

where I0(x) is the modified Bessel function of the first kind and 0th

order, given by

I0(x) = 1 +∞∑

k=1

[(x

2

)k 1

k!

]2

α = N−12 determines the size of the window.

β controls the shape of the window.If β = 0, w(n) is the rectangular window.If β = 5.44, w(n) is the Hamming window.


Choosing β:Let

δ = min{δp, δs}A = −20 log δ

Then

β =

0.1102(A− 8.7), A > 50

0.5842(A− 21)0.4 + 0.07886(A− 21), 50 ≥ A ≥ 21

0, A < 21

Given the transition width ∆f =ws−wp

2π , N is found as

N ≥

{A−7.9514.36∆f , A ≥ 210.9∆f , A < 21

Example: Design a linear phase lowpass filter with ωp = 0.3π,ωs = 0.5π, and a stopband ripple of δs = 0.01, using the Kaiserwindow.


FIR filter design using frequency sampling

The desired frequency response Hd(e jω) is sampled at Nequidistant points in the interval [0, 2π].

H(k) ≡ Hd

(2πk

N

)k = 0, 1, . . . ,N − 1

The filter coefficients h(n) are obtained as the N-point IDFT ofH(k).

h(n) =1

N

N−1∑k=0

H(k)e j 2πN

nk n = 0, 1, . . . ,N − 1

In order to obtain real h(n), we consider only conjugate symmetricresponses:

H(N − k) = H∗(k), k = 0, 1, . . . ,N − 1

2

and H(0) is real.


In this case, h(n) can be expressed as

h(n) =1

N

H(0) + 2

N−12∑

k=1

Re{

H(k)e j 2πN

nk}

Examples:

Obtain a linear phase lowpass filter with ωc = π2 and N = 7,

using the frequency sampling approach.


Filter Structures

Cascade structure:

H(z) =K∏

k=1

Hk(z)

where

Hk(z) = bk0 + bk1z−1 + bk2z−2 k = 1, 2, . . . ,K

for FIR systems, and

Hk(z) =bk0 + bk1z−1 + bk2z−2

1 + ak1z−1 + ak2z−2Parallel structure:

H(z) = C +K∑

k=1

Hk(z)

where

Hk(z) =bk0 + bk1z−1

1 + ak1z−1 + ak2z−2


Lattice Structure

The filter is realized as a cascade of two-port networks.

Useful in adaptive filtering, speech coding, wavelet filterbanksetc.

Robust to coefficient quantization.

Consider the first order filter A1(z) = 1 + a1(1)z−1.For the two-port network, we have

f1(n) = x(n) + K1x(n − 1)

g1(n) = K1x(n) + x(n − 1)

f1(n) = y(n) if K1 = a1(1).Now consider A2(z) = 1 + a2(1)z−1 + a2(2)z−2.From the lattice structure,

f2(n) = f1(n) + K2g1(n − 1)

g2(n) = K2f1(n) + g1(n − 1)


f2(n) = x(n) + K1(1 + K2)x(n − 1) + K2x(n − 2)

= x(n) + a2(1)x(n − 1) + a2(2)x(n − 2)

Thus,

K2 = a2(2)

K1 =a2(1)

1 + a2(2)

In general, with f0(n) = g0(n) = x(n), we have

fm(n) = fm−1(n) + Kmgm−1(n − 1) m = 1, 2, . . . ,M − 1

gm(n) = Kmfm−1(n) + gm−1(n − 1)

Ki are called reflection coefficients


With am(0) = 1, we have

fm(n) =m∑

k=0

am(k)x(n − k)

Fm(z) = Am(z)X (z)

Similarly,

g2(n) = K2f1(n) + g1(n − 1)

= a2(2)x(n) + a2(1)x(n − 1) + x(n − 2)

With bm(m) = 1, we have

gm(n) =m∑

k=0

bm(k)x(n − k)

Gm(z) = Bm(z)X (z)

where, if {1, am(1), . . . , am(m)} are the coefficients of Am(z),{am(m), am(m − 1), . . . , am(1), 1} are the coefficients of Bm(z)


It can be shown that

Bm(z) = z−mAm(z−1)

=⇒ The zeros of Bm(z) are reciprocal of those of Am(z).Step-up recursion:A0(z) = 1.

Am(z) = Am−1(z) + Kmz−1Bm−1(z)

= Am−1(z) + Kmz−mAm−1(z−1)

Step-down recursion:

Am(z) = Am−1(z) + Kmz−1Bm−1(z)

= Am−1(z) + Km[Bm(z)− kmAm−1(z−1)]

Am−1(z) =Am(z)− KmBm(z)

1− K 2m

m = M − 1, . . . , 1

Km = am(m)Vijaya Krishna A EC-305: DSP

The procedure breaks down if any Km = 1.

Roots of Am(z) lie inside the unit circle if and only if |Ki | < 1for i = 1, 2, . . . ,m

Examples

Find H(z) if K1 = K2 = 13 , and K3 = 1.

Draw the lattice structure for H(z) = 8 + 4z−1 + 2z−2 + z−3.

All-pole lattice filters:

H(z) =1

1 +∑N

k=1 z−k=

1

AN(z)

Or,X (z) = AN(z)Y (z)

Can be obtained by interchanging the input and the output in theFIR lattice.


Set

fN(n) = x(n)

f0(n) = y(n)

The recursive relations are

fN(n) = x(n)

fm−1(n) = fm(n)− Kmgm−1(n − 1)

gm(n) = Kmfm−1(n) + gm−1(n − 1)

y(n) = f0(n) = g0(n)


Documents

FIR Design