DSP Lecture 24

7/30/2019 DSP Lecture 24

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Lectures 23-24 EE-802 ADSP SEECS-NUST

EE 802-Advanced Digital SignalProcessing

Dr. Amir A. Khan

Office : A-218, SEECS

9085-2162; [email protected]

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Lecture Outline

Filter Design (From Oppenheim)

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Introduction

The term Filters originate from frequency-selective

filtering

modify certain frequencies relative to others

Analysis vs. Design

Analysis Pole-zero distribution System properties (Causality/Stability)

Frequency response

System structural realizations

Study system behavior

Design

Practical realization poses causality and stability constraints

coefficients ak have strict constraints

Additionally, a linear phase requirement forH(z) may be desirable

Find system coefficients, i.e. M;N;a1; ;aN; b0; ;bM,

s.t.

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Filter Design Overview

Design of Discrete-Time IIR Filters FromContinuous-Time (Analog) Filters Impulse Invariance

Bilinear Transformation

Common Discrete-time filters Butterworth Chebyshev

Elliptic

Design of FIR Filters Windowing Method Frequency Sampling Method

Optimum Filter Design (Park-McClellan)

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Typical low-pass specifications

Parameters

[0;wp] = pass-band

[ws;p] = stop-band

[wp;ws] = transition-band

dp1; dp2 = pass-band ripple; Often expressed in dB via 20log10(1+dp)

ds = stop-band ripple; Often expressed in dB via 20log10(ds)

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FIR or IIR ?

The choice between IIR and FIR usually based on phase considerations

GLP (generalized linear phase) constraint

FIR with GLP

IIR with GLP

IIR filters cannot be stable, causal and GLP at same time

If GLP desired : choose FIR

If GLP not desired : IIR can be chosen

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7/38Lectures 23-24 EE-802 ADSP SEECS-NUST

IIR Filter Design

Based on established methods for analog filter design

Approach usually restricted to low pass and band pass filters

Basic filter type is low pass

High pass or band pass filters generally created using

frequency transformations

Design Steps

Choose prototype analog filter

Butterworth, Chebyshev, Elliptic

Choose analog to digital transformation method

Impulse Invariance, Bilinear Transformation Transform digital filter specs to equivalent analog filter specs

Design analog filter

Transform analog filter to digital filter

Perform Frequency transformation if necessary7



IIR Filter Design Steps

1. Start with filter specifications in digital frequency2. Choose a design methodology (impulse invariance, bilinear

transformation)

3. Convert digital frequency specs. into analog frequency specs (i.e.

to depending on design method chosen

4. Select a prototype analog filter (Butterworth, Chebyshev, Elliptic)

5. Design the analog filter according to the analog frequency specs

involves selecting the filter cut-off frequencies and order6. Obtain the corresponding filter transfer function (H(s) Laplace Tr.)

7. Convert this analog filter (H(s)) to the digital filter (H(z))

according to the design methodology initially chosen8



Impulse Invariance Roundup !

Based on choosing a discrete time impulse responseh[n] that is similar to

continuous-time impulse responseh(t) (Step 7)

Motivated by desire to maintain the shape of frequency response

Frequency axis mapping is linear

Major problem is aliasing

TWw

Wj

0

Im(z)

Re(z)

unit-circle

T

p

T

p

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continuous-time impulse responseh(t)



Major problem is aliasingWj

0

Im(z)

Re(z)

unit-circle

T

p

T

p

TWw

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continuous-time impulse responseh(t)



Major problem is aliasing

TWw

Wj

0

Im(z)

Re(z)

unit-circle

T

p

T

p

Many-one mapping = source of aliasing

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Bilinear Transformation (B.T.)

Algebraic transformation b/w variables sand z

One-one mapping between the s and z-planes

Given a C.T. prototype filterHc(s), the corresponding D.T. filterH(z) is

The parameter Td has no influence whatsoever in the filter design, it is

cancelled out as we start with digital specs. and return to digital filter while

passing by analog filter

1

1

2 1( )1

d

zs

T z

1

1

2 1( ) ( ( ))

1c

d

zH z H

T z

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Properties of Bilinear Transformation

1 ( / 2)1 ( / 2)

d

d

T szT s

Substitute s j W

1 2 2

1 2 2

d d

d d

T j Tz

T j T

W

W

/ /

/ /

W(1) If0, then |z|1 for any

similarly, if0, then |z|1 for all W

(2) If0, then1 2

1 2

d

d

j Tz

j T

W W

/

/1z

1 2

1 2

j d

d

j Te

j T

w W W

/

/

2 1

1

j

j

d

es

T e

w

w

)

) )

/2

/2

2 sin / 22 2tan / 2

2 cos / 2

j

j

d d

e j js j

T e T

w

w

w w

w

W

2tan( )

2dT

wW 2arctan( / 2)dTw W

Since 0

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2tan( )

2dT

wW

2arctan( / 2)dTw W

Whole of left-half s-plane mapped to inside the unit circle in z-plane

The whole of imaginary axis on s-plane mapped to unit circle, no aliasing problem

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2tan( )

2dT

wW

2arctan( / 2)dTw W

Bilinear transformation avoids problem of aliasing through complete mapping but

everything comes at a cost ? Whats the cost here?

Non-linear frequency mapping as opposed to impulse invariance where we had a

linear mapping

Non-linear compression of frequency axis has to be compensated in B.T.

This non-linear phenomenon is called Frequency Warping

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Frequency Warping Effect of Bilinear

Transformation

Note that thecritical frequencies

are pre-warped

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Phase Warping and Non-linearity of Phase

by B.T.

Dashed line is linear phase and

solid line is phase resulting from bilinear transformation

Effect more pronounced at higher frequencies 17



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FIR Filter Design

Windowing Method Truncate (window) the ideal response to

make it FIR

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FIR Filt D i R t l



FIR Filter Design-RectangularWindowing Problems

Magnitude spectra of Rectangular windows of increasing lengths (samples)

Reduction in width of main lobe as M increases

Area under sidelobes remain the same (note Normalized Amplitude plotted here)20

FIR Filter Design Rectang lar



FIR Filter Design-RectangularWindowing Problems

Gibbs phenomenon occurs whenever there is truncation

Increasing M does not yield significant result as oscillations

increase without reducing in amplitude

Transition region becomes smaller with increasing M

Ripples continue to exist especially at discontinuities

Problem is due to sharp discontinuity (in time domain) of rectangular

windows

Solution : Use windows with tapered ends in time domain instead of

sharp discontinuities

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G tti A d Sh Di ti it f



Getting Around Sharp Discontinuity of

Rectangular Windows

No abrupt discontinuity in time-domain response of windows translates to

low amplitude side lobes in frequency domain

Advantage is the reduced number of ripples

On the hindsight, tapered window results in a wider transition band

(frequency domain)

Wider transition band can always be compensated by using larger length

windows (higher order filter, remember filter order = Window length - 1)

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Properties of Commonly Used Windows

Rectangular

1, 0

0, otherwise

n Mnw

Bartlett (triangular)

2 / , 0 / 2, even

2 2 / , / 2

0, otherwise

n M n M M

n n M M n M w

)0.5 0.5cos 2 / 0

0, otherwise

n M n M n

pw

)0.54 0.46cos 2 / 0

0, otherwise

n M n M n

pw

) )0.42 0.5cos 2 / 0.08cos 4 / 0

0, otherwise

n M n M n M n

p pw

Hanning

Hamming

Blackman

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Rectangular

Bartlett (triangular)

Hanning

Hamming

Blackman

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Highest amplitude

high oscillations at discontinuity

Smallest width

the sharpest transition

Wider transition region (wider main-lobe) compensated by much lower side-

lobes and thus less ripples 26


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Example : Window Comparison

Rectangular window

Hanning Window

Less ripples in Hanning Window at the cost of larger transition band

Hanning Window

M= 40

Transition band reduced by increasing window length

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Specifications of Window Design Method

Filter responseH(ej ) should not the shaded regions

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Properties of Window Design Method (1)

Equal transition bandwidth on both sides of desired (ideal) cut-off

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Equal peak approximation error in passband and stopband

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Main lobe of window is wider than the width of transition band

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Peak approximation error depends on window shape and

independent of window size (filter order) 32


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Design Example

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Step 1-Design Example : Choice of Window

dB40)01.0log(20;01.02

d

Hanning, Hamming and Blackman all satisfy the criterion, we can

chose between Hanning and Hamming to have a smaller transition

band as compared to Blackman for same order

Suppose we choose the Hanning Window 34


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Step 2-Design Example : Filter Order

dB40)01.0log(20;01.02

dSuppose we choose the Hanning Window

Width of main lobe = s p = 0.3 0.2 0.1

pw sw

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Step 3 Design Example : Specify Ideal


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Step 3-Design Example : Specify Ideal

Response Hd(w)

pw sw

Ideal Filter cut-off frequency

Ideal low-pass filter

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Step 4 Design Example : Specify Ideal


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Step 4-Design Example : Specify Ideal

Impulse Response hd [n]

Non-causal

Make it causal: Delay by M/2

) )

)40

405.0sin][

n

nnh

dp

p

Ideal Filter Coefficients

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Step 5-FIR filter coefficients

) ) )40

405.0sin][

n

nnhd

p

p

Ideal Filter Coefficients

)0.5 0.5cos 2 / 0

0, otherwise

n M n M n

pw

Hanning

][].[][ nwnhnh d

FIR Filter Coefficients

Find frequency response H(ejw) and verify if it meets specifications

Otherwise, repeat the process by changing either filter order,

window type, or by slightly moving the ideal filter band edge freq.

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DSP Lecture 24

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