iir digital filter structures filter design

7/22/2019 IIR Digital Filter Structures Filter Design

1/35

1

Digital Signal Processing

IIR digital filter structuresFilter design


2/35

2

Basic IIR Digital Filter Structure

An N-th order IIR digital transfer function is characterised by 2N+1 unique

coefficients, and in general, requires 2N+1 multipliers and 2N two input adders forimplementation

Direct form IIR filter: Filter structures in which the multiplier coefficients areprecisely the coefficients of the transfer function

Consider, a 3rdorder IIR filter with transfer function

We can implement H(z) as a cascade

where


3/35

3

Basic IIR Digital Filter Structure

The filter section H1(z) can be seen to be an FIR filter and can be realised:

Time domain representation of H2(z) is given by


4/35

4

Direct Form IIR Digital Filter Structure

Cascade of the two structures leads to H(z) and is known asdirect form I

structure

It is noncanonic as it uses 6 delays to realise 3rdorder transfer function

Its transpose (direct form It)


5/35

5

Direct Form IIR Digital Filter Structure

Various other noncanonic directform structures can be derived bysimple block diagrammanipulations

Observe in the direct formstructure (on the right), signalvariables in nodes 1 and 1 are thesame, so the two top delays canbe shared

Likewise, the middle and bottomtwo delays can be shared

We get direct form II and itstranspose (direct form IIt) isalso shown

Now, it is canonic


6/35

6

Cascade form IIR Digital Filter Structure

By expressing the numerator and the denominator polynomials of the

transfer function as a product of polynomials of lower degree, a digitalfilter can be realised as a cascade of low-order filter sections

)(

)(

)(

)(

)(

)(

)(

)()(

3

3

2

2

1

1

zD

zP

zD

zP

zD

zP

zD

zPzH

Consider, for example,H(z)=P(z)/D(z) expressed as

Examples of cascade realisationsobtained by different pole-zeropairings

Examples of cascade realisationsobtained by different ordering ofsections


7/35

7


There are altogether a total of 36 different cascade realisations of H(z)based on pole-zero pairings and orderings

Due to finite wordlength effects, each such realisation behavesdifferently

Usually, the polynomials are factored into a product of 1storder and 2ndorder polynomials:

For a first order,


8/35

8

Consider an example:

The direct form II and cascaderealisations are


How?

21

21

1

1

5.08.01

0455.08227.01

4.01

44.0)(

zz

zz

z

zzH

Hint: try to remember a general form for direct form II and then reuse it!


9/35

9

Parallel Form IIR Digital Filter Structures

A partial-fraction expansion of the transfer function in z-1leads to the

parallel form Istructure

Assuming simple poles, the transfer function H(z) can be expressed as

In the above for a real pole,

A direct partial fraction expansion of the transfer function in z leads tothe parallel form IIstructure

In the above for a real pole,

Well see some examples.


10/35

10

A partial fraction expansion of

in z-1yields


1.04.01

6.0

)5831.04.0(1

25.0

)5831.04.0(1

25.0)( 111

zzjzjzH

)1)(1(

)1()1(

11 BA

AbBa

B

b

A

aUse

Parallel form I


11/35

11

A partial fraction expansion of

in z yields


1

1

1

1

1

1

)5831.04.0(1

)1458.01.0(

)5831.04.0(1

)1458.01.0(

4.01

24.0)(

zj

zj

zj

zj

z

zzH

)1)(1(

)1()1(

11 BA

AbBa

B

b

A

aUse

Parallel form II


12/35

FIR filter design

We have briefly explored the design of FIR filters in Lecture 5

Well look at it again here using MATLAB

Well use the window method to design FIR filters

12


13/35

13

Ideal filter and impulse response

An ideal LPF Its impulse

What is the problem we cant have infinite coefficient length

So, we need to truncate

13

Gain

freqwc

1LPF

Inverse DFT

Similar to


14/35

1414

Let us use a rectangular window

We can see that the shape of the LPF has changed

This is know as Gibbs phenomenon oscillatory behaviour in the magnituderesponses caused by truncating the ideal impulse response function (i.e. the

rectangular window has an abrupt transition to zero)

Gibbs phenomenon can be reduced by

Using a window that tapers smoothly at each end

Providing a smooth transition from passband to stopband in the magnitudespecifications

14

Ideal filter and impulse response

DFT


15/35

151515

Filter length

The filter length (i.e. order) affects the

magnitude response

Length increase, number of ripples in bothpassband and stopband increases but witha corresponding decrease in the ripplewidths

That is as N increase, it gets closer to theideal LPF

But height of largest ripples remain thesame independent of length

Similar oscillatory behaviour could beobserved in the magnitude responses ofthe truncated versions of other types ofideal filters

15

Infinite N

N=25

N=13


16/35

1616

Common window functions

Some common window functions

Low pass filter frequency response

N=51 and2

cw

Magnitude response of windowfunctions with M=25


17/35

171717

FIR filter design

Step 1 - Specify the ideal or required frequency response, HD(w)

Step 2 - Obtain impulse response, hD(n) of the desired filter by evaluating theinverse Fourier transform

Important note: well look at hD(n) for standard filter designs in the next slide

Step 3 - Select a window function and then the number of filter coefficients

Step 4 - Obtain values of w(n) for the chosen window function and the values ofthe actual FIR coefficients, h(n), by multiplying hD(n) with w(n)

h(n)=hD(n) w(n)


18/35

FIR filter design (Steps 1 and 2)

18

We can makeuse of availablehD(n)


19/35

19

FIR filter design (step 3) window?

First problem in step 3 - which window function?

The ratio of the main lobe/side lobe can be used

Example, the frequency response of Hamming and Blackman windows

Main lobe centred at =0

Other ripples are called side lobes


20/35

FIR filter design (step 3) window?

To ensure a fast transition from passband to stopband, window should

have a very small main lobe width

To reduce the passband and stopband ripple , the area under thesidelobes should be small so to increase the stopband attenuation, weneed to decrease the sidelobe amplitude

Most of the time, these two requirements are contradictory

Example: LPF with length=51 and cut-off at /2

20


21/35

21

FIR filter design (step 3) length?

Second problem in step 3 number of filter coefficients?

To obtain good (i.e. sharper) transition band reduce main lobe widthincrease the filter length

But increased filter length means increased computational complexity

Eg: Frequency response of Hamming window with length 101 and 201are shown below

21

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200

Normalized Frequency (rad/sample)

Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50


Magnitude(dB)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-200

0

200


Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50


Magnitude(dB)


22/35

FIR filter design (step 3) length?

Using the same window as in the previous slide

LPF with Hamming window, with normalised cut-off=0.3

22

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4000

-3000

-2000

-1000

0


P

hase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0

50


M

agnitude(dB)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-6000

-4000

-2000

0


Ph

ase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0

50


Magnitude(dB)


23/35

23

To select the suitable order i.e. the length

The lowest order that can meet the requirements

There are several methods:

Kaiser, Bellanger, Hermann

Kaisers formula:

Actual location of transition band is immaterial

Bellangersformula:

Hermanns formula is more complex so we will not consider here (but generally itgives slightly more accurate value for the order)

23

FIR filter design (step 3) order selection

)(6.14

13log20 10

ps

sp

ffN

1)(3

10log2 10

ps

sp

ffN

2/)(6.14

13log20 10

ps

spN


24/35

2424

Example Kaisers formula

Estimate the order of a linear-phase lowpass FIR filter with the following

specifications: passband edge Fp=1.8 kHz, stopband edge Fs=2 kHz,peak passband ripple p=0.1 dB, minimum stopband attenuation s=35dB and sampling rate FT=12 kHz.

Get the peak ripple values: p=0.0115, s=0.0178

Using the Kaisers formula:

which gives 98.27

Since N must be an integer, we round up the value to give N=99

As the order is odd, a Type 2 FIR filter can be designed. To design aType 1 FIR filter, N should be increased to 100

In Matlab, you can use kaiserord function but it only gives anapproximation

Normally, you should check to see if the filter specifications are met, ifnot change N

24

)12000/180012000/2000(6.14

13)0178.0(0115.0log20 10

N


25/35

25

Another example

A requirement exists for an FIR digital filter to meet the following specifications:

Passband: 150-250 Hz Transition width: 50 Hz

Passband ripple: 0.1 dB

Stopband attenuation: 60 dB

Sampling frequency: 1 kHz

Obtain the filter coefficients

Solution - The peak ripple values are p=0.0115, s=0.001

Using Kaisers formula:

which give N=49.85. So, we could take N=50 (type 1, odd length)

Note that the passband information is not used to obtain N, so it will be the sameeven if we shift the passband or even if we use it for highpass, stopband etc

If the transition bands are not same then we use the smaller transition band

bigger transition band requires smaller N, so the smaller transition bandrequires a higher order and is more important for the order selection

)1000/50(6.14

13)001.0(0115.0log20 10 N)(6.14

13log20 10

ps

sp

ffN


26/35

A common mistake

Example: Obtain the FIR coefficients for a highpass filter with

Fc=600 Hz with FT=2000 Hz

Assume we start with N=100

Use the known impulse response for HPF

Problem

The phase is completely non-linear

The magnitude response is not good

26

0 20 40 60 80 100 120-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

20

40

60

80

Normalized Frequency ( rad/sample)

Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-15

-10

-5

0

5

Normalized Frequency ( rad/sample)

Magnitude(dB)


27/35

27

To obtain linear phase response

Coefficients must be symmetry/antisymmetry

So say, change N=201

Linear phase response achieved

But there are still ripples (Gibbs phenomenon)

A common mistake solution 1

0 50 100 150 200 250-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10000

-5000

0

5000


Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0

50


Magnitude(dB)


28/35

28

Let us use a window, say Hamming

Less ripple in the magnituderesponse (in the passband)

Why is the cut-off frequency in theplot at 0.6 rad/sample?

We used 0.3 to obtain the impulseresponse function, right?

A common mistake solution 2

0 50 100 150 200 2500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 50 100 150 200 250-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10000

-5000

0

5000


Phase(degrees)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0

50


Magnitude(dB)


29/35

Taking the example further

29

Let us actually use the filterafter designing it!

It could be improved using higher N and improved window such asKaiser window

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-80

-60

-40

-20

0

20

Frequency

PowerSpectrumM

agnitude(d

B)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-120

-100

-80

-60

-40

-20

0

20

Frequency

PowerSpectrumM

agnitude(dB)


30/35

Using MATLAB fir1 function

30

Using MATLAB, filter coefficients can easily obtained using fir1 function!

Hamming window is used by default For the same input x in the previous slide, let us design a bandpass filter

to extract only the component at 650 Hz

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-80

-60

-40

-20

0

20

Frequency

PowerSpectrumM

ag

nitude(dB)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-120

-100

-80

-60

-40

-20

0

20

Frequency

PowerSpectrumM

ag

nitude(dB)

0 50 100 150 200 250-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08


31/35

Improved FIR filter design

31

The windows such as Hanning, Hamming etc are known as fixed window designs as

the

ripple values are fixed)

FIR filter design can also be improved using adjustable window functions such asDolph-Chebyshev and Kaiser that provide control over by means of an additionalparameter characterising the window

The basis of FIR filter design is to obtain the set of minimal number of coefficients

for the required response

So, computer based optimisation methods have been invented

Parks-McClellan algorithm can be used to obtain the FIR coefficients (in MATLAB,we can design this filter using remez function)

Normally, improved design over the standard window based method

Known as computer based FIR filter design

The only other FIR design method is frequency sampling method

But knowledge of the window based method is sufficient for this course

Note: Kaiser order and Kaiser window are two different matters, though we normally use themtogether


32/35

32

IIR filter design

32

The most common approach to IIR filter design:

Convert the digital filter specifications into an analogue prototypelow-pass filter specifications

Determine the analogue low-pass filter transfer function, Ha(s)

Transform Ha(s) in the desired digital transfer function G(z)

The most widely used transformation is the bilinear transformation that

maps the imaginary axis in the s-plane (j) onto the unit circle of the z-plane

So, relation between G(z) and Ha(s)

However, this design requires some knowledge of analogue low-passfilters and also s-plane and therefore, we will skip this method andstraightaway utilise IIR filter design using MATLAB


33/35

3333

IIR filter design using MATLAB

33

We have done IIR filter design using MATLAB in Lecture 5

There are several classical IIR filters

Butterworth, Chebyshev Types I and II, Elliptic

MATLAB functions

butter Butterworth filter with flat passband (no ripples)

ellip Elliptic filter with ripples (but normally requiring lower order thanButterworth for same transition band)

cheby1 - Chebyshev filter controlling peak-to-peak ripple in the passband

cheby2 - Chebyshev filter controlling the amount of stopband ripple

First step is to estimate the order of the filter given the specifications usingfunctions: buttord, cheb1ord, cheb2ord, ellipord etc

Next, obtain the coefficients, BandAusing functions: buttord, besself, cheby1,cheby2, ellip, etc

Finally, do the filtering using filtfilt function

The frequency response of the filter can be obtained using function freqz(B,A)


34/35

34

IIR filter design using MATLAB -example

Design an elliptic IIR low-pass filter with the specifications: Fp=0.8 kHz,

Fs=1 kHz, FT=4 kHz, p=0.5 dB, s=40 dB

MATLAB code:

[N,Wn]=ellipord(0.4,0.5,0.5,40)

[B,A]=ellip(N,0.5,40,Wn)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-400

-300

-200

-100

0


Phase(degrees

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0


Magnitude(dB)


35/35

3535

IIR filter design using MATLAB -example

To filter a signal, use filtfilt function

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-120

-100

-80

-60

-40

-20

0

20

Frequency

PowerSpectrumM

agnitude(dB)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-120

-100

-80

-60

-40

-20

0

20

Frequency

PowerSpectrumM

agnitude(dB)

iir digital filter structures filter design

Documents