Filter design techniques ch7 iir

Filter design process

In general the design of filter is composed

from three different steps

1

Filter design techniques

In the design of frequency selective filters,

the desired filter characteristics are

specified in the frequency domain in terms

of the desired magnitude and phase

response of the filter

In the filter design process, we determine

the coefficients of a causal FIR or IIR filter

that closely approximate the desired

frequency response specifications

2

Which type of filter is to be used

In general FIR filters are used in filtering

problems where a linear phase is required

IIR filters can be used if linear phase is not

required

In general IIR filter has a lower side-lobes

in the stop band than an FIR having the

same number of parameters

3

Which type of filter is to be used

If some phase distortion is tolerable then

IIR filter is preferable because its

implementation requires fewer

parameters, less memory and lower

computational complexity

A comparison between FIR and IIR filters

is made in the next slide

4

IIR versus IIR

5

Relation between IIR filter and

the difference equationsThe input output relations of the IIR filter are

governed by the difference equation

𝑘=0

𝑁

𝑎𝑘𝑦 𝑛 − 𝑘 =

𝑘=0

𝑀

𝑏𝑘𝑥 𝑛 − 𝑘

The above equations means that we can

compute the filter output if the filter coefficients

𝑎𝑘 and 𝑏𝑘 are known

The goal of this section is to determine these

coefficients

6

Relation between IIR filter and

the difference equationsIf we apply the z transform to the LCCD

difference equation which describes the

IIR filter we got the discrete filter system

function given by

𝐻 𝑧 = 𝑘=0𝑀 𝑏𝑘𝑧

−𝑘

1 + 𝑘=1𝑁 𝑎𝑘𝑧

−𝑘

It can be noticed that 𝐻(𝑧) is a rational

function

7

Causality and its implications

All ideal filters are non causal

Non causal means that these filters are

not physically realizable

To illustrate this concept let us consider an

ideal low pass filer

The frequency response of the filter is

given by 𝐻 𝜔 =1 𝜔 ≤ 𝜔𝑐0 𝜔𝑐 < 𝜔 < 𝜋

8


The impulse response of the filter is given

by ℎ 𝑛 =𝜔𝑐

𝜋

𝑠𝑖𝑛𝜔𝑐𝑛

𝜔𝑐𝑛

Typical plot of the ideal LPF filter for 𝜔𝑐 =𝜋

4

is shown below

9


From the plot of ℎ[𝑛] it is clear that ℎ 𝑛 ≠0 for 𝑛 < 0 which means that the filter is

not causal

If the filter is not causal this means that the

filter output samples 𝑦[𝑛] depends on the

future values of the input 𝑥[𝑛 + 1], 𝑥[𝑛 +2] which is impossible to estimate

10

Processing continuous time

signal with digital filterConsider a discrete-time filter is to be used

to low pass filter a continuous time signal

The complete system may be presented

as shown below

11

Determining specifications for a

discrete time filter

When designing a LPF we need its

magnitude frequency response to be as

shown

12

Design of discrete time IIR filters

from continuous time filters

The traditional approach to design IIR

filters involves the transformation of a

continuous-time filter into a discrete-time

filter meeting prescribed specifications

This design procedure can be achieved by

finding the system function 𝐻𝑐(𝑠) or the

impulse response of the continuous filter

ℎ𝑐(𝑡)

13



There are four well known analog filters

used in the design of IIR digital filters

these are

Butterworth

Chebyshev I

Chebyshev II

Elliptic

14



The system function 𝐻(𝑧) or the impulse

response ℎ 𝑛 for the discrete-time filter is

obtained by using either

Impulse invariance method

Bilinear transformation

15

IIR filter design by Impulse

invarianceIn the impulse invariance method we can

design an IIR filter by sampling the

impulse response of analog filter

according to

ℎ 𝑛 = 𝑇𝑑ℎ𝑐 𝑛𝑇𝑑Where 𝑇𝑑 represents the sampling interval

16


invarianceIf the continuous time filter is band limited,

so that

𝐻𝑐 𝑗Ω = 0 Ω ≥𝜋

𝑇𝑑

𝐻 𝑒𝑗𝜔 = 𝐻𝑐(𝑗𝜔

𝑇𝑑) 𝜔 ≤ 𝜋

17


invariance

The discrete-time filter and the continuous

time filter frequency responses are related

by a linear scaling of the frequency axis

which is given by 𝜔 = Ω𝑇𝑑 for 𝜔 ≤ 𝜋

Where 𝜔 is the discrete frequency axis

and Ω is the continuous frequency axis

18

Practical IIR filter and aliasing

Any practical continuous filter can not be

exactly band limited

This means that the sampling process will

cause interference between successive

samples which causes aliasing as shown

19

Practical IIR filter and aliasing

The aliasing problem makes the impulse

invariance method not a suitable method for

the design of high pass filters

However if the continuous time filter

approaches zero at high frequencies, the

aliasing is negligible and can be ignored

20

IIR filter and continuous to discrete

system function transformation

Consider the system function of the

continuous-time filter expressed in terms of

partial fractions as shown below

The corresponding impulse response is

(inverse Laplace transform)

𝑁 represents the number

of poles

21



The impulse response of the discrete time filter

obtained by sampling 𝑇𝑑ℎ𝑐(𝑡) as shown

The system function of the discrete time filter is

22



If we compare the system function for both

continuous time 𝐻𝑐 𝑠 and discrete time 𝐻(𝑧)system function, we observe that a pole at

𝑠 = 𝑠𝑘 in the S- plane transforms to a pole at

𝑧 = 𝑒𝑠𝑘𝑇𝑑 in the z plane

Also the coefficients in the partial fraction for

both 𝐻𝑐 𝑠 and 𝐻(𝑧) are equal, except for the

scaling multiplier 𝑇𝑑

23



Note that 𝑠𝑘 is acomplex number which can be

written as 𝑠𝑘 = 𝜎 + 𝑗𝜔 this means that 𝑧 =𝑒𝜎𝑇𝑑𝑒𝑗𝜔𝑇𝑑

If the real par of 𝑠𝑘 is less than zero (𝜎 negative),

then 𝑧 < 1 , the corresponding pole in the

discrete time filter falls inside the unit circle

This means that the causal discrete time filter is

stable

24

Example 1

Example 1 convert the analog filter defined

by the system function 𝐻𝑎 𝑠 =1

(𝑠+0.1)2+9into

a digital IIR filter by means of impulse

invariance method

25

Example1 solution

Solution

Since the filter system function 𝐻𝑐 𝑠 for the

continuous filter is given we can find

𝐻(𝑧) directly through the partial fraction (no

need to go back to the time domain and then

use sampling)

From 𝐻𝑐 𝑠 we can find that the filter has two

poles located at

𝑠1 = −0.1 + 𝑗3 and 𝑠2 = −0.1 − 𝑗3

26

Example1 solution

Solution

If we analyze 𝐻𝑐 𝑠 by the partial fraction

we get

To find 𝐻(𝑧) we use the expression

developed in slide 23

27

Example1 solution

If we plot the poles and zeros of the

previous example we can see that 𝜎 < 0and both poles are located in the left side

of the s plane as shown below

28

Example 1 solution

If we manipulate 𝐻(𝑧) expression we find that

𝐻(𝑧) can be written as

If we compare this expression for 𝐻(𝑧) with


−𝑘

1+ 𝑘=1𝑁 𝑎𝑘𝑧

−𝑘 we can find the filter

coefficients are

𝑏0 = 1, 𝑏1 = −𝑒−0.1𝑇𝑑cos(3𝑇𝑑), 𝑎0 = 1, 𝑎1 =

− 2𝑒−0.1𝑇𝑑 cos 3𝑇𝑑 , 𝑎2 = 𝑒−0.2𝑇𝑑,

29

2

Example 2

The system function of a discrete time

system is 𝐻 𝑧 =1

1−𝑒−0.2𝑧−1+

1

1−𝑒−0.4𝑧−1,

determine the filter coefficients

30

Example 2 solution

Note the discrete time system can be

rewritten as 𝐻 𝑧 =2− 𝑒−0.4+𝑒−0.2 𝑧−1

1− 𝑒−0.4+𝑒−0.2 𝑧−1+𝑒−0.6𝑧−2

If we compare this expression for 𝐻(𝑧) with


−𝑘


−𝑘 we can find the filter

coefficients are

𝑏0 = 2 , 𝑏1 = 𝑒−0.4 + 𝑒−0.2 , 𝑎0 = 1 , 𝑎1 =

− 𝑒−0.4 − 𝑒−0.2, 𝑎2 = 𝑒−0.6

31

Example 3

Design a Butterworth filter whose continuous

frequency response is given by

𝐻𝑐(𝑗Ω)2 =

1

1+Ω

Ω𝑐

2𝑁

using the impulse invariance method

The filter specifications are

32

Example 3 solution

Solution

Since the parameter 𝑇𝑑 cancels in the impulse

invariance procedure we can select 𝑇𝑑 = 1 ,

therefore Ω = 𝜔

In order to deign the filter we need to transform

the discrete filter specifications to specifications

on the continuous time filter

33

Example 3 solution

Solution

The continuous time filter specifications became

The filter magnitude at the pass band and stop

band frequencies is given by

34

Example 3 solution

Solution

From the above magnitude conditions, the

magnitude squared of the continuous filter at

both the pass and stop bands is given by

0.89125 2 =1

1+0.2𝜋

Ω𝑐

2𝑁

0.17783 2 =1

1+0.3𝜋

Ω𝑐

2𝑁

35

Example 3 solution

Solution

If we solve the above two equations for 𝑁 and

Ω𝑐 we got 𝑁 = 5.8858 ≈ 6 and Ω𝑐 = 0.7032

Since 𝑁 = 6 then the filter has 12 poles of the

magnitude squared function 𝐻𝑐 𝑠 𝐻𝑐 −𝑠 =1

1+𝑆

Ω𝑐

2𝑁

The poles can easily be found by using the

following matlab commands

36

Example 3 solution

𝑎 = 1

b=[68.399 0 0 0 0 0 0 0 0 0 0 0 1]

[𝑧 𝑝 𝑘] = 𝑡𝑓2𝑧𝑝(𝑎, 𝑏)

37

Example 3 solution

Solution

The poles are uniformly

distributed in angle of360°

12= 30° on a circle of

radius Ω𝑐 = 0.7032 as

shown

38

Example 3 solution

Solution

The poles that describe a stable system are

those poles located in the left side of the s-plane

The number of these poles is 6 and one solution

for these poles is on the assumption that the first

pole located a at an angle of 105°

39

Example 3 solution

Solution

Now the system function of the continuous time

filter is given by

To find 𝐻(𝑧) we use the expression developed in

slide 23

40

Example 3 solution

Solution

In order to find the filter coefficients again we need

to write 𝐻(𝑧) as a rational function and then find

the coefficients as we have did in the previous two

examples

41

Impulse invariance summery

To summarize, the impulse invariance method maps

the s-plane to the z-plane by sampling the impulse

response of continuous (analog) time filter

Impulse invariance suffers from aliasing problem

and therefore is not suitable for HPF design

42

Impulse invariance summery

In order to find the filter coefficients follow these steps

1. find 𝐻𝑐 𝑠 , expand it in partial fraction, find the poles

of 𝐻𝑐 𝑠

2. Next find the discrete system function according to

𝐻 𝑧 = 𝑘=1𝑁 𝑇𝑑𝐴𝑘

1−𝑒𝑠𝑘𝑇𝑑𝑧−1

3. Convert 𝐻(𝑧) to rational function

4. Compare the rational expression for 𝐻(𝑧) with


−𝑘


−𝑘 and determine the coefficients

accordingly

43


The bilinear transformation is an algebraic

transformation between the 𝑠 and 𝑧 planes

Bilinear transformation maps the entire 𝑗Ω-axis in the s-plane to one revolution of the

unit circle in the z-plane

Bilinear transformation avoids the aliasing

problem associated with time invariant

approach

44


The bilinear transformation corresponds to

replacing 𝑠 by

𝑠 =2

𝑇𝑑

1 − 𝑧−1

1 + 𝑧−1

This means that the discrete and

continuous system functions are related by

𝐻 𝑧 = 𝐻𝑐2

𝑇𝑑

1 − 𝑧−1

1 + 𝑧−1

45


To develop the properties of the algebraic

transformation we solve for 𝑧 to obtain

𝑧 =2

𝑇𝑑

1 +𝑇𝑑2𝑠

1 −𝑇𝑑2𝑠

Substituting 𝑠 = 𝜎 + 𝑗𝜔, we got

𝑧 =1 + 𝜎𝑇𝑑2+ 𝑗Ω𝑇𝑑2

1 − 𝜎𝑇𝑑2− 𝑗Ω𝑇𝑑2

46


Again as in the time invariant procedure, if 𝜎 <0, 𝑧 < 1, for any value of Ω

This means the poles fall in the left half of the s-

plane or inside the unit circle of the z-plane,

which means that the system is stable

Similarly if of 𝜎 > 0, 𝑧 > 1, for any value of Ω

The poles fall in the right half of the s-plane or

outside the unit circle of the z-plane, which

means that the system is unstable

47


In order to show that the 𝑗Ω-axis of the s-plane

maps onto the unit circle, substitute 𝜎 = 0 into

𝑧 =1+𝜎𝑇𝑑2+𝑗Ω𝑇𝑑2

1−𝜎𝑇𝑑2−𝑗Ω𝑇𝑑2

and simplify the expression we

got

𝑧 =1 + 𝑗Ω

𝑇𝑑2

1 − 𝑗Ω𝑇𝑑2

Recall that 𝑧 = 𝑒𝜎𝑒𝑗𝜔 = 𝑒𝑗𝜔

48


The expression for 𝑠 =2

𝑇𝑑

1−𝑧−1

1+𝑧−1can be

rewritten as s =2

𝑇𝑑

1−𝑒−𝑗𝜔

1+𝑒−𝑗𝜔or equivalently

𝑠 = 𝜎 + 𝑗Ω =2

𝑇𝑑

2𝑒−𝑗𝜔2 sin

𝜔

2

2𝑒−𝑗𝜔2 cos

𝜔

2

=2𝑗

𝑇𝑑tan

𝜔

2

From which we can conclude that

Ω =2

𝑇𝑑tan

𝜔

2or 𝜔 = 2𝑡𝑎𝑛−1

Ω𝑇𝑑

2

49

Mapping between the z-plane

and the s-planeThe properties of the bilinear transformation as

mapping from the s-plane to the z-plane are

illustrated graphically as shown below

50

Axis mapping notes

Note that the Bilinear transformation maps

the range of continuous frequencies 0 ≤Ω ≤ ∞ maps to 0 ≤ 𝜔 ≤ 𝜋 on the discrete

frequency axis

Also the range of continuous frequencies

−∞ ≤ Ω ≤ 0 maps to −𝜋 ≤ 𝜔 ≤ 0 on the

discrete frequency axis as illustrated in the

next slide

51

Axis mapping notes

52

Example 1

Design a digital Butter worth low pass

filter whose specifications are

Using the bilinear transformation

53

Example 1 solution

The design procedure starts by reflecting the

specifications of the discrete filter to

specifications of the digital filter as shown

Let 𝑇𝑑 = 1

54

Example 1 solution

The magnitude of the filter at the pass

band frequency Ω𝑝 = 2 tan0.2𝜋

2is given

by

Also note that the magnitude of the filter at

the stop band Ω𝑠 = 2 tan0.3𝜋

2is given by

55

Example 1 solution

By substituting the values of Ω𝑝 and Ω𝑠 in

the magnitude squared of the Butterworth

filter we got the following equations

By solving these two equations for Ω𝑐 and

𝑁 we got 𝑁 = 5.305 ≈ 6 and Ω𝑐 = 0.766

56

Example 1 solution

If we plot the poles of 𝐻𝑐 𝑠2 of the filter

we got the following graph

57

Example 1 solution

From the pole plot it can be shown that

58

Example 2

Design a single pole low pass digital filter

with a 3-dB bandwidth of 0.2𝜋, using the

bilinear transformation applied to the

analog filter defined by 𝐻 𝑠 =Ω𝑐

𝑠+Ω𝑐where

Ω𝑐 is the 3-dB bandwidth of the filter

59

Example 2 solution

By using the linear transformation we have

Ω𝑐 =2

𝑇𝑑tan

0.2𝜋

2=0.65

𝑇𝑑

The system function of the analog filter can

be written as 𝐻 𝑠 =

0.65

𝑇𝑑

𝑠+0.65

𝑇𝑑

By substituting 𝑠 =2

𝑇𝑑

1−𝑧−1

1+𝑧−1, we got the

discrete system function as

60

Example 2 solution

𝐻 𝑧 =

0.65

𝑇𝑑

21−𝑧−1

1+𝑧−1+0.65

𝑇𝑑

= 0.2451+𝑧−1

1−0.509𝑧−1

61

IIR filter design in MATALB

We have seen from the previous discussion both

the impulse invariance and bilinear

transformations needs too many computations

A simpler way to design an IIR filter can be done

easily with the aid of MATLAB

It is easy to show this by an example

62

IIR filter design in MATLAB

Design a band pass filter with the following

specifications

Pass band frequencies from 1000 to 2000 Hz

The stop bands starting 500 Hz away on either side

A 10 kHz sampling frequency

At most 1 dB of pass band ripple and at least 60 dB of

stop band attenuation

63


Solution

In order to design the filter we need to specify its

coefficients

This can be achieved by using the following

statements in MATLAB

𝑓𝑠 = 10 𝑘𝐻𝑧 % Sampling frequency

𝑓𝑝1 = 1 𝑘𝐻𝑧 % The first band pass frequency

𝑓𝑝2 = 2 𝑘𝐻𝑧 % The second band pass frequency

𝑓𝑠1 = 500 𝐻𝑧 % The first stop band frequency

𝑓𝑠2 = 2500 𝐻𝑧 % The second stop band frequency

64


𝑅𝑝 = 1 𝑑𝐵 % The ripple in the pass band

𝑅𝑆 = 60 𝑑𝐵 % The attenuation in the stop band

The filter order and normalized frequencies can be

computed by using the following statement

[𝑛,𝑊𝑛] = 𝑏𝑢𝑡𝑡𝑜𝑟𝑑𝑓𝑝1 𝑓𝑝2

𝑓𝑠,𝑓𝑠1 𝑓𝑠2

𝑓𝑠, 𝑅𝑝, 𝑅𝑠

The filter coefficients can be computed from the

following statements

𝑏, 𝑎 = 𝑏𝑢𝑡𝑡𝑒𝑟 𝑛,𝑊𝑛 ;

65


More examples to design IIR filter is by the use of these

statements [𝑏, 𝑎] = 𝑏𝑢𝑡𝑡𝑒𝑟(𝑛,𝑊𝑛, 𝑜𝑝𝑡𝑖𝑜𝑛𝑠)

[𝑏, 𝑎] = 𝑏𝑢𝑡𝑡𝑒𝑟(5,0.4); % Lowpass Butterworth

[𝑏, 𝑎] = 𝑐ℎ𝑒𝑏𝑦1(𝑛, 𝑅𝑝,𝑊𝑛, 𝑜𝑝𝑡𝑖𝑜𝑛𝑠)

[𝑏, 𝑎] = 𝑐ℎ𝑒𝑏𝑦1(4,1, [0.4 0.7]); % Bandpass Chebyshev Type I

[𝑏, 𝑎] = 𝑐ℎ𝑒𝑏𝑦2(𝑛, 𝑅𝑠,𝑊𝑛, 𝑜𝑝𝑡𝑖𝑜𝑛𝑠)

[𝑏, 𝑎] = 𝑐ℎ𝑒𝑏𝑦2(6,60,0.8, ′ℎ𝑖𝑔ℎ′); % Highpass Chebyshev Type II

[𝑏, 𝑎] = 𝑒𝑙𝑙𝑖𝑝(𝑛, 𝑅𝑝, 𝑅𝑠,𝑊𝑛, 𝑜𝑝𝑡𝑖𝑜𝑛𝑠)

[𝑏, 𝑎] = 𝑒𝑙𝑙𝑖𝑝(3,1,60, [0.4 0.7], ′𝑠𝑡𝑜𝑝′); % Bandstop elliptic

66

Filter design in MATLAB

We can design IIR or FIR filters in MATALB

using the filter design and analysis tool

FDATOOL

In the command window of MATLAB type

FDATOOL

A window the filter design window will appear as

shown in the next slide

67

68

Using FDATOOL

To design a filter, you need to specify the

sampling rate, pass and stop band frequencies

You can also determine the filter type,

implementation, FIR or IIR

Once the filter is designed you can view or

export the filter coefficients to workspace or file

using the command file as illustrated in the next

slide

69

Exporting filter coefficients

70

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Filter design techniques ch7 iir

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