session 8 i i r filter (infinite impulse...

Session 8 : IIR Filter

I I R Filter(Infinite Impulse Response)

Session 8

Ir. Dadang Gunawan, Ph.DElectrical Engineering

University of Indonesia


The Outline8.1 State-of-the-art8.2 Coefficient Calculation Method for IIR Filter8.2.1 Pole-Zero Placement Method8.2.2 Impulse Invariant Method8.2.3 Matched-z-transform (MZT) Method 8.2.4 Bilinear z-transform (BZT) Method8.3 Classical Analog Filter8.3.1 Butterworth Filter8.3.2 Chebysev Filter8.3.3 Elliptic Filter8.4 Review


State of the art• The basic IIR filter is characterized by the follo-wing

equation :

• Where h(k) is the impulse response of the filter which is theoretically infinite in duration

∑∑

∑

==

∞

=

−−−=

−=

M

kk

N

kk

k

knxaknxbny

knxkhny

00

0

)()()(

)()()(

8.1


State of the art (cont’d)

• bk and ak are the coefficients of the filter • x(n) and y(n) are the input and output to the filter• Transfer function for the IIR filter is :

∑

∑

=

−

=

−

−−

−−

=++++++

= M

k

kk

N

k

kk

MM

NN

za

zb

zazaazbzbbzH

0

01

10

110

......)(

8.1



• The important thing is to find suitable values for the coefficients bk and ak

• Note that the current output y(n) is a function of the past outputs y(n-k) . So that it show the feedback system of some sort

• The strength of IIR filters comes from the flexibility the feedback arrangement provides

• Remember that the transfer function of IIR filter can be shown as the pole and zero equations

8.1



• Here is an example tolerance scheme for an IIR bandpass filter

8.1

Figure 8.1



ε2 : passband ripple parameterδp : passband deviationδs : stopband deviationfp1 and fp2 : passband edge frequencyfp1 and fp2 : stopband edge frequencyAp : passband ripple = 10. log10(1+ ε2)

= 20. log10(1- δp)As : stopband attenuation = -20. log10(δp)

8.1


Coefficient calculation methods for IIR filters

There are 4 methods to calculate the coefficients :1. Pole-zero placement2. Impulse invariant3. Matched z-transform4. Bilinear z-transform

Learn carefully

8.2


Pole-zero placement Method

• The idea is : when a zero is placed at a given point on the z-plane, the frequency response will be zero at the corresponding pointwhile a pole produces a peak at the corresponding frequency point

• Note that for the coefficient of the filter to be real, the poles and zeros must either be real

Watch the figure 8.2 below

8.2.1


Pole-zero placement Method (cont’d)

8.2.1

Figure 8.2



• Here is an example to make a bandpass digital filter which is required to meet the following specifications :- complete signal rejection at dc and 250 Hz- a narrow passband centered at 125 Hz- a 3dB bandwidth of 10 Hz

Fist we must determine where to place the poles and zeros on the z-plane . Watch the frequency on 250 Hz and 125 Hz

8.2.1



These are at angles of 0o and 360o x 250/500 = 180o

and the place poles at +- 360o x 125/500 = +-90o

The radius, r, of the poles is determined by the desi-red bandwidth. An approximate relationship betweenr , for r > 0.9 and bandwidth bw is given by :r = 1 – (bw/Fs).πSo that, by substituting the value of bw=10 Hz and Fs=500 Hz , giving r = 0.937After it, we can draw the pole-zero diagram below :

8.2.1



8.2.1

Figure 8.3



From the pole-zero diagram, the transfer function can be written as follow :

2

2

2

2

2/2/

877969.011

877969.01

))(()1)(1()(

−

−

+−

=

+−

=

−−+−

=

zz

zz

rezrezzzzH jj ππ

8.2.1



So that, the difference equation is :

Look at again the transfer function. It shows filter which is a second-order section, with coefficients :

b0 = 1 a1 = 0b1 = 0 a2 = 0.877969b2 = -1

)2()()2(877969.0)( −−+−−= nxnxnyny

8.2.1


Impulse Invariant Method

• First, consider these component :- H (s) : a suitable analog transfer function - h (t ) : the impulse response- h (nT) : z transforming with T sampling interval - H (z) : desired transfer function

• Those component are useful and obtained by using Laplace Transform and also z-transformation

• Look at the example on DSP textbook

8.2.2


Impulse Invariant Method (cont’d)

Here are the steps in Impulse Invariant Method :1. Determine a normalized analog filter H(s) that

satisfies the specifications for the desired digital filter2. If necessary, expand H(s) using partial fraction to

simplify the next step3. Obtain the z-transform of each partial fraction to

obtain :

∑∑=

−= −

→−

M

KTp

KM

K K

K

zeC

psC

k1

11 1

8.2.2


Impulse Invariant Method (cont’d)

4. Obtain H(z) by combining the z-transforms of the partial fractions into second-order terms and possibly one-first-order term. If the actual sampling frequency is used then multiply H(z) by T

8.2.2


Matched z-transform (MZT) method

• It provides a simple way to convert an analog filter into an equivalent digital filter

• The idea is : each of the poles and zeros of the analog filter is mapped directly from the s-plane to the z-plane using the following equation :

• It maps a pole or zero at the location s=a in the s-plane, onto a pole or zero in the z-plane at z=eaT

)1()( 1 aTezas −−→−

8.2.3


Matched z-transform (MZT) method (cont’d)

• Here is an example for having a filter with a 3 dB cutoff frequency of 150 Hz in sampling frequency of 1.28 kHz. The normalized of transfer function of an analog filter is given by :

121)(

2 ++=

sssH

To obtain the transfer function, watch the answer below

8.2.3



The cutoff frequency may be expressed as ωc=2π x 150 = 942.4778 rad/s . The transfer function ofthe denormalized analog filter is obtained by repla-cing s by s/ωc :

22

2

2

)()('

cc

c

css

ss

sHsH

ωωωω

++=

==

Find poles byabc formula

8.2.3



Remember : so that :

We have the real and imaginary poles :

jjp

p

ci

cr

43.66622

43.66622

=+=

−=−=

ω

ω

acbbs 42

212 −±

−=

8.2.3



Then, prT = -0.52065 cos (piT) = 0.8675piT = +0.52065 e prT = 0.5941

The transfer function become :

21

5

594134.0030818.11108876.8)( −− +−

×=

zzzH

8.2.3


Bilinear z-transform (BZT) Method

• It is the most important method• The idea is: to convert an analog filter H(s) into an

equivalent digital filter is to replace s as follow:

• That transformation maps the analog transfer function, H(s), from the s-plane into the discrete transfer function, H(z), in the z-plane

Tkork

zzks 21,

11

==+−

=

8.2.4


BZT Method (cont’d)

• Look at the figure below. It shows the transforma-tion using BZT method

8.2.4

S-plane Z-planeFigure 8.4



• Here are the steps for using BZT1. Use the digital filter specifications to find suitable

normalized, prototype, analog low pass filter H(s)2. Determine and prewarpe the bandedge or critical

frequencies of the desired filterwhen : ωp = specified cutoff frequency

ωp’ = prewarped cutoff frequency

8.2.4



Remember that in bandpass and bandstop filter, there are the lower and upper passband edge frequencies or we can say ωp1’ and ωp2’ .

3. Denormalize the analog prototype filter by replacing s in the transfer function, H(s), using these following transformation :

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2tan'

Tpp

ωω

8.2.4



lowpass to lowpass

lowpass to highpass

lowpass to bandpass

lowpass to bandstop2

02

20

2

'

'

ω

ω

ω

ω

+=

+=

=

=

sWss

Wsss

ss

ss

p

p

8.2.4



where :

4. Apply the BZT to obtain the desired digital filter transfer function, H(z), by replacing s in the frequency-scaled (i.e. denormalized) transfer function, H’(s) as follows

11

+−

=zzs

121220 '''' pppp W ωωωωω −==

8.2.4


Example of BZT Method

Learn in DSP textbook [Ifeachor and Jervis] pages 474 – 477 . Its very urgent !

Lowpass filter

Bandpass filter

Highpass filter

8.2.4


Classical Analog Filter

• There are four types of Classical Analog filter :1. Butterworth filter2. Chebysev type I3. Chebysev type II4. Elliptic• All types of filter are derived from lowpass prototype

filter

8.3


Butterworth Filter

Here is sketch of frequency response on Butterworth filter

8.3.1

Figure 8.5


Butterworth Filter (cont’d)

• The important equations on Butterworth filter are :

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−

≥

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

pp

ps

Ap

As

N

pp

NH

ωω

ωω

ωlog2

110

110log

1

1)(10

10

22

Magnitude squareFrequency response Filter order

8.3.1


Chebysev Filter

• Chebysev Type I : equal ripple in the passband,monotonic in the stopband

• Chebysev Type II : equal ripple in the stopband,monotonic in the passband

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−

≥−

−

pp

ps

Ap

As

N

ωω1

10

101

cosh

110

110cosh

8.3.2


Chebysev Filter

Here is sketch of frequency response on Chebysev

8.3.2

Type 1 Type 2Figure 8.6


Elliptic Filter

• The elliptic filter exhibits equiripple behavior in both the passband and the stopband

• This is the following magnitude-squared response:

• GN(ω’) is a Chebysev rational function

)(1)'( 222

2

ωεω

NGKH

+=

8.3.3


Elliptic Filter

Here is sketch of frequency response on Elliptics

8.3.3

Figure 8.7


Preparation for the review

IT IS THE END OF THIS SESSIONPREPARE FOR THE REVIEW

Open your DSP textbook and read more

ARE YOU READY ? IT WILL TAKE ONLY A FEW MINUTES

8.4


Review

1. What is the meaning of Infinite Impulse Response and its effect in digital filter ?

2. Compare the performance of three filter : Butterworth, Chebysev, and Elliptic

3. Find out several applications of IIR filter in our daily life. Refer to internet, magazine, journal.

8.4

Not so hard… Isn’t it ?

session 8 i i r filter (infinite impulse...

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