학습에앞서

§ 학습목표– IIR 필터의동작을이해한다.– IIR 필터의 pole, zero 개념을이해한다.– IIR 필터의주파수응답을학습한다.

2

The General IIR Difference Equation

§ The general difference equation of digital filters is

– {al}: feedback coefficients– {bk}: feedforward coefficients– N+M+1: Number of coefficients– If {al} are all zero, the difference equation reduces to the difference

equation of an FIR system.

3

åå==

-+-=M

kk

N

ll knxblnyany

01][][][

One Feedback Term

§ First-order case where M = N = 1, i.e.,

§ Example

§ Initial Rest Condition: y[n] = 0, for n<0 because x[n] = 0, for n<0

4

y[n] = a1y[n -1]+ b0x[n] +b1x[n -1]

y[n] = 0.8y[n -1]+ 5x[n]

y[0] = 0.8y[-1] + 5x[0]Need y[-1] to get started

Time-Domain Response (1/2)

§ Compute y[n]

§ Continue the recursion

5

]3[2]1[3][2][][5]1[8.0][

-+--=+-=

nnnnxnxnyny

ddd

Time-Domain Response (2/2)

§ Plot y[n]

6

Impulse Response of a First-Order IIR System

§ Consider the first-order recursive difference equation with b1=0.

§ Impulse Response

7

][]1[][ 01 nxbnyany +-=

][]1[][ 01 nbnhanh d+-=

Example: Impulse Response

§ Example

8

][3]1[8.0][ nxnyny +-=

][)8.0(3][)(][ 10 nunuabnh nn ==

System Function

9

111

0

0

110

01010

if1

)(][)(

azza

b

zabzabznuabzHn

n

n

nn

n

nn

>-

=

===

-

¥

=

-¥

=

-¥

-¥=

- ååå

Another First-Order IIR System

§ Another First-Order IIR Filter

Because the system is linear and time-invariant, it follows

10

y[n] = a1y[n -1]+ b0x[n] +b1x[n -1]

H(z) =b0

1- a1z-1 +

b1z-1

1- a1z-1 =

b0 + b1z-1

1 - a1z-1

]1[)(][)(][ 11110 -+= - nuabnuabnh nn

shifta is1-z

Step Response of a First-Order Recursive System (1/2)

11

][]1[][ 01 nxbnyany +-=

Step Response of a First-Order Recursive System (2/2)

§ Plot Step Response

12

y[n] = 0.8y[n -1]+ 3u[n]

( ) ][8.0115][1

1][ 1

1

11

0 nunuaabny nn

++

-=--

=

Delay Property

§ Delay in Time-Domain <-->Multiply X(z) by z-1

13

x[n]« X(z)

x[n -1]« z -1X(z)

Proof: x[n -1]z -nn= -¥

¥

å = x[l]z- (l+1)

l=-¥

¥

å

= z-1 x[l]z -l

l= -¥

¥

å = z-1X(z)

System Function of an IIR Filter

§ System function of the first-order difference equation

§ Example

14

y[n] = a1y[n -1]+ b0x[n] +b1x[n -1]

Y (z) = a1z-1Y(z) + b0X(z) + b1z

-1X(z)

H(z) =Y (z)X(z)

=b0 +b1z

-1

1- a1z-1 =

B(z)A(z)

(1 - a1z-1)Y (z) = (b0 + b1z

-1 )X(z)

]1[2][2]1[8.0][ -++-= nxnxnyny

)(8.01

22)( 1

1

zXzzzY ÷÷

ø

öççè

æ-+

= -

-

Poles and Zeros (1/2)

§ Roots of Numerator & Denominator

– Zeros at H(z)=0

– Poles at H(z)à∞

15

H(z) =b0 + b1z

-1

1 - a1z-1 ® H (z) =

b0z + b1

z - a1

b0z + b1 = 0 Þ z = -b1

b0

z - a1 = 0 Þ z = a1

16

1

1

8.0122)( -

-

-+

=zzzH

Zero at z = -1

Pole at z = 0.8

Poles and Zeros (2/2)

Frequency Response

§ Example

17

w

ww

ˆ

ˆˆ

1

1

8.0122)(

8.0122)( j

jj

eeeH

zzzH -

-

-

-

-+

=®-+

=

== )()()( ˆ*ˆ2ˆ www jjj eHeHeH ww

w

w

w

w

ˆcos6.164.1ˆcos88

8.0122

8.0122

ˆ

ˆ

ˆ

ˆ

-+

=-+

×-+

-

-

j

j

j

j

ee

ee

0ˆ@ 40004.0

88)(2ˆ ==

+= wwjeH

Frequency Response Plot

18

w

ww

ˆ

ˆˆ

8.0122)( j

jj

eeeH -

-

-+

=

Three-Dimensional Plot of a System Function

19

Three Domains

§ Relationship among the n-, z-, and domains.

20

Use H(z) to getFreq. Response wjez =

-w

21

z-Transform Tables

Summary

§ This lecture introduced a new class of LTI systems that have infinite duration impulse responses, i.e., a IIR system.– IIR digital filters involve previously computed values of the output

signal as well as values of the input signal in the computation of the present output.

§ The z-transform system functions for IIR filters are rational functions that have poles and zeros.– Poles of the system function H(z) are important because properties

such as the shape of the frequency response or the form of the impulse response can be inferred quickly from the pole locations.

22

åå==

-+-=M

kk

N

ll knxblnyany

01][][][