Lectures 9 IIR Systems: First Order System · 2008. 2. 29. · 1 EE3054 Signals and Systems...

1

EE3054

Signals and Systems

Lectures 9

IIR Systems: First Order System

Yao Wang

Polytechnic University

Some slides included are extracted from lecture presentations prepared by McClellan and Schafer

2/29/2008 © 2003, JH McClellan & RW Schafer 2

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2

FIR system: Review

� Described by a feedforwarddifference equation

� Impulse response is finite duration (finite impulse response or FIR)

� Characterized by impulse response h[n], system function H(z) (Z-transform of h[n]) and frequency response H(e^jw)

∑=

−==

=

M

k

n

knxkhnhnxny

bnh

0

][][][*][][

][

∑=

−=M

k

k knxbny

0

][][

)()(][ ω̂jeHzHnh ↔↔

IIR System: General

∑∑==

−+−=M

k

k

N

k

k knxbknyany

01

][][][

Weighted average of

input samples

Weighted average of

past output samples

(feedback)

Still a linear time-invariant system

Impulse response is infinitely long generally

Called Infinite Impulse Response (IIR) system

3

Roadmap

� First discuss first order system

� Time domain: output for given input, impulse response

� Z-domain: transfer function, characterization by poles, how to compute output using Z-domain

� Frequency response

� Next discuss second order system

� Finally to general IIR system

∑=

−+−=M

k

k knxbnyany

0

1 ][]1[][

∑=

−+−+−=M

k

k knxbnyanyany

0

21 ][]2[]1[][

2/29/2008 © 2003, JH McClellan & RW Schafer 6

ONE FEEDBACK TERM (First

Order System)

� CAUSALITY

� NOT USING FUTURE OUTPUTS or INPUTS

y[n] = a1y[n −1]+ b0x[n] +b1x[n −1]FIR PART of the FILTER

FEED-FORWARDPREVIOUS

FEEDBACK

� ADD PREVIOUS OUTPUTS

4

2/29/2008 © 2003, JH McClellan & RW Schafer 7

FILTER COEFFICIENTS

� ADD PREVIOUS OUTPUTS

� MATLAB

� yy = filter([3,-2],[1,-0.8],xx)

y[n] = 0.8y[n −1]+ 3x[n] −2x[n −1]

SIGN CHANGEFEEDBACK COEFFICIENT

2/29/2008 © 2003, JH McClellan & RW Schafer 8

COMPUTE OUTPUT

5

2/29/2008 © 2003, JH McClellan & RW Schafer 9

COMPUTE y[n]

� FEEDBACK DIFFERENCE EQUATION:

y[n] = 0.8y[n −1]+ 5x[n]

y[0] = 0.8y[−1] + 5x[0]

� NEED y[-1] to get started

2/29/2008 © 2003, JH McClellan & RW Schafer 10

AT REST CONDITION

� y[n] = 0, for n

6

2/29/2008 © 2003, JH McClellan & RW Schafer 11

COMPUTE y[0]

� THIS STARTS THE RECURSION:

� SAME with MORE FEEDBACK TERMS

y[n]= a1y[n −1]+ a2y[n − 2] + bk x[n − k]k=0

2

∑

2/29/2008 © 2003, JH McClellan & RW Schafer 12

COMPUTE MORE y[n]

� CONTINUE THE RECURSION:

7

2/29/2008 © 2003, JH McClellan & RW Schafer 13

PLOT y[n]y[n] has infinite duration!

Is IIR system LTI?

� If x[n]=0, y[n]=0 for n

8

Properties of LTI system:

Review

� Any LTI system can be characterized by

its impulse response h[n]=T(δ[n])

� Output to any input is related by

� y[n]=x[n]*h[n]

2/29/2008 © 2003, JH McClellan & RW Schafer 16

y[n]= a1y[n −1]+ b0x[n]

IMPULSE RESPONSE

u[n] =1, for n ≥ 0

h[n]= a1h[n −1]+ b0δ[n]

][)(][ 10 nuabnhn=

h[n] has infinite duration!

9

2/29/2008 © 2003, JH McClellan & RW Schafer 17

PLOT IMPULSE RESPONSE

h[n] = b0 (a1)nu[n] = 3(0.8)nu[n]

� Show that for the example system� y[n]=0.8 y[n-1] + 5 x[n]

y[n] = x[n]* h[n] yields same result as the direct computation using recursion

h[n]=5 * 0.8^n u[n]

x[n]=2δ[n]-3δ[n-1]+2 δ[n-3]

By linearity and time invariance:

y[n] = 2 h[n] – 3 h[n-1] + 2 h[n-3]

10

� When x[n] and h[n] are both infinite duration,

numerical computation of convolution is

generally infeasible.

� But we can still use the recursion based on the

difference equation, although this is tedious.

� Z-transform comes to rescue!

� Y(z)= X(z) H(z)

� Determine H(z), X(z), Y(z)

� From Y(z), determine y[n] (inverse Z-transform)

2/29/2008 © 2003, JH McClellan & RW Schafer 20

CONVOLUTION PROPERTY

� MULTIPLICATION of z-TRANSFORMS

� CONVOLUTION in TIME-DOMAIN

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

h[n]y[n] = h[n]∗ x[n]x[n]

H(z)

IMPULSE

RESPONSE

11

2/29/2008 © 2003, JH McClellan & RW Schafer 21

System Function of First

Order System

� Impulse response:

� Infinite duration!

� Z-transform (System

Function) H(z) = h[n]z−n

n=−∞

∞

∑

∑∑∞

=

−−∞

−∞=

==0

1010 ][)()(n

nnn

n

n zabznuabzH

][)(][ 10 nuabnhn=

2/29/2008 © 2003, JH McClellan & RW Schafer 22

Derivation of H(z)

� Recall Sum of Geometric Sequence:

� Yields a COMPACT FORMrn

n=0

∞

∑ =1

1− r

11

1

0

0

1

10

0

10

if1

)()(

azza

b

zabzabzHn

n

n

nn

>−

=

==

−

∞

=

−∞

=

− ∑∑If |r|

12

2/29/2008 © 2003, JH McClellan & RW Schafer 23

Recap:

� FIRST-ORDER IIR FILTER:

y[n]= a1y[n −1]+ b0x[n]

H(z) =b0

1− a1z−1

][)(][ 10 nuabnhn=

Transform pair

2/29/2008 © 2003, JH McClellan & RW Schafer 24

Another first order system

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

13

Can we determine H(z) more

easily

� Can we determine H(z) w/o determining

h[n] first?

� YES: by apply Z-transform to the

difference equation!

2/29/2008 © 2003, JH McClellan & RW Schafer 26

DELAY PROPERTY of X(z)

� DELAY in TIMEMultiply X(z) by z-1

x[n]↔ X(z)

x[n −1]↔ z −1X(z)

Proof: x[n −1]z −n

n= −∞

∞

∑ = x[ℓ]z− (ℓ+1)ℓ=−∞

∞

∑

= z−1 x[ℓ]z −ℓ

ℓ= −∞

∞

∑ = z−1X(z)

14

2/29/2008 © 2003, JH McClellan & RW Schafer 27

y[n] = a1y[n −1]+ b0x[n] +b1x[n −1]

Z-Transform of IIR Filter

� DERIVE the SYSTEM FUNCTION H(z)

� Use DELAYDELAY PROPERTY

� Apply transform on both sides

Y (z) = a1z−1Y(z) + b0X(z) + b1z

−1X(z)

2/29/2008 © 2003, JH McClellan & RW Schafer 28

H(z)=Y (z)

X(z)=b0 +b1z

−1

1− a1z−1 =

B(z)

A(z)

Y (z) − a1z−1Y(z) = b0X(z) + b1z

−1X(z)

(1 − a1z−1)Y (z) = (b0 + b1z

−1)X(z)

Y (z) = a1z−1Y(z) + b0X(z) + b1z

−1X(z)

15

2/29/2008 © 2003, JH McClellan & RW Schafer 29

Example

� DIFFERENCE EQUATION:

�� READREAD the FILTER COEFFS:

y[n] = 0.8y[n −1]+ 3x[n] −2x[n −1]

)(8.01

23)(

1

1

zXz

zzY

−

−=

−

−

H(z)

2/29/2008 © 2003, JH McClellan & RW Schafer 30

POLES & ZEROS

� ROOTS of Numerator & Denominator

POLE: H(z) ���� inf

ZERO:

H(z)=0

H(z) =b0 + b1z

−1

1 − a1z−1 → H (z) =

b0z + b1z − a1

b0z + b1 = 0 ⇒ z = −b1

b0

z − a1 = 0 ⇒ z = a1

16

2/29/2008 © 2003, JH McClellan & RW Schafer 31

EXAMPLE: Poles & Zeros

� VALUE of H(z) at POLES is INFINITEINFINITE

POLE at z=0.8

ZERO at z= -1

∞→=−

+=

=−−

−+=

−

+=

−

−

−

−

0)(8.01

)(22)(

0)1(8.01

)1(22)(

8.01

22)(

29

1

54

1

54

1

1

zH

zH

z

zzH

2/29/2008 © 2003, JH McClellan & RW Schafer 32

POLE-ZERO PLOT

2 + 2z −1

1−0.8z−1

ZERO at z = -1

POLE at

z = 0.8

17

Stability of the System

� FIRST-ORDER IIR FILTER:

y[n]= a1y[n −1]+ b0x[n]

H(z) =b0

1− a1z−1

][)(][ 10 nuabnhn=

Pole at z=a_1

When |a_1| < 1

h[n] = b0 (a1)nu[n] = 3(0.8)nu[n]

18

When |a_1| >1

� Show h[n]

� System produce unbounded output for

finite input!

� Unstable!

� BIBO stability

� BIBO: bounded input bounded output

Stability from Pole Location

� A causal LTI system with initial rest conditions is stable if all of its poles lie strictly inside the unit circle!� Our example is for 1st order system with 1 pole only. Above statement is true for systems of any order, which can be decomposed into sum of first order systems.

� Zero locations do not affect system stability

� FIR systems are always stable (poles at zeros only)

19

2/29/2008 © 2003, JH McClellan & RW Schafer 37

FREQUENCY RESPONSE

� SYSTEM FUNCTION: H(z)

� H(z) has DENOMINATOR

� FREQUENCY RESPONSE of IIR

� We have H(z)

� THREE-DOMAIN APPROACH

H(ej ?ω ) = H(z ) z = e j ?ω

h[n]↔ H (z)↔ H(e j?ω )

2/29/2008 © 2003, JH McClellan & RW Schafer 38

FREQUENCY RESPONSE

� EVALUATE on the UNIT CIRCLE

H(ej ?ω ) = H(z ) z = e j ?ω

20

2/29/2008 © 2003, JH McClellan & RW Schafer 39

FREQ. RESPONSE FORMULA

ω

ωω

ˆ

ˆˆ

1

1

8.01

22)(

8.01

22)(

j

jj

e

eeH

z

zzH

−

−

−

−

−

+=→

−

+=

=2ˆ)( ωjeH ω

ω

ω

ω

ω

ω

ˆ

ˆ

ˆ

ˆ2

ˆ

ˆ

8.01

22

8.01

22

8.01

22j

j

j

j

j

j

e

e

e

e

e

e

−

+⋅

−

+=

−

+−

−

−

−

=−−+

+++−

−

ωω

ωω

ˆˆ

ˆˆ

8.08.064.01

4444jj

jj

ee

ee

ω

ωˆcos6.164.1

ˆcos88

−

+

?ˆ@,40004.0

88)(,0ˆ@2ˆ πωω ω ==

+== jeH

2/29/2008 © 2003, JH McClellan & RW Schafer 40

Frequency Response Plot

ω

ωω

ˆ

ˆˆ

8.01

22)(

j

jj

e

eeH

−

−

−

+=

freqz(b,a)

b=[2,2]

a=[1, -0.8]

21

2/29/2008 © 2003, JH McClellan & RW Schafer 41

UNIT CIRCLE

� MAPPING BETWEEN

z = e j?ω

z = 1 ↔ ?ω = 0

z = −1 ↔ ?ω = ±π

z = ± j ↔ ?ω = ± 12π

z and ?ω

2/29/2008 © 2003, JH McClellan & RW Schafer 42

SINUSOIDAL RESPONSE

� x[n] = SINUSOID => y[n] is SINUSOID

� Get MAGNITUDE & PHASE from H(z)

ω

ω

ωω

ω

ˆ)()( where

)(][ then

][ if

ˆ

ˆˆ

ˆ

jez

j

njj

nj

zHeH

eeHny

enx

==

=

=

22

2/29/2008 © 2003, JH McClellan & RW Schafer 43

POP QUIZ

� Given:

� Find the Impulse Response, h[n]

� Find the output, y[n]

� Whenx[n] = cos(0.25πn)

H(z) =2 + 2z−1

1− 0.8z −1

2/29/2008 © 2003, JH McClellan & RW Schafer 44

Evaluate FREQ. RESPONSE

2 + 2z−1

1 − 0.8z −1at ?ω = 0.25π

zero at ω=πω=πω=πω=π?ω = 0.25π

0ˆ is 1 == ωz

23

2/29/2008 © 2003, JH McClellan & RW Schafer 45

POP QUIZ: Eval Freq. Resp.

� Given:

� Find output, y[n],

when

� Evaluate at

x[n] = cos(0.25πn)

H(z) =2 + 2z−1

1− 0.8z −1

z = e j0.25π

y[n]= 5.182cos(0.25πn − 0.417π )

309.1

25.0

22

22

182.58.01

)(22)(

j

je

e

jzH

−

−=

−

−+=

π

2/29/2008 © 2003, JH McClellan & RW Schafer 46

THREE DOMAINS

Use H(z) to get

Freq. Response

z = e j?ω

24

READING ASSIGNMENTS

� This lecture focuses on First Order

System

� Chapter 8, Sects. 8-1, 8-2, 8-3, 8-4, 8-5

� 8-3.2 Block diagram structure: study by

yourself

� 8-5: Frequency response of first order

system: study by yourself (Slides 36-45)