Chapter 6 Discrete-Time System

Chapter 6Chapter 6 Discrete-Time System Discrete-Time System

2/90

Operation of discrete time system

1. Discrete time system1. Discrete time system

1 0( ) ( 1) ( )y n b y n a x n

where and are multiplier

D is delay element

0a

0a

1b

1b

Fig. 6-1.

3/90

2. Difference equation2. Difference equation

Difference equation

0 0

( ) ( ) 0M N

k kk k

b y n k a x n k

0 0

( ) ( ) ( )M N

k kk k

y n a x n k b y n k

where and is constant or function of n ka kb

4/90

Example 6-1– Moving average

Example 6-2– Integration

0 1 2( ) ( ) ( 1) ( 2)y n a x n a x n a x n

If 0 1 2 1/ 3a a a

1( ) ( ) ( 1) ( 2)

3y n x n x n x n

1 0 1( ) ( 1) ( ) ( 1)y n b y n a x n a x n

If 1 0 11, 1/ 2b a a

1( ) ( 1) ( ) ( 1)

2y n y n x n x n

5/90

Functional relationship of discrete time system

3. Linear time-invariant system3. Linear time-invariant system

( ) ( ) ( )k

y n h k x n k

where is impulse response of system( )h n

Fig. 6-2.

6/90

– The system is said to be linear if

– The system is said to be time-invariant if

1 1 2 2 1 1 2 2( ) ( ) ( )a x n a x n a y n a y

( ) ( )

( ) ( )

x n y n

x n k y n k

7/90

Form and transfer function– Difference equation of discrete time system

0 0

( ) ( ) ( )M N

k kk k

y n a x n k b y n k

After z-transform

0 0

( ) ( ) ( )N M

k kk k

k k

Y z a z X z b z Y z

8/90

– Transfer function

0

0

( )( )

( ) 1

Nk

kk

Mk

kk

a zY z

H zX z b z

( )H z

If 0kb

0

( ) ( )N

kk

y n a x n k

0

( )( )

( )

Nk

kk

Y zH z a z

X z

where is impulse response of systemka

9/90

– Impulse response of discrete time system

1( ) ( ) , 0,1,2,h k H z k

If is power series( )H z

1 2

( ) ( )

= (0) (1) (2)

k

n k

H z h k z

h h z h z

In this case, ( ) ( )x n n

0

0

( ) ( ) ( )

= ( ) ( )

= (0) ( ) (1) ( 1) (2) ( 2)

= ( ), 0,1,2,

k

k

y n h k x n k

h k n k

h n h n h n

h n n

10/90

Example 6-31

1

1( )

1 0.5

zH z

z

Using power series

1 2 3( ) 1 1.5 0.75 0.375H z z z z

(0) 1, (1) 1.5, (2) 0.75, (3) 0.375,h h h h

11/90

Using other method

1 1( ) 0.5 ( ) ( ) ( )Y z z Y z X z z X z

( ) 0.5 ( 1) ( ) ( 1)y n y n x n x n

( ) ( ) ( 1) 0.5 ( 1)y n x n x n y n

Substitute , and initial value ( ) ( )x n n ( 1) 0y

1 0( )

0 0

nn

n

(0) 1y

(1) (1) (0) 0.5 (0) 0 1 0.5 1.5y x x y

(2) (2) (1) 0.5 (1) 0.5 ( 1.5) 0.75y x x y

(3) (3) (2) 0.5 (2) 0.5 0.75 0.375y x x y

(0) 1, (1) 1.5, (2) 0.75, (3) 0.375,h h h h

12/90

Example 6-4

1 0( ) ( 1) ( )y n b y n a x n

Using z-transfrom

11 0( ) [ ( ) ( 1)] ( )Y z b z Y z y a X z

11 0 1( )(1 ) ( ) ( 1)Y z b z a X z b y

0 11 1

1 1

( ) ( ) ( 1)( )

1 1

a X z b yY z

b z b z

Using inverse z-transfrom

10 1 1

0

( ) ( ) ( ) ( ) ( 1)n

n k n

k

y n a b x k b y

13/90

If initial value ( 1) 0y

01

1

( )( )

1

a X zY z

b z

01

1

( )1

aH z

b z

Using inverse z-transfrom

0 1( ) ( )nh n a b

14/90

System stability– BIBO(bounded input, bounded output)

[ ]k

h k

Bounded )()(then

Bounded; )( if

)()(

)()()(

kx

x

k

k

khBny

Bnx

knxkh

knxkhnypf)

15/90

Raible tabulation1 2

1 2 1 0( ) 0n nn nD z a z a z a z a z a

Table 6-1. Raible’s tabulation

16/90

– If part or all factor is 0 in the first row, then this table is ended

• Singular case

– Using substitution

»

(1 )z z

(1 ) (1 )n n nz n z

n th order case

17/90

Example 6-52( ) 0.25 0D z z z

18/90

Example 6-6

3 2( ) 3.3 3 0.8 0D z z z z

19/90

– Singular case3 2((1 ) ) (1 3 ) 3.3(1 2 ) 3(1 ) 0.8 0D z z z z

0 0 00 : 0, 0, 0b c d

0 0 00 : 0, 0, 0b c d

20/90

Example 6-73 2( ) 1 0D z z z z

3 2[(1 ) ] (1 3 ) (1 2 ) (1 ) 1 0D z z z z

21/90

Discrete time system transfer function H(z)

4. Description of Pole-Zero 4. Description of Pole-Zero

1 2

1 2

1 20 1 2

20 1 2

( )( )

( )

( )( ) ( ) =

( )( ) ( )

=

N

N

N N NN

N N NN

Y zH z

X z

K z z z z z z

z p z p z p

a z a z a z a

b z b z b z b

where K is gain

Zeros of at :

Poles of at : ( )H z

( )H z1 2, , , Nz p p p

1 2, , , Nz z z z

22/90

– Description of pole and zero in z-plane

( 1)( )

( 0.5 0.5)( 0.5 0.5)( 0.75)

K zH z

z j z j z

Fig. 6-3.

23/90

Example 6-81 2

1 2 3

1 2( )

1 1.75 1.25 0.375

z zH z

z z z

( 2)( 1)( )

( 0.5 0.5)( 0.5 0.5)( 0.75)

z z zH z

z j z j z

Fig. 6-4.

24/90

Example 6-9

2

2

2

1 2

( )( )( )

( 0.5 0.5)( 0.5 0.5)

( 1) =

0.5

(1 ) =

1 0.5

K z j z j zH z

z j z j

K z

z z

K z

z z

K=0.2236

Fig. 6-5.

25/90

Frequency response of system

– Method for calculation

– Method for geometric calculation

5. Frequency response5. Frequency response

( ) ( )

= ( ) ( )

j T

n

z en

j T j T

n

H z h n z

H e h n e

26/90

Example 6-10

• Method for calculation

– Substitution of and using Euler’s formular

1( )

0.5

zH z

z

j Tz e

1( )

0.51 cos( ) sin( )

=cos( ) 0.5 sin( )

j Tj T

j T

eH e

eT j T

T j T

27/90

– Substitution of

–

0

0( ) 2 / 0.5 4 0jH e

/ 8, 8 4

ss T T

41 cos( / 4) sin( / 4)

( )cos( / 4) 0.5 sin( / 4)

=2.51 51.2

j jH e

j

4 2( ) ( )=1.26 71.6sj T j

H e H e

3 3

8 4( ) ( )=0.55 82.1sj T j

H e H e

2( ) ( )=0 0sj T jH e H e

28/90

Fig. 6-6.

29/90

• Method for geometric calculation

0 1 2( )( )j T jAH e e

B

0( )j T AH e

B

Magnitude response

01 2( )j TH e

Phase response

Fig. 6-7.

30/90

6. Realization of system6. Realization of system

( ) ( 1)y n x n

Unit delay

Adder/subtractor

( ) ( ) ( )w n x n y n

Constant multiplier

( ) ( )y n Ax n

Branching

1

2

( ) ( )

( ) ( )

y n x n

y n x n

Signal multiplier

( ) ( ) ( )w n x n y n

Fig. 6-8.

31/90

Direct form

– Direct form 1

0

1

( )( )

( ) 1

Ni

ii

Ni

ii

a zY z

H zX z b z

0 0

( ) ( ) ( )N N

i ii i

y n a x n i b y n i

32/90Fig. 6-9.

(a)

(b)

33/90

– Direct form 2

( )( )

( )

N zH z

D z

( ) ( )( ) ( ) ( )

( )

N z X zY z H z X z

D z

( )( )

( )

X zW z

D z

( ) ( ) ( )Y z N z W z

1

( ) ( ) ( )N

ii

w n x n b w n i

0

( ) ( )N

ii

y n a w n i

Inverse transform

poles

zeros

34/90Fig. 6-10.

(a)

(b)

35/90

Example 6-11

– Direct form 1

– Direct form 2

1 2

1 2

3 3.6 0.6( )

1 0.1 0.2

z zH z

z z

( ) 3 ( ) 3.6 ( 1) 0.6 ( 2) 0.1 ( 1) 0.2 ( 2)y n x n x n x n y n y n

( ) ( ) 0.1 ( 1) 0.2 ( 2)w n x n w n w n

( ) 3 ( ) 3.6 ( 1) 0.6 ( 2)y n w n w n w n

36/90

Fig. 6-11.

(a)

(b)

37/90

Quantization effect of parameters– Quantization error of parameters

• Input signal quantization

• Accumulation of arithmetic roundoff errors

• Coefficient of transfer function quantization

38/90

Cascade and parallel canonic form– Cascade canonic form or series form

0 1 2

01

( ) ( ) ( ) ( )

= ( )

l

l

ii

H z a H z H z H z

a H z

11

11

1( )

1i

ii

a zH z

b z

1 21 2

1 21 2

1( )

1i i

ii i

a z a zH z

b z b z

first order

second order

39/90

Fig. 6-12.

Fig. 6-13.

(a)

(b)

40/90

– Parallel canonic form

1 2

1

( ) ( ) ( ) ( )

= ( )

r

r

ii

H z A H z H z H z

A H z

first order

second order

01

1

( )1

ii

i

aH z

b z

10 1

1 21 2

( )1

i ii

i i

a a zH z

b z b z

41/90

Fig. 6-14.

42/90

Fig. 6-15.

(a)

(b)

43/90

Example 6-12

– Cascade canonic form

1 2

1 2

3 3.6 0.6( )

1 0.1 0.2

z zH z

z z

3( 1)( 0.2)( )

( 0.5)( 0.4)

z zH z

z z

1

1 1

1( )

1 0.5

zH z

z

1

2 1

1 0.2( )

1 0.4

zH z

z

1 2

1 1

1 1

( ) ( ) ( )

1 1 0.2 =

1 0.5 1 0.4

H z H z H z

z z

z z

Quantization error of parameter is decreased

44/90

– Parallel canonic form

31 2( ) 3( 1)( 0.2)

( 0.5)( 0.4) 0.5 0.4

AA AH z z z

z z z z z z z

1 2 33, 1, 7A A A

1 1

1( )

1 0.5H z

z

2 1

7( )

1 0.4H z

z

45/90

Fig. 6-16.

(a)

(b)

46/90

Example 6-132

2

2( 1)( 1.4 1)( )

( 0.5)( 0.9 0.81)

z z zH z

z z z

1

1 1

1( )

1 0.5

zH z

z

1 2

2 1 2

1 1.4( )

1 0.9 0.81

z zH z

z z

Fig. 6-17.

47/90

Example 6-142

2

2( 1)( 1.4 1)( )

( 0.5)( 0.9 0.81)

z z zH z

z z z

2

2

3 41 22

( ) 2( 1)( 1.4 1)

( 0.5)( 0.9 0.81)

=0.5 0.9 0.81

H z z z z

z z z z z

A z AA A

z z z z

1 24.94, 2.19A A

3 4 3.17A A

3 4 6.26A A Substitute 1, 1z z

3 44.72, 1.55A A

1 1

2.19( )

1 0.5H z

z

1

2 1 2

4.72 1.55( )

1 0.9 0.81

zH z

z z

48/90

Fig. 6-18.

49/90

FIR system– Direct form

• Tapped delay line structure or transversal filter

1

( ) ( )N

ii

y n a x n i

Fig. 6-19.

50/90

– Cascade canonic form

1 20 1 2

0 1

( ) ( ) ( )sNN

nk k k

n k

H z h n z a a z a z

where ( 1) / 2sN N

Fig. 6-20.

51/90

– Linear phase FIR system structure

• N is even : type 1 and type 3

( ) ( ) 0,1, ,h n h N n n N

( ) ( ) 0,1, ,h n h N n n N or

0

( ) ( ) ( )N

k

y n h n x n k

/2 1

0 /2 1

/2 1 /2 1

0 0

( ) ( ) ( / 2) ( / 2) ( ) ( )

( ) ( ) ( / 2) ( / 2) ( ) ( )

N N

k k N

N N

k k

h k x n k h N x n N h k x n k

h k x n k h N x n N h N k x n N k

52/90

– Type 1

– Type 3

• N is odd : type 2 and type 4– Type 2

– Type 4

/2 1

0

( ) ( ) ( ) ( ) ( / 2) ( / 2)N

k

y n h k x n k x n N k h N x n N

/2 1

0

( ) ( ) ( ) ( )N

k

y n h k x n k x n N k

( 1)/2

0

( ) ( ) ( ) ( )N

k

y n h k x n k x n N k

( 1)/2

0

( ) ( ) ( ) ( )N

k

y n h k x n k x n N k

53/90

Fig. 6-21.

(a)

(b)

54/90

Lattice structure– Lattice structure of FIR filter

1 21 2( ) 1 N

NH z a z a z a z

1 2( ) ( ) ( 1) ( 2) ( )Ny n x n a x n a x n a x n N

Difference equation

1

ˆ( ) ( ) ( ) ( ) ( )N

kk

y n x n x n x n a x n k

Error between and( )x n ˆ( )x n

FIR filter use linear predictor

1

ˆ( ) ( )N

kk

x n a x n k

where is prediction coefficient ka

55/90

• N=1, output of FIR filter

– Single-stage FIR lattice structure

1( ) ( ) ( 1)g n k x n x n

1( ) ( ) ( 1)y n x n a x n

1( ) ( ) ( 1)y n x n k x n

where is reflection coefficient 1k

Fig. 6-22.

56/90

• N=2

2 1 2 1

1 2 1

1 1 2 2

( ) ( ) ( 1)

= ( ) ( 1) ( 1) ( 2)

= ( ) ( ) ( 1) ( 2)

y n y n k g n

x n k x n k k x n x n

x n k k k x n k x n

2 2 1 1

2 1 2 1

2 1 1 2

( ) ( ) ( 1)

= ( ) ( 1) ( 1) ( 2)

= ( ) ( ) ( 1) ( 2)

g n k y n g n

k x n k k x n k x n x n

k x n k k k x n x n

Fig. 6-23.

57/90

• N=3

3 1 1 2 2 3

2 1 3 1 2 3 3

( ) ( ) ( ) ( 1)

+( ) ( 2) ( 3)

y n x n k k k k k x n

k k k k k k x n k x n

3 3 2 1 3 1 2 3

1 1 2 2 3

( ) ( ) ( ) ( 1)

+( ) ( 2) ( 3)

g n k x n k k k k k k x n

k k k k k x n x n

Substitution 1 1 1 2 2 3 2 2 1 3 1 2 3 3 3, , b k k k k k b k k k k k k b k

3 1 2 3( ) ( ) ( 1) ( 2) ( 3)y n x n b x n b x n b x n

3 3 2 1( ) ( ) ( 1) ( 2) ( 3)g n b x n b x n b x n x n

Fig. 6-24.

58/90

• M-stage lattice structure

0

( ) ( )M

M ii

y n b x n i

0

( ) ( )M

M M ii

g n b x n i

If ( ) ( )x n n

0

( )M

iM i

i

Y z b z

0

( )M

iM M i

i

G z b z

1

( ) ( )MM MG z z Y

z

59/90

– Calculating filter coefficient

» m-stage

1

0

( )M

M ii

Y z b z

11 1( ) ( ) ( )m m m mY z Y z k z G z

11 1( ) ( ) ( )m m m mG z k Y z z G z

1 11 1

( ) ( )( ) m m m

m m m

G z k Y zY z Y k z

z

1 2

( ) ( )( ) , 1

1m m m

m mm

Y z k G zY z k

k

1 2

( ) (1/ )( )

1

mm m m

mm

Y z k z Y zY z

k

1( ) ( ),M

M MG z z Yz

M m

60/90

Fig. 6-25.

61/90

Example 6-15

(1)

1 22 ( ) 1 0.9 0.8Y z z z

2M

22 2

2 2

1 2

1( ) ( )

= (1 0.9 0.8 )

=0.8 0.9

G z z Yz

z z z

z z

62/90

(2) M Mk b

2 2 0.8k b

1 1 2 1k k k b

1 10.8 0.9k k

1 0.5k

Fig. 6-26.

63/90

Example 6-16

– in example 6-15

1 22 ( ) 1 0.9 0.8Y z z z

22 2 2

1 22

1 2 2 2

2

1

( ) (1/ )( )

1

(1 0.9 0.8 ) 0.8 (1 0.9 0.8 ) =

1 (0.8)

=1 0.5

Y z k z Y zY z

k

z z z z z

z

2 0.8k

1 0.5k

64/90

– Generalization of calculating filter coefficient

» Substitute for

» Substitute for

0

( )M

iM Mi

i

Y z b z

m

0

( )m

im mi

i

Y z b z

1/ z z

M

0

(1/ )m

im mi

i

Y z b z

,0

(1/ )m

m im m m i

i

Y z b z

65/90

» At 3m

1 2 33 30 31 32 33( )Y z b b z b z b z

2 33 30 31 32 33(1/ )Y z b b z b z b z

1 2

( ) (1/ )( )

1

mm m m

mm

Y z k z Y zY z

k

0

( )m

im mi

i

Y z b z

0

(1/ )m

im mi

i

Y z b z

66/90

,10 0

1, 20

,0 0

2

1

=1

m mi m m i

mi m m m imi i i

m ii m

m mi i

mi m m m ii i

m

b z k z b zb z

k

b z k b z

k

Divided by iz

,1, 2

,1

mi m m m im i

m

b k bb

k

0,1, , 1i m

, 1, , 2,1, 1mm M M k

m mmk b

67/90

Example 6-17

–

1 2 33( ) 1 0.5 0.2 0.5Y z z z z

3m 1 2 3

3 31 32 33( ) 1Y z b z b z b z

3 33 0.5k b

68/90

– Calculating of coefficient

• At

,1, 2

,1

mi m m m im i

m

b k bb

k

2 ( )Y z

3, 0m i

30 3 3320 2

3

2

1

1 ( 0.5)( 0.5) = 1

1 ( 0.5)

b k bb

k

69/90

• At

• At

3, 1m i

31 3 3221 2

3

2

1 ( )

0.5 ( 0.5)(0.2) = 0.8

1 ( 0.5)

b k bb

k

3, 2m i

32 3 3122 2

3

2

1 ( )

0.2 ( 0.5)(0.5) = 0.6

1 ( 0.5)

b k bb

k

2 22k b

70/90

– Calculation of

• Go first stage

2 ( )Y z

1 22 21 22

1 2

( ) 1

1 0.8 0.6

Y z b z b z

z z

21 2 2111 2

2

2

1 ( )

0.8 (0.6)(0.8) = 0.5

1 (0.6)

b k bb

k

1 11k b

71/90

Fig. 6-27.

72/90

– Lattice structure of IIR filter• All-pole system

1 21 2

1( )

1 NN

H za z a z a z

Difference equation

1 2( ) ( ) ( 1) ( 2) ( )Ny n x n a y n a y n a y n N

1 2( ) ( ) ( 1) ( 2) ( )Nx n y n a x n a x n a x n N

1

( ) ( ) ( )N

kk

y n x n a x n k

73/90

– N=1

1( ) ( ) ( 1)y n x n k y n

1( ) ( ) ( 1)g n k x n y n

Fig. 6-28.

74/90

– N=2

1 2 2( ) (1 ) ( 1) ( 2) ( )y n k k y n k y n x n

2 2 1 2( ) ( ) (1 ) ( 1) ( 2)g n k y n k k y n y n

Fig. 6-29.

75/90

– N th order

( ) ( )Ny n x n

1 1( ) ( ) ( 1), , 1, , 2,1m m m my n y n k g n m N N

1 1( ) ( ) ( 1), , 1, , 2,1m m m mg n k y n g n m N N

0 ( ) ( )g n y n

Fig. 6-30.

76/90

• General IIR system function

0

1

1

( )( )

( )1

Mi

ii

N

ii

b zB z

H zA za z

All-pole lattice structure

Ladder structure

1/ ( )A z

( )B z

77/90

– Lattice-ladder structure

» System output

» System transfer function

0

( ) ( )M

m mm

y n d g n

0

( )( )

( )

( ) ( ) =

( ) ( )

Mm

mm

Y zH z

X z

G z B zd

X z A z

78/90

0

0

0 0

0

( )( )

( ) ( )

( ) ( ) =

( ) ( )

1 = ( )

( )

Mm

mm

Mm

mm

M

m mm

G zB zd

A z X z

G z G zd

G z X z

d G zA z

where 0

( )( )

( )m

m

G zG z

G z

0

( ) ( )M

m mm

B z d G z

79/90

Fig. 6-31.

80/90

Example 6-18

1 2 3

1( )

1 0.5 0.2 0.5H z

z z z

3 2 10.5, 0.6, 0.5k k k

Fig. 6-32.

81/90

Example 6-19

– Equation for each node

1 2 3

1 2 3

1( )

1 0.5 0.2 0.5

z z zH z

z z z

1 2 33

1 22

11

0

( ) 0.5 0.2 0.5

( ) 0.6 0.8

( ) 0.5

( ) 1

G z z z z

G z z z

G z z

G z

82/90

• Comparison of coefficients

1 2 3

0 0 1 1 2 2 3 3

1 1 20 1 2

1 2 33

0 1 2 3

1 2 3

( ) 1

= ( ) ( ) ( ) ( )

= (0.5 ) (0.6 0.8 )

+ ( 0.5 0.2 0.5 )

=( 0.5 0.6 0.5 )

+( 0.8 0.2 )

B z z z z

d G z d G z d G z d G z

d d z d z z

d z z z

d d d d

d d d z

1 2 32 3 3( 0.5 )d d z d z

0 1 2 3

1 2 3

2 3

3

0.5 0.6 0.5 1

0.8 0.2 1

0.5 1

1

d d d d

d d d

d d

d

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Fig. 6-33.

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– Advantage of digital filter compare with analog filters• Digital filter can have a truly linear phase response

• The performance of digital filters does not vary with environmental change

• The frequency response of a digital filter can be automatically adjusted if it is implemented using a programmable processor

• Several input signals or channels can be filtered by one digital filter without the need to replicate the hardware

• Both filtered and unfiltered data can be saved for further use

7. Introduction to digital filter 7. Introduction to digital filter

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• Advantage can be readily taken of the tremendous advancement in VLSI technology to fabricate digital filters and to make them small in size, to consume low power, and to keep the cost down

• In practice, the precision achievable with analog filters is restricted

• The performance f digital filters is repeatable from unit to unit

• Digital filters can be used at very low frequencies

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– Disadvantage of digital filters compare with analog filters• Speed limitation

• Finite wordlength effect

• Long design and development times

– Block diagram of digital filter with analog input and output

Fig. 6-34.

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Types of digital filters : FIR and IIR filters– IIR filter

– FIR filter

0

( ) ( ) ( )k

y n h k x n k

0

( ) ( ) ( )N

k

y n h k x n k

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– IIR filtering equation is expressed in recursive form

– Alternative representations for FIR and IIR filters

0 0 1

( ) ( ) ( ) ( ) ( )N M

k kk k k

y n h k x n k b x n k a y n k

where and are coefficient of filterska kb

IIR is feedback system of some sort

0

( ) ( )N

k

k

H z h k z

0

1

( )1

Nk

kk

Mk

kk

b zH z

a z

These transfer functions are very useful in evaluating their frequency responses

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Choosing between FIR and IIR filters– Relative advantages of the two filter type

• FIR filters can have an exactly linear phase response

• FIR filters are always stable. The stability of IIR filters cannot always be guaranteed

• The effect of using a limited number of bits to implement filters such as roundoff noise and coefficient quantization errors are much less severe in FIR than in IIR

• FIR requires more coefficients for sharp cutoff filters than IIR

• Analog filters can be readily transformed into equivalent IIR digital filters meeting similar specifications

• In general, FIR is algebraically more difficult to synthesize

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– Guideline on when to use FIR or IIR• Use IIR when the only important requirements are sharp cutoff filters

and high throughput

• Use FIR if the number of filter coefficients is not too large and, in particular, if little or no phase distortion is desired