Lecture Digital signal processing: Lecture 3 - Zheng-Hua Tan

Digital Signal Processing, Fall 2006
Lecture 3: The z-transform
Zheng-Hua Tan
Department of Electronic Systems
Aalborg University, Denmark

1

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Course at a glance
MM1

Discrete-time
signals and systems

MM2
Fourier-domain
representation

Sampling and
reconstruction

System

System
structures

System
analysis
MM6

MM5
Filter

MM4
z-transform
MM3
2

DFT/FFT

Filter structures

MM9,MM10

MM7

Filter design
MM8

Digital Signal Processing, III, Zheng-Hua Tan, 2006

1


Part I: z-transform
z-transform
Properties of the ROC
Inverse z-transform
Properties of z-transform

„
„
„
„

3

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Limitation of Fourier transform
Fourier transform

„

∞

∑ x[n]e − jωn

X ( e jω ) =

n = −∞

1
x[ n] =
2π

π

∫−π X (e

jω

)e jωn dω

Condition for the convergence of the infinite sum

„

| X ( e jω ) | = |

∞

∞

∞

n = −∞

n = −∞

n = −∞

∑ x[n]e − jωn | ≤ ∑ | x[n] ||e − jωn | ≤ ∑ | x[n] | < ∞

If x[n] is absolutely summable, its Fourier transform exists
(sufficient condition).
1
x[n] = a n u[n]
| a |< 1 : X (e jω ) =

Example
1 − ae − jω

„

„

a = 1 : X ( e jω ) =

∞
1
+ ∑ πδ (ω + 2πk )
1 − e − jω k = −∞

| a |> 1 : 4

Digital Signal Processing, III, Zheng-Hua Tan, 2006

2


z-transform
Fourier transform

„

X ( e jω ) =

∞

∑ x[n] e − jω n

n = −∞

z-transform

„

∞

X ( z) =

∑ x[n]

z−n

Z

x[n] ↔ X ( z )

n = −∞

The complex variable z in polar form z = re jω

„

∞

∞

−∞

−∞

X ( z ) = X ( re jω ) = ∑ x[ n](re jω ) − n = ∑ ( x[n]r − n )e − jωn

|z| = r = 1,
5

X ( z ) = X (e jω )

Digital Signal Processing, III, Zheng-Hua Tan, 2006

z-plane
z-transform is a function of a complex
variable Æ using the complex z-plane

„

Z-transform on unit circle
<-> Fourier transform
Linear frequency axis in
Fourier transform
ÆUnit circle in z-transform
(periodicity in freq. of
Fourier transform)

6

Digital Signal Processing, III, Zheng-Hua Tan, 2006

3


Region of convergence – ROC
Fourier transform does not converge for all
∞
sequences
jω
− j ωn

„

X (e ) =

∑ x[n]e

n = −∞

z-transform does not converge for all sequences or
for all values of z.

„

X(z) = X ( re jω ) =

∞

∑ x[n]

z −n =

n = −∞

∞

∑ ( x[n]r − n )e − jωn

n = −∞

ROC – for any given seq., the set of values of z for
which the z-transform converges

„

∞

∑ | x[n]r − n |< ∞

n = −∞

7

∞

∑ | x[n] | | z | − n < ∞

ROC is ring!

n = −∞

Digital Signal Processing, III, Zheng-Hua Tan, 2006

ROC
Outer boundary is a circle (may extend to infinity)
Inner boundary is a circle (may extend to include the
origin)
If ROC includes unit circle, Fourier transform
converges

„
„

„

8

Digital Signal Processing, III, Zheng-Hua Tan, 2006

4


Zeros and poles
The most important and useful z-transforms –
rational function:

„

P( z )
Q( z )

P ( z ) and Q ( z ) are polynomials in z
X ( z) =

Zeros: values of z for which X(z)=0.
Poles : values of z for which X(z) is infinite.
Close relation between poles and ROC

„
„
„

9

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Right-sided exponential sequence
x[n] = a n u[n]
∞

∑ a n u[n]z − n

X ( z) =

n = −∞
∞

= ∑ (az −1 ) n
n=0

„

ROC
∞

∑ | az −1 |n < ∞

n=0

„

z-transform
∞

X ( z ) = ∑ ( az −1 ) n =
n=0

Z

u[n] ↔
10

1
,
1 − z −1

1
z
=
,
1 − az −1 z − a

| z |>| a |

| z |> 1

Digital Signal Processing, III, Zheng-Hua Tan, 2006

5


Left-sided exponential sequence
x[n] = − a n u[− n − 1]
∞

X ( z ) = − ∑ a n u[− n − 1]z − n
n = −∞
−1

∞

n = −∞

n=0

= − ∑ a n z − n = 1 − ∑ (a −1 z ) n
„

ROC
∞

∑ | a −1 z |n < ∞

n=0

„

z-transform
X ( z) =

11

1
z
=
,
1 − az −1 z − a

| z |<| a |

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Sum of two exponential sequence
1
1
x[n] = ( ) n u[n] + (− ) n u[n]
3
2
∞
1 −1 n ∞
1

X ( z ) = ∑ ( z ) + ∑ (− z −1 ) n
3
n =0 2
n =0

„

1
2z( z − )
1
1
12
=
+
=
1 −1
1
1
1 −1
1+ z
( z − )( z + )
1− z
3
2
3
2

ROC

1

1 −1
z |< 1 and | (− ) z −1 |< 1
3
2
1
1
| z |> and | z |>
3
2
1
| z |>
2

|

12

Digital Signal Processing, III, Zheng-Hua Tan, 2006

6


Sum of two exponential sequence
Another way to calculate:
1
1
x[n] = ( ) n u[n] + (− ) n u[n]
2
3
Z

1
1
1
( ) n u[n] ↔
, | z |>
1
2
2
1 − z −1
2
Z
1
1
1
, | z |>
(− ) n u[n] ↔
1
3
3
1 + z −1
3
Z
1
1
1
1
+
( ) n u[n] + (− ) n u[n] ↔
1
1

2
3
1 − z −1 1 + z −1
3
2
13

| z |>

1
2

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Two-sided exponential sequence
1
1
x[n] = (− ) n u[n] − ( ) n u[−n − 1]
3
2
Z
1
1
1
(− ) n u[n] ↔
, | z |>
1
3
3
1 + z −1

3
Z
1
1
1
, | z |<
− ( ) n u[ − n − 1] ↔
1
2
2
1 − z −1
2
1
1
1
1
, | z |> , | z |<
X ( z) =
+
1 −1
1 −1
3
2
1+ z
1− z
3
2

1
)

12
=
1
1
( z − )( z + )
2
3
2z( z −

14

Digital Signal Processing, III, Zheng-Hua Tan, 2006

7


Finite-length sequence
⎧ an , 0 ≤ n ≤ N −1
x[n] = ⎨
otherwise.
⎩0,
N −1

N −1

X ( z ) = ∑ a n z − n = ∑ (az −1 ) n
n =0

=

1 − (az )
1 zN − aN
= N −1
−1
1 − az
z
z−a
N −1

ROC

n =0

−1 N

∑ | az

−1 n

| <∞

n =0

| a |< ∞ and z ≠ 0

z k = ae j ( 2πk / N ) ,

k = 0,1,..., N − 1

pole at z = a

z k = ae j ( 2πk / N ) ,
15

k = 1,..., N − 1

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Some common z-transform pairs

16

Digital Signal Processing, III, Zheng-Hua Tan, 2006

8


Part II: Properties of the ROC
„
„
„
„

17

z-transform
Properties of the ROC
Inverse z-transform
Properties of z-transform

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Properties of the ROC

18

Digital Signal Processing, III, Zheng-Hua Tan, 2006

9


Properties of the ROC
„

„

19

The algebraic expression
or pole-zero pattern does
not completely specify
the z-transform of a
sequence Æ the ROC
must be specified!
Stability, causality and
the ROC

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Part III: Inverse z-transform
„

„
„
„

20

z-transform
Properties of the ROC
Inverse z-transform
Properties of z-transform

Digital Signal Processing, III, Zheng-Hua Tan, 2006

10


Inverse z-transform
„

„

Needed for system analysis: 1) z-transform,
2) manipulation, 3) inverse z-transform.
Approaches:
Inspection method
Partial fraction expansion
Power series expansion

‰
‰

‰

21

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Inspection method
„

By inspection, e.g.
X ( z) = (

„

1
),
1 −1
1− z
2

1
2

Make use of
Z

a n u ( n) ↔ (
∴
if
22

| z |>

1
),
1 − az −1

| z |>| a |

1
x[n] = ( ) n u[n]
2
1
of course,
| z |< ?
2

1
x[n] = −( ) n u[−n − 1]
2

Digital Signal Processing, III, Zheng-Hua Tan, 2006

11


Partial fraction expansion
„

23

For rational function, get the format of a sum
of simpler terms, and then use the inspection
method.

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Second-order z-transform
X ( z) =
1−

X ( z) =

1
1
, | z |>
3 −1 1 − 2
2
z + z
4
8

1
1
1 −1
(1 − z )(1 − z −1 )
2
4

A1

A2
+
1 −1
1
(1 − z ) (1 − z −1 )
4
2
1 −1
A1 = (1 − z ) X ( z ) | z =1/ 4 = −1
4
1
A2 = (1 − z −1 ) X ( z ) | z =1/ 2 = 2
2
−1
2
X ( z) =
+
1 −1
1 −1
(1 − z ) (1 − z )
4
2
X ( z) =

M

X ( z) =

∑b z

−k

∑a z

−k

k =0
N

k =0

k

k

M

b
X ( z) = 0
a0

∑ (1 − c z
k =1
N

k

∑ (1 − d
k =1

k

−1

)

z −1 )

N

Ak
−1
1
−
d
k =1
kz

X ( z) = ∑

Ak = X ( z )(1 − d k z −1 ) | z = d k

1
1
x[n] = 2( ) n u[n] − ( ) n u[n]
2
4

24

Digital Signal Processing, III, Zheng-Hua Tan, 2006

12


What about M>=N?
X ( z) =

1 + 2 z −1 + z −2
, | z |> 1
3
1
1 − z −1 + z −2
2
2

X ( z) =

A1
A2
+
1 −1 (1 − z −1 )
(1 − z )
2
Found by long division.

2
1 − 2 3 −1
−2
−1

z − z +1 z + 2z +1
2
2
z − 2 − 3 z −1 + 2

X ( z ) = B0 +
B0 = 2

− 5 z −1 − 1

1
A1 = (1 − z −1 ) X ( z ) | z =1/ 2 = −9
2
A2 = (1 − z −1 ) X ( z ) | z =1 = 8
X ( z) = 2 −

25

8
9
+
−1
1
(1 − z −1 ) (1 − z )
2

(1 + z −1 ) 2
1
(1 − z −1 )(1 − z −1 )
2

1
x[n] = 2δ [n] − 9( ) n u[n] + 8u[n]
2

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Power series expansion
„

By long division
X ( z) =

1
, | z |>| a |
1 − az −1

1 + az −1 + a 2 z −2 + ...
1 − az −1 1
1 − az −1
az −1
az −1 − a 2 z − 2
a 2 z − 2 ...
1
= 1 + az −1 + a 2 z − 2 + ...
1 − az −1

x[n] = a n u[n]
26

Digital Signal Processing, III, Zheng-Hua Tan, 2006

13


Finite-length sequence
1 −1
z )(1 + z −1 )(1 − z −1 )
2
1
1
X ( z ) = z 2 − z − 1 + z −1
2
2
X ( z ) = z 2 (1 −

1
1
x[n] = δ [n + 2] − δ [n + 1] − δ [n] + δ [n − 1]
2
2

27

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Part IV: Properties of z-transform
„
„
„

„

28

z-transform
Properties of the ROC
Inverse z-transform
Properties of z-transform

Digital Signal Processing, III, Zheng-Hua Tan, 2006

14


Linearity
Z

x1[n] ↔ X 1 ( z ), ROC = Rx1

X ( z) =

∑ x[n] z

−n

n = −∞

Z

x2 [n] ↔ X 2 ( z ), ROC = Rx2

„

∞

Linearity
Z

ax1[n] + bx2 [n] ↔ aX 1 ( z ) + bX 2 ( z ), ROC contains Rx1 ∩ Rx2
at least
Z
1
, | z |> 1
1 − z −1
Z
z −1
u[n − 1] ↔
, | z |> 1
1 − z −1
Z 1 − z −1
u[n] − u[n − 1] = δ [n] ↔
= 1, All z
1 − z −1
u[n] ↔

29

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Time shifting
Z

x[n − n0 ] ↔ z − n0 X ( z ),

X ( z) =

ROC = Rx1 (except for 0 or ∞ )
„

30

Example

∞

∑ x[n] z

−n

n = −∞

1
, | z |>
1
4
z−
4
z −1
1
X ( z) =
= z −1 (

)
1 −1
1 −1
1− z
1− z
4
4
Z
1 n
1
( ) u[n] ↔
1
4
1 − z −1
4
Z
1 n −1
1
( ) u[n − 1] ↔ z −1 (
)
1
4
1 − z −1
4
X ( z) =

1

Digital Signal Processing, III, Zheng-Hua Tan, 2006

15


Multiplication by exponential sequence
Z

z0n x[n] ↔ X ( z / z 0 ), ROC = |z 0|Rx

∞

∑ x[n] z

X ( z) =

−n

n = −∞

F

e jω0 n x[n] ↔ X (e j (ω −ω0 ) )

Examples

„

Z

1
,

| z |> 1
1 − z −1
Z
1
1
a nu[n] ↔
=
,
| z |> a
−1
1 − ( z / a)
1 − az −1
1
1
x[n] = cos(ω0 n)u[n] = (e jω ) n u[n] + (e − jω ) n u[n]
2
2
1
1
(1 − cos ω0 z −1 )
2
2
X ( z) =
+
=
,
1 − e jω z −1 1 − e − jω z −1 1 − 2 cos ω0 z −1 + z − 2

u[n] ↔

0

0

31

0

| z |> 1

0

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Differentiation of X(z)
dX ( z )
nx[n] ↔ − z
, ROC = Rx
dz
Z

„

Example

X ( z) =

∞

∑ x[n] z

−n

n = −∞

X ( z ) = log(1 + az −1 ), | z |>| a |
dX ( z ) − az − 2
=
dz
1 + az −1
Z
dX ( z )
az −1
nx[n] ↔ − z
=
, | z |>| a |
dz
1 + az −1
nx[n] = a (− a ) n −1 u[n − 1]
x[n] =

32

a (− a) n −1
u[n − 1]
n

Digital Signal Processing, III, Zheng-Hua Tan, 2006

16

Conjugation of a complex sequence
Z

x [ n] ↔ X ( z ), ROC = Rx
*

33

*

*

X ( z) =

∞

∑ x[n] z

−n

n = −∞

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Time reversal
Z

x[ − n] ↔ X (1 / z ), ROC = 1 / Rx

„

∞

∑ x[n] z

−n

n = −∞

Example
x[n] = a − nu[−n]
X ( z) =

34

X ( z) =

1
, | z |<| a −1 |
1 − az

x[n] = a nu[n]
1
X ( z) =
, | z |>| a |
1 − az −1

Digital Signal Processing, III, Zheng-Hua Tan, 2006

17


Convolution of sequences
Z

x1[n] * x2 [n] ↔ X 1 ( z ) X 2 ( z ),

X ( z) =

1
x[n] = ( ) n u[n]
2
h[n] = u[n]
y[n] ?

35

1
1
, | z |>
1 −1
2
1− z
2
1
H ( z) =
, | z |> 1
1 − z −1
1

1
Y ( z) =
, | z |>
1 −1
2
−1
(1 − z )(1 − z )
2
1
1
1
2
)
=
−
(
1 −1
1 (1 − z −1 )
(1 − z )
1−
2
2
X ( z) =

Example

y[n] =

−n

n = −∞

ROC contains Rx1 ∩ Rx2
„

∞

∑ x[n] z

1

1
(u[n] − ( ) n+1 u[n])
1
2
1−
2

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Initial-value theorem
X ( z) =

x[0] = lim X ( z )
z →∞

lim X ( z ) = lim
z →∞

=

z →∞

∞

∑ x[n] lim

n = −∞

z →∞

∞

∑ x[n] z

−n

n = −∞

∞

∑ x[n] z

−n

n = −∞

z− n

= x[0]

36

Digital Signal Processing, III, Zheng-Hua Tan, 2006

18


Properties of z-transform

37

Digital Signal Processing, III, Zheng-Hua Tan, 2006

Summary
„
„
„
„

38

z-transform
Properties of the ROC
Inverse z-transform
Properties of z-transform

Digital Signal Processing, III, Zheng-Hua Tan, 2006

19

Course at a glance
MM1

Discrete-time
signals and systems

MM2
Fourier-domain
representation

Sampling and
reconstruction

System

System
structures

System
analysis
MM6

MM5
Filter

MM4
z-transform
MM3

39

DFT/FFT

Filter structures

MM9,MM10

MM7

Filter design
MM8

Digital Signal Processing, III, Zheng-Hua Tan, 2006

20