Introduction to Discrete-time Systems: Z-Transform and its Applications in EE 510, Lecture notes of Teaching method

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EE 510: Lumped Systems Theory

Fall 2016

Handout 3:

Introduction to Discrete-time Systems

Prof. Mohamed Zribi

Updated 2 October 2016

Outline of the Handout

1. Overview of the Z-Transform
2. The Inverse Z-transform
3. Properties of the Z-Transform
4. Transfer Functions
5. State Space Representations of DT Systems
6. Solving Difference Equations using the Z-transform
7. Solutions of Discrete-time State Space Equations
8. Modal Decomposition of the solution of
9. Derivation of Discrete Models from Continuous Ones (^2)

x(k +1) = Ax(k) + Bu(k)

1. Overview of the Z-transform

 The one-sided Z-transform is given by:

where z is a complex number.

 The Z-transform has an associated region of convergence for z ,

which is determined by when the infinite sum converges.

 The Z -transform has several applications. For example, it is

useful for solving difference equations.

 Often X ( z ) is evaluated using an infinite sum.

n

n 0

X (z) Z{x[n]} x[n]z

−

∞

=

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 4

Z-Transform of a Signal

f [k]

Z

f (0)

f (1)

f (2)

f (3)

f (4)

…

F(z)

Z

f (0) · z 0

• f (1) · z -

• f (2) · z -

• f (3) · z -

• f (4) · z -

…

5

-n

n=

F(z) = f(n)× z

∞

∑

EE 510 Lumped Systems Theory Dr. Mohamed Zribi

Example 2:

Determine the Z-transform of the signal

[ ]

( ) [ ]

( ) [ ] [ ]

( ) [ ]

( ) [ ]

( )

1 2

0 1 2

2

0

0

2

X z 0 1 2

Hence

z

Weknow that:

0 , otherwise

1 , 2

1 , 1

2 , 0

− −

− − −

=

−

−

∞

=

= − +

= = + +

=















=

− =

=

=

∑

∑

X z z z

xn z x z x z x z

X xnz

n

n

n

xn

n

n

n

n

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 7

Solution:

Example 3: Z Transform of Unit Impulse Signal

f (^) impulse(k) Fimpulse(z)

Z

Z

f (0) = 1

f (1) = 0

f (2) = 0

f (3) = 0

f (4) = 0

…

1 · z 0

+0 · z -

+0 · z -

+0 · z -

+0 · z -

…

(^0) -1 0 1 2 3 4 5 6 7 8 9

1

F (z) 1 impulse

=

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 8

Example 4: Delayed Unit Impulse Signal

f (^) delay (k) Fdelay (z)

Z

Z

f(0) = 0

f(1) = 1

f(2) = 0

f(3) = 0

f(4) = 0

…

0 · z 0

+1 · z -

+0 · z -

+0 · z -

+0 · z -

…

1 Fdelay (z) z

−

(^0) -1 0 1 2 3 4 5 6 7 8 9

1

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 10

Therefore, the Z-transform of f [n]= δ [ n - 1] is such that:

1

0

( ) { [ 1]} [ 1]

with an ROC consisting of the entire -plane except 0.

n

n

F z Z n n z z z

z z

∞ − −

=

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 11

Example 6: Z-Transform of Unit Step Signal

u (^) step (k) Ustep (z)

Z

Z

u(0) = 1

u(1) = 1

u(2) = 1

u(3) = 1

u(4) = 1

…

1 · z 0

+1 · z -

+1 · z -

+1 · z -

+1 · z -

…

U (z) 1 z z z ...

1 2 3

step =^ + + + +

− − −

(^0) -1 0 1 2 3 4 5 6 7 8 9

1

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 13

n → ∞, |a |<1,

1 a

1 a

1 a

(1 a)(1 a a ... a ) 1 a a ... a

n 1

2 n 2 n

−

−

−

− + + + +

• • • • =

−

Recall that,

assuming

1 a

1

1 a

1 a

1 a

(1 a)(1 a a ... a ) 1 a a ...

n 1

2 n 2

−

=

−

−

−

− + + + +

• • • =

−

→∞

→∞

n

n

lim

lim

z- 1

z

1 - z

U (z) 1 z z z ...

• 1

1 2 3

step =^ + + + + = =

− − −

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 14

16

Example 7:

Given that f [k] =e -akT, find F(z).

E(z) can be written in power series form as

In this example, f [k] may be generated by sampling the

exponential function f (t)=e

-at .

, 1 1

1

( ) 1 1 ( ) ( )

1 1

1 2 2 1 1 2

< −

= −

=

= + + + = + + +

− − − − −

− − − − − − − −

e z z e

z

e z

F z e z e z e z e z

aT aT aT

aT aT aT aT  

EE 510 Lumped Systems Theory Dr. Mohamed Zribi

Solution:

17

Example 8:

Find the Z-transform of a sampled unit ramp.

A sampled unit ramp can be written as f (kT)=kT. By the

definition of the Z-transform

and

In order to find a closed form of F(z), we multiply the above

equation by z,

( ) (^12233 ) 0

= ⋅ = − + − + − + 

∞

=

− F z ∑ kT z T z z z k

k

zF ( z )= T ( 1 + 2 z −^1 + 3 z −^2 + )

1 1

( ) ( ) ( 1 1 2 ) 1 −

= −

− = + − + − + = − z

Tz

z

T zF z F z T z z 

Thus

( 1 )^2

( ) −

= z

Tz F z

EE 510 Lumped Systems Theory Dr. Mohamed Zribi

Solution:

19

• 1 2

jk jk - 1

1 2

1

(^21111)

1

(^2) j - 1 j 1

1

jk jk

• j - 1
• jk

j - 1

jk

1 - 2cos z z

sin z F(z) 2j

e e (k) sin(k )

1 2cos z z

1 cos z

) 1 cos z jsin z

1

1 cos z jsin z

1 (

) 1 e z

1

1 - e z

1 F(z) ( 2

e e (k) cos(k )

1 - e z

1 (k) e F(z)

1 - e z

1 [ ] e F(z)

−

−

− −

−

− − − −

− −

−

⋅ +

⋅ ←→ =

− = =

− ⋅ +

− ⋅

− ⋅ + ⋅

− ⋅ − ⋅

=

− ⋅

⋅

←→ =

= =

⋅

= ←→ =

⋅

= ←→ =

θ

θ θ

θ

θ

θ θ θ θ

θ

θ θ

θ θ

θ θ

θ

θ

θ

θ

Z

Z

Z

Z

f

f

f

f k

Example 10:

EE 510 Lumped Systems Theory Dr. Mohamed Zribi

Example 11: Z-Transform of an Exponential Signal

f exp (k) Fexp (z)

Z

Z

f (0) = 1

f (1) = a

f (2) = a 2

f (3) = a 3

f (4) = a 4

…

1 · z 0

+a · z -

+a^2 · z -

+a^3 · z -

+a^4 · z -

…

z- a

z

1 - az

1

F (z) 1 az a z a z ...

• 1

1 2 2 3 3 exp

= =

= + + + +

− − −

(^0) -1 0 1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

a=1.

EE 510 Lumped Systems Theory Dr. Mohamed Zribi 20