z-Transform1The z-TransformpIntroductionpz-TransformpProperties of the Region of Convergence for the z-

TransformpInverse z-Transformpz-Transform PropertiespUnilateral z-TransformpSolving the Difference EquationspZero-Input ResponsepTransfer Function representationpSummary

z-Transform21. IntroductionpRole in Discrete-Time Systems

l z-Transform is the discrete-time counterpart of the Laplace transform.

pResponse of Discrete-Time Systemsl If a system is lumped, it can be described as a difference equation

2y[k] + 3y[k-1] + y[k-2] = u[k] + u[k-1] - u[k-2] for k = 0, 1, 2l The response of the system is excited by an input u[k] and some initial

conditions.l The difference equations are basically algebraic equations, their solutions

can be obtained by direct substitution.l The solution however is not in closed form and is difficult to develop

general properties of the system.l A number of design techniques have been developed in the z-Transform

domain.

z-Transform31. The z-TransformpPositive and Negative Time Sequence

l A discrete-time signal f[k] = f[kT], where k is an integer ranging (-∗<t<∗), is called a positive-time sequence if f[k]=0 for k < 0; it is called a negative-time sequence if f[k] = 0 for k > 0.

lWe mainly consider the positive-time sequences.

pz-Transform Pairl The z-transform is defined as

where z is a complex variable, called the z-transform variable.

pExamplel x[n] = { 1, 2, 5, 7, 0, 1}; x[n] = (1/2)n u{n}

X z Z x n x n z n

n

( ) [ [ ] ] [ ]≡ ≡ −

= − ∞

∞

∑

Difference with the Fourier Transform

Difference with the Fourier Transform

z-Transform40. The z-Transform (c.1)pExample

l f[k] = bk for all positive integer k and b is a real or complex number.

l If , then the infinity power series converges and

l The region is called the region of convergence.pUnit Step Sequence

l The unit sequence is defined as

l The z-Transform is pExponential Sequence

l f[k] = e akT

F z f n z b z b zn

n

n n

n

n

n

( ) [ ] ( )= = =−

=

∞−

=

∞−

=

∞

∑ ∑ ∑0 0

1

0

b z − <1 1

F zb z

zz b

( ) =−

=−−

11 1

b

Q z zz

n

n

( ) = =−

−−

=

∞

∑ 11 1

0

F z e ze

a n T

n

na T( ) = =

−=

∞−∑

0

11

q nf o r nf o r n

[ ], , ,

==<

1 0 1 20 0

b z<

z-Transform51. The z-Transform (c.2)pRegion of Convergence

l For any given sequence, the set of values of z for which the z-transform converges is called the region of convergence.

pViewpointsl The representation of the complex variable z

l Consider the z-transform

l Convergent Condition

z re j= ω

X r e x n r ej j n

n

( ) [ ] ( )ω ω= −

= − ∞

∞

∑

x n r n

n

[ ] −

= − ∞

∞

∑ < ∞

ROC includes the unit circle ==> Fourier Transform converges

Convergence of the z-Transform==> The z-transform and its derivatives

must be continuous function of z.

z-Transform61. The z-Transform (c.3)

X z P zQ z

( ) ( )( )

=

pRational Function

pEx.x n a u nn[ ] [ ]= − − − 1x n a u nn[ ] [ ]=

z-Transform7

2. Properties of the Region of Convergence for the z-Transformp Properties

l The ROC is a ring or disk in the z-plane centered at the origin, i.e., l The Fourier transform of x[n] converges absolutely if and only if the ROC of the z-

transform of x[n] includes the unit circle.l The ROC cannot contain any poles.l If x[n] is a finite-duration sequence, i.e. a sequence that is zero except in a finite

interval , then the ROC is the entire z-plane except possibly z=0 or z= .

l If x[n] is a right-sided sequence, i.e. a sequence that is zero for n<N1< , the ROC extends outward from the outermost finite pole in X(z) to z= .

l If x[n] is a left-sided sequence, i.e., a sequence that is zero for n>N2>- , the ROC extends inward from the innermost (smallest magnitude) nonzero pole in X(z) to (and possibly including) z=0.

l A two-sided sequence is an inifinite-duration sequence that is neither right-sided nor left-sided. If x[n] is a two-sided sequence, the ROC will consist of a ring in the z-plane, bounded on the interior and exterior by a pole, and, consistent with property 3, not containing any poles.

l The ROC must be a connected region.

0 ≤ < < ≤ ∞r z rR L

− ∞ < ≤ ≤ ≤ ∞N n N1 2

∞∞

∞

∞

z-Transform8

2. Properties of the Region of Convergence for the z-TransformpExample

l ROC is a Ring

l ROC is the interior of a circle

l ROC is the outerior of a circle

Unit circle

No common ROC case ?No common ROC case ?

z-Transform93. The Inverse z-TransformpMethods

l Direct Divisionl Partial Fraction Expansion

pDirect Divisionl F(z) = -2z2 +3z +3z-2 + 3z-3 + 9z-4

f[k] = {-2, 3, 0, 0, 3, 3, 9, ...}

pEx. 3/(z2 - z -2)

z z z z z zz z z

z z zz z z

z z z z z2 4 3 2

4 3 2

3 2

3 2

2 2 3 4

2 2 5 6 32 2 4

3 3 6 33 3 6

3

2 3 3 3 9− − − + + − +

− + +− − +

− −+

− + + + +− − −

F zz z z z

z z( ) =

− + + − +− −

2 5 6 32

4 3 2

2

z-Transform103. The Inverse z-Transform (c.1)pPartial Fraction Expansion and Table Lookup

l If M<N and the poles are all first order

l If M>= N and the poles are all first order, the complete partial fraction expression can be

l If X(z) has multiple-order poles and M>=N

X zb

a

c z

d z

kk

M

kk

N( )

( )

( )

=−

−

−

=

−

=

∏

∏0

0

1

1

1

1

1

1

X zb

a

A

d zk

kk

N

( )( )

=− −

=∑0

01

1 1A d z X zk k z d k

= − −=( ) ( )1 1

X z B zA

d zr

r

r

M Nk

kk

N

( )( )

= +−

−

=

−

−=

∑ ∑0

11 1

A d z X zk k z d k= − −

=( ) ( )1 1

X z

c z

d z

kk

M

kk

N( )

( )

( )

=−

−

−

=

−

=

∏

∏

1

1

1

1

1

1

X z B zA

d z

C

d zr

r

r

M Nk

kk k i

Nm

im

m

s

( )( ) ( ),

= +−

+−

−

=

−

−= ≠

−=

∑ ∑ ∑0

11

111 1

Cs m d

d

dwd w X wm

is m

s m

s m is

w d i

=− −

−

−

−

−−

= −

11 1

1( )!( )[( ) ( )]

z-Transform113. The Inverse z-Transform (c.2)

X zz z

z z

z z

z z

BA

z

A

z( )

( )( )

=+ +

− +=

+ +

− −= +

−+

−

− −

− −

− −

− − −−

1 2

13

2

1

2

1 2

11

21 1

1

2

1

1 2

1 2

1 2

1 10

1

1

2

1

pExamples

l ROC:

l ROC:

l ROC:

z > 1

z <1

2

1

21< <z

z-Transform124. The Unilateral z-TransformpDefinition

pTime Delay

X z Z x n x n z n

n

( ) [ [ ] ] [ ]≡ ≡ −

=

∞

∑0

x n[ ] X z( )

x n k[ ]− z X z x n zk k n

n

k− − +

=

+ −∑( ) [ ]1

x n k[ ]+ z X z x n zi n

n

k

( ) [ ]−

−

=

−

∑0

1

z-Transform135. Solving the Difference Equations

pGoalsl Solving LTIL differential equations using Laplace transform.

pExample2y[k] + 3y[k-1] + y[k-2] = u[k] + u[k-1] - u[k-2]where u(t) = q(t), y(-1) = -1 and y(-2) = 1.

pExerciseFind the response of

y[k+1] - 2y[k] = u[k] and y[k+1] - 2y[k] = u[k+1] y[-1]=1 and u[k] = 1, for k =0, 1, 2, ...

z-Transform14

6. Zero-Input Response-- Characteristic Polynomial

pConsider the zero-input response

l The denominator of Yzi, is called the characteristic polynomial.l The roots of Yzi, is called the modes of the system.

pWhy the name, "mode" ?l The zero-input response of the system excited by any initial

conditions can always be expressed as

l The zero-input response is for t>=0

l The zero-input response is always a linear combination of the two time functions (-0.5)k and (-1)k.

Y zy y y z

z z

y y z y z

z zzi ( )

[ ] [ ] [ ] ( [ ] [ ]) [ ]=

− − − − − −+ +

=− − − − − −

+ +

−

− −

3 1 2 1

2 3

3 1 2 1

2 3 1

1

1 2

2

2

Y zy y z y z

z z

k z

z

k z

zzi ( )

( [ ] [ ]) [ ]

.=

− − − − − −+ +

=+

++

3 1 2 1

2 3 1 0 5 1

2

2

1 2

y k k kzik k( ) ( . ) ( )= − + −1 20 5 1

The form of the zero-input response excited by any initial conditions is completely determined by the modes of the system

The form of the zero-input response excited by any initial conditions is completely determined by the modes of the system

z-Transform156. Zero-State Response-- Transfer Function

The Transfer FunctionsThe ratio of the z transforms of the output and input with all initial conditions zero or

The transfer function of the above example is (z2 + z -1)/(2z2 + 3z +1)

H z Y zU z i n i t i a l c o n d i t i o n s

( ) ( )( )

== 0

pConsider2y[k] + 3y[k-1] + y[k-2] = u[k] + u[k-1] - u[k-2]l If all initial conditions are zero, we have

pWays to Find Transfer FunctionsThe transfer function is the z transform of the impulse response.The function can be obtained from the zero-state response excited by any input, in particular, step or sinusoidal functions.The function can be obtained from the difference equation description.

l y[k]-y[k-1]+2y[k-2]-3y[k-3] = u[k]

Y z z zz z

U z z zz z

U z( ) ( ) ( )=+ −+ +

=+ −+ +

− −

− −

12 3

12 3 1

1 2

1 2

2

2

z-Transform16

6.1 Poles and Zeros of Proper Transfer FunctionspFor a proper rational functions

l where N(z) and D(z) are polynomials with real coefficients. If N(z) and D(z) have no common factors, then the roots of D(z) and N(z) arerespectively the poles and zeros of the system.uexamples

pDefinitionl A finite real or complex number λ is a pole of H(z) if the absolute value

of H(λ ) = ∗. It is a zero of H(z) if H(λ ) = 0.

pExamplesl Find the zero-state response of a system with transfer function, H(z) =

(z2+z-1)/(2z2 + 3z +1) excited by unit step function.

H z N zD z

( ) ( )( )

=

Y z zz z

U z( ) ( )( . ) ( )

( )= =+

+ +3 4

2 0 5 1

z-Transform17

6.1 Poles and Zeros of Proper Transfer Functions (c.1)

H zz z j z j

z z z

1

3 2

0 5510 9 08 05 08 050551

0 7 055 0801

( ) .( . )( . . )( . . )

.. . .

=+ − + − −

=− − +

H z zz z z2 3 2

10 7 055 0801

( ). . .

=−

− − +

H z zz z z3 3 2

5 5 1 0 90 7 0 5 5 0 8 0 1

( ) . ( . ). . .

=−

− − +

H z zz z z4 3 2

5 5 1 1 10 7 0 5 5 0 8 0 1

( ) . ( . ). . .

=− −

− − +

H z z zz z z5

2

3 2

5 5 1 1 9 10 7 0 5 5 0 8 0 1

( ) . ( . ). . .

=− +

− − +

pConsider the following systems

Positions of zeros and poles

z-Transform186.2 Time Responses of Modes and PolespRemarks

l The zero-input response is essentially dictated by the modes; the zero-state response is essentially dictated by the poles.

pThree parts in z-planel The unit circlel The region outside the unit circle.l The region inside the unit circle.

pObservationsl The poles σ +- jω or re+-jθ

l The response rkcos kθ or rk sinkθl θ determines the frequency of the oscillation.l The highest frequency is determined by (-π /T, π /T].

z-Transform196.2 Time Responses of Modes and Poles(c.1)Summaryu The time response of a pole (mode), simple or repeated, approaches zero as

k -->∗ if and only if the pole (mode) lies inside the unit circle or its magnitude is smaller than 1.

u The time response of a pole (mode) approaches a nonzero constant as k -->∗if and only if the pole is simple and located at the z=1.

pExamplesl y[k+3] - 1.6 y[k+2] - 0.76y[k+1] - 0.08 y[k] = u [k+2] - 4u[k].

yzi [k], H(z), yzs[k] as k -> *

z-Transform20

7. Transfer-Function Representation-- Complete Characterization

pSystem Description for LTIL systemsl Convolutionl Difference equationl Transfer function.

pSystem Connectionl The transfer function of

connection can be easily derived from algebraic manipulation of transfer functions.

H(z)H(z)U(z) Y(z)

= H(z)U(z)

?h(k)h(k)

y k

h k i u ii

k

[ ]

[ ] [ ]= −=∑0

u k[ ]

H1(z)H1(z) H2(z)H2(z)H1(z)H1(z)

H2(z)H2(z)+

H1(z)H1(z)

H2(z)H2(z)

+-+

H1(z)H1(z)

H2(z)H2(z)

+++

u y

u y u y

u y

z-Transform21

7. Transfer-Function Representation-- Complete Characterization (c.1)

p Questionsl Transfer functions represent the input-output relation when initial conditions are

set to zeros.l What is the representation of the transfer function for a system

p Consider the difference equationsl D(z) Y(z) = N(z) U (z)Ex. D(z) = (z+0.5)(z-1)(z-2); N(z)= z2 - 3z +2

yzs[k] = k1(-0.5)k + (terms due to the poles of U(z))p Missing Poles

l If N(z) and D(z) have common factors, R(z), then the roots of R(z) are modes but not poles of the system and {the set of the poles} {the set of the modes}

l The root of R(z) are called the missing poles.p Completed Characterization

l A system is completed characterized by its transfer function if N(z) and D(z) arecoprime.

H z z zz z z z

( ) ( )( )( . )( )( ) .

=− −

+ − −=

+1 2

05 1 2105