DISCRETE-TIME SYSTEM REPRESENTATION

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DISCRETE-TIME SYSTEM REPRESENTATION

LECTURE 4

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OUTLINE

• Obtain discrete-time models (State-space or transfer functions) from continuous-time models

• Continuous TF with sampler -> Discrete TF

• Continuous TF with ZOH -> Discrete TF

• Continuous TF with delay -> Discrete TF

• Continuous SS -> (ZOH augmented) Discrete SS

• Discrete TF -> Discrete SS

• Discrete SS -> Discrete TF

• Properties of discrete models (poles/zeros/eigenvalues…)

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REVIEW OF MATHEMATICAL FACT

• Z-Transform is nothing but the Laplace Transform of sampled signals, i.e., the starred Laplace transform

• Convolution of two functions in time becomes

multiplication of two functions in the s- (or z-) domain.

(linked by Laplace or z-transform)

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ANALYZING HYBRID SYSTEMS

ME561 Lecture 5- 3

Analyzing Hybrid Systems

• To analyze computer-controlled system, a unified approach is need to look at both continuous-time signals and sampled (discrete-time) signals

⇒ Laplace transform is still the tool of choice (with some added consideration for sampled signals)

C(z)

D/A u(kT) e(kT) A/D

T

G(s) ⁺ r

–

y u(t) e(t)

Critical elements to analyze: sampler, ZOH, delay

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CONTINUOUS TF -> DISCRETE TF

• Commonly, some signals in a system are impulse sampled and

others are continuous. Continuous signals can be transformed into discrete ones by sampling

ME561 Lecture 5- 4

Continuous TF -> Discrete TF

• Commonly, some signals in a system are

impulse sampled and others are continuous.

Continuous signals can be transformed into discrete ones by sampling

d_T

x(t) x*(t)

X(s) X*(s)

Recall the Laplace transform of the sampled signal (z-tranform) is:

[ ]

₀

0

( ) *( ) *( ) *( ) ^s ( ) ^kTs

k

X z X s x t ^∞ x τ e d⁻ ^τ τ ^∞ x kT e⁻

=

= = ^L =

∫

⋅ =

∑

⋅

X(kT)

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CONTINUOUS-TIME SYSTEM WITH (INPUT) SAMPLER

ME561 Lecture 5- 5

Continuous-time System with (Input) Sampler

( )

G s x t^*( ) x t( ) ( )

y t

*( ) y t

( )

*( ) X s X s

*( )

Y s Y s( )

Y ( ) s = G ( ) s X ⋅ *( ) s

Fact: _{Y s} _{( )} ₌ _{G s X} _{( )} _⋅ _{*( )} _s Clear

( ) ( ) ( ) Y z = G z X z ⋅

Equivalently Fact:

Not readily clear (property of

Starred Laplace Transform to be proven)

ME561 Lecture 5- 5

Continuous-time System with (Input) Sampler

( )

G s x t^*( ) x t( ) ( )

y t

*( ) y t

( )

*( ) X s X s

*( )

Y s Y s( )

Y *( )s = G *( )s X⋅ *( )s

Fact:

_{Y s}_{( )} ₌ _{G s X}_{( )} _⋅ _{*( )}_s

Clear

( ) ( ) ( ) Y z = G z X z⋅

Equivalently Fact:

Not readily clear

(property of

Starred Laplace Transform to be proven)

Fact:

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ME561 Lecture 5- 6

Proof:

0

0 0

0

( ) ( ) *( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

t

t t

k k

k

y t g t x d

g t x kT d g t x kT d

g t kT x kT

τ τ τ

τ ^∞ τ δ τ τ ^∞ τ τ δ τ τ

= =

∞

=

= − ⋅

= − ⋅ − = − ⋅ −

= − ⋅

∫

∑ ∑

∫ ∫

∑

( )

G s x t^*( ) x t( ) ( )

y t

*( ) y t

( )

*( ) X s X s

*( )

Y s Y s( )

Y ( ) s = G ( ) s X ⋅ *( ) s

Or

Y z ( ) = G z X z ( ) ⋅ ( )

[ ]

0 0

( )

0 0 0

0 0

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) , where

( ) ( ) ( ) ( )

n

n k

n k m

k n k k m

m k

Y z Z y t g nT kT x kT z

g nT kT x kT z g mT x kT z m n k

g mT z x kT z G z X z

∞ ∞

−

= =

∞ ∞ ∞ ∞

− − +

= = = =

∞ ∞

− −

= =

= = − ⋅

= − ⋅ ⋅ = ⋅ ⋅ = −

= ⋅ ⋅ ⋅ = ⋅

∑ ∑

∑∑ ∑∑

∑ ∑

j

k k = j

j

k k = j

ME561 Lecture 5- 6

Proof:

0

0 0

0

( ) ( ) *( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

t

t t

k k

k

y t g t x d

g t x kT d g t x kT d

g t kT x kT

τ τ τ

τ ^∞ τ δ τ τ ^∞ τ τ δ τ τ

= =

∞

=

= − ⋅

= − ⋅ − = − ⋅ −

= − ⋅

∫

∑ ∑

∫ ∫

∑

( )

G s x t^*( ) x t( ) ( )

y t

*( ) y t

( )

*( ) X s X s

*( )

Y s Y s( )

Y *( )s = G *( )s X⋅ *( )s Or Y z( ) = G z X z( ) ⋅ ( )

[ ]

0 0

( )

0 0 0

0 0

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) , where

( ) ( ) ( ) ( )

n

n k

n k m

k n k k m

m k

Y z Z y t g nT kT x kT z

g nT kT x kT z g mT x kT z m n k

g mT z x kT z G z X z

∞ ∞

−

= =

∞ ∞ ∞ ∞

− − +

= = = =

∞ ∞

− −

= =

= = − ⋅

= − ⋅ ⋅ = ⋅ ⋅ = −

= ⋅ ⋅ ⋅ = ⋅

∑ ∑

∑∑ ∑∑

∑ ∑

j

k k = j

j

k k = j

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PROPERTY OF STARRED LAPLACE TRANSFORM

ME561 Lecture 5- 7

Property of Starred Laplace Transform

Y *( )s = G s X( )⋅ *( ) *s = G s( ) *⋅X *( )s = G *( )s X⋅ *( )s

( )

G s x t^*( ) x t( ) ( )

y t

*( ) y t

( )

*( ) X s X s

*( )

Y s Y s( )

Y s( ) = G s X( ) ⋅ *( )s

x ( t )

X ( s ) G ( s )

y ( t )

Y ( s )

Y z( ) = Z Y s( ) = Z G s X s( ) ( ) = Z GX s( ) = GX z( ) ≠ G z X z( ) ( )

( ) ( ) ( )

Y z = G z X z ⋅

Unless constant or pure delay (Ex4.3 to be explained later)

Given

UNLESS CONSTANT

OR PURE DELAY

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SUMMARY — CONTINUOUS-TIME SYSTEM WITH   INPUT SAMPLER

ME561 Lecture 5- 8

Summary-- Continuous-time System with Input Sampler

• If the input to a continuous-time system G(s) is an impulse-sampled signal (i.e., G(s) is preceded by a sampler), then it makes sense to calculate the z-

transform of this transfer function, and the

corresponding pulse transfer function of G(s) is given by

[ ]

¹

[ ]

( ) ( ) ( )

G z Z G s

⁻

G s

t kT

= = Z L

=

( )

G s _{x t}^*_{( )} x t( ) ( )

y t

*( ) y t

( )

*( ) X s X s

*( )

Y s Y s( )

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ME561 Lecture 5- 9

Ex4.1 Calculating Pulse Transfer Function (Continuous-time system with sampler)

( )

G s x t^*( ) x t( ) ( )

y t

*( ) y t

( )

*( ) X s X s

*( )

Y s Y s( ) G s

( ) = s a

+ 1

Method I: (Table lookup)

From the z-transform table (Table 3.1), we have

G z s a

z

z e ^aT e ^aT z ( ) =

L

+

NM O

QP

⁼ ⁻ ⁻ ⁼ ⁻ ⁻ ⁻

Z 1 1

1 ¹

Method II: (Direct Calculation)

The impulse response for G(s) is

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ME561 Lecture 5- 10

Ex4.1 (cont.)

[ ]

( )

1

( )

^at

g t = L

⁻

G s = e

⁻

Method II: (Direct Calculation)

The impulse response for G(s) is

Hence

[ ]

¹

[ ]

( ) ( ) ( )

G z Z G s

⁻

G s

t kT

= = Z L

=

g kT( ) = e⁻^akT , k = 0 1 2, , ,...

G z g kT e z e z

e z e z

akT k

k

aT k

k aT aT

( ) = ( ) = ⋅ = =

− =

−

− −

=

∞ −

=

∞

− − −

∑ ∑

Z

0 0

1 1

1

c h c h

1

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PULSE TRANSFER FUNCTION OF CASCADED SYSTEMS

ME561 Lecture 5- 11

Pulse Transfer Function of Cascaded Systems

d_T x*(t) x(t)

X*(s) X(s) d_T G(s)

u*(t) u(t) U*(s) U(s) d_T H(s)

y*(t) y(t) Y*(s) Y(s)

dT

x*(t) x(t) X*(s) X(s) u(t) G(s)

H(s) U(s) dT

y*(t) y(t) Y*(s) Y(s)

(a)

(b)

Two similar configurations, systems in cascade with/without

a sampler in between

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CASE (A)

ME561 Lecture 5- 12

Case (a)

d_T x*(t) x(t)

X*(s) X(s) d_T G(s)

u*(t) u(t) U*(s) U(s) d_T H(s)

y*(t) y(t) Y*(s) Y(s)

U s( ) = G s X( )⋅ *( )s and Y s( ) = H s U( ) ⋅ *( )s

U *( )s = G *( )s X⋅ *( )s and Y *( )s = H *( )s U⋅ *( )s

Y *( )s = H *( )s G⋅ *( )s X⋅ *( )s

Y z ( ) = H z G z X z ( ) ⋅ ( ) ⋅ ( )

H z( ) = Z H s( ) G z( ) = Z G s( )

PSLT

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CASE (B)

ME561 Lecture 5- 13

Case (b)

d_T x*(t) x(t)

X*(s) X(s) u(t) G(s)

H(s) U(s) d_T

y*(t) y(t) Y*(s) Y(s)

Y s( ) = H s G s X( ) ( )⋅ *( )s = HG s X( ) ⋅ *( )s

Y *( )s = H s G s( ) ( ) *⋅X *( )s = HG s( ) *⋅X *( )s

Y z

( ) =

HG z X z

( ) ⋅ ( )

PSLT

( ) ( ) H z G z

≠

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EX4.3 PULSE TRANSFER FUNCTION OF CASCADED SYSTEMS

ME561 Lecture 5- 14

Ex4.3 Pulse Transfer Function of Cascade Systems

d_T x*(t) x(t)

X*(s) X(s) d_T G(s)

u*(t) u(t) U*(s) U(s) d_T H(s)

y*(t) y(t) Y*(s) Y(s)

G s s a H s

( ) = ( ) s b

+ =

+

1 1

and

[ ] [ ]

( )( )

2

( ) ( ) ( ) ( ) ( )

( )

1 1

bT aT bT aT

Y z H z G z Z H s Z G s X z

z z z

Z Z

s b s a z e⁻ z e⁻ z e⁻ z e⁻

= = ⋅

= ⋅ = ⋅ =

+ + − − − −

Solution:

Case (a)

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ME561 Lecture 5- 15

Ex4.3 (cont.)

d_T x*(t) x(t)

X*(s) X(s) u(t) G(s)

H(s) U(s) d_T

y*(t) y(t) Y*(s) Y(s)

G s s a H s

( ) = ( ) s b

+ =

+

1 1

and

Solution:

Y z

X z HG z HG s H s G s

s b s a a b s b s a a b

z z e

z

bT z e aT

( )

( ) ( ) ( ) ( ) ( )

( ) ( )

= = =

= + ⋅

L

+

NM O

QP

⁼ ⁻

L

⁺ ⁻ ⁺

NM O

L QP

NM O

QP

⁼ ⁻

L

⁻ ⁻ ⁻

NM O

QP

− −

Z Z

Z 1 1 Z 1 1 1 1

( )

( )( )

1

( )

bT aT

z e e

a b z e z e

− −

= −

− − −

( )( )

2

bT aT

z

z e− ⁻ z e− ⁻

vs.

Case (b)

Case (a)

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PULSE TRANSFER FUNCTION OF CLOSED-LOOP SYSTEMS

ME561 Lecture 5- 16

Pulse Transfer Function of Closed-Loop Systems

dT

e(t) E(s) e*(t)

E*(s) y(t) G(s)

Y(s)

H(s)

r(t) R(s)

+ –

E s R s H s Y s Y s G s E s

( ) ( ) ( ) ( ) ( ) ( ) *( )

= −

= ⋅

E s( ) = R s( ) − H s G s E( ) ( )⋅ *( )s

E *( )s = R *( )s − H s G s( ) ( ) *⋅E *( )s = R*( )s − HG *( )s E⋅ *( )s PSLT

E s

HG s R s

*( ) = *( ) *( )

+ 1 ⋅

1 ^Y ^s

G s

HG s R s

*( ) *( )

= + ⋅

1

Y z G z

HG z R z

( ) ( )

= + ⋅

1

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OTHER

CONFIGURATIONS

ME561 FALL 2001

Table 4.1 Pulse Transfer Function of Five Typical Closed-Loop System Configurations From Figure 4.6, we see that

Y s( )= G s G( )⋅ _C*( )s E⋅ *( )s or Y s*( )= G s G*( )⋅ _C*( )s E⋅ *( )s Using the z-transform notation, we get

Y z( )= G z G z E z( )⋅ _C( )⋅ ( )

where G z( )= Z G s( ) . Since E z( ) = R z( )−Y z( ), we have

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OUTLINE

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CONTINUOUS-TIME PRECEDED BY ZOH

ME561 Lecture 5- 18

Continuous-time System Preceded by ZOH

[ ]

* ( ) _P ( ) _ZOH ( ) * * ( )

Y s = G s G⋅ s ⋅ U s G s e

ZOH s

Ts

( ) = 1− ⁻ Y z

U z( ) G s G_P _ZOH s G z ( ) = Z ( )⋅ ( ) = ( )

G z e

s G s G s

s e G s

s

G s

s z G s

s

Ts

P P Ts P P P

( ) ( ) ( ) ( ) ( ) ( )

=

L

−

NM O

QP

⁼

L

NM O

QP

⁻

L

NM O

QP

⁼

L

NM O

QP

⁻ ^⋅

L

NM O

QP

− − −

Z 1 Z Z Z ₁ Z

G z z G s

s

z z

G s s

P P

( ) ( ) ( ) ( )

= −

L

NM O

QP

⁼ ⁻

L

NM O

−

QP

1 ₁ 1

Z Z

Pure delay comes out directly

G_P(s) y(t)

Y(s) G_ZOH(s) u(k) or u*(t)

U(z) or U*(s) u(t)

Plant U(s) ZOH y(k) or y*(t)

Y(z) or Y*(z) G(z) or G*(s)

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REVIEW OF ZOH

ME561 Lecture 5- 19

Review of ZOH

f t ( ) = f kT ( ) for kT ≤ < t ( k + 1 ) T

Unit Impulse Response

X s x t t t T

s s e e s

R R

sT

( ) = ( ) = ( ) − ( − )

= −

−

L L 1 L 1

1 1 1

G_ZOH(jω)

x*(kT) X*(jω) x_R(t)

X_R(jω)

Time 0

1

Time 0

1

T

G s e

ZOH s

sT

( ) = 1− ⁻

0 Time 1

T

“An integrator that is automatically reset to zero after one sampling period”

ME561 Lecture 5- 19

Review of ZOH

f t( ) = f kT( ) for kT ≤ <t (k +1)T

Unit Impulse Response

X s x t t t T

s s e e s

R R

sT

( ) = ( ) = ( ) − ( − )

= −

−

L L 1 L 1

1 1 1

G_ZOH(jω)

x*(kT) X*(jω) x_R(t)

X_R(jω)

Time 0

1 Time

0 1

T

G s e

ZOH s

sT

( ) = 1− ⁻

0 Time 1

T

“An integrator that is automatically reset to zero after one sampling period”

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ME561 Lecture 5- 20

Ex4.2

( ) G s

( ) x t

*( ) y t

( ) X s

*( )

Y s zoh

*( )

x t ₁

( ) ( 1) G s = s s

+

Delays that are integer multiple of sample period can be taken out and replaced with the proper z^-1terms

[ ] [ ]

( )

2

1 1

1

( ) ( ) ( ) ( )

( 1) 1

( 1)

Ts

P ZOH

Ts

G z G s G s G s e

s e

s s

s s

−

− −

= = ⋅ ⋅

+

= ⋅

− +

Z Z = Z

Z

(

¹

)

2

1 1 (

z 1)

s s

= − ⋅

− Z +

( )

2

( ) 1 1 1 1

1 G z z 1

s s

s

∴ = − − ⋅ − +

Z +

(

¹

)

( 1 2

( 1 ) (1 )

1 ) 1 ( 1)( )

T T

T

T e z e Te

z z

z

z e e

T z

z z z

z

− − −

−

− + + −

−

= − − + = −

−

− −

−

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ME561 Lecture 5- 21

Ex4.4

( ) G s

( ) x t

*( ) y t

( ) X s

*( )

Y s zoh

*( ) x t

G s K

P( ) = s a

+

1 ( ) 1

( ) (1 ) (1 )

( )

G sP k

G z z z

s s s a

− −

= − ⋅ = − ⋅

Z Z +

(

¹ ^z⁻¹

)

^K_a ¹_s _s ¹ _a ^z ⁻_{z a}¹ ^K ⁻¹ ¹_s _s ¹ _a

= − ⋅ − = ⋅ −

+ +

Z Z L

(

¹

)

¹ ¹

1 _at

t kT

e akT

z K z K

z a ⁻ z a e⁻

=

− −

= ⋅ Z − = ⋅ Z −

1 1

aT

z z e a z

z z

z K

−

= − −

−

(

¹ ^aT

)

aT

K e

a z e

−

= − −

− C.T. pole: ^Note −a D.T. pole: e ^–aT ( recall z = e^sT )

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AN IMAGINED APPLICATION

ME561 Lecture 5- 22

An Imagined Application

r(t) R(s) r(k)

R

^*

(s) C(z)

ZOH G(s)

U

^*

(s)

Y

^*

(s)

Y

^*

(s) u(k)

y(k)

e(k) E

^*

(s)

What techniques do we need for the analysis of this Control problem?

What techniques do we need for the analysis of

this control problem?

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TABLE FOR SYSTEMS PRECEDED BY ZOH

ME561 Lecture 5- 23

Table for Systems Preceded by ZOH

-- M. Tomizuka, ME232 Notes, UC Berkeley

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ZOH EQUIVALENCE OF SYSTEM WITH TIME DELAY

ME561 Lecture 5- 24

ZOH Equivalence of System with Time Delay

G s_P

( ) =

e⁻^{T s}^D

⋅

G s

( )

TD = nT + TL

(

¹

)

1 ¹

( ) 1 ( )

1 ^L ^L

T T

n

s s

k n

t T

e G s z e G s

z z

s z s

− −

− − −

+

=

⋅ − ⋅

= − Z = Z L

(

¹

)

^{( )}

(

¹

)

^{( )}

( ) 1 1 ^L

T s p P

n s T

G s e G s

G z z Z z Z e

s s

− − − − ⋅

= − = −

(

¹

)

1

( ) 1 ( )

( ) 1 ^L^s ^L^s

p

n T

n

T G s z G s

G z z z e e

s z s

− −

+

⇒ = − ⋅ ⋅Z ⋅ = − ⋅Z ⋅

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EX4.5 ZOH EQUIVALENT WITH TIME DELAY

ME561 Lecture 5- 25

Ex4.5 ZOH Equivalent with Time Delay

G s e

P s

s

( )

= .

+

−0 5

1

Find the equivalent ZOH transfer function with a sampling time of 0.2 sec.

Since T_D = 0.5 sec, we can select n = 2 and T_L = 0.1

Solution:

0.1

2 1 3

1 1 1 1

( ) ( 1) 1

s

z e z s

G z e

z s s z s s

− −

+

− −

= = −

+ +

Z Z

Use the following facts:

1

1 1

1 ^T

Z z

s z

Z z

s z e⁻

= −

+ = −

G

_p

(s) = e

^−0.5s

s + 1

(28)

ME561 Lecture 5- 26

Ex4.5 (cont.)

0.1 1 1

1 Z e s

s z

− =

−

0.1

0.1 0.1 0.1 0.2 0.1

1 1 0.2

1

1 ^k

s t T k

k

k e

Z e Z e Z e Z e

s z e

− − + − + ∞ − + −

= −

∞

= = = = =

+ −

( )

0.1

3 0.2 3

1 1 0.0952 0.0861

( ) 1 0.8187

z e z

G z z z z e z z

−

− +

= − =

− − −

0.1 1 2

1

1 1

1 0 1

s k

k

Z e z z z

s z

− ∞ − − −

=

= ⋅ = + + + =

∑

−

0.1

0.1 1 2

0

0.1 0.3

.2

1 0

1

s e

Z e z z

s e e z

e

− − − − − −

= + + + = −

+ −

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OUTLINE

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STATE SPACE REPRESENTATION

ME561 Lecture 6- 3

x(t) or x(k) n ¥ 1 vector State Vector u(t) or u(k) r ¥ 1 vector Input Vector

y(t) or y(k) m ¥ 1 vector Output Vector F or A n ¥ n matrix System Matrix

G or B n ¥ r matrix Input Matrix C(t) or C(k) m ¥ n matrix Output Matrix

D(t) or D(k) m ¥ r matrix Direct Transmission Matrix

State Space Representation

( ) ( ) ( )

t t t

= ⋅ + ⋅

x F x G u y C x D u

Continuous-Time

( 1) ( ) ( )

( ) ( ) ( )

k k k

+ = ⋅ + ⋅

= ⋅ + ⋅

x A x B u

y C x D u

Discrete-Time

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STATE SPACE AND I/O

ME561 Lecture 6- 4

State Space and I/O

• Data needed to represent a LTI system:

( ) ( ) ( )

t t t

= ⋅ + ⋅

x F x G u y C x D u

( 1) ( ) ( )

( ) ( ) ( )

k k k

+ = ⋅ + ⋅

= ⋅ + ⋅

x A x B u

y C x D u

Continuous-Time Discrete-Time

F G C D

x x

y u

F G

C D u y

x ^{1 s} x

F G C D

A B C D

( )k x

( )k u

(k +1) x

( )k y

A B C D A B C D

( )k ( )k u

y

1 z x( )k (k +1)

x

Often referred to as system (A,B,C,D)

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ZOH EQUIVALENT SYSTEM

ME561 Lecture 6- 5

ZOH Equivalent System

Given

the dynamics

and the selected sampling time T (for the sampling and ZOH),

can we figure out the equivalent discrete-time representation

x A x B u

y C x D u

( ) ( ) ( )

k k k

+ = ⋅ + ⋅

= ⋅ + ⋅

1

i.e., what is the relationship between (F,G) and (A,B)?

C and D remain unchanged.

F G C D

F G

C D u y

x 1 s x

( ) ( ) ( )

t t t

= ⋅ + ⋅

x F x G u y C x D u

A B

C D A B

C D u( )k ( )k

y

1 z x( )k (k +1)

x

ZOH ^u^{( )}^k

( )k

y F G

C D u

x 1 s x

y

ME561 Lecture 6- 5

ZOH Equivalent System

Given the dynamics

and the selected sampling time T (for the sampling and ZOH), can we figure out the equivalent discrete-time representation

x A x B u

y C x D u

( ) ( ) ( )

k k k

+ = ⋅ + ⋅

= ⋅ + ⋅

1

i.e., what is the relationship between (F,G) and (A,B)?

C and D remain unchanged.

F G C D

F G

C D u y

x 1 s x

( ) ( ) ( )

t t t

= ⋅ + ⋅

x F x G u y C x D u

A B

C D A B

C D u( )k ( )k

y

1 z x( )k (k +1)

x

ZOH ^u^{( )}^k

( )k

y F G

C D u

x 1 s x

y

DISCRETE-TIME SYSTEM REPRESENTATION