11. discrete-time control: background1,2pcsl.ecs.umass.edu/cbs/handouts/ch11_discrete_time... ·...

3/22/2005 11-1 Copyright ©2005 (Yossi Chait)

11. Discrete-Time Control: Background1,2

Discrete-time control system are hybrid: part continuous time and part continuous time.

∑ PCDI

Y-FR

DOUE

∑ ∑

In studying how to analyze such systems we’ll visit: • Impulse sampling and zero-order hold

• z transform

• Stability

• Design 1. Ogata, K., Discrete-Time Control Systems, Prentice-Hall, Inc., 1987.2. Houpis, CH., and Rasmussen, SJ., Quantitative Feedback Theory Fundamentals and Applications, Marcel

Dekker AG, 1999.


11.1. Sampling and Hold

An ideal sampler comprises of a switch that closes to admit an input signal every sampling period T. The finite duration of the sampling is assumed infinitesimal. This is called analog-to-digital(A/D) conversion. A digital-to-analog (A/D) conversion converts to sampled-data signal back to a continuous-time signal – typically via a zero-order hold (ZOH) circuit.

tt kT


For a signal zero at t < 0, h(t) is related to x(t) as follows

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

(0) 1 1 ( ) 1 1 2

(2 ) 1 2 1 3

h t x t t T x T t T t T

x T t T t T

= − − + − − −

+ − − − +…

The Laplace transform of a delayed step is

and for the ZOH we jave

( )[ ] ( )( )1

0

( )k TskTs

k

e eh t H s x kT

s

∞ − +−

=

−= = ∑L


We define

The sampled signal is a train of impulses (the strength of each equals x(t) at kT)

where ( ) 0 unless .t kT t kTδ − = =

Nyquist (Shannon’s) sampling theorem says:

A function f(t) which contains no frequency components greater than ωc (i.e., band limited) can be represented by x(kT) with T < π/ωc.


In addition to assuming no aliasing, we assume that quantization, saturation, and conversion errors in the A/D conversion are negligible.

11.2. The z Transform

Define

so we can write

The machinery developed for Laplace transform domain, such frequency response, root locus, and stability analysis is readily applicable in the z domain for discrete-time control system having sampling as discussed above.


Examples of z transforms:

( ) ( )1, 0

00, 0

kx t

k=⎧

= δ = ⎨ ≠⎩

so

The unit step function

( ) ( )1x t t=

has a z transform of


A table of z transforms is provided.

Important Properties.

•

•

• Final Value Theorem: if X(z) has only stable poles (inside the unit circle) with the exception of one at z = 1, then the final value is

A table of important z transform properties is provided.


Inverse z transform.

• direct division

• partial fraction expansion

• other methods

( )( ) ( )

1 2

1 210 5 10 5

.1 .2 1 1.2 0.2

z z zX z

z z z z

− −

− −+ +

= =− − − +

Example. Find x(k) for

1 2 1 2

1 2 3

2 3

2 3 4

1 1.2 0.2 10 5 10 12 2 17 2 17 20.4 3.4

z z z zz z z

z zz z z

− − − −

− − −

− −

− − −

− + +− +

−− +

3 4

3 4 5

4 5

18.4 3.4 18.4 22.08 3.68 18.68 3.68

z zz z z

z z

− −

− − −

− −

−− +

−4 5 6 18.68 22.416 3.736z z z− − −− +

1 2 3 410 17 18.4 16.86z z z z− − − −+ + + +


Since

( )0

( ) ,k

k

X z x kT z∞

−

== ∑

direct comparison of this infinite series with the long division gives

Example. Find x(k) for ( )( ) ( )

101 .2

zX z

z z=

− −

If X(z) has one or more zeros at the origin, use this trick


From z transform tables

Note: if we did not divide by z, the expansion of x(z) would yield terms not appearing in the table.


11.3. The Pulse Transfer Function

Here we study transfer functions in the z domain. Consider the system shown below.

The continuous-time convolution integral becomes a convolution integral in discrete time

∞

( ) ( ) ( )

( ) ( ) ( ) ( )

0

0

, 0.

h

h

y kT g kT T x hT

x kT hT g hT x k h g k h

=∞

=

= −

= − < = < =

∑

∑The z transform of y(kT) is

( ) ( ) ( ) ( )0 0 0

k

k k h

Y z y kT z g kT hT x hT∞ ∞ ∞

−

= = == = −∑ ∑ ∑


( ) ( ) ( ) ( ) ( )0 0 0 0

( ) ( ), .

m h m h

m h m h

g mT x hT z g mT z x hT z

G z X z m k h

∞ ∞ ∞ ∞− + − −

= = = == =

= = −

∑ ∑ ∑ ∑

The pulse transfer function is

( )( )

.Y z

GX z

=

Back to the original system.

To relate sampled signals, we note that

( ) ( )* * , 0,1,2,sX s X s j k k= ± ω = …

so


and the starred Laplace transform with becomessTz e=

( ) ( ) ( ).Y z G z X z=

Note that in this system

That is

How do we obtain a pulse TF for a system? Typically, we use basic block diagram algebra and z-Transform Tables (see attachments).

Example. Find the PTF of

( )1 1

( )1

TseG s

s s s

−−=

+


( ) ( )( )

( )

( )( )

( ) ( )( )( )

2 2

1 2

1 12 1 11

1 1 1 1 1 11 11 11 11

1 11 11 1 .1 11 11

( )

T T T

TT

Tsez z

s s s s ss s s

T e z e Te zTzz

z e zz e zz

G z

− − − − −

− − −− − −−

− ⎡ ⎤⎡ ⎤− ⎡ ⎤− −− = − + +⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦+ ++⎣ ⎦ ⎣ ⎦

⎡ ⎤ − + + − −⎢ ⎥−− + + =⎢ ⎥ − −− −−⎢ ⎥⎣ ⎦

=

=

=Z Z Z

Example. Find the PTF of a closed-loop discrete-time control system shown below.

( )e t∑

( )r t ( )e kT

T( )G s

( )c t

−

( )H s

*E R HC R HGE= − = −

Starring both sides

[ ]

*

* *1

1

ER HG

=+


Combining the last 2 equations gives

Note: we have assumed all samples have same sampling period and are synchronized.

Some systems do not have a PTF. This occurs when the input signal dynamics cannot be decoupled from the dynamics of the system.

∑

( )r t

T

( )G s( )c t

−

( )H s


Example.

E∑

R *ET

KC

−( )ZOH s 1

1s+

plantzero-order

holddigital

controller

*C kGEE R C== −

( ) ( ) ( ) 11 1

1 1 1

11 1

1 1 1 1

T

T T

k e zk k k kz z

s s z e z e z

− −− −

− − − − −

−⎡ ⎤ ⎡ ⎤= − − = − − =⎢ ⎥ ⎢ ⎥+ − − −⎣ ⎦ ⎣ ⎦Z

( )( )( )

1

1

1.

1 1

T

T

k e zCz

R k k e z

− −

− −

−⇒ =

⎡ ⎤− − +⎣ ⎦


Example. E∑

R *ET

PIDC

−( )ZOH s ( )

11s s+

( ) ( ) ( )1 111

11

1d

p i d psi

T KPID s K K K s PID z K z

K Tz− −

−= + + ⇒ = + + −−

( ) ( ) ( )[ ] ( ) ( ) ( ) ( )1 2

12 1 1

1 .3697 .26421 .

1 1 .3697 1z z

G z ZOH s G s zs s z z

− −−

− −

⎡ ⎤ += = − =⎢ ⎥+ − −⎣ ⎦Z Z

1 0, 1.d i pK K K−= = =Let

( ) ( )( )

1 2 1

1 2 2 1.3697 .2642 .3697 .2642

.1 1 .6321 .6321

p

p

K G zC z z zz

R K G z z z z z

− −

− −+ +

= = =+ − + − +


Step Scope

1

Gain

num(z)

([1 -.3679],[1 -Discrete

Transfer Fcn1

0 5 10 15 200

0.2

0.4

0.6

0.8

1

1.2

1.4

sec

c(kT

)


For a unit step

( )1 2

1 2 1

1 2 3 4 5 6

.3697 .2642 11 .6321 1

.3679 1.3996 1.3996 1.1469 .8944

z zC z

z z zz z z z z z

− −

− − −

− − − − − −

+=

− + −= + + + + + +…

Since z-1 implies time shift by one sampling period, taking inverse z transform gives

Note: it is possible to compute the response between sampling instances.


11.4. s-plane to z-plane Mapping

Stability and performance of continuous-time (CT) systems depends on pole location. Since s and z are related by z = eTs, we can study these discrete-time properties of a PTF by related z domain pole location via the map.

Poles and zeros in the s plane are mapped to the z plane via z = eTs. Denote the complex number s = σ+jω so

( ) ( )2 , 0, 1,T j j T kT jT Tz e e e e e kσ+ ω ω+ πσ ω σ= = = = ± …

Note that s plane frequencies with integer multiple of ωs difference are mapped into the same z plane location.

Stable systems have al their poles in the open left-half s plane, or in the z plane

The jω-axis in the s plane maps into |z| = 1 circle.


The phase of z, ωT, varies from -∞ to ∞ as ω varies from -∞ to ∞. In particular, along the jω axis, as ω varies from -½ωs to -½ωs we have |z|=1 and ∠z varies CCW from –π to π. From -½ωs to -½ωs the phase ∠z varies again CCW from –π to π. This is depicted below.

planez

2sj ω−

jω

σ

planes

2sj ω

2sj ω−

sjω

32

sj ω

Im

Re

primary2

sj ω

−∞

jω

σ

Im

Re


planezplanes

ω

1−σ 2σ0

j Im

1Reσ

0 1

Im

Reσ

jω2

sj ω

2sj ω−

2jω1jω

-1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0

jω

2sj ωsjω

σ


-1 -0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

planez


11.4. Stability Analysis

Discrete-time stability is determined from the roots of the characteristic equation 1+L(z)=0.


11.4.1 Stability tests

• Jury Stability Test an algebraic test using the characteristic equation.

• Routh test via bilinear transformation.

jω

σ

planesIm

planez planew

j∞

Re

Im

j− ∞

2sj ω

2sj ω−

Re11−

( ) 1 11 1 00 1 n n

n nL z a z a z a z a−−= + = + + + +1 1

1 1 01 1 1

01 1 1

n n

n nw w w

a a a aw w w

−

−+ + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + − + − +⎝ ⎠ ⎝ ⎠ ⎝ ⎠

11. discrete-time control: background1,2pcsl.ecs.umass.edu/cbs/handouts/ch11_discrete_time... ·...

Documents