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Exercises

24.3. Suppose that.

1 - 0.2Z-1F(z) -

- (1 + 0.6Z-1)(1 - 0.3Z-1)(1 - Z-1)

(a) Calculate the corresponding time-domain responsef*(t).

f*(t).

24.4. Determine

the inverse transform of

z(z + 1)

(z - 1)(z2 - z + 1)

by the following methods:

(a) Partial fraction expansion.

(b) Long division.24.5Calthe

I-trofth

repsit

dthat the s ampling period is t1t = 2 min. The pulse is f = 3 for 2 ::; t < 6.

6.-

f 31-

00 2

.14 6

.1

24.6. The pulse transfer function of a process is given by

Y(z) - 5(z + 0.6)

(a) Calculate the response y(nM) to a unit step change in x using the partial fraction

method.(b)Cheyouansinpa

(a)byuslo

di(c) What is the steady-state value of y?24.7Thedesite

trajT(foa

bareissit

d(a)Deran

expforth

Latro

thtetT(s).

(b) Determine the corresponding I-transform T(z) for s ampling periods of t1t = 4and 8 min.

0 20

J

40

24.8. The dynamic behavior of a temperature sensor and transmitter can be described by

the first-order transfer function,

T(s) 8s + 1

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588 RESPONSE

If the actual temperature changes as follows (t in seconds):

{

a

s

mv

o(b) If samples of the measured temperature are automatically logged in a digital

coevtwm

at=

tthe logged temperature? .

24.9. The transfer function for a process model and a zero-order hold can be written as

(1 - e-SM

)3.8e-zs

H(s)Gp(s) = s (lOs + 1)(5s + 1)

Derive an expression for the pulse transfer function ofH(s)Gp(s) when ~t = 2.24.

Thpu

trafuoap

isgbY(z) - 2.7z-!(z + 3)

Ii

(a) Calculate the response y(n~t) to a unit step change in x using the partial fractionmethod.

(b) Check your answer in part (a) by using long division.

(c) What is the steady-state value of y?24.Agachis

ustop

c

feedbackcolo

Thoptrfi

gG(s)

and is

I

B(s)

(l

)(l-e-Sl!.t

)(10

)-21

G(s) = E(s) = 2 1 + 8s s 12s + 1 e

(a) Suppose that a sampling period of~t = 1 min is selected. Calculate HG(z), the

putrafu

ofGwZ(b) If a unit step change in the controller error signal e(t) is made, calculatethesa

opreb

u

24.Deth

putr

fu

z

tprocess Gp(s) = K/[(5s + 1)(3s + 1)] using partial fraction expansion in the

s-dC

yorew

thi

S2

there.

24.13. Find HG(z) if G(s) = (1 - 9s)/[(3s + 1)(15s + 1)] for ~t = 4 (use partial fractionex

Wisthcod

e

resinth

ouYnfoastcit

i

t24.14. Verify the z-transform in Table 24.1 for f(t) = t2. What is the z-transform

for

f(t) = 1 - e-at?

I

24.FintheresYnfoth

dieY-Y

+0.

=

LeXo=1,Xn=0fon2:1.U

lod

a

a

ithe results.

ill...

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Exercises 589.0.8rl

F(z) = (1 - 0.8z-l)Z

. . oro

eaea

rau

+ 1)/('Tzs

the steady-state gain of the pulse transfer function.

.18. Determine the sampled function f(nf:..t) corresponding to the z-transform

0.5z-1( z )

-- 1 - 1.5z-1 + 0.5z-zUsepartfracex

(f:.=1)ancothrewt

ldmethod for the first six sampled values (n = 0, 1, . . . , 5)._..1For

G(s=1/[(+1)(+2)]obG(fo

M=1.Dt

rI

to a unit step change in the input. Repeat using Tustin's method (approximate

z-transform) and compare the step responses for the first five samples.~.20Todetetheeffof

polanzeloc

astI step responses of the pulse transfer functions shown below for the first six sampling

instants, n = 0 t o n = 5. What conclusions can you make concerning the effect of

pole and zero locations?

1

(a) i"""-=- Z-I

1

(b) 1 + 0.7z-1

1(c) 1 - 0.7z-1

1

(d) (1 + 0.7z-I)(1 - O.3z-l)

1 - 0.5rl

(e) (1 + 0.7rl)(1 - 0.3z-l)

f 1 - 0.2rl( ) (1 + 0.6rl)(1 - 0.3z-l)

:4.21. For the transfer functions shown below, determine the corresponding pulse transfer

function HGp(z) for the system and a zero-order hold.

1

(a) Gp(s) = (s + 1)36(1 - s)

(b) Gp(s) = (s + 2)(s + 3)ForsamperofM=1an

f:..=2,dewa

pozof HGp(z) lie outside the unit circle for either process. Discuss the significance ofthese results.

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it has been determined that the closed-loop system is stable when '1"/

-10

.following instrumentation changes? Justify your answers using qualitative arguments.

11.2. The block

diagram of a feedback control system is shown in the drawing. Determine

11.Anop

unprht

trfvp

m('+1-

1Can this process be made closed-loop stable by using a proportional feedback con-

troller, Gc(s) = Kc? Can closed-loop stability be achieved using an ideal proportional-

derivative controller, Gc(s) = Kc(1 + TDS)? Justify your answers.

11.4. For the liquid-level control system in Figure 10.22, determine the numencal valuesofKcan'1"thre

inastcls

lne

dyw

thc

vhaticnumerical values are available:

A=3ft2

ih = 10 gal/min

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Exercises 269.and the following transfer functions:

Se -2s

Gp(s)G,(s) = 0.8

4e-s

!GL(s) = 15s + 1

o e

oc

wa

ms

eo

smore

c osysu

yoana e

prot me

srer

to.e

met

cos

rer

to.

11.Itisdestocontheexite

Tzofthhe

sitdrabyadj

thestefloratwsU

dioiitemperature Tt. The dynamic behavior of the heat exchanger can be approximated

by the transfer functions:If!

n(s) 2.Se -s of

T2(s) O.ge-ZS

[ ]. .

---;--( )

=5 1

= dimensIOnless

whethetim

conanti

dehauos

Tchas the following steady-state characteristics:

whepistheconoutexinm

At

noconp=12m

Afasuchinthc

o

rane

steavalin20s

(astotafitic

tature transmitter has negligible dynamics and is designed so that its output signalvarilinefro4to20m

asTzvafr1

to1oI p

Liquidout

Liquidin

[2] = Steam trap

y

(a) If a proportional feedback controller is used, what is Kern?What is the frequencyoftheres

oscw

Ke=K(

Ut

dsmetandEulide(b) Estimate Kern using the Routh criterion and a 1/1 P ade approximation for the

time-delay term. Does this analysis provide a satisfactory approximation?

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lIIil:"'I'

270 STABILITY OF CLOSED-LOOP

I

II

".

.p (2s + 1)(lOs + 1) L 5s + 1

1

Does the presence of a "right half plane zero" (i.e., a > 0) in the procefunction affect the stability of the closed-loop system? (Hint: Consider th

uations where a = 0 and a > 0.) .

seprp

e

troasshinthd!ilIII1illI,II1/i~ .I!1111",I',I:Ililll!I, I -P 4.5

2

(10s + 1)2

c

(a)Fow

ras)o

T

wt

rs(b) What practical arguments might be used to restrict the range(s) of acT

evfu(c)Ifaprcanswers to (a) and (b) be affected?

11.A

opu

pisd

tt, B(s) e-2sCaapr

fec

ssof

Kcre

inastcls11.Aprinc

vat

witi

coo

1,2,2,a5

sra

ofcog

thri

ascisus

wiT=2

m11.Abldi

ofafecsofco

gaKa

Kist

csL

1

s + 1II

s-2

(8 + 10)(s + 5)

1

s + 1

II