Essentials of Process Control phần 9 ppt

Therefore, the real sampler can be closely approximated by an impulse sampler: An impulse sampler is a device that converts a continuous input signal to a sequence of impulses or delta f

crrwrr% 14: Sampling, z Transforms, and Stability 4’79

i.e., the output of the hold is maintained at a constant value over the sampling p&o&The hold converts the discrete signal, which is a series of pulses, into a continuoussignal that is a stairstep function The equivalent block diagram of this system isshown at the bottom of Fig 14.2 The transfer function of the hold is H(,) The transferfunction of the computer controller is D&,

Figure 14.3 shows a digital control computer The process output variables yt,

y2, * *t yIy are sensed and converted to continuous analog signals by transmitters

TI,T'L,.-., TN These data signals enter the digital computer through a multiplexed

analog-to-digital (A/D) converter The feedback control calculations are done in the

t Digital computer

t D/A 1

FIGURE 14.3

Computer control.

480 PART WI:: Sampled-Data Systems

computer using some algorithm, and the calculated controller output signals are sent I

to holds associated with each control valve through a multiplexed digital-to-analog

(D/A) converter A block diagram of one loop is shown in the bottom of Fig 14.3

The sampling rate of these digital control computers can vary from several times

a second to only several times an hour.‘The dynamics of the process dictate the

sam-pling time required The faster the process, the smaller the samsam-pling period Ts must

be for good control One of the important questions that we explore in this chapter

and the following one is what the sampling rate should be for a given process For

a given number of loops, the smaller the value of T, specified, the faster the

com-puter and the input/output equipment must be This increases the cost of the digital

hardware

14.2

IMPULSE SAMPLER

A real sampler (Fig 14.1) is closed for a finite period of time This time of closure

is usually small compared with the sampling period T, Therefore, the real sampler

can be closely approximated by an impulse sampler: An impulse sampler is a device

that converts a continuous input signal to a sequence of impulses or delta functions

Remember, these are impulses, not pulses The height of each of these impulses is

infinite The width of each is zero The area or “strength” of the impulse is equal to

the magnitude of the input function at the sampling instant

,

t

If the units of fit, are, for example, kilograms, the units of &, are kilograms/minute

The impulse sampler is, of course, a mathematical fiction; it is not physically

realizable But the behavior of a real sampler-and-hold circuit is practically

identi-cal to that of the idealized impulse sampler-and-hold circuit The impulse sampler

is used in the analysis of sampled-data systems and in the design of sampled-data

controllers because it greatly simplifies the calculations

Let us now define an infinite sequence of unit impulses 6tt) or Dirac delta

func-tions whose strengths are all equal to unity One unit impulse occurs at every

sam-pling time We call this series of unit impulses, shown in Fig 14.4, the function I(+

41, = $f) + &T,) + S(t-27-,) + &3T,) + I(f) = i: $-nT,)

(14.3)

II =oThus, the sequence of impulses 4;) that comes out of an impulse sampler can be

expressed as

n TT FIGURE 14.4Impulse sampler.

Laplace-transforming Eq (14.4) gives

(14.5)

482 PART FIVE : Sampled-Data Systems

Equation (14.4) expresses the sequence of impulses that exits from an impulse pler in the time domain Equation (14.5) gives the sequence in the Laplace domain.Substituting io for s gives the impulse sequence in the frequency domain

n=OThe sequence of impulses &, can be represented in an alternative manner The

It,) function is a periodic function (see Fig 14.4) with period T.y and a frequency o,

in radians per minute

So Laplace-transforming Eq (14.11) gives

(14.12)

CIIAI~IK IJ: Sampling, z Transforms, and Stahiliti 483

Substituting io for s gives

F;iw) = f ‘y ~~i(“+n,.,), (14.13)

s n=-mEquation (14.4) is completely equivalent to’ Eq (14.11) in the time domain

Equation (14.5) is equivalent to Eq (14.12) in the Laplace domain Equation (14.6)

is equivalent to Eq (14.13) in the frequency domain We use these alternative forms

of representation in several ways later

14.3

BASIC SAMPLING THEOREM

A very important theorem of sampled-data systems is:

To obtain dynamic information about a plant from a signal that contains ponents out to a frequency omax, the sampling frequency o, must be set at a rate greater than twice urnax.

high-It is therefore always recommended that signals be analog filtered before they are

sampled This eliminates the unimportant high-frequency components Trying to ter the data after it has been sampled using a digital filter does not work

fil-To prove the sampling theorem, let us consider a continuous fit, that is a sine

wave with a frequency 00 and an amplitude Ao.

+ sin[(oo - os)t] + sin[(oo - 20,)t] + * * *)

(14.17)

Thus, the sampled function 4;) contains a primary component at frequency wo plus

an infinite number of complementary components at frequencies 00 + os, oo+2%, * * ,mo - o,, 00 - 20,, The amplitude of each component is the amplitude

of the original sine wave fttJ attenuated by l/T, The sampling process produces asignal with components at frequencies that are multiples of the sampling frequencyplus the original frequency of the continuous signal before sampling Figure 14.52illustrates this in terms of the frequency spectrum of the signal This is referred to

by electrical engineers as “aliasing.”

Now suppose we have a continuous function fit) that contains components over

a range of frequencies Figure 14% shows its frequency spectrum f&j If this signal

is sent through an impulse sampler, the output f(T) has a frequency spectrum &, asshown in Fig 14% If the sampling rate or sampling frequency os is high, there is nooverlap between the primary and complementary components Therefore, &, can befiltered to remove all the high-frequency complementary components, leaving justthe primary component This can then be related to the original continuous function.Therefore, if the sampling frequency is greater than twice the highest frequency inthe original signal, the original signal can be determined from the sampled signal

If, however, the sampling frequency is less than twice the highest frequency

in the original signal, the primary and complementary components overlap Thenthe sampled signal cannot be filtered to recover the original signal, and the sampledsignal predicts incorrectly the steady-state gain and the dynamic components of theoriginal signal

Figure 143 shows that J:, is a periodic function of frequency o Its period

is ws

This equation can also be written

Going into the Laplace domain by substituting s for io gives

(14.18)

(14.19)

( 14.20)Thus Ft., is a periodic function of s with period io, We use this periodicity property

to develop pulse transfer functions in Section 14.5

If*1 h

z TRANSFORMATION

14.4.1 Definition

Sequences of impulses, such as the output of an impulse sampler, can be z

trans-formed For a specified sampling period T,, the z transformation of an

impulse-sampled signal & is defined by the equation

‘[-fi;,] = ho, + hz-’ + &r,,Z-* + fi~~,z-~ + * * * + j&)z-‘* + - (14.2 1)

The notation ‘%[I means the z transformation operation The j&T,rj values are the

magnitudes of the continuous function ftl, (before impulse sampling) at the sampling

periods We use the notation that the z transform of 4;; is F(,)

The z variable can be considered an “ordering” variable whose exponent represents

the position of the impulse in the infinite sequence j&

Comparing Eqs (14.5) and (14.22), we can see that the s and z variables are

related by

We make frequent use of this very important relationship between these two complex

variables

Keep in mind the concept that we always take z tran’sforms of impulse-sampled

signals, not continuous functions We also use the notation

(14.24)This means exactly the same thing as Eq (14.22) We can go directly from the time

domain 4;) to the z domain Or we can go from the time domain hT, to the Laplace

domain FTsl and on to the z domain Fez).

14.4.2 Derivation of z Transforms of Common Functions

Just as we did in learning Russian (Laplace transforms), we need to develop a small

German vocabulary of z transforms

(:IIAPW~ IJ: Sampling, z Transforms, and Stability 487

Passing the step function through an impulse sampler gives J;;, = K~~(~jlt,), whereltI) is the sequence of unit impulses defined in Eq (14.3) Using the definition of ztransformation [Eq (14.22)J gives

~[.I$)1 = 2 J&7$-n = f(O) + hT,)P + A2w-2 + *

s must be large enough to keep the exponential less than 1

The z transform of the impulse-sampled step function is

You should now be able to guess the z transformation oft* We know there will

be an s3 term in the denominator of the Laplace transformation of this function So wecan extrapolate our results to predict that there will be a (z - 1)3 in the denominator

of the z transformation

We find later in this chapter that a (2 1) in the denominator of a transfer function

in the z domain means that there is an integrator in the system, just as the presence

of an s in the denominator in the Laolace domain tells us there is an integrator

488 PART PIVI:: Satnplcd-Data Syskms

(14.27)

Remember that the Laplace transformation of the exponential was Kl(s + a)

So the (s + a) term in the denominator of a Laplace transformation is similar to the(z - ewaTs) term in a z transformation Both indicate an exponential function In the

s plane we have a pole at s = -a In the z plane we find later in this chapter that wehave a pole at z = e?Tv So we can immediately conclude that poles on the negativereal axis in the s plane “map” (to use the complex-variable term) onto the positivereal axis between 0 and + 1

ID Exponential multiplied by time

In the Laplace domain we found that repeated roots l/(s + ~2)~ occur when wehave the exponential multiplied by time We can guess that similar repeated rootsshould occur in the z domain Let us consider a very genera1 function:

This function can be expressed in the alternative form

: w e

we

oots

1 1

z- ’ (2i) sin(o Ts)

1

z sin(o r,)YE-

2i 1 + zw2 - 22-l cos(oT.y) = z2 + 1 - 2zcos(oT,)

F Unit impulse function

By definition, the z transformation of an impulse-sampled function is

(14.32)

4) = ho, + &)z-’ + fi27,)z-2 +

If f& is a unit impulse, putting it through an impulse sampler should give an &T, that

is still just a unit impulse a([) But Eq (14.4) says that

But if& must be equal to just au), the term ho) in the equation above must be equal

to 1 and all the other terms f(T,), &T,), must be equal to zero Therefore, the ztransformation of the unit impulse is unity

where k is an integer Consider an arbitrary function f&o) The original function

j&) before the time delay is assumed to be zero for time less than zero Running thedelayed function through an impulse sampler and z transforming give

~[J;:-D)] = >: &T,-kT.,)Z-n

II =o

Now we let x = y1- k.

.u=-k

since ftxr) = 0 for .X < 0 The term in the brackets is just the I mnsform of (I,

since x is j dummy variable of summation

%[e-af A;)] = -g e-anT.y fi,&)z-n = 2 finTs)(ZeuT’)-”

Now substitute zl = zeaT’ into the equation above

3e-“‘f;l‘,l = 2 fin7dzr = F(z,)

n = O Q

E XAMPLE 14.3 Suppose we want to take the z transformation of the function ft, =

Kte-“‘Z(,, Using Eqs (14.26) and (14.37) gives

cE(Kte-“‘z(,)] = %,[CKff(,)] = ,zyf)2Substituting ~1 = zeaT,- gives

KT,zeaTs KTsD?-aT’

~‘[Kfe-“‘z(~)l = (ze”Ts - 1)2 = (z _ ,-UT,)2 (14.38)

This is exactly what we found in Example 14.2

C Final-value theorem

lim h,, = iirr( t x

n

(14.39)

The steady-state value of fit, or the limit of At, as time goes to infinity is K Running

j&J through an impulse sampler and z transforming give

Multiplying both sides by (z - 1)/z and letting z + 1 give

The definition of the z transform of f(;, is

F(z) = fro, + j&z-’ + j&52 +

Letting z go to infinity (for Iz-‘j < 1) in this equation gives fro,, which is the limit

also unique; i.e., there is only one fCnT,) that corresponds to a given FtI,.

However, keep in mind that more than one continuous function At, gives thesame impulse-sampled function & The sampled function f(‘;, contains informa-tion about the original continuous function ftt, only at the sampling times Thisnonuniqueness between fCT, (and Fc,)) and fct, is illustrated in Fig 14.6 Both contin-

uous functions f,(,) and j$r) pass through the same points at the sampling times butare different in between the sampling instants They would have exactly the same ztransformation

ml ^ 1 , .I e- : 1 ,.-e 7 t ,-,, n‘>f,\,-mc

492 PAW WI:.: Sampled-Data Systems

The classical mathematical method for inverting a z transform is to use the

lin-earity theorem [Eq (14.36)] We expand the function FCzj into a sum of simple terms

and invert each individually This is completely analogous to Laplace

transforma-tion inversion Let F(,, be a ratio of polynomials in z, Mth-order in the

numera-tor and Nth-order in the denominanumera-tor We facnumera-tor the denominanumera-tor into its N roots:

(z - Pl>(z - p2k - p3> k - pN>

(14.42)

where Zcz) = Mth-order numerator polynomial Each root pi can be expressed in

terms of the sampling period:

z - e-a~T.s z - e-azT.s z - e-ajT, + ‘ + z - ~-NNT.> (14.45)

The coefficients A, B, C, , W are found and F(,, is inverted term by term to give

3-1 [F,,,] = hnT,) = Ae-“16 + &-“NT., + + we-wT.~ ( 14.46)

EXAMPLE 14.4 We show in Example 14.8 that the closedloop response to a unit step

change in setpoint with a sampled-data proportional controller and a first-order process

is

5:) = (i: - l)[z - b + K, K, K,,( I - 1~): K,,(I - 6)1 (14.47)

n u

I1

it

n

KC = feedback controller gain

K, = process steady-state gain7,) = process time constantFor the numerical values of K, = T(, = 1, K, = 4.5, and T, = 0.2, Y(,, becomes

Y(z) = (z - l)(z - 0.003019)0.81592Expanding in partial fractions gives

Y(z) = 0.81592z - l -z - 0.0030190.8159~

The pole at 0.003019 can be expressed as

0 003019 = y5.803 = p’T.sThe value of the term nT, is 5.803

y(,$,) = y(~.~,~) = 0.8159 - 0.8159e-“uT.Y = 0.8159(1 - e-5.803n) (14.51)Table 14.1 gives the calculated results of y(jzr,) as a function of time n

B Long division

An interesting z transform inversion technique is simple long division of the merator by the denominator of F(, The ease with which z transforms can be invertedwith this technique is one of the reasons z transforms are often used

nu-By definition,

F(z) = fro, + hr,,z-’ + h27;)z-2 + jj3T,,z-3 +

If we can get F(:) in terms of an infinite series of powers of z-t, the coefficients

in front of all the terms give the values of fCllr,,, The infinite series is obtained bymerely dividing the numerator of F,,, by the denominator of Fczj

494 p,4Kr I:IVE: Sampled-Dalia Systems

where Ztzj and f’(;) are polynomials in z The method is easily understood by looking

fro, = 0f(~,y) = fio.2) = 0.8 1%

Ycz) = 2.17562-l - 0.8098~-~ + 3.254~-~ - 2.310~-~ + (14.53)The system is unstable for this value of gain (KC = 12), as we show later in this chapter.Notice that this example demonstrates that a first-order process controlled by a sampled-data proportional controller can be made closedloop unstable if the gain is high enough

With the use of an analog controller, the first-drder process can never be closedloop

unstable Thus, there is a very important difference between continuous and discreteclosedloop systems Analog continuous controllers have an inherent advantage over dis-crete sampled-data controllers because they know what the output is doing at all points

in time The discrete controller knows only what the output is at the sampling times wInversion of z transforms by long division is very easily accomplished numeri-cally by a digital computer The FORTRAN subroutine LONGD given in Table 14.2performs this long division The output variable Y is calculated for NT sampling

(IWI i,,~( IJ: Sampling, I Transform, and Stability 495

c does long-division using subroutine longd dimension a( IO), h( IOj,y( 100)

c Case for kc=12 aO=O.

a(Ij=2.1756 a(2)=0.

b(lj=0.3722 b(2)= - 1.357 n=2

m=I nt=6 yo=o.

call longd(aO,a,b,yO.y,n,m,nt)

do 10 k=l,nt write(6,l)k,y(k)

if{m.gt.n) nmax=m

do 10 k=l,nmax d(kj=a(kj if(k.gt.mj d(k)=O.

10 continue d(nmax+ Ij=O.

iflaO.eq.O )go to 30 yO=aO

do 20 k=l,nmav

20 d(k)=d(k)-yO*b(k) y(I)=d(l)

496 PART WE: Sampled-Data Sysretns

% F o r r n numerutor atrd denominator polytwmials

num=[2 I756 O];

times, given the coefficients AO, Ăl), Ă2), , ĂM) of the numerator and the

co-efficients B( 1), B(2), , B(N) of the denominator

Y(z) = YO + Y(l)z -’ + Y(2)z-2 + Y(3)zC3 + *

A0 + Ă l)z-’ + Ẵ)z-~ + + ĂM)z-~ (14.54)

= 1 + B(l)z-’ + B(2).@ + * * * + B(N)z-N

C Use of MATLAB for inversion

Now that we have discussed the classical inversion methods, we are ready tosee how inversion of z transforms can be easily accomplished using MATLAB soft-warẹ Specific numerical values of parameters must be specified Table 14.3 gives

a program that solves for the values of the output at the sampling periods for theYcz) considered in Example 14.6 First the numerator and denominator polynomialsare formed The number of sampling periods (ntotul) is specified, and the samplingperiod is set Then the [y,x]=nimpulse(num,den,r~totnl) command is used to gener-ate the output sequence ytn~,) at each value of n The results are the same as thoseobtained by long division

14.5

PULSE TRANSFER FUNCTIONS

We know how to find the : transformations of functions Let us now turn to theproblem of expressing input/output transfer function relationships in the z domain.Figure 14.7 shows a system with samplers on the input and on the output of the

Pulse transfer functions.

process Time-, Laplace-, and z-domain representations are shown Gtz) is called a

pulse transferfunction It is defined below

A sequence of impulses z.$, comes out of the impulse sampler on the input ofthe process Each of these impulses produces a response from the.process Considerthe kth impulse L(TkT, )( Its area or strength is u(kr,Y) Its effect on the continuous output

of the plant y(,) is

where yk([) = response of the process to the kth impulse

g([) = unit impulse response of the process = 2-l [Gel]

Figure 14.7 shows these functions

The system is linear, so the total output y(,) is the sum of all the yk's

498 PART FIVE: Sarnpkd-thta Systems

The continuous function yt,) coming out of the process is then impulse sampled,producing a sequence of impulses yt, If we z-transform yf,), we get

my;,,1 = >7 Y(nT,$)z-‘* = Y(z)

of Eq (14.59)

p=oDefining G(,) in this way permits us to use transfer functions in the z domain[Eq (14.60)] just as we use transfer functions in the Laplace domain Gtz) is the ztransform of the impulse-sampled response g;;, of the process to a unit impulse func-tion 6~~) In z-transforming functions, we used the notation %[&I = %[F;“,,] = F~,J

In handling pulse transfer functions, we use similar notation

where G& is the Laplace transform of the impulse-sampled response g&, of the cess to a unit impulse input

pro-(14.63)G& can also be expressed, using Eq (14.12), as

A hold device is always needed in a sampled-data process control system The

zero-order hold converts the sequence of impulses of an impulse-sampled function ,f;l;, to

FIGURE 14.8

Zero-order hold.

a continuous stairstep function fHtI) The hold must convert an impulse f* of area.w

or strength f&,) at time t = rzT, to a square pulse (not an impulse) of height fcnr,$,and width T, See Fig 14.8 Let the unit impulse response of the hold be defined ashi) If the hold does what we want it to do (i.e., convert an impulse to a step up andthen a step down after Ts minutes), its unit impulse response must be

We are now ready to use the concepts of impulse-sampled functions, pulse transferfunctions, and holds to study the dynamics of sampled-data systems Consider thesampled-data system shown in Fig 14.9a in the Laplace domain The input entersthrough an impulse sampler The continuous output of the process Yc,) is

Y(.\) = G(s) u;.)

Yes) is then impulse sampled to give Y;‘r, Equation (14.13) says that YFs, is

ys, = ; ‘y Y(.s+inw,) \ ,,=-”

(14.67)

5 0 0 PART FIVE : Sampled-Data Sj slcins

(b) Series elements with intermediate sampler

(c) Series elements that are continuous

Without sampler

(d) System of Example 14.7

FIGURE 14.9

Openloop sampled-data systems

Substituting for Yts+inw,y), using Eq (14.67), gives

We showed [Eq (14.20)] that the Laplace transform of an impulse-sampled function

is periodic

4.68)

ction

cttwtm IA: Sampling, z Transforms, and Stability 501

‘l:s) = u&+iw,) = I/*-. <s WJ,~) = u~v+iZw,) = “* ( 14.69)Therefore, the U~~+i,lw,~~ terms can be factored out of the summation in Eq (14.68)

The term in the parentheses is Gt, according to Eq (14.64), and therefore the output

of the process in the Laplace domain is

In the z domain, this equation becomes

Y2(z) = G u:+%(z) u(z) (14.74)Thus, the overall transfer function of the process can be expressed as a product ofthe two individual pulse transfer functions if there is an impulse sampler betweenthe elements

Consider now the system shown in Fig 14.9c, where the two continuous ments GltSJ and G2CsI do not have a sampler between them The continuous outputY2(.v) is

ele-Y?(s) = G2c.r) Cc,, = Gxs$h(.s,U;, (14.75)