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CHAPTER

7

7.1 SECOND-ORDER SYSTEM

Transfer Function

This section introduces a basic system called a second-order system or a quadratic lag.

Second-order systems are described by a second-order differential equation that relates

the ouput variabley in Chap. 6. Some systems are inherently second-order, and they do not result from a

series combination of two first-order systems. Inherently second-order systems are notextremely common in chemical engineering applications. Most second-order systems

that we encounter will result from the addition of a controller to a first-order process.

Lets examine an inherently second-order system and develop some terminology that

will be useful in our analysis of the control of chemical processes.

Consider a simple manometer as shown in Fig. 71. The pressure on both legs of

the manometer is initially the same. The length of the fluid column in the manometer

isL. At time t 0, a pressure difference is imposed across the legs of the manometer.

Assuming the resulting flow in the manometer to be laminar and the steady-state fric-

tion law for drag force in laminar flow to apply at each instant, we will determine the

transfer function between the applied pressure difference P and the manometer read-

ing h. If we perform a momentum balance on the fluid in the manometer, we arrive at

the following terms:

( ) (Sum of for

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138 PART 2 LINEAR OPEN-LOOP SYSTEMS

where

Sum of forces

causing fluid to move

Un

bbalanced pressure forces

causing motion

Frictional forces

opposing motion

Unbalanced pressure forcescausing motion

( ) P P D gh D1 22 2

4 4p r p

Frictional forces

opposing motion

Skin

friction

at wall

Shear stress

at wall

Area in contact

with wall

Frictional forces

opposing motionWal

t ll pm

pm

pDLV

DDL

D

dh

dtDL( ) ( )

8 8 1

2(

After (Final)Before (Initial)

L

h h/2

h/2

DReference level

t= 0

P1= 0 P1 P2P2= 0

FIGURE 71

Manometer.

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CHAPTER 7 HIGHER-ORDER SYSTEMS: SECOND-ORDER AND TRANSPORTATION LAG 139

The rate of change of momentum of the fluid [the right side of Eq. (7.2)] may be

expressed as

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2

3

162

2 2

1 2L

g

d h

dt

L

D g

dh

dth

P P

g

P

g

m

r r r

(7.4)

(A more detailed version of the analysis of the manometer can be found in Bird et al.,

1960). Note that as with first-order systems, standard form has a coefficient of 1 on the

dependent variable term, h in this case. Second-order systems are described by a second-

order differential equation. We may rewrite this Eq. (7.4) in general terms as

t zt22

22

d Y

dt

dY

dtY X t ( )

(7.5)

where

t22

3

L

g(7.6)

216

2zt

m

r

L

D g(7.7)

X tP

gY h( )

r

and

(7.8)

Solving for tand zfrom Eqs. (7.6) and (7.7) gives

t 2

3

L

gs

(7.9)

zm

r

8 3

22D

L

gdimensionless

(7.10)

By definition, both tand zmust be positive. The reason for introducing tand zin theparticular form shown in Eq. (7.5) will become clear when we discuss the solution of

Eq. (7.5) for particular forcing functionsX(t).

Equation (7.5) is written in a standard form that is widely used in control theory.

If the fluid column is motionless (dY/dt 0) and located at its rest position (Y 0)

before the forcing function is applied, the Laplace transform of Eq. (7.4) becomes

t zt2 2 2s Y s sY s Y s X s( ) ( ) ( ) ( )

(7.11)

From this, the transfer function follows:

Y ( ) 1

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denominator oft2s2 2zts 1. All such systems are defined as second-order. Notethat it requires two parameters, tand z, to characterize the dynamics of a second-order

system in contrast to only one parameter for a first-order system. We now discuss theresponse of a second-order system to some of the common forcing functions, namely,

step, impulse, and sinusoidal.

Step Response

If the forcing function is a unit-step function, we have

X s s( )

1

(7.13)

In terms of the manometer shown in Fig. 71, this is equivalent to suddenly applying a

pressure difference [such that X(t) P/rg 1] across the legs of the manometer attime t 0.

Superposition will enable us to determine easily the response to a step function of

any other magnitude.

Combining Eq. (7.13) with the transfer function of Eq. (7.12) gives

Y ss s s

( )

1 12 12 2t zt

(7.14)

The quadratic term in this equation may be factored into two linear terms that contain

the roots

sa z

t

z

t

2 1

(7.15)

sb z

t

z

t

2 1

(7.16)

Equation (7.14) can now be written

Y ss s s s sa b

( )

1 2/t

( )( ) (7.17)

The response of the system Y(t) can be found by inverting Eq. (7.17). The roots saand

sb will be real or complex depending on value of the parameter z. The nature of the

roots will, in turn, affect the form ofY(t). The problem may be divided into the threecases shown in Table 7.1. Each case will now be discussed.

TABLE 71

Step response of a second-order system

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CASE I STEP RESPONSE FOR y< 1. For this case, the inversion of Eq. (7.17) yields

the result

Y t ett( ) tan

11

1

1

2

2 12

zz

t

z

zz t/ sin 1

(7.18)

To derive Eq. (7.18), use is made of the techniques of Chap. 3. Since z< 1, Eqs. (7.15)to (7.17) indicate a pair of complex conjugate roots in the left half-plane and a root at

the origin. In terms of the symbols of Fig. 31, the complex roots correspond to s2 and

s2* and the root at the origin to s6.The reader should realize that in Eq. (7.18), the argument of the sine function is in

radians, as is the value of the inverse tangent term.

By referring to Table 3.1, we see that Y(t) has the form

Y t C e C t

Ctt( ) 1 2

23

21 1z t zt

zt

/ cos sin

(7.19)

The constants C1, C2, and C3 are found by partial fractions. The resulting equation is

then put in the form of Eq. (7.18) by applying the trigonometric identity used in Chap. 4,

Eq. (4.26). The details are left as an exercise for the reader. It is evident from Eq. (7.18)

that Y(t) 1 as t.

The nature of the response can be understood most clearly by plotting the solution

to Eq. (7.17) as shown in Fig. 73, where Y(t) is plotted against the dimensionless vari-

able t/t for several values ofz, including those above unity, which will be consideredin the next section. Note that for z< 1 all the response curves are oscillatory in nature

and become less oscillatory as z is increased. The slope at the origin in Fig. 73 iszero for all values ofz. The response of a second-order system for z< 1 is said to beunderdamped.

What is the physical significance of an underdamped response? Using the manom-

eter as an example, if we step-change the pressure difference across an underdamped

manometer, the liquid levels in the two legs will oscillate before stabilizing. The oscil-

lations are characteristic of an underdamped response.

CASE II STEP RESPONSE FOR y 1. For this case, the response is given by theexpression

Y tt

e t( ) / 1 1t

t

(7.20)

This is derived as follows: Equations (7 15) and (7 16) show that the roots s and s are

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Using MATLAB/Simulink to Determine the Step Response of the Manometer

Consider a manometer as illustrated in Fig. 71. The manometer is being used to determine the

pressure difference between two instrument taps on an air line. The working fluid in the manometer

is water. Determine the response of the manometer to a step change in pressure across the legs of

the manometer.

Data

L 200 cm

g 980 cm/s2

Pg

ttr

0 010 0

forcm for

D 0.11 cm, 0.21 cm, 0.31 cm (Three Cases)

Solution. From Eq. (7.4), we have the governing differential equation for the manometer:

m

r

1 0 01

1 0 3

cP g/ cm s

g/cmfor th

. ( )

.

ee working fluid water,

m

r

1 0 01

1 0 3

cP g/ cm s

g/cmfor th

. ( )

.

ee working fluid water,

Actually, the response for z> 1 is not new. We saw it previously in the discussionof the step response of a system containing two first-order systems in series, for which

the transfer function is

Y s

X s s s

( )

( )

1

1 11 2t t( )( ) (7.22)

This is true for z> 1 because the roots s1 and s2 are real, and the denominator of Eq.(7.12) may be factored into two real linear factors. Therefore, Eq. (7.12) is equivalent to

Eq. (7.22) in this case. By comparing the linear factors of the denominator of Eq. (7.12)

with those of Eq. (7.22), it follows that

t z z t 12 1 ( ) (7.23)

t z z t 22 1 ( ) (7.24)

Note that ift1t2, then tt1t2 and z 1. The reader should verify these results.

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The Simulink model for simulating the transfer functions is shown in Fig. 74, and the response is

shown in Fig. 75.

FIGURE 74

Simulink diagram for manometer simulation.

0.136s2 + 2.70s + 1

0.136s2 + 0.738s + 1

1

Critically damped manometer

Overdamped manometer

Pressure forcingfunction

10/s

Scope

1

0.136s2 + 0.340s + 1

1

Underdamped manometer

0

2

4

6Height

Underdamped

Overdamped

Criticallydamped

8

10

12

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The presence ofs on the right side of Eq. (7.33) implies differentiation with respect to t

in the time response. In other words, the inverse transform of Eq. (7.31) is

Y td

dtY t( ) ( ) impulse step ( ) (7.34)

Application of Eq. (7.34) to Eq. (7.18) yields Eq. (7.32). This principle also yields the

results for the next two cases.

CASE II IMPULSE RESPONSE FOR y 1. For the critically damped case, the response

is given by

Y t te t( ) / 12t

t

(7.35)

which is plotted in Fig. 78.

CASE III IMPULSE RESPONSE FOR y> 1. For the overdamped case, the response is

given by

Y t ett( ) /

1 1

11

2

2

t zz

tz t sinh

(7.36)

which is also plotted in Fig. 78.

To summarize, the impulseresponse curves of Fig. 78 show the same general

behavior as the step response curves of Fig. 73. However, the impulse response always

returns to zero. Terms such as decay ratio, period of oscillation, etc., may also be used

to describe the impulse response. Many control systems exhibit transient responsessuch as those of Fig. 78.

Sinusoidal Response

If the forcing function applied to the second-order system is sinusoidal

X t A t( ) sin w

then it follows from Eqs. (7.12) and (4.23) that

Y sA

s s s( )

w

w t zt 2 2 2 2 2 1( )( ) (7.37)

The inversion of Eq. (7.37) may be accomplished by first factoring the two quadratic

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units of time. The transportation lag parameter tis simply the time needed for a particleof fluid to flow from the entrance of the tube to the exit, and it can be calculated from

the expression

t volume of tube

volumetric flow rate

or t AL

q

(7.43)

It can be seen from Fig. 710 that the relationship betweeny(t) andx(t) is

y t x t( ) t( )

(7.44)

Subtracting Eq. (7.42) from Eq. (7.44) and introducing the deviation variablesXxxsand Yyysgive

Y t X t ( ) t( )

(7.45)

If the Laplace transform ofX(t) isX(s), then the Laplace transform ofX(tt) is e

stX(s).

This result follows from the theorem on translation of a function, which was discussed

in App. 3A. Equation (7.45) becomes

Y s e X ss( ) ( ) t

orY s

X se s

( )

( )

t

(7.46)

Therefore, the transfer function of a transportation lag is est.

The transportation lag is quite common in the chemical process industries where

a fluid is transported through a pipe. We shall see in a later chapter that the presence of

a transportation lag in a control system can make it much more difficult to control. In

general, such lags should be avoided if possible by placing equipment close together.

They can seldom be entirely eliminated.

APPROXIMATION OF TRANSPORT LAG. The transport lag is quite different from the

other transfer functions (first-order, second-order, etc.) that we have discussed in that it

is not a rational function (i.e., a ratio of polynomials.) As shown in Chap. 13, a system

containing a transport lag cannot be analyzed for stability by the Routh test. The trans-

port lag can also be difficult to simulate by computer. For these reasons, several approx-

imations of transport lag that are useful in control calculations are presented here

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Keeping only the first two terms in the denominator gives

e s

s

t

t1

1 (7.47)

This approximation, which is simply a first-order lag, is a crude approximation of a

transport lag. An improvement can be made by expressing the transport lag as

ee

e

ss

s

t

t

t

/

/

2

2

Expanding numerator and denominator in a Taylor series and keeping only terms of

first-order give

es

s

s

t t

t

1 2

1 2

/

/first-order Pad

(7.48)

This expression is also known as afirst-order Padapproximation.

Another well-known approximation for a transport lag is the second-order Pad

approximation:

es s

s s

s

t t t

t t

1 2 12

1 2 12

2 2

2 2

/ /

/ / second-ordeer Pad(7.49)

Equation (7.48) is not merely the ratio of two Taylor series; it has been optimized to

give a better approximation.

The step responses

of the three approximations

of transport lag presented

here are shown in Fig. 711.

The step response of e

ts isalso shown for comparison.

Notice that the response for

the first-order Pad approxi-

mation drops to1 before ris-

ing exponentially toward 1.

The response for the second-

order Pad approximation

jumps to

1 and then descendsto below 0 before returning

gradually back to1.

Although none of the

approximations for ets is

very accurate, the approxima-

i f t s i f l h i i l i li d b l fi d d d

FIGURE 711

Step response to approximation of the transport lag ets:

(1)1

1ts ; (2) first-order Pad; (3) second-order Pad ; (4) ets.

01.0

0.6

0.2

0.2

0Y

0.6

1.0

(2)

(3)

(1)

(4)

1 2t/

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