dynamic behavior of first_second order systems

7/31/2019 Dynamic Behavior of First_Second Order Systems

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Chapter 4: Dynamic Behavior of First-order & Second-Order Systems 1

Dynamic Behavior of First-order and Second-order Systems

1. Analysis of first-order systemsConsider the system shown in Figure 1 which consists of a tank of uniform cross

sectional area A to which is attached a flow resistance R. Assume that qo is related to the

head by linear relationship

R

hqo = (4. 1)

qi

h

q

V

A

Figure 1 Liquid level system

The change in liquid height inside the tank can be described by the following equation;

R

hq

dt

dhA = (4. 2)

At steady state, this equation can be written as:

R

hq s

s =0 (4. 3)

Subtracting Equation (4.3) from Equation (4.2) gives

R

hhqq

dt

hhdA s

ss )()()(

=

(4. 4)

Defining the deviation variables as

s

s

hhH

qqQ

=

=(4. 5)

Equation (4.4) can be written as

QR

H

dt

dHA =+ (4. 6)


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Taking the L.T. of this equation gives

)()(

)( sAsHR

sHsQ += (4. 7)

Equation (4.7) can be rearranged into the standard form of the first-order system to give

ARwhere

sR

sQsH

=

+=

1)()(

(4. 8)

In a general form, Equation (4.8) can be written in a general form:

1)(

)(

)(

+==

s

KsG

sU

sY

(4. 9)

Where

)]([)(

)]([)(

tusU

tysY

l

l

=

=(4. 10)

G is called transfer function. It is a convenient way to represent linear dynamic models. It

relates one input and one output:

Systemu(t)

U(s)

y(t)

Y(s)

Figure 2 General system representation

The term (time constant) and K (steady state gain) characterize the first-order system.

Note that the both parameters depend on operating conditions of the process and that the

transfer function does not contain the initial conditions explicitly.

Properties of transfer functions

Time constantof a process is a measure of the time necessary for the process to adjust to

a change in the input.

Steady state gain is the steady state change in output divided by the sustained change in

the input. It characterizes the sensitivity of the output to the change in input. The steady-

state of a transfer function can be used to calculate the steady-state change in an output

due to a steady-state change in the input. For example, suppose we know two steady

states for an input, u, and an output, y. Then we can calculate the steady-state gain, K,

from:


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ssss

ssss

uu

yyK

,1,2

,1,2

= (4. 11)

Another important property of the transfer function is that the orderof the denominator

polynomial (in s) is the same as the order of the equivalent differential equation. A

general nth-order differential equation has the form

ubdt

dub

dt

udb

dt

udbya

dt

dya

dt

yda

dt

yda

om

m

mm

m

mon

n

nn

n

n ++++=++++

11

1

111

1

1 LL (4. 12)

Where u and y are input and output deviation variables respectively. The transfer function

obtained by L.T. is

o

n

n

n

n

o

m

m

m

m

n

i

i

i

m

i

i

i

asasa

bsbsb

sa

sb

sU

sYsG

++++++

===

=

=

L

L

1

1

1

1

0

0

)(

)()( (4. 13)

The steady state gain of G(s) in Equation (4.13) is bo/ao, obtained by setting s = 0 in G(s).

Definition: The order of the transfer function is defined to be the order of the

denominator polynomial

Multiplicative rule

UGGY = 21 (4. 14)

Additive rule

G2

G1

+

+

U1

U2

2211 UGUGY += (4. 15)

G1U G2

Y


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Dynamic response

The dynamic response for a step change in u(t) of magnitude M: U(s) = M/s

)1()( t

eKMty

= (4. 16)

The transient response for a step change is shown in Figure 3.

0 t

Mu(t)

y(t)KM

Figure 3 Response of a first-order system to a step change in the input.

Dynamic Characteristics

A first-order process is self-regulating. The process reaches a new steady state.

The ultimate value of the output is K for a unit step change in the input, or KMfor a step of size M. This can be seen from Equation (4.16), which

yields tasKMy . This characteristic explains the name steady state or

static gain given for the parameter K, since for any step change in the input the

resulting change in the output steady state is given by

)()( inputKoutput = (4. 17)

This Equation tells us by how much we should change the value of the input in

order to achieve a desired change in the output, for a process with given K. Thus,to effect the same change in the output, we need:

A small change in the input if K is large (very sensitive systems)

A large change in the input if K is small. See Figure 5.

The value of y(t) reaches 63.2 % of its ultimate value when the time elapsed is

equal to one time constant . When the time elapsed is 2, 3, 4, the percent


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response is 86.5, 95 and 98 respectively. From these facts, one can consider the

response essentially completed in three to four time constants.

The slope of the response at t = 0 is equal to 1.

1)(

)(

]/)([0

0

== =

=

t

t

t

e

td

KMtyd

(4. 18)

This implies that if the initial rate of change of y(t) were to be maintained, the

response would reach its final value in one time constant (see the dashed line in

Figure 4). The corollary conclusions are:

The smaller the value of time constant, the steeper the initial response of the

system (Figure 5).

10

KM

y

t1

1.0

0.632

00.0

Figure 4 Response of a first-order system to a step change in the input.

y(t)

0

y(t)

0

K2

< K1

K1

K2

Time Time

t1

< t2

t2

t1

KM

K1M

K2M

Figure 5 Effect of static gain, time constant on the response of first-order lag system.


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2. Pure capacitive system (Integrator)

Consider the first order system described by Equation (4.19)

)(1 tbuyadt

dya o =+ (4. 19)

If ao = 0, then from Equation (4.19) we take

)()(1

tKutua

b

dt

dy== (4. 20)

Which gives a transfer function

s

K

sU

sYsG ==

)(

)()( (4. 21)

In this case, the process is calledpurely capacitive orpure integrator.

Dynamic response

For a step change in u of magnitude M: U(s) = M/s. Equation (4.21) yields

2)(

s

KMsG = (4. 22)

After inversion we find that

Ktty =)( (4. 23)

Notice that the output grows linearly with time in unbounded fashion (Figure 6). Thus,

tasy

Note that a pure capacitive process causes serious control problem because it can not

balance itself. For small change in the input, the output grows continuously. This attributeis known as non-self-regulating process.

y(t)

t

K

Figure 6 Unbounded response of pure capacitive process


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3. Analysis of second-order systems

A second-order system is one whose output, y(t), is described by a second-order

differential equation. For example, the following equation describes a second-order linear

system:

)(12

2

2 tbuyadt

dya

dt

yda o =++ (4. 24)

If ao 0, then Equation (4.24) yields

)(22

22 tKuy

dt

dy

dt

yd=++ (4. 25)

Equation (4.25) is in the standard form of a second-order system, where

= natural period of oscillation of the system

= damping factor

K = steady state gainThe very large majority of the second- or higher-order systems encountered in a

chemical plant come from multicapacity processes, i.e. processes that consist of two or

more first-order systems in series, or the effect of process control systems.

Laplace transformation of Equation (4.25) yields

12)(

22 ++=

ss

KsG

(4. 26)

Dynamic response

For a step change of magnitude M, U(s) = M/s, Equation (4.26) yields

)12()(

22 ++=

sss

KMsY

(4. 27)

The two poles of the second-order transfer function are given by the roots of the

characteristic polynomial,

01222 =++ ss (4. 28)

and they are

112

2

2

1

=

+= pandp (4. 29)

Therefore, Equation (4.27) becomes

))(()(

21

2

pspss

KsY

=

(4. 30)

The form of the response of y(t) will depend on the location of the two poles in the

complex plane. Thus, we can distinguish three cases:


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Case A: (over-damped response), when > 1, we have two distinct and real poles.

In this case the inversion of Equation (4.30) by partial fraction expansion yields

+=

tteKty

t1sinh

11cosh1)( 2

2

2(4. 31)

Where cosh(.) and sinh(.) are the hyperbolic trigonometric functions defined by

2cosh

2sinh

+

=

=ee

andee

(4. 32)

Case B: (critically damped response), when = 1, we have two equal poles (multiple

pole).

In this case, the inversion of Equation (4.30) gives the result

+=

tet

Kty 11)( (4. 33)

Case C: (Under-damped response), when < 1, we have two complex conjugate poles.

The inversion of Equation (4.30) in this case yields

=

=

+

=

21

2

2

1tan

1

)sin(1

11)(

and

where

teKtyt

(4. 34)

Several remarks can be made concerning the responses shown in Figure 7:

1. Responses exhibiting oscillation and overshoot (y/KM > 1) are obtained only for

values of less than one (Under-damped response). This response is initially

faster than the other two responses; however, it does not reach steady-state earlier

due to oscillations. Note that almost all under-damped responses are caused by the

interactions of the controllers with the process units they control.

2. Large values of yield a sluggish (slow) response (Over-damped response). Thelarger the value of the more sluggish the system. Note that system response

resembles a little the response of a first-order system to a unit step input. But

when compared to a first-order response we notice that the system initially delays

to respond and then its response is rather sluggish (Figure 8). Over-damped are

the responses of multicapacity systems, which result from the combination of

first-order systems in series.


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3. The fastest response without overshoot is obtained for the critically-damped case

( =1).

y(t)/KM

t/tau

Figure 7 Dimensionless response of second-order system to input step change.


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0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time

y/KM

Second-Order system

First-order system

Figure 8 Comparison between first-order and second-order responses.

Characteristics of under-damped systems

Figure 9 Characteristics of an under-damped response.

- Overshoot: Is the ratio ofa/b, where b is the ultimate value of the response and a is the

maximum amount by which the response exceeds its steady state value. It can be shown

that it is given by the following expression:


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=

21exp

OS (4. 35)

- Decay ratio: Is the ratio of the amount above the stead state value of two successive

peaks, c/a. it can be shown that it can be calculated by the following equation:

=

21

2exp

DR (4. 36)

- Rise time: tris the the process output takes to first reach the new steady state value.

- Time to first peak: tp is the time required for the output to reach its first maximum

value.

- Settling time: ts is defined as the time required for the process output to reach and

remain inside a band whose width is equal to 5 % of the total change in the output.

- Period: Equation (4.34) defines the radian frequency, to find the period of oscillation

P (i.e. the time elapsed between two successive peaks), use the well-known

relationship = 2/P; thus:

21

2P

= (4. 37)

4. Dynamic systems with dead time

Time delay is inherent property of chemical processes. The transfer function of a time

delay of units is given by

se)s(G = (4. 38)

Consider a first-order system with a dead time between the input and the output. This

system can be represented by a series of two systems as shown in Figure 10. For the first-

order system, we have the following transfer function:

1s

K

)]t(u[

)]t(y[

+=

l

l(4. 39)

while for the dead time we have:

se)]t(y[

)]t(y[ =

l

l(4. 40)

Therefore, the transfer function between the input u(t) and the delayed output y(t-) is

given by:

1s

Ke

)]t(u[

)]t(y[ s

+=

l

l(4. 41)

Same argument can be used with second-order systems.


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Figure 10 Block diagram for first-order process with dead time

Dynamic characteristics

The fitting of the first order plus time-delay model to the step response, using the

tangent method, requires the following steps, as shown in Figure 11:

Figure 11 Graphical analysis of the process reaction curve to obtain parameters of a

first-order plus time-delay model.

1. The process gain for the model is found by calculating the ratio of the change in the

steady-state value of y to the size of step change M in x.

2. A tangent is drawn at the inflection point of the step response; the intersection of the

tangent line and the time axis (where y = 0) is the time delay.

3. If the tangent is extended to intersect the steady-state response line (where y = KM),the point of intersection corresponds to time t = + .

This method suffers from using only a single point to estimate the time constant. Use

of multiple points may provide a better estimate.

1s

K

+ se

)]t(u[l )]t(y[l )]t(y[ l


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Polynomial approximations to e-s

Processes with dead time are difficult to control because the output does not contain

information about current events.

The introduction of exponent form results in a non-rational transfer function, which

can not be put in the form of a ratio of two polynomials in s. Quite often the exponentialterm is approximated by the first- (Equation 4.42) or second-order Pad approximations

(Equation 4.43)

s2

1

s2

1e s

+

(4. 42)

22

22

s

s12s21

s12

s2

1e

+

+

+

(4. 43)

Approximation of higher-order systems

Systems with order that is higher than one can be represented by

=

= +==

n

1i

i

n

1i

i

)1s(

K)s(G)s(G (4. 44)

It can be approximated by low-order transfer function with dead time as follows

1sKe)s(G

1

s

+=

(4. 45)

where 1 is the dominant time constant, i.e. dominated implies that the overall process

transient process is determined primarily by a few large time constants; that is response

modes that are much slower than the others. The dead time is defined as

=

=n

2i

i (4. 46)

Dynamic systems with inverse response

The dynamic behavior of certain processes shows an initial response which opposes in

direction to what it ends up, Figure 12. Such behavior is called inverse response. Such

behavior is the net result of two opposing effects. Systems show an inverse response if

one of the roots of the numerator (i.e. one of the zeros of the transfer function) has

positive real part.


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Figure 12 Inverse response

5. Examples

Non-interacting system

Two tanks are connected in series as shown in Figure 13. The time constants are 2 = 1

and 1 = 0.5; R1 = R2 = 1. Sketch the response of the level in tank 2 if a unit-step change

is made in the inlet flow rate to tank 1.

Solution

q(t)

R1, q

1(t)

R2

q2(t)

h1

h2

Figure 13 Two-tank liquid level system: Noninteracting


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A balance on tank 1 gives

dt

dhAqq 1

11 = (4. 47)

A balance on tank2 gives

dtdhAqq 2221 = (4. 48)

The flow head relationships are given by the following expressions

2

22

1

11

R

hq

R

hq

=

=

(4. 49)

Equation (4.47) can be rearranged to give

ARs

R

sQ

sH

=+= ,1)(

)(

1

11

(4. 50)

1

1

)(

)(

1

1

11

+=

=

ssQ

sQ

R

HQ

(4. 51)

In the same manner, the following equations van be obtained for tank2

1)(

)(

2

2

1

2

+=

s

R

sQ

sH

(4. 52)

Having the transfer function for each tank, we can obtain the overall transfer functionH2(s)/Q(s) by multiplying Equations (4.51) and (4.52) to eliminate Q1

11

1

)(

)(

2

2

11

2

+

+=

s

R

ssQ

sH

(4. 53)

Note that all roots here are real; thus, this is an over-damped system. By comparing this

equation with the general form of the over-damped system, it follows that

)1(

)1(

2

2

2

1

=

+=

(4. 54)

It can be proved that H2(s) is

)2(1)( 22tt

eetH = (4. 55)

Generalization for several noninteracting systems in series

Considern noninteracting first-order systems as represented by the block diagram of

Figure 14. The block diagram is equivalent to the relationships


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1)(

)(

1)(

)(

1)(

)(

1

2

2

1

2

1

11

+=

+=

+=

s

k

sX

sX

s

k

sX

sX

s

k

sX

sX

n

n

n

n

o

LLLLLL

(4. 56)

To obtain the overall transfer function, we simply multiply together the individual

transfer functions; thus

= +

=n

i i

i

o

n

s

k

sX

sX

1 1)(

)(

(4. 57)

Xo Xn1

1

1

+sk

12

2

+sk

1+sk

n

n

X2X1 Xn-1

Figure 14 Noninteracting first-order systems

Interacting system

q(t)

R1, q

1(t) R2 q2(t)

h1

h2

Figure 15 Two-tank liquid level system: Interacting

Solution

The balances on tanks 1 and 2 are the same as before and are given by Equations (4.47)

and (4.48). However, the flow-head relationship for tank I is now

1

211

)(

R

hhq

= (4. 58)


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The flow-head relationship for R2 is the same as before. Expressing Equation (4.58) in

terms of deviation variables gives

1

211

)(

R

HHQ

= (4. 59)

Transforming model equations gives

)()(

)()()(

)()()(

)()()(

222

2111

2221

111

sHsQR

sHsHsQR

ssHAsQsQ

ssHAsQsQ

=

=

=

=

(4. 60)

The combination of these equations gives

1)()(

)(

2121

2

21

22

++++=

sRAs

R

sQ

sH

(4. 61)

Notice that the difference between the transfer function for noninteracting system and the

interacting system is the presence of the term A1R2 in the coefficient of s.

To understand the effect of interaction on the transient of a system, consider a two-tank

system for which the time constants are equal, if the tanks are noninteracting, the transfer

function relating inlet flow to outlet flow is

2

2

1

1

)(

)(

+

=ssQ

sQ

(4. 62)

The unit-step response for this transfer function is

tt et

etQ =1)(2

(4. 63)

If the tanks are interacting, the overall transfer function, according to Equation (4.61), is

(assuming A1 = A2)

13

1

)(

)(2

2

++=

ssQ

sQ

(4. 64)

Notice that R2A2 = 2 = . By application of the quadratic formula, the denominator of

this transfer function can be written as

)162.2)(138.0(

1

)(

)(2

++=

sssQ

sQ

(4. 65)

For this example, we see that the effect of interaction has been to change the effective

time of the interacting system. One time constant has become considerably larger and the

other smaller than the time constant of either tank in the noninteracting system. The

response of Q2(t) to a unit step change in Q(t) for interacting case [Equation (4.65)] is

62.238.0

2 17.117.01)(tt

eetQ += (4. 66)


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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

Q2

Noninteracting

Interacting

Figure 16 Effect of interaction on step response of two-tank system

From Figure 16, it can be seen that interaction slows up the response. This is

understandable, when inlet to the first tank is stepped, the flow q1 will be reduced by the

build-up of level in tank 2.

In general, the effect of interaction on a system containing two first-order lags is to

change the ratio of effective time constants in the interacting system. In terms of transientresponse, this means that the interacting system is more sluggish than the noninteracting

system.

Fitting of a first order system

Figure 17 gives response of the temperature T in a continuous stirred-tank reactor to a

step change in feed flow rate w from 120 to 125 kg/min. Find an approximate first-order

model for the process for these operating conditions.

SolutionFirst note that w = M = 125-120 = 5 kg/min. Since T = T() T(0) = 160 140 =

20 C, the process gain is

min4

min5

20

kg

C

kg

C

w

TK

oo

==

=


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The time constant obtained from the graphical construction shown in Figure 17 is 5 min.

note that this result agrees with the time constant when 63.2% of the response is

complete, that is

T = 140+0.632*20 = 152.6 C

Consequently, the desired process model is

15

4

)(

)(

+=

ssW

sT

0 t

w (kg/min)

u(t)

y(t)160

120

125

140

5

152.6

Figure 17 Temperature response of a stirred-tank reactor for a step change in feed

flow rate.

dynamic behavior of first_second order systems

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