Bryan L. Programmable controllers. Theory and implementation

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CLOSED-LOOP PROPORTIONAL CONTROL

Figure 15-31a illustrates a typical open-loop process control system where

the process variable and transfer function in Laplace form are represented by

the equation:

PV SP Hc Hp

Hc Hp

ssss

() () () ()

()

() ()

()( )( )

Figure 15-31b shows this same system in a closed-loop configuration. For

the closed-loop system, the process variable is defined by:

PV E Hc Hp

ssss() () () ()

()( )( )

Replacing E

(s)

with SP

(s)

– PV

(s)

yields:

PV SP PV Hc Hp

PV SP Hc Hp PV Hc Hp

PV Hc Hp SP Hc Hp

sssss

ssss sss

ssssss

() () () () ()

() () () () () () ()

() () () () () ()

=−

()

−

()

Solving for PV

(s)

over SP

(s)

yields the closed-loop transfer function of this

process control system:

Hc Hp

()

() ()

Figure 15-31. (a) Open-loop and (b) closed-loop process control systems.

HpHc

–

PV PVSP

–

(b) Closed-loop system

(a) Open-loop system

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The term Hc

(s)

represents the controller’s transfer function, while the term

(s)

represents the process’s transfer function. The process’s transfer func-

tion may take the form of a first-order response or a second-order response

(overdamped, underdamped, or critically damped). As we will discuss later,

the ideal controller transfer function for a system with a second-order

process plus lag and dead time is one with proportional, integral, and

derivative (PID) components.

A closed-loop system’s response to a proportional controller creates an error

that cannot be eliminated. This error is referred to as offset. Figure 15-32

shows the graph of a proportional controller’s transfer function, in which a

50% CV output keeps the process variable at the set point. If a load

disturbance occurs, the error will increase and the controller will change the

output variable to try to bring the error back to zero. However, if the load

disturbance requires a permanent output change in the controller (CV

new

), the

one-to-one relationship between the controller and error will prohibit a zero

error value, because the original function curve is changed. For instance, if

the process variable in Figure 15-33a increases due to a load disturbance,

the control variable (assuming a direct-acting controller) will increase pro-

portionally to try to bring the error back to zero. If the load disturbance causes

a permanent change in the required controller output, the new output level

(CV

new

) will cause the process variable to return to the set point value.

However, as PV begins to approach the set point (see Figure 15-33b), the

controller will reduce its output because the error is diminishing. The reduced

CV output will cause the error to increase, because the process requires a

control variable output at the level of CV

new

to maintain the process variable

at the set point. Thus, error will always be present in the system.

100%

50%

new

min

max

Change in output due

to load disturbance

–

Load

Disturbance

Figure 15-32. An offset caused by a load disturbance in a closed-loop system.

In process applications, the need for a permanent change in CV is typical,

thus proportional controllers always produce a small amount of error. This

error limits the use of proportional controllers to applications that include a

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manual reset, allowing the operator to change the bias, or operating point, of

the controller. This changes the level of controller output associated with

E = 0 from CV to CV

new

. Another way to minimize the system error is to

increase the gain, K

, thus reducing the proportional band. This method is

precarious, however, because too much gain will make the system oscillate

like an ON/OFF controller with a small deadband. The reason for this is that

if a small error occurs around the set point, the controller will have a large

output swing (CV) to correct for the process variable (PV). This will push the

error in the opposite direction. When the error goes the opposite direction of

the set point, the controller will quickly respond with another large output

change, thus forcing the process variable to return to its original direction.

Hence, the system will behave like an ON/OFF system if the proportional gain

is too large.

The following example illustrates the effect of error in a proportional

controller controlling a first-order system. Note that the step change in set

point is permanent, simulating a permanent disturbance or change.

Figure 15-33. (a) Desired response of a proportional controller to a load disturbance and

(b) its actual response.

(b)

new

(a)

50% 50%

new

A load disturbance causes

to deviate from the set point.

A load disturbance causes

to deviate from the set point.

The controller increases the control

variable to

new

to compensate for

the disturbance. The increase to

new

causes a decrease in the value of

PV,

instigating a decrease in the control

variable. The decrease in the control

variable causes the process variable

to increase again.

The controller increases the

control variable to

new

compensate for the disturbance,

bringing the error to zero.

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EXAMPLE 15-5

The closed-loop system shown in Figure 15-34 has a first-order

process (

) with a gain of 5 and a time constant of τ = 30 seconds.

The proportional controller (

) has a proportional gain of 8. (a) Find

the closed-loop transfer function, (b) calculate the response to a unit

step

, and (c) plot the response, indicating the system time constant

(

sys

)

and the steady-state value.

Figure 15-34. Closed-loop system.

SOLUTION

(a) The transfer function of a closed-loop system is expressed as:

Hc Hp

()

() ()

As discussed in Chapter 14, the Laplace transfer function of a first-

order process with lag is:

()

So the process’s transfer function is:

()

30 1

A proportional controller’s Laplace transfer function is simply the value

of its gain, so:

()

Therefore, the closed-loop transfer function of the entire system is:

HpHc

EPVSP

–

(

)

(

)

= 8

+ 1

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()

(

)

(

)

(

)

(

)

[]

(

)

(

)

(

)

(

)

40 30 1

30 41

30 1

40 30 1

30 1

This transfer function indicates that this is a first-order system. To

express it in the form of a first-order system, we must divide the

numerator and denominator by 41 to obtain:

()

(

)

(

)

0 976

0 732 1

(b) The response to a unit step change in the set point is given by:

()

()=

unit step

Therefore, using the previous equation, the process variable response

will be:

PV SP

() ()













0 976

0 732 1

Thus:

()

(. )

























1 0 976

0 732 1

0 976

0 732 1

According to Table 14-1, this response is in the form of the inverse

Laplace transform of a first-order response to a step input with lag:

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−













=−













()

sec

Hence, in the time domain, the process variable response will be equal to:

PV e

()

. sec

.=−













−

0 976 1

0 732

system to a unit step change in the set point. Note that the gain of the

system is 0.976, meaning that the process variable in the system will

not reach the value of the unit step input. Instead, the system will

respond only 0.976 to the unit step change of 1. The process variable

steady-state value (

), which is the final value of

PV,

can be

computed using the final value theorem:

PV e

=−

























=−

→∞

−

→∞

lim .

lim[ . ( )]

0 976 1

0 9761 0

0 976

0 732

Figure 15-35. System response to step change.

Therefore, the system will always have a residual error of 2.4% (1.0 –

0.976 = 0.024). The system time constant,

sys

= 0.732, indicates that

the system will take 0.732 seconds to reach 0.615, 63% of the steady-

state value.

The open-loop response of the process to a step change would have

a steady-state value of 5 (see Figure 15-36a), where 1

τ would occur

at 30 seconds. The open-loop response of the controller and process

to a unit step would have a gain of 40 with this same τ

constant (30 sec).

τ = 0.732

234

(sec)

0.615

0.976

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The closed-loop response of the controller and process, however,

would have a much smaller gain (in this case, 0.976) due to the

negative feedback in the control loop.

Figure 15-36. Open- and closed-loop responses shown in (a) detail and

(b) 10× detail.

The value of the gain in a proportional controller influences the response of

a second-order closed-loop system as shown in Figure 15-37. The higher the

gain, the faster the process responds; however, cycling and overshoot occur.

Lowering the gain makes the response much slower and the value at steady

state smaller. For example, the value of K

= 1 in Figure 15-37 will cause a

slow, closed-loop response to a unit step change in the set point with a final

steady-state value of 0.5.

Gain

234

(sec)

0.615

Open Loop

HcHp

(Gain of 40)

Hc Hp

EPV

–

Open Loop

(Gain of 5)

(see graphic below)

1τ 3τ

10 20 30 40 50 60 70 80 90 100

Open Loop

HcHp

Value of 25.2 at 1τ

63% (3.15)

95% (4.75)

Open-Loop

Closed Loop

HcHp

(Final value of 0.976)

Closed Loop

Gain

(sec)

(a)

(b)

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15-6 INTEGRAL CONTROLLERS (I MODE)

An integral controller provides an output whose rate of change is propor-

tional to the error deviation. This means that the larger the error, the faster the

controller’s output changes and vice versa. An integral controller will stop

adjusting its output once the error becomes zero. When used in conjunction

Figure 15-37. Responses of a second-order closed-loop system to different values of

proportional gain.

HpHc

EPVSP

–

Hc = K

Hp =

(10

+ 1)(2.5

+ 1)

Second-Order Process

(

= 10 min,

= 2.5 min)

1.5

1.0

0.5

(min)

0 10.0 20.0

= 4

= 10

= 2

= 1

Proportional Gain

Proportional

Gain =

Reponses for various values of

to the closed-loop response

where

(

)

= (unit step)

(

)

(

)

(

)

(

)

+ 1

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with a proportional controller, an integral controller will bring the system’s

residual error to zero. An integral controller’s output (CV) is represented by:

dCV

where:

dCV

the rate of change in controller output in % over seconds

the integral gain in % of the controller output per second

per % error

the error in %

This differential equation indicates that the controller’s output CV

(t)

can

be obtained by taking the integral over time of both sides of the equation

so that:

dCV

dCV K Edt

CV CV K Edt

CV K Edt CV

tt I

−=

∫∫

∫

() ( )

where the term CV

(t=0)

is the value of the output at t = 0. When an integral

controller is used in a closed-loop system, it calculates CV

(t)

for every change

in error. So, if the value of the error changes after the controller has calculated

a previous value CV

(t)

, then it will use this previous value of CV

(t)

as the CV

(t=0)

value and calculate a new CV

(t)

output based on the new error.

The integral gain K

(see Figure 15-38) indicates the sensitivity of the

output’s rate of change to the percentage of error that occurs over time. A

large value of K

means that a small error will produce a large rate of change

in the controller output. Conversely, a small value of K

means that a small

error produces a small rate of change in the controller output. In Figure 15-

38, the rate of change of K

is greater than that of K

Figure 15-39 illustrates the reaction of an integral controller’s output to a

change in the process variable due to a load disturbance. Note that, at the

moment the error occurs, the controller starts the integration of the error

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Figure 15-38. Integral gain.

Figure 15-39. Integral controller’s response to a step change in the process variable.

value, meaning that the control variable begins to increase as a function of

the magnitude of the error. The error in Figure 15-39 is constant, creating a

ramp integration of the controller’s output. That is, the amount of error

remains constant over time, so the control variable increases at a steady rate.

SP PV

SPPV

= SP

= 0

dCV

= –

dCV

= 0

dCV

= +

100%

50%

Error