PV () s ( 1 + Hc Hp () s () s ) = SP Hc Hp () s () s () s
PV () s ( 1 + Hc Hp () s () s ) = SP Hc Hp () s () s () s
Solving for PV (s) over SP (s) yields the closed-loop transfer function of this process control system:
(a) Open-loop system
SP
Hc Hp PV
(b) Closed-loop system
Figure 15-31. (a) Open-loop and (b) closed-loop process control systems.
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The term Hc (s) represents the controller’s transfer function, while the term Hp (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.
CV Change in output due
to 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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A load disturbance causes PV
A load disturbance causes PV
to deviate from the set point.
to deviate from the set point.
The controller increases the
The controller increases the control
control variable to CV new to
variable to CV new to compensate for
compensate for the disturbance,
the disturbance. The increase to CV new
bringing the error to zero.
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.
Figure 15-33. (a) Desired response of a proportional controller to a load disturbance and (b) its actual response.
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 P , 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.
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E X AM PLE 1 5 -5
The closed-loop system shown in Figure 15-34 has a first-order process ( Hp ) with a gain of 5 and a time constant of τ = 30 seconds.
The proportional controller ( Hc ) has a proportional gain of 8. (a) Find the closed-loop transfer function, (b) calculate the response to a unit step 1 s , and (c) plot the response, indicating the system time constant
( τ sys ) and the steady-state value.
Figure 15-34. Closed-loop system.
S OLU T I ON
(a) The transfer function of a closed-loop system is expressed as:
As discussed in Chapter 14, the Laplace transfer function of a first- order process with lag is:
Hp () s =
τ s + 1 So the process’s transfer function is:
Hp () s =
30 s + 1
A proportional controller’s Laplace transfer function is simply the value of its gain, so:
Hc () s = 8
Therefore, the closed-loop transfer function of the entire system is:
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( )( ) 8 s + = 30 1
PV 5
() s
[ ( )( ) s + 30 1 ]
SP () s
40 () s + = 30 1
( s + 30 1 + 1 )
( s + 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:
(b) The response to a unit step change in the set point is given by:
SP () s =
( unit step )
Therefore, using the previous equation, the process variable response will be:
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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Hence, in the time domain, the process variable response will be equal to:
(c) Figure 15-35 illustrates the time response of the closed-loop 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 ( PV ss ), which is the final value of PV, can be computed using the final value theorem:
= lim[ . t →∞ 0 976 1 0 ( − )] = . 0 976
t (sec) 1 2 3 4
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, τ 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).
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PV
Open Loop HcHp
Gain
(Gain of 40)
Hc Hp PV
Open Loop Hp
Hp PV
(Gain of 5) (see graphic below)
Closed Loop
t (sec) τ 1 2 3 4
PV (a) Gain
10 9 Open Loop HcHp 8 Value of 25.2 at 1 τ
95% (4.75) 4 Open-Loop Hp
2 Closed Loop HcHp (Final value of 0.976)
t (sec) 10 20 30 40 50 60 70 80 90 100
3 τ Figure 15-36. Open- and closed-loop responses shown in (a) detail and
(b)
(b) 10 × detail.
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.
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 P = 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.
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Proportional Gain = K P
(10 s + 1)(2.5s + 1) Second-Order Process
( τ 1 = 10 min, τ 2 = 2.5 min)
PV 1.5
Proportional Gain K P = 10
K P =4 0.5 K P =2 K P =1
0 t (min) 0 10.0 20.0
Reponses for various values of K P in Hc
to the closed-loop response
1 where SP ( s) = (unit step) s
Figure 15-37. Responses of a second-order closed-loop system to different values of
proportional gain.