8. DISEÑO METODOGICO
8.4 INSTRUMENTOS PARA LA RECOLECCION DE DATOS Y RESULTADOS
We now walk through the stirred-tank heater system once again. This time, we will take a closer look at the transfer functions and the units (Fig. 5.5).
16 Can an open loop be still a loop? You may wonder, what is an open loop? Often, we loosely refer to
elements or properties of part of a system as open loop, as opposed to a complete closed-loop system. You will see more of this language in Chap. 7.
17 In real life, we expect probable simultaneous reference and disturbance inputs. As far as analysis goes,
the mathematics is much simpler if we consider one case at a time. In addition, either case shares
the same closed-loop characteristic polynomial. Hence they should also share the same stability
and dynamic response characteristics. Later when we talk about integral error criteria in controller design, there are minor differences, but not sufficient to justify analyzing a problem with simultaneous reference and load inputs.
5.2. Closed-Loop Transfer Functions T Stirred-tank heater (mV) (°C) TH(°C) Tsp (°C) − + + Gc Ga GL Gp Gm Km T i(°C) T(°C) + (mV) (mV)
Figure 5.5. Block diagram of a simple SISO closed-loop system with phys-
ical units.
Process Model
The first item on the agenda is “process identification.” We either derive the transfer functions of the process based on scientific or engineering principles, or we simply do a step-input experiment and fit the data to a model. Either way, we need to decide what the controlled variable is. We then need to decide which should be the manipulated variable. All remaining variables are delegated to become disturbances.
With the stirred-tank heater, we know quite well by now that we want to manipulate the heating-coil temperature to control the tank temperature. The process function Gpis defined based on this decision. In this simple illustration, the inlet temperature is the only disturbance, and the load function is defined accordingly. From Subsection 2.8.2 and Eq. (2.49b), we have the first-order process model:
T = GLTi+ GpTH = KL τps+ 1 Ti+ Kp τps+ 1 TH. (5.13)
From Subsection 2.8.2, we know that the steady-state gain and the time constant are dependent on the values of flow rate, liquid density, heat capacity, heat transfer coefficient, and so on. For the sake of illustration, the heat transfer analysis is skipped. Let’s presume that we have done our homework, substituted in numerical values, and have found Kp = 0.85◦C/◦C andτp = 20 min.
Signal Transmitter
Once we know what to control, we need to find a way to measure the quantity. If the transducer (sensor and transmitter packaged together) is placed far downstream or is too well insulated and the response is slow, the measurement transfer function may appear as
Tm
T = Gm =
Kme−tds τms+ 1
, (5.14)
where Kmis the measurement gain of the transducer,τmis the time constant of the device, and td accounts for transport lag. In the worst case, the sensor may be nonlinear, meaning that the measurement gain would change with the range of operating conditions.
With temperature, we can use a thermocouple, which typically has a resolution of the order of 0.05 mV/◦C. [We could always use a resistance temperature detector (RTD) for better resolution and response time.] That is too small a change in output for most 12-bit analog-digital converters, so we must have an amplifier to boost the signal. This is something we do in a laboratory, but commercially, we should find off-the-shelf transducers with the
sensor and amplifier packaged together. Many of them have a scaled output of, for example, 0–1 V or 4–20 mA.
For the sake of illustration, let’s presume that the temperature transmitter has a built-in amplifier that allows us to have a measurement gain of Km= 5 mV/◦C. Let’s also presume that there is no transport lag and that the thermocouple response is rapid. The measurement transfer function in this case is simply
Gm= Km= 5 mV/◦C.
This so-called measurement gain is really the slope of a calibration curve – an idea that we are familiar with. We do a least-squares fit if this curve is linear and find the tangent at the operating point if the curve is nonlinear.
Controller
The amplified signal from the transmitter is sent to the controller, which can be a computer or a little black box. Not much can be said about the controller function now, except that it is likely a PID controller or a software application with a similar interface.
A reminder is that a controller has a front panel with physical units such as degrees celsius. (Some also have relative scales of 0–100%.) Therefore, when we dial a change in the set point, the controller needs to convert the change into electrical signals. That’s why Kmis part of the controller in the block diagram (Fig. 5.5).
Actuator Control Valve
Last but not least, designing a proper actuator can create the most headaches. We have to find an actuator that can drive the range of the manipulated variable. We also want the device to have a faster response than the process. After that, we have to find a way to interface the controller to the actuator. A lot of work is masked by the innocent-looking notation Ga.
For the stirred-tank heater example, several comments should be made here. We need to consider safety. If the system fails, we want to make sure that no more heat is added to the tank. Thus we want a fail-close valve – meaning that the valve requires energy (or a positive signal change) to open it. In other words, the valve gain is positive. We can check the thinking as follows: If the tank temperature drops below the set point, the error increases. With a positive proportional gain, the controller output will increase, hence opening up the valve. If the process plant has a power outage, the valve closes and shuts off the steam. But how can the valve shut itself off without power?
This leads to the second comment. One may argue for emergency power or a spring- loaded valve, but to reduce fire hazards, the nominal industrial practice is to use pneumatic (compressed-air-driven) valves that are regulated by a signal of 3–15 psi. The electrical signal to and from the controller is commonly 4–20 mA. A current signal is less susceptible to noise than a voltage signal is over longer transmission distances. Hence, in a more applied setting, we expect to find a current-to-pressure transducer (I/P) situated between the controller output and the valve actuator.
Finally, we have been sloppy in associating the flow rate of steam with the heating-coil temperature. The proper analysis that includes a heat balance of the heating medium is in the Review Problems. To sidestep the actual calculations, we have to make a few more
5.3. Closed-Loop System Response
assumptions for the valve gain to illustrate what we need to do in reality:
(1) Assume that we have the proper amplifier or transducer to interface the controller output with the valve, i.e., converting electrical information into flow rate.
(2) We use a valve with linear characteristics such that the flow rate varies linearly with the opening.18
(3) The change in steam flow rate can be “translated” to changes in heating-coil tem- perature.
When the steady-state gains of all three assumptions are lumped together, we may arrive ata valve gain Kvwith units of degrees celsius per millivolt. For this illustration, let’s say the valve gain is 0.6◦C/mV and the time constant is 0.2 min. The actuator controller function would appear as Gv = Kv τvs+ 1 = 0.6◦C/mV 0.2 s + 1 .
The closed-loop characteristic equation of the stirred-tank heater system is hence
1+ GcGvGpGm = 1 + Gc
(0.6)(0.85)(5)
(0.2s + 1)(20s + 1) = 0.
We will not write the entire closed-loop function C/R or, in this case, T/Tsp. The main
reason is that our design and analysis will be based on only the characteristic equation. The closed-loop function is handy to do only time-domain simulation, which we can easily compute by using MATLAB. That being said, we need to analyze the closed-loop transfer function for several simple cases so we have a better theoretical understanding.
5.3. Closed-Loop SystemResponse
In this section, we will derive the closed-loop transfer functions for several examples. The scope is limited by how much sense we can make out of the algebra. Nevertheless, the steps that we go through are necessary to learn how to set up problems properly. The analysis also helps us to better understand why a system may have a faster response, why a system may become underdamped, and when there is an offset. When the algebra is clean enough, we can also make observations as to how controller settings may affect the closed-loop system response. The results generally reaffirm the qualitative statements that we have made concerning the characteristics of different controllers.
The actual work is rather cookbook-like:
(1) With a given problem statement, draw the control loop and derive the closed-loop transfer functions.
(2) Pick either the servo or the regulator problem. Reminder: the characteristic polyno- mial is the same in either case.
18 In reality, the valve characteristic curve is likely nonlinear and we need to look up the technical
specification in the manufacturer’s catalog. After that, the valve gain can be calculated from the slope of the characteristic curve at the operating point. See Homework Problem I.33 and the Web Support.
− +
R E
Gc Gp C
Figure 5.6. Simple unity feed-
back system.
(3) With the given choices of Gc(P, PI, PD, or PID), Gp, Ga, and Gm, plug their transfer functions into the closed-loop equation. The characteristic polynomial should fall out nicely.
(4) Rearrange the expressions such that we can redefine the parameters as time constants and steady-state gains for the closed-loop system.
All analyses follow the same general outline. What we must accept is that there are no handy-dandy formulas to plug and chug. We must be competent in deriving the closed-loop transfer function, steady-state gain, and other relevant quantities for each specific problem. In our examples, we will take Gm= Ga = 1 and use a servo system with L = 0 t o highlight the basic ideas. The algebra tends to be more tractable in this simplified unity feedback system with only Gcand Gp(Fig. 5.6), and the closed-loop transfer function is
C R =
GcGp 1+ GcGp,
(5.15)
which has the closed-loop characteristic equation 1+ GcGp = 0.
Example 5.1: Derive the closed-loop transfer function of a system with proportional control
and a first-order process. What is the value of the controlled variable at steady state after a unit-step change in set point?
In this case, we consider Gc= Kc and Gp = [Kp/(τps+ 1)], and substitution into Eq. (5.15) leads to19 C R = KcKp τps+ 1 + KcKp. (E5.1)
19 You may wonder how transfer functions are related to differential equations. This is a simple illus-
tration. We use y to denote the controlled variable. The first-order process function Gp arises from
Eq. (3.6):
τp
dy
dt + y = Kpx.
In the unity feedback loop with Gc= Kc, we have x= Kc(r− y). Substitution for x in the ODE
leads to τp dy dt + y = KcKp(r− y) or τp dy dt + (1 + KcKp)y= KcKpr.
It is obvious that Eq. (E5.1) is the Laplace transform of this equation. This same idea can be applied to all other systems, but, of course, nobody does that. We all work within the Laplace transform domain.
5.3. Closed-Loop System Response
We now divide both the numerator and denominator with (1+ KcKp) to obtain C R = KcKp/(1 + KcKp) [τp/(1 + KcKp)]s+ 1 = K τs + 1, (E5.2) where K = KcKp 1+ KcKp, τ = τp 1+ KcKp
are the closed-loop steady-state gain and time constant.
Recall Eq. (5.11); the closed-loop characteristic equation is the denominator of the closed- loop transfer function, and the probable locations of the closed-loop pole are given by
s= −(1 + KcKp)/τp.
There are two key observations. First, K< 1, meaning that the controlled variable will change in magnitude less than a given change in set point, the source of offset. The second is thatτ < τp, meaning that the system has a faster response than the open-loop process. The system time constant becomes smaller as we increase the proportional gain. This is consistent with the position of the closed-loop pole, which should “move away” from the origin as Kcincreases.
We now take a formal look at the steady-state error (offset). Let’s consider a more general step change in set point, R= M/s. The eventual change in the controlled variable, by means of the final-value theorem, is
c(∞) = lim s→0s K τs + 1 M s = M K .
The offset is the relative error between the set point and the controlled variable at steady state, i.e., (r− c∞)/r: ess = M− M K M = 1 − K = 1 − KcKp 1+ KcKp = 1 1+ KcKp. (E5.3)
We can reduce the offset if we increase the proportional gain.
Let’s take another look at the algebra for evaluating the steady-state error. The error that we have derived in the example is really the difference between the change in controlled variable and the change in set point in the block diagram (Fig. 5.6). Thus we can write
E = R − C = R 1− GcGp 1+ GcGp = R 1 1+ GcGp .
Now if we have a unit-step change R= 1/s, the steady-state error by means of the final-value theorem is (recall that e= e)
ess = lim s→0s 1 1+ GcGp 1 s = 1 1+ lim s→0GcGp = 1 1+ Kerr , (5.16)
where Kerr= lim
s→0GcGp. We call Kerr the position error constant.
20 For the error to
20 In many control texts, we also find the derivation of the velocity error constant (by using R= s−2) and
approach zero, Kerr must approach infinity. In Example 5.1, the error constant and the
steady-state error are
Kerr= lim s→0GcGp = KcKp τps+ 1 = K cKp, and again ess = 1 1+ KcKp. (5.17)
Example 5.2: Derive the closed-loop transfer function of a system with proportional control
and a second-order overdamped process. If the second-order process has time constants 2 and 4 min and process gain 1.0 (units), what proportional gain would provide us with a system with damping ratio of 0.7?
In this case, we consider Gc= Kc, and Gp = {Kp/[(τ1s+ 1)(τ2s+ 1)]}, and substitution
into Eq. (5.15) leads to C R = KcKp (τ1s+ 1)(τ2s+ 1) + KcKp = KcKp/(1 + KcKp) τ1τ2 1+KcKp s2+ τ1+τ2 1+KcKp s+ 1. (E5.4)
The key is to recognize that the system may exhibit underdamped behavior even though the open-loop process is overdamped. The closed-loop characteristic polynomial can have either real or complex roots, depending on our choice of Kc. (This is much easier to see when we work with a root locus in Chap. 7.) For now, we rewrite the closed-loop function as
C R =
K
τ2s2+ 2ζτs + 1, (E5.4a)
where the closed-loop steady-state gain is K = [(KcKp)/(1 + KcKp)], and the system natural time period and damping ratio are
τ =! τ1τ2 1+ KcKp, ζ = 1 2 (τ1+ τ2) τ1τ2(1+ KcKp) . (E5.5)
If we substituteζ = 0.7, Kp= 1, τ1 = 2, and τ2= 4 into the second expression, we should
find the proportional gain Kcto be 1.29.
Last, we should see immediately that the system steady-state gain in this case is the same as that in Example 5.1, meaning that this second-order system will have the same steady-state error.
In terms of controller design, we can take an entirely analytical approach when the system is simple enough. Of course, such circumstances are not common in real life. Furthermore, we often have to compromise between conflicting criteria. For example, we cannot require a system to have both a very fast rise time and a very short settling time. If we want to provide a smooth response to a set-point change without excessive overshoot, we cannot also expect a fastand snappy initial response. As engineers, itis our job to decide.
In terms of design specification, it is not uncommon to use the decay ratio (DR) as the design criterion. Repeating Eq. (3.29), we know that the DR [or the overshoot (OS)] is a function of the damping ratio:
DR= (OS)2= exp −2πζ 1− ζ2 . (5.18)
5.3. Closed-Loop System Response
From this equation we can derive
ζ2= (ln DR)2
4π2+ (ln DR)2. (5.19)
If we have a second-order system, we can derive an analytical relation for the controller. If we have a proportional controller with a second-order process, as in Example 5.2, the solution is unique. However, if we have, for example, a PI controller (two parameters) and a first-order process, there are no unique answers as we only have one design equation. We mustspecify one more design constraintin order to have a well-posed problem.
Example 5.3: Derive the closed-loop transfer function of a system with PI control and a
first-order process. What is the offset in this system?
We substitute Gc= Kc[(τIs+ 1)/τIs] and Gp = [Kp/(τps+ 1)] into Eq. (5.15), and we find that the closed-loop servo transfer function is
C R = KcKp(τIs+ 1) τIs(τps+ 1) + KcKp(τIs+ 1) = (τIs+ 1) τIτp KcKp s2+τI(1+KcKp) KcKp s+ 1 . (E5.6)
There are two noteworthy items. First, the closed-loop system is now second order. The integral action adds another order. Second, the system steady-state gain is unity and will not have an offset. This is a general property of using PI control. [If this is not immediately obvious, try taking R= 1/s and apply the final-value theorem. We should find the eventual change in the controlled variable to be c(∞) = 1.]
With the expectation that the second-order system may exhibit underdamped behavior, we rewrite the closed-loop function as
C R =
(τIs+ 1)
τ2s2+ 2ζ τs + 1, (E5.6a)
where the natural time period and DR of the system are
τ =! τIτp KcKp , ζ = 1 2(1+ KcKp) ! τ I KcKpτp . (E5.7)
Although we have the analytical results, it is not obvious how choices of the integral time constant and the proportional gain may affect the closed-loop poles or the system DR. (We may get a partial picture if we consider circumstances under which KcKp 1.) Again, the analysis is deferred until we cover root locus; we should find that to be a wonderful tool in assessing how controller design may affect system response.
Example 5.4: Derive the closed-loop transfer function of a system with PD control and a
first-order process.
Closed-loop transfer function (5.15) with Gc= Kc(1+ τDs) and Gp= [Kp/(τps+ 1)] is C R = KcKp(τDs+ 1) (τps+ 1) + KcKp(τDs+ 1) = KcKp(τDs+ 1) (τp+ KcKpτD)s+ 1 + KcKp. (E5.8)
The closed-loop system remains first order and the function is that of a lead–lag element. We can rewrite the closed-loop transfer function as
C R =
K (τDs+ 1)
τs + 1 , (E5.8a)
where the system steady-state gain and time constant are
K = KcKp
1+ KcKp, τ =
τp+ KcKpτD 1+ KcKp .
The system steady-state gain is the same as that with proportional control in Example 5.1. We, of course, expect the same offset with PD control too. The system time constant depends on various parameters. Again, this analysis is deferred until we discuss root locus.
Example 5.5: Derive the closed-loop transfer function of a system with proportional control
and an integrating process. What is the offset in this system?
Let’s consider Gc= Kcand Gp = 1/As; substitution into Eq. (5.15) leads to C R = Kc As+ Kc = 1 ( A/Kc)s+ 1. (E5.9)
We can see quickly that the system has unity gain and there should be no offset. The point is that integral action can be introduced by the process and we do not need PI control under such circumstances. We come across processes with integral action in the control of rotating bodies and liquid levels in tanks connected to pumps (e.g., Example 3.1).
Example 5.6: Provide illustrative closed-loop time-response simulations. Most texts have
schematic plots to illustrate the general properties of a feedback system. This is something that we can do ourselves by using MATLAB. Simulate the observations that we have made in previous examples. Use a unity feedback system.
We consider Example 5.3 again; let’s pickτpto be 5 min and Kpbe 0.8 (unit). Instead of using the equation that we derived in Example 5.3, we can use the following statements in MATLAB to generate a simulation for the case of a unit-step change in the set point. This approach is much faster than using Simulink.
kc=1; % The two tuning parameters to
taui=10; % be varied
% The following statements are best saved in an M-file
Gc=tf(kc*[taui 1],[taui 0]); % The PI controller function
Gp=tf(0.8,[5 1]); % The process function
Gcl=feedback(Gc*Gp,1) % Unity closed-loop function
% GcGp/(1 + GcGp)
step(Gcl); % Personalize your own plotting
% and put a hold for additional % curves
5.3. Closed-Loop System Response
In these statements, we have usedfeedback()to generate the closed-loop function C/R. The unity feedback loop is indicated by the “1” in the function argument. Try first with Kc= 1 andτI with values of 10, 1, and 0.1. Next, selectτI = 0.1 and repeatwith Kc= 0.1, 1, 5, and 10. In both cases, the results should follow the qualitative trends that we anticipate.