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(6.22)
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Subtracting Eq. (6.22) from Eq. (6.21) and introducing the deviation variables
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give qx-qY = vg
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FIGURE 6-4 Mixing process.
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Taking the Laplace transform of this expression and rearranging the result give 1 Y(s) -=7s + 1 X(s) (6.23)
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where T = V/q. This mixing process is, therefore, another first-order process, for which the dynamics are now well known. We next bring in an example from DC circuit theory. RC Circuit Consider the simple RC circuit shown in Fig. 6.5 in which a voltage source v(t) is applied to a series combination of a resistance R and a capacitance C. For t < 0, v(t) = v,. Determine the transfer function relating et(t) to v(t), where et(t) is the voltage across the capacitor. Applying Kirchhoff s law, which states that in any loop the sum of voltage rises [v(t) in this example] must equal the sum of the voltage drops, gives i dt I Recalling that the current is the rate of change of charge with respect to time (coulombs per second), we may replace i by dqldt in Eq. (6.24) to obtain v(t) = Ri(t) + $
MO 1 v(t) = R- + p(O dt
Since the voltage across the capacitance is given by the relationship
(6.25)
4 e, = c
the initial charge on the capacitor is simply
(6.26)
qs = Ce,, Initially, when the circuit is at steady state and the capacitor is fully charged, the voltage across the capacitor is equal to the source voltage v,; therefore, Eq. (6.25) can be written for these steady-state conditions as VS = $4. = e,,
RC circuit.
LINEAR OPEN-LOOP SYSTEMS
Subtracting Eq. (6.27) from Eq. (6.25) and introducing the deviation variables v=v-vvs e=4-4s
E, = e, - e,, = g (6.28)
we obtain the result
V,Rdp,rs
(6.29)
V =RC%+E, (6.30)
Taking the transform of Eq. (6.30) and rearranging the result give 1 E,(s) -=7s + 1 V(s) (6.31)
where G- = RC. Again we obtain a first-order transfer function. The three examples that have been presented in this section are intended to show that the dynamic characteristics of many physical systems can be represented by a first-order transfer function. In the remainder of the book, more examples of first-order systems will appear as we discuss a variety of control systems.
Summary
In each example of a first-order system, the time constant has been expressed in terms of system parameters; thus
mC r=hA
7=AR
V Q-Z-.T=RC
for thermometer, Eq. (5.5) - . for liquid-level process, Eq. (6.8) for mixing process, Eq. (6.23)
for RC circuit, Eq. (6.31)
LINEARIZATION
Thus far, all the examples of physical systems, including the liquid-level system of Fig. 6.1, have been linear. Actually, most physical systems of practical importance are nonlinear. Characterization of a dynamic system by a transfer function can be done only for linear systems (those described by linear differential equations). The conve-
PHYSICAL EXAMPLES OF FIRST-ORDER SYSTEMS
nience of using transfer functions for dynamic analysis, which we have already seen in applications, provides significant motivation for approximating nonlinear systems by linear ones. A very important technique for such approximation is illustrated by the following discussion of the liquid-level system of Fig. 6.1. We now assume that the resistance follows the square-root relationship
q. = Ch 12 (6.32)
where C is a constant. For a liquid of constant density and a tank of uniform cross-sectional area A, a material balance around the tank gives
s(t) - qo(t) = A$
Combining Eqs. (6.32) and (6.33) gives the nonlinear differential equation
q - Ch Q = f$
(6.33)
(6.34)
At this point, we cannot proceed as before and take the Laplace transform. This is owing to the presence of the nonlinear term h *, for which them is no simple transform. This difficulty can be circumvented as follows.X By means of a Taylor-series expansion, the function q.(h) may be expanded around the steady-state value h,; thus
q. = qO(h,) + q;(h,)(h - h,) + q: (h );j - hS)2 + ...
where qA(h,) is the first derivative of q. evaluated at h S, qi(h ,) the second derivative, etc. If we keep only the linear term, the result is qo = q&s) + q:(k)(h - h,)
(6.35)
Taking the derivative of q. with respect to h in Eq. (6.32) and evaluating the derivative at h = h, gives
q;(h;) = (1/2)Ch;
Introducing this into Eq. (6.35) gives qo = qo, + $P - hs)
where qo, = q&s)
(6.36)
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