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.--0.1s 1 H(s) = 10 - - ~ (6.13) s(s + 1) i i s(s + 1) The first term in Eq. (6.13) can be inverted as shown in Eq. (5.12) to give lO( 1 -em ). The second term, which includes e -O.ls, must be inverted by use of the theorem on translation of functions given in Chap. 4. According to this theorem, the inverse of e-s of(s) is f(t - to) with f(t) = 0 for t - to < 0 or t < to. The inverse of the second term in Eq. (6.13) is
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The complete solution to this problem, which is the inverse of Eq. (6.13), is H(f) = lO(1 - e- ) H(f) = lo((1 - e- ) - [l - e-( -O.l)]} Simplifying the expression for H(t) for t > 0.1 gives H(r) = l.O52e- t >O.l t co.1 t >O.l (6.14)
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From Eq. (5.16), the response of the system to an impulse of magnitude 9 is given by H(t)(hpd, = (9)$e- = eer In Fig. 6.2, the pulse response of the liquid-level system and the ideal impulse response are shown for comparison. Notice that the level rises very rapidly during the 0.1 min that additional flow is entering the tar& the level then decays exponentially and follows very closely the ideal impulse response.
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The responses to step and sinusoidal forcing functions are the same for the liquid-level system as for the mercury thermometer of Chap. 5. Hence, they need
0 0.1 0.2
(b)
FIGURE 6-2 Approximation of an impulse function in a liquid-level system. (Example 6-l) (n) pulse input; (b) response of tank level.
PHYSICAL EXAMPLES OF FIRST-ORDER SYSTEMS
not be rederived. This is the advantage of characterizing all first-order systems by the same transfer function. Liquid-Leml Process with Constant-flow Outlet An example of a transfer function that often arises in control systems may be developed by considering the liquid-level system shown in Fig. 6.3. The resistance shown in Fig. 6.1 is replaced by a constant-flow pump. The same assumptions of constant cross-sectional ama and constant density that were used before also apply here. For this system, Eq. (6.2) still applies, but q(t) is now a constant; thus
q(t) - qo = A%
At steady state, Eq. (6.15) becomes (6.16) 4s - 40 = 0 Subtracting Eq. (6.16) from Eq. (6.15) and introducing the deviation variables Q = q - qs and H = h - h, gives Q=Ag Taking the Laplace (6.17)
transform of each side of Eq. (6.17) and solving for H/Q gives
H(s)
Q(s) = As
(6.18)
Notice that the transfer function, l/As, in Eq. (6.18) is equivalent to integration. One realizes this from the discussion on the transform of an integral presented in Chap. 4. Therefore, the solution of Eq. (6.18) is (6.19) If a step change Q(r) = u(t) were applied to the system shown in Fig. 6.3 the result is (6.20) h(t) = h, + t/A The step response given by Eq. (6.20) is a ramp function that grows without limit. Such a system that grows without limit for a sustained change in input is
q. = Constant
FIGURE
Liquid-level system with constant flow outlet.
LINEAR OPEN-LOOP SYSTEMS
said to have nonregulation. Systems that have a limited change in output for a sustained change in input are said to have regulation. An example of a system having regulation is the step response of a first-order system, which is shown in Fig. 5.6. The transfer function for the liquid-level system with constant outlet flow given by Eq. (6.18) can be considered as a special case of Eq. (6:!), as R + Q). The next example of a first-order system is a mixing process. l., _.
Mixing Process
Consider the mixing process shown in Fig. 6.4 in which a stream of solution containing dissolved salt flows at a constant volumetric flow rate q into a tank of constant holdup volume V. The concentration of the salt in the entering stream, x (mass of salt/volume), varies with time. It is desired to determine the transfer function relating the outlet concentration y to the inlet concentration X. Assuming the density of the solution to be constant, the flow rate in must equal the flow rate out, since the holdup volume is fixed. We may analyze this system by writing a transient mass balance for the salt; thus Flow rate of salt in - flow rate of salt out = rate of accumulation of salt in the tank Expressing this mass balance in terms of symbols gives
qx -
4Y =
d(b)
We shall again introduce deviation variables as we have in the previous examples. At steady state, Eq. (6.21) may be written qxs - 4Ys = 0 x = x - x ,
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