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FIGURE 5-6 Response of a first-order system to a step input.
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we should expect the flow of heat to commence immediately, with the result that the mercury temperature rises, causing a corresponding rise in the column of mercury. As the mercury temperature rises, the driving force causing heat to flow into the mercury will diminish, with the result that the mercury temperature changes at a slower rate as time proceeds. We see that this description of the response based on physical grounds does agree with the response given by Eq. (5.12) and shown graphically in Fig. 5.6. Several features of this response, worth remembering, are 1. The value of Y(t) reaches 63.2 percent of its ultimate value when the time elapsed is equal to one time constant T. When the time elapsed is 27,37, and 47, the percent response is 86.5, 95, and 98, respectively. From these facts, one can consider the response essentially completed in three to four time constants. 2. One can show from Eq. (5.12) that the slope of the response curve at the origin in Fig. 5.6 is 1. This means that, if the initial rate of change of Y(t) were maintained, the response would be complete in one time constant, (See the dotted line in Fig. 5.6.) 3. A consequence of the principle of superposition is that the response to a step input of any magnitude A may be obtained directly from Fig. 5.6 by mult lplying the ordinate by A. Figure 5.6 actually gives the response to a unit-step function input, from which all other step responses am derived by superposition. These results for the step response of a first-order system will now be applied to the following example.
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RESPONSE OF FIRST-ORDER SYSTEh4S
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Example 5.1. A thermometer having a time constant* of 0.1 min is at a steadystate temperature of 90% At time t = 0, the thermometer is placed in a temperature bath maintained at 100 F. Determine the time needed for the thermometer to read 98 . In terms of symbols used in this chapter, we have 7 = 0.1 min xs = 90 A = loo
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The ultimate thermometer reading will, of course, be loo , and the ultimate value of the deviation variable Y(m) is IO . When the thermometer reads 98 , Y(t) = 8 . Substituting into Eq. (5.12) the appropriate values of Y, A, and T gives 8 = lO(1 - c ).~) Solving this equation for t yields t = 0.161 min The same result can also be obtained by referring to Fig. 5.6, where it is seen that Y/A = 0.8 at t/r = 1.6.
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Impulse Response
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The impulse response of a first-order system will now be developed. Anticipat-
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ing the use of superposition, we consider a unit impulse for which the Laplace transform is X(s) = 1 (5.13)
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Combining this with the transfer function for a first-order system, which is given by Eq. (5.7), results in
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This may be rearranged to
l/r Y(s) = -
s + l/7
(5.15)
The inverse of Y(s) can be found directly from the table of transforms and can
he written in the form 7Y(t) = f - *
(5.16)
A plot of this response is shown in Fig. 5.7 in terms of the variables t/T and -rY(t). The response 50 an impulse of magnitude A is obtained, as usual, by / multiplying TY(t) from Fig. 5.7 by A/r.
*The time constant given in this problem applies to the thermometer when it is located in the temperature bath. The time constant for the thermometer in air will be considerably different from that given because of the lower heat-transfer coefficient in air.
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2 VT
FIGURE 5-7 Unit-impulse response order system.
o f a first-
Notice that the response rises immediately to 1.0 and then decays exponentially. Such an abrupt rise is, of course, physically impossible, but as we shall see in Chap. 6, it is approached by the response to a finite pulse of narrow width, such as that of Fig. 5.4.