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If the forcing function is a unit-step function, we have X(s) = ; In terms of the damped vibrator shown in Fig. 8.1 this is equivalent to suddenly applying a force of magnitude K directed toward the right at time t = 0. This follows from the fact that X is defined by the relationship X(t) = F(t)IK . Super-
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HIGHER-ORDER SYSTEMS: SECOND-ORDER AND TRANSPORTATION LAG
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position will enable us to determine easily the response to a step function of any other magnitude. Combining Eq. (8.12) with the transfer function of Eq. (8.11) gives 1 Y(s) = 1 (8.13) s T2S2 + 257s + 1 The quadratic term in this equation may be factored into two linear terms that contain the roots
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s1 = -r+ 7 7
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,y2 = -g- m 7 7
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(8.14) (8.15)
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Equation (8.13) can now be written 1172 y(s) = (s)(s - Sl)(S - s2) (8.16)
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The response of the system Y(t) can be found by inverting Eq. (8.16). The roots sr and s2 will be real or complex depending on the parameter 6. The nature of the roots will, in turn, affect the form of Y(t). The problem may be divided into the three cases shown in Table 8.1. Each case will now be discussed.
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CASE I STEP RESPONSE FOR 6 < 1. For this case, the inversion of Eq. (8.16)
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yields the result (8.17) i 1 To derive Eq. (8.17), use is made of the techniques of Chap. 3. Since 6 < 1, Eqs. (8.14) to (8.16) indicate a pair of complex conjugate roots in the left-half plane and a root at the origin. In terms of the symbols of Fig. 3.1, the complex roots correspond to s:! and S; and the root at the origin to Sg. By referring to Table 3.1, we see that Y(t) has the form Y(f) = Cl + e --5f 7 C2 cos \/1_52$ + C3 sin i Y(f) = 1 - J&e-ct sin J-t + tan- C-F f
4-7;)
(8.18)
The constants C 1, C2, and C3 are found by partial fractions. The resulting equation is then put in the form of Eq. (8.17) by applying the trigonometric identity used
TABLE 8.1
CaSe
I II III
<I = 1 >l
Nature of roots Complex Real and equal Real
Description of response Underdamped or oscillatory Critically damped Overdamped or nonoscillatory
LINEAR OPEN-LOOP SYSTEMS
in Chap. 5, Eq. (5.23). The details are left as an exercise for the reader. It is evident from Eq. (8.17) that Y(t) --f 1 as t + ~0. The nature of the response can be understood most clearly by plotting Eq. (8.17) as shown in Fig. 8.2, where Y(t) is plotted against the dimensionless variable t/r for several values of {, including those above unity, which will be considered in the next section. Note that, for 4 < 1, all the response curves are oscillatory in nature and become less oscillatory as c is increased. The slope at the origin in Fig. 8.2 is zero for all values of t. The response of a second-order system for 6 < 1 is said to be under-damped.
CASE II STEP RESPONSE FOR 5 = 1. For this case, the response is given by
expression
Y(t) = 1 - (1 + ;,,-1,
(8.19)
This is derived as follows: Equations (8.14) and (8.15) show that the roots st and s2 are real and equal. Referring to Fig. 3.1 and Table 3.1, it is seen that
FIGURE 8-2 Response of a second-order system to a unit-step forcing function.
HIGHER-ORDER SYSTEMS: SECOND-ORDER AND TRANSPORTATION LAG
Eq. (8.19) is in the correct form. The constants are obtained, as usual, by partial fractions. The response, which is plotted in Fig. 8.2, is nonoscillatory. This condition, b = 1, is called critical dumping and allows most rapid approach of the response to Y = 1 without oscillation.
CASE III STEP RESPONSE FOR 5 > 1. For this case, the inversion of Eq.
(8.16) gives the result Y(t) = 1 - e-@ i where the hyperbolic functions are defined as p - e-a sinh a =
2 ea + eMa cash a = 2
sinh ,@ $ i
(8.20)
The procedure for obtaining Eq. (8.20) is parallel to that used in the previous cases. The response has been plotted in Fig. 8.2 for several values of 5. Notice that the response is nonoscillatory and becomes more sluggish as 4 increases. This is known as an over-dumped response. As in previous cases, all curves eventually approach the line Y = 1. Actually, the response for f > 1 is not new. We met it previously in the discussion of the step response of a system containing two first-order systems in series, for which the transfer function is 1 Y(s) -= (71s + l)(r*s + 1) X(s) (8.21)
This is true for f > 1 because the roots sr and ~2 are real, and the denominator of Eq. (8.11) may be factored into two real linear factors. Therefore, Eq. (8.11) is equivalent to Eq. (8.2 1) in this case. By comparing the linear factors of the denominator of Eq. (8.11) with those of Eq. (8.21), it follows that 71 = (l + &T)r
5-2 = ({ - JG)r (8.22) (8.23)
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