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G, = K
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The process is now equivalent to the sampled-data control of a first-order process with proportional control. For this specific example, we obtain for G(s) =
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K(l - e- ) G(s) = s (7s + 1) = K(1 - e- )x LS+$ I (
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TABLE 23.1
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system C(z) and C(z, m) c(z) = G(z)R(z) c(z, m) = G(z, m)R(z)
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( ) = 1 + GH(z) c(z, m) = RG(z, m) _ GH(z, mW(d 1 + GH(z)
CR(z)
C(z) =
G(z)JVz) 1 + GH(z)
G(z, m)Wz) ( m, = 1 + GH(z)
C(z) =
1 + GlG2Zf(z)
GzU(z)
C(Z, m) = UG2(z, m) _
UG2(z)GlGH(z, ml
1 + GlG2H(z)
Gl(z)G2(zW(z) ( ) = 1 + GI(z)GzH(z) cl(z)G2(z, m)Wz) c(zsm) = 1 + GI(z)GzH(z)
6. c
( ) = 1 +
C(Z, m) =
G2W)
Gl(z)G2H(z)
UG2(z9
_ G~(z)uG~H(z)G~(z~ m) I + G~If(z)Gl(z)
SAMPLED-DATA
CONTROL
SYSTEMS
G(s) may now be written as two terms: G(s) = K l/r (l/r)emTs s[s + (l/T)] - s[s + (l/r)]
To obtain G(z) for Eq. (23.25), the Z-transform of each term on the right must be found. The first term can be transformed easily by using the following transform pair from Table 22.1: z 1 - eeaT i a !
(23.25)
-S(S + a> (Z - l)[z - e-UT)
The second term can be transformed with the use of the following important theorem. Theorem. If g(r) is Z-transformable and has the Z-transform G(z), then the Ztransform of the delayed function g(t - n T) is given by Z{g(t - nT)) = z--G(z) (23.26) This theorem applies when the delay time nT is an integral number of sampling periods. This theorem applies only when g(t) = 0 for t < 0, a condition that will always apply in this book. The proof of this theorem can be found in other mferences (IOU, 1959). Note that this theorem is similar to the one for the transform of a delayed continuous function (i.e., L{g(t - 7)) = e-rSG(s)). Applying the transform pair and the theorem just given to the terms on the right side of Eq. (23.25) gives
z 1 - e-T!~ z-lz 1 - e-T!~ 1 ( ( 1 G(z) = K cz _ 1) (z - ,-T/T] - K cz _ 1) (z _ ,-T/T)
(23.27)
or G(z) = Kz (1 - z- )(l - e-nT) (2. - l)(z - e- ) Simplifying gives
(23.29)
(23.28)
where b = ewTiT.
To obtain G,G,R(z) for a unit-step change in R, we proceed as fokws:
G,GpR(s) Taking the Z-transform. G,G&z) = Kz(l - b) (z - l)(z - b) (23.31) 117 = K = K S(TS + 1) s[s + (l/r)] (23.30)
For a unit-step change in R, Eq. (23.21) becomes C(z) = 1 + G(z) =
G&+R(z)
Kz(1 - b)l(z - l)(z - b) 1 + [K(l - b)/(z - b)]
(23.32)
OPEN-LOOP
CLOSED-LOOP
RESPONSE
K(l - b)z ( ) = (z - l)[z - b + K(1 - b)]
(23.33)
INVERSION. The inversion of C(z) may be obtained by long division or by partial fraction expansion. Using the latter method, we proceed as follows C(z) - =
K(l - b) (z - l){z - [b - K(1 - b)l) B
Z-a!
(23.34)
or C(z) PC
K(l - b) A (z - l)(z - (Y) = - + z - l
(23.35)
where (Y = b - K(l - b). Solving for the constants A and B gives A = Kl(1 + K), B = -Kl( 1 + K). Inverting by means of the table of transforms gives c(nT) = & (1 - (u ) = & { 1 - [b - K(l - b)] ] (23.36)
Stab&y. From the result given by Eq. (23.36), the stability of this system can be studied as follows. For c(nZ ) to converge, lb - K(l - b)l < 1 Note that b < 1 since b = eTT and T/T is positive. The inequality may be written in two ways: I. b - K(l - b) < 1 II. b-K(l-b)>-1 For I.: -b+K(l-b)>-1 K(l- b) > -1+ b K > -(l - b)/(l - b) K > -1 or Since K is always positive, this result is of no practical value. For II.: b-K(l-b)>-1 -b+K(l-b)<l K(l - b) < (1 + b) l+b KCl - b (requirement for stability) (23.37)
This is a useful result and shows how a sampled-data system differs from a continuous system. For a continuous system, proportional control of a first-order system is always stable. For the sampled-data case, them is a value of controller gain K, above which the system goes unstable. hznsient response. For this example, the transient response consists of connected arcs of exponentials. A typical response is shown in Fig. 23.5. The sampled-data
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