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c(O) = 0 ,-(T) = 1 - e-!T-aT)/T
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c(2T) = c(3T) = *.. c(nT) = 0 The desired response Cd is therefore
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Cd(Z) zz 0 + [I - f+*-nrqz-- + oz-2 + . .
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DESIGN OF SAMPLED-DATA CONTROLLERS
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FIGURE 27-6
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Minimal prototype response for Example
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or C,(z) = (1 - d)z- (27.32)
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For a load algorithm, we use Q. (27.14) to obtain D(z). Notice that the peaks in Fig. 27.6 occur at time ur into each sampling interval. For this problem, we may use Eq. (26.11) with K = 1 and n = 0 to obtain, after simplification
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G(z) = &$)[z+$j
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(27.33)
For this example,
e-am UC,(s) = -am
S(TS + 1) = e
l/r s + (l/T)
To obtain UG,(z), we shall make use of the modified Z-transform as was done in Chap 25. With this in mind,
let ,-hTs = e-am
or AT = ar orh =arlT We can now write the m parameter in the modified Z-transforms as
m = 1 - A = 1 - (w/T)
The Z-transform of UGP(s) becomes
U%(z) = z[ ,;;I$] = z ( s(Tsl+ ,)
e-(l-adT)T/~
with
1 - (u/T).
From the table of transforms (Table 22. l), we obtain UG,(z)
= 1 -
z - l
z - b
SAMPLED-DATA CONTROL SYSTEMS
Cl- b)
0 0 c
FIGURE 27-7 z - 1k -2 0 2 tlT 6
Response under slow sampling, load, minimal prototype algorithm to unit-step in load. a = 0.5, T/r = 1.
This may be simplified to U, $(z) = ( ;zTkj; +$
Introducing Eqs. (27.32), (27.33), and (27.34) into Eq. (27.14) gives D(z) = z(1 - db) z. - w] 1
(1-d)2(z-1)[z-e]
(27.35)
It is instructive to examine the continuous response for this example as shown in Fig. 27.7. Although the response is zero at sampling instants after t = T, there is intersample ripple. Furthermore, one can show that the manipulated variable does not settle down, as was the case in Example 27.1, but continues to oscillate with decreasing amplitude as shown in Fig. 27.7. The reason for this unsatisfactory behavior is that the minimal prototype response is too demanding in returning the process variable to the set point. If the designer selects a nonminimal prototype response, which permits the response to return to the set point at 3T or later, the intersample ripple will be eliminated. A possible nonminimal prototype response is shown in Fig. 27.8.
Ir\__ FIGURE 27-8 T 2T 3T 4T t Possible nonminimal a5 21.2.
prototype response for Example
DESIGN OF SAMPLED-DATA CONTROLLERS
Response of Sampled-Data System to Other Disturbances
As stated in item 7 of the design specifications in this chapter, it is important to test the control algorithm for inputs that differ from the input for which the algorithm was designed. The computation for other inputs is straightforward, but it can be quite tedious. One uses Eq. (27.13) with the appropriate R(z) or UG,(z). For the algorithm developed in Example 27.1, the response to two different inputs will be considered: a step change in set point, and a ramp change in load.
STEP CHANGE IN SET POINT. For a set-point change, Eq. (27.13) becomes:
G(zP(z)R(z)
( ) = 1 + G(z)D(z) For a unit-step change in R
(27.36)
Substituting this expression for R(z) and those for G(z) and D(z) from Eqs. (27.19) and (27.21), respectively, into Eq. (27.36) gives
l - b cfz(z - Y)
z(z - b) (z - l)[z + (1 + b)] z 1 C(z) = az(z - Y) 1+ 1-b z(z - b) (z - l)[z + (1 + b)]
(27.37)
The inversion of this expression gives a result, shown graphically in Fig. 27.9.
FIGURE 27.9 Response under fast sampling, load, minimal prototype algorithm to a unit-step change in set point (a = 0.5).
SAMPLED-DATA CONTROL SYSTEMS
Notice that the response gives an overshoot with a settling time of 3T and the response is free of intersample ripple. From this result, one concludes that the response is satisfactory for a step change in set point.
RAMP CHANGE IN LOAD. For the algorithm developed in Example 27.1, consider a ramp change in load, for which u(r) = t or U(S) = l/s2. For a load
change, Eq. (27.13) becomes
UG,k)
( I = 1 + G(z)D(z) For this case, one can show that z(1 - b) WT)(Z - l)(z - b)
(27.38)
(27.39)
Introducing this expression for UG,(z) and G(z) and D(z) from Eqs. (27.19) and (27.2 l), respectively, into Eq. (27.38) gives, aker considerable algebraic manipulation C(z) = --&z + (1 + b)l{z[T
- ~(1 - b)] - [bT - (1 - b)~]}
Inverting this expression by the method of long division gives c(O) = 0
c(T) = 0 c(2T) = T - (1 - b)~ c(3T) = 2T - (1 - b2)7 c(nT) = ~(1 - b)(2 + b)
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