barcode reader integration with asp.net Physical variables State variables are called physical variables when they in Software

Generator ANSI/AIM Code 128 in Software Physical variables State variables are called physical variables when they

1. Physical variables State variables are called physical variables when they
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are readily measured and observed (level, temperature, composition, etc.). Physical variables were discussed at the beginning of this chapter and illustrated for a liquid-level system in Example 28.1 where x I = h t and x 2 = h2. 2. Phase variables State variables that are chosen to be the dependent variable and its successive derivatives are called phase variables. Phase variables were
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*Although the analog computer will not be discussed in this book, simulation software, such as TUTSIM or ACSL (Advanced Computer Simulation Language), will be discussed in a later chapter. These simulation languages contain integrator blocks that are equivalent to the response of an integrator in an analog computer
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STATE-SPACE REPRESENTATION OF PHYSICAL SYSTEMS
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selected at the beginning of this chapter in Eqs. (28.3~) and (28.3b) where xl = y and x2 = j. 3. Canonical variables If the state variables are selected to be canonical variables, the result is that the matrix A is diagonal. At this point, it is sufficient to say that canonical variables are selected as state variables for ease in matrix computation. In general, the canonical variables are not readily identified with physical variables. In addition to the types of state variables listed above, any other legitimate set of variables can be selected. In Example 28.1, we used physical variables, namely the levels in the tanks of the liquid-level system. In the next two examples, the method for selecting state variables will be shown. Example 28.2. For the two-tank liquid-level system of Example 28.1, shown in Fig. 28.1, obtain the state-space description as expressed by Qs. (28.11) and (28.12) when phase variables are selected for the state variables. To simplify the problem, let ~2 = 0, i.e., there is only one input ~1. For the system shown in Fig. 28.1, one can show that
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H2h) Ul@> _ (71s + R2 1x72s + 1)
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(28.18)
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where ~1 = AlRl and 72 = AzR2 Introducing the parameters in Fig. 28.1 into Eq. (28.18) gives
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_ HZ(S) Ul(S)
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213 (Is + l)($ + 1) 4 2)(s
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(28.19)
_ H2(s) Ul(S) (s + +
(28.20)
To obtain the differential equation corresponding to Eq. (28.20), we cross-multiply to obtain (S + 2)(s + 3)H2 = 4u1
(s2+5s +6)H2 = 4U1
This may be expressed as the following differential equation: h2+5h2 +6h2
= 4~1
(28.21)
Let the state variables be the following phase variables: Xl = h2
(28.22) (28.23)
STATE-SPACE
METHODS
We may now write ii 1 = x2[= i2] i2 = 62 Equation (28.2 1) becomes i2 + 5x2 + 6xl = 4ut The system can be described by Eqs. (28.24) and (28.26): ii1 = x2 i2 = -6xl -5x2 + 4ul In terms of a matrix expression, Eqs. (28.27) may be written: i = Ax+bul where A = [-i -i] b = [j (28.26) (28.24) (28.25)
(28.27~) (28.276)
If the output y is to be the level in tank 2, y = cx where c = [l O] Example 28.3. For the PI control system shown in Fig. 28.2, obtain a state-space representation in the form of Eq. (28.7); thus i=Ax+br where r is a scalar. Let x1=c n2 = 2 = fl With this choice of state variables, we have selected phase variables. From Fig. 28.2, we may write C(s) -=
M(s) (71s + KP 1x72s + 1)
(28.28) (28.29)
or A C(s) -= (s + a)(s + b) M(s) (28.30)
FIGURE 28-2 PI Control System for Example 28.3.
STATE-SPACE REPRESENTATION OF PHYSICAL. SYSTEMS
where A = K&Q u = l/r1 b = 11~ Cross-multiplying Eq. (28.30) gives
[s* + (a + b)s + ab]C(s) = AM(s) or . % + (a + b)k + abc = Am From Eqs. (28.28), (28.29), and (28.31) we obtain .il = x* i2 = -abxl - (a + b)x2 + Am (28.32) (28.33) (28.31)
We must now obtain the state variables associated with the PI controller. From Fig. 28.2, we obtain
M(s) = K (71s + 1) E(s) c 71s
or T~sM(s) = K,qsE(s) + K,, (s) In terms of the time domain, this expression becomes riz = K,i + (K,/q)e From the signals entering and leaving the comparator, we may write e = r - c or, since x 1 = c, we may write e=r-xl and k=F-fl Combining Eqs. (28.34) and (28.35) gives tit = K,(i - i1) + (KJq)(r - xl) or rh = K,f - Kcx2 + (K&)r - (KcIq)xl (28.36) (28.35b) (28.35~) (28.34)
At this stage, we are faced with the difficulty of having a derivative term on the right side of Eq. (28.36). In state-space representation, all variables on the right side must be state variables, not derivatives of state variables. One way to handle the present difficulty is to define a new state variable xg; let xg = m-Kg or
(28.37) (28.38)
= h - K,;
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