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e*( - )Bu(T)d
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(29.4)
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Equation (29.4) is a well known result that has been derived in many books on the solution of ordinary differential equations.
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i = Ax
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(29.5)
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Let us now turn our attention to the solution of the matrix differential equation
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This is Eq. (29.1) for the case of no inputs (i.e., u = 0). The initial conditions for Eq. (29.5) may be expressed as x(O). One can show that the solution to Eq. (29.5) with initial conditions x(O) is given by
x(t) = I + At + g2 + . . .
+ $tk} x(O)
The infinite series of matrix terms within the braces is given the symbol ek. This symbol is chosen to recall that the infinite series of the scalar term ear is 1 + at + gt2 + . . . + Etk Using the symbol ek, we may write Eq. (29.6) as
x(t) = eA x(0) (29.7) ak
The symbol eAr is an IZ X n matrix in which each element contains a power series of t. The solution to Eq. (29.1) can be shown to be
. x(t) = e* x(O) + e*( - )Bu(T)dT i0 (29.8)
Notice that Eq. (29.8) resembles Eq. (29.4), which is the solution for the scalar differential equation. Since e Ar is awkward and perhaps misleading as to its nature, e is sometimes replaced by +(t); thus
t)(t)
= eAr
(transition
matrix)
(29.9)
Either of the terms +(t) and eAr can be used for the transition matrix. In this book, we shall use eAr.
Example 29.1. Solution of a matrix differential equation.
Solve the following
matrix differential equation i= - 1 0 where u(t) is a unit-step function and
-2 x+
x(O) = [ -:, I
STATE-SPACE
METHODS
One can show that
,*r =
emr 0
,-t _ ,-2t
e-2r
In the next section, the method used to obtain the elements of this matrix will be developed. Applying Eq. (29.8) gives
x(t) = [- o-,] +[
x(t) =
e-(t--7) - (33e-2( -7)
0*5e-2( -7)
]I: 1 f,
2e- + 0.5ee2 - 0.5ee2
Determining e * t One method for determining the elements of the transition matrix e Ar is to use Laplace transforms. Consider the matrix differential equation of JZq. (29.1) i = Ax + Bu If we take the Laplace transform of each side, we obtain sX(s) - x(O) = AX(s) + BU(s) or 1 sX(s) - AX(s) = x(O) + BU(s) Solving for X(s) gives (~1 - A)X(s) = x(O) + BU(s) (29.10)
To obtain an expression for X(s), pre-multiply both sides of Eq. (29.10) by (81 - A)- ; thus (d- A)- @1 - A)X(s) = @I- A)- x(O) + (d- A)- BU(s) This equation becomes X(s) = (sI - A)- x(O) + (s1 - A)- BU(s) To obtain x(t) from Eq. (29.1 l), we may take the inverse transform; thus x(t) = L-l{ (s1 - A)- x(O)} + L-l{ (s1 - A)- BU(s)} By comparing Eqs. (29.8) and (29.12), we see that eAt = L-l{ (s1 - A)- } (29.13) (29.12) (29.11)
TRANSFER
FUNCTION MATRIX
e*( -T)~U(T)dT = L- { ($1 - A)- BU(s)}
TRANSFER FUNCTION MATRIX
(29.14)
When x(O) = 0, a case frequently used in control applications, we obtain from Eq. (29.11) . (29.15) X(s) = (sI- A)- BU(s) This may be written X(s) = G(s)U(s) where G(s) = @I- A)- B (transfer function matrix) (29.17) (29.16)
The term G(s) is called the transfer &nction matrix and serves the same purpose as the transfer function for the scalar case; namely, it relates a set of state variables X(s) to a set of inputs U(s). If we prefer to relate the output to the input as expressed by Eq. (29.2), we may proceed as follows. Taking the Laplace transform of both sides of Eq. (29.2) gives Y(s) = CX(s) Combining Eqs. (29.15) and (29.18) gives * Y(S) = C(sI- A)- BU(s) We may now write Y(s) = Gl(s)W) where Gl(s) = C(sIA)- B (29.20) The term Gt(s) in Eq. (29.20) is also a transfer function matrix that relates the output vector Y to the input vector U.
Example 29.2. Determine the transfer function matrix for the 2-tank liquid-level system shown in Fig. 29.1. As developed in Example 28.1 [Eq. (28.17)] of the previous chapter, this system is described by li = Ah+Bu where A=
(29.18)
(29.19)
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