To convert a transfer function into state equations in phase variable form, we first convert the transfer function to a differential equation by cross-multiplying and taking the inverse Laplace transform, assuming zero initial conditions. Then we represent the differential equation in state space in phase variable form. An example illustrates the process.
Find the state-space representation in phase-variable form for the transfer function shown in Figure (1):
Step 1. Find the associated differential equation:
The corresponding differential equation is found by taking the inverse Laplace Transform, assuming zero initial conditions:
Step 2. Select the state variables. Choosing the state variables as successive derivatives, we get:
Using this notation, we can rewrite equation (1) as:
Step 3. Differentiating both sides of the last equations, we must find _x1 and _x2, then we use Eq. (2) to find x3. Proceeding in this way we obtain the state equations. Since the output is c=x1, the combined state and output equations are:
Step 4. Expressing the last equations in vector-matrix form, we get the state-space representation of the system as:
At this point, we can create an equivalent block diagram of the systemof Figure 1(a) to help visualize the state variables.We draw three integral blocks as shown in Figure 1(b) and label each output as one of the state variables, xi(t), as shown.
A transfer function with a polynomial in s in the numerator
The transfer function of the previous Example has a constant term in the numerator. If a transfer function has a polynomial in s in the numerator that is of order less than the polynomial in the denominator, as shown in Figure 2(a), the numerator and denominator can be handled separately. First separate the transfer function into two cascaded transfer functions, as shown in Figure 2(b); the first is the denominator, and the second is just the numerator. The first transfer function with just the denominator is converted to the phase-variable representation in state space as demonstrated in the last example. Hence, phase variable x1 is the output, and the rest of the phase variables are the internal variables of the first block, as shown in Figure 2(b).
The second transfer function with just the numerator yields:
Where, after taking the inverse Laplace transform with zero initial conditions, we obtain:
But the derivative terms are the definitions of the phase variables obtained in the
first block. Thus, writing the terms in reverse order to conform to an output equation, we obtain:
Hence, the second block simply forms a specified linear combination of the state
variables developed in the first block.
From another perspective, the denominator of the transfer function yields the
state equations, while the numerator yields the output equation. The next example
demonstrates the process.
Find the state-space representation of the transfer function shown in
Step 1. Separate the system into two cascaded blocks, as shown inFigure 3(b).The
first block contains the denominator and the second block contains the numerator.
Step 2. Find the state equations for the block containing the denominator. We notice that the first block’s numerator is 1/24 that of Example 1. Thus, the state equations are the same except that this system’s input matrix is 1/24 that of Example 1.
Step 3. Introduce the effect of the block with the numerator. The second block of
Figure 3(b) yields:
Taking the inverse Laplace transform with zero initial conditions, we get:
Thus, the last box of Figure 3(b) ‘‘collects’’ the states and generates the output equation:
Although the second block of Figure 3(b) shows differentiation, this block was implemented without differentiation because of the partitioning that was applied to the transfer function. The last block simply collected derivatives that were already formed by the first block.
Thus, the full state-space representation of the system is:
Once again we can produce an equivalent block diagram that vividly represents
our state-space model:
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- Sistemas de Control Automatico, Kuo
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