The time response of a linear system can be obtained using the state-variable approach. State space analysis is an important area of systems theory. In this page, the state-variable approach is introduced with the aim of using this approach to solve second-order systems. The description of state-variable theory is kept simple, keeping in mind the depth of coverage needed for circuit analysis.
A state variable is similar to the variable selected for solving a differential equation. Normally a state variable should be continuous, implying that it should reflect energy stored in a system. Hence the voltage across a capacitor and the current through an inductor are suitable to be selected as state variables.
Using state variables, state equations can be formed. We can obtain the solution, using either differential equations or Laplace transforms. State space analysis is a suitable technique for program-based solution using a computer. On the other hand, it tends to be rather tedious for solution by hand. But in view of its importance as an analytical tool, it is appropriate to have a brief look at state space analysis. First, we start off with a first-order system, and then extend the technique to higher-order systems.
Equation (1) presents the differential equation of a first-order linear system, where y(t) is the state variable and x(t) is the excitation function. In this equation, k and b are constants. Equation (2) is the same as equation (1), excepting that the derivative of y(t) is represented differently in manner suitable for state space analysis.
Equation (3) is obtained from equation (1). Multiply both sides of equation (3) by e-kt and get equation (4). Equation (5) is obtained from equation (4).
Integrate equation (5), with respect to t. Let the constant of integration be y(0), the initial value of y(t). In this page, the initial instant of time t is set to be zero, to make the equations more easily understandable. We can set the initial instant of time t to be t0, and then we need t to replace by (t - t0). In addition, y(0) is replaced by y(t0).
Equation (6) is the integral of equation (5).Equation (7) is obtained from equation (6) and it is the solution of differential equation, presented by equation (2).
Equation (7) has two parts, one representing the zero-input response and the other representing the zero-state response.
We apply what we have learned to an RC circuit, shown in Fig. 1
Equations (12), (13), (14) and (15) show how we can obtain a differential equation for the circuit in Fig. 1.
Based on equation (7), the solution is presented by equation (17). The values of components and the current source are specified in equation (18). On substituting these values into equation (17), we obtain equation (19). The time response of the capacitor voltage is expressed by equation (19).
Next, we extend this technique to higher-order systems.
Equation (20) presents the state equation. For a higher-order system, the derivatives, the state variables are column vectors. For a second-order system, the number of elements in these vectors is two. For a third-order system, the number of elements in these vectors is three, and so on. In this page, the scope is restricted to single-single output systems, and hence the system has only one excitation signal. The theory can be extended to multiple-input, multiple output systems. In equation (20), A is a square matrix and B is a column vector. The rank of A equals the order of the system, and the number of elements in B is the same as that of y(t). In order to obtain an explicit form of solution of equation (20), it is necessary to introduce a state transition matrix, φ(t) . Equation (21) presents the state transition matrix. If A is an n × n matrix, then φ(t) is also an n × n matrix. The derivative of φ(t) is expressed by equations (22) and (23). In equation (21), eAt is represented by a time-series.
Next we obtain solution for the state variables. Equation (20) is the state equation. It can be presented as shown by equation (24). Multiply both sides of equation (24) by e-At, and then we get equation (25).
As in the case of the first-order system, equation (25) can be represented, as shown by equation (26). Integral of equation (26) is presented by equation (27). The solution is presented by equation (28).
Next we need to determine eAt. Equation (29) presents the characteristic equation of the system. The roots of the characteristic equation are either real or complex numbers. In order to determine eAt, we make use of equation (30), which is obtained based on the Cayley-Hamilton theorem. The roots are called eigenvalues and the number of eigenvalues or poles equal the order of the system. For an nth order system, there are n eigenvalues, and they are defined as shown by equation (31).
Each eigenvalue satisfies equation (30) and hence we get the set of equations, defined by equation (32).
Next two examples are presented to show how eAt is determined.
Given matrix A, obtain eAt.
Solution:
From equation (33), form the characteristic equation, as shown by equation (34). The eigenvalues are presented by equation (35).
Since the rank of matrix is two, the expansion of eAt contains only two terms, as shown by equation (36). Each of the two eigenvalues satisfies equation (36) and (37) and hence equation (38) is obtained. The solution is presented by equation (39).
From the values of α0 and α1, eAt can be determined, as shown by equation (40).
When the multiplicity of an eigenvalue is more than one, we get the coefficients, as shown below. Given that the multiplicity of an eigenvalue is m, the first m equations are obtained as shown by equation (41). The remaining equations are obtained as before.
Now an example is presented.
Given the matrix as stated by equation (43), determine eAt.
Solution:
The eigenvalues are obtained, as shown by equations (44) and (45).
Based on the Cayley-Hamilton theorem we obtain equations (46) and (47). By substituting each of the eigenvalues, we get equation (48). Since the multiplicity of the eigenvalue at - 1 is two, we get the equation in the middle of equation (48). Since there are three equations and three unknowns, we can solve the set of simultaneous equations and obtain the values of α0, α1, and α2. From equation (46), it can be seen that we need the matrix A2. Equation (50) presents A2.
Based on the values of I, A, and A2, we can determine eAt, as shown below.
By substituting the values of α0, α1, and α2, into equation (51), we get equation (52).
It is possible to use Laplace transforms to obtain eAt. Given the state equation as presented by equation (53), we can express both sides in terms of Laplace transforms and obtain (54). Each element of the derivative vector column is the first-order derivative of the corresponding state variable. We know how to get the Laplace transform of the derivative of a function. Hence we get equation (54). Equation (55) is obtained from equation (54).
The state variable vector can be presented, as shown by equation (56). The inverse transform of equation (56) yields the state variables in time domain. By comparing equations (28) and (57), we get equation (58).
The inverse of a matrix can be obtained as shown below. Given M, obtain its cofactor matrix CF(M). Each element of CF(M) is the cofactor of the corresponding element in M. For a 2 × 2 matrix, the inverse is obtained, as illustrated by equations (59), (60), (61) and (62). The transpose of the cofactor matrix is the adjoint matrix. Divide each element by the determinant of M, and the result is the inverse of M.
We now obtain eAt of a 2 × 2 matrix. The matrix defined by equation (33) is chosen, so that we can verify the results obtained earlier. Equation (63) presents A and (sI - A). The inverse of (sI - A) is obtained and it is presented by equation (63). We get eAt from equation (64). It can be seen that equation (65) is the same as equation (40). It is easier to use the Laplace transforms.
The inverse of a 3 × 3 matrix can be obtained, as shown below.
Next we take up a 3 × 3 matrix. The matrix defined by equation (43) is chosen, so that we can verify the results obtained earlier. Equation (67) presents A and (sI - A). The inverse of (sI - A) is obtained and it is presented by equation (68). The inverse transform of equation (68) can be obtained and we get equation (52).
For a single-input, single-output system, we can obtain the transfer function of the system as shown below. Equation (69) presents the state equation and the equation for output q(t). Here output q(t) is expressed as a function of state variables and the input function x(t). When a transfer function is to be determined, the initial conditions are set to be of zero value. Hence we get equation (70). Equation (71) is obtained from equation (70). Replace the state variables in the equation for output by the expression in equation (71) and the result is equation (72). A transfer function is the ratio of Laplace transform of output to the Laplace transform of input and hence the transfer function can be obtained, as shown by equation (73).
The examples for obtaining a transfer function based on the state variable approach are presented in the next chapter.
This page has presented a brief introduction to state space analysis, restricting the scope to what is needed for circuit analysis. Proof of Cayley-Hamilton theorem is presented in text books on modern control theory. The next topic for study is the time response of second-order circuits.