We give two more examples to demonstrate the ideas of input/output description of a dynamical system.

Example 1 A system with three state variables (order = 3):

We must first calculate the adjoint of the determinant of the matrix (sI-A) (see matrix inverse under tools in the supplementary section):

We next solve for det(sI-A) :

Finally, we have

Example 2: Consider the shown spring-mass system.

m > 0, b > 0 and k > 0. Let the state variables be the position and velocity of the mass.
The output of the system is the position and the input is the external force f .

(a) Is the system stable?
(b) Find the transfer function.

For parts (c), (d) and (e) let m = 1 kg, b = N.s/m, and k = 1 N/m.
(c) Find the state variables (position and velocity of the mass) as functions of time if the initial conditions are equal to zero and f(t) = 1.
(d) Find the output y(t) if the initial conditions are equal to zero and f(t) = sint .
(e) Find the output y(t) if the initial position is equal to 1 m, the initial velocity is equal to zero and f(t) = sint .


Solution:

Let the position of the mass be z. Let us first find the state space representation of the system. We previously found the differential equation that governs the dynamics of this system:

The two state variables of the system are

The input u = f and the output y = z . Thus, the state space equations are written as:

(a) In order to test the stability of the system we find the eigenvalues of the matrix A:

Note that for m > 0, b > 0 and k > 0 the real part of the eigenvalues is always negative. Therefore, the system is stable. For any initial condition the mass will go to the equilibrium point and stops if it is left to move without applying the external force (input).

(b) Transfer function

Note that the transfer function of the system can lso be found by taking the Laplace transform of both sides of the differential equation that governs the system dynamics and setting all the initial conditions to zero.

(c) To find the state variables as functions of time we use the variation of parameters formula. Let us first compute the state transition matrix with m = 1 kg, b = N.s/m, and k = 1 N/m:

Exercise: Complete the integration and get the answer for part (c). Is there an easier way to solve this part of the example?

(d) Since all the initial conditions are equal to zero we can use the transfer function found in part (b) to find y(t).

(e) This part is the same as part (d) except for the initial condition. In part (d) the initial condition is assumed to be zero and the output is the forced response. Thus, if we find the unforced response and add it to the one found in part (d) we get the answer.

The unforced response (or response due to the initial condition) is given as:

The output is equal to the unforced response found here plus the forced response found in part (d), that is (variation of parameters formula)