Modal Control System

This system has only one feedback loop, but this loop doesn't have a regulator as such. Regulation occurs due to the input system coefficient and the state variable coefficients. Therefore, the calculation of these coefficients will give us the result for the system. 

The modal control system should provide a monotonic transient function without overshoot (which corresponds to a binomial pole distribution). 

The control object is covered by feedback on selected state variables with some transmission coefficients. The matrix of these coefficients is a modal regulator. The principal quantity z may be one of the state variables, but this is not necessary.

Using additional corrective devices, it is necessary to provide a single static transmission coefficient between the control action and the main value z.

The development of a modal control system begins with the transformation of the source data. We are given a system of equations:

x = 20v,

2y' + y = x - 2z,

z'' + z' + 3z = 15y

Value of the transition time tpp = 2,4 s.

Now we introduce the following notation:

x1 = y        x2 = z          r = 1         n = 3

y1 = z        x3 = z'         m = 1

u1 = u

We substitute the replaced variables in the system of equations, and also make the change: v = 47u. Now we write the obtained equations in the normal Cauchy form, expressing the variables x. We get:

From here we can write the following matrices:

Now let's look at the diagram shown on the demo sheet and, in accordance with it, we will write a system of two equations in matrix form:

We calculate the matrix A-BK. We call it the matrix E.

Subtract the diagonal matrix P from the resulting matrix, calculate the determinant of the resulting matrix and equate it to zero to obtain the characteristic equation of the system.

Since we use the binomial distribution of poles, the third-order characteristic equation should have the form:

Here A is the geometric mean pole equal to: A = τpp/tpp = 6,3/2,4 = 2.625 s. 

We equate the values of p of the same degree. From here we find all the coefficients.

The next and last point is to find the coefficient of the system km. To do this, recall the system of equations written according to the scheme.

Static Mode:

We describe these equations in more detail.

From the first equation we obtain the following three equalities:

We write what the coefficient of the system is equal to:

The resulting coefficient of the system will stand at the input of the system and change the input signal. The scheme will looks like this: