Question

2.10-2. Consider the system described by
x(k + 1) = c
0 1
0 3
d x(k) + c
1
1
d u(k)
y(k) = [-2 1]x(k)
(a) Find the transfer function Y(z)/U(z).
(b) Using any similarity transformation, find a different state model for this system.
(c) Find the transfer function of the system from the transformed state equations.
(d) Verify that A given and Aw derived in part (b) satisfy the first three properties of similarity transformations.
The fourth property was verified in part (c).

Answer 1

PROBLEM A:

a) A system is described by,

…… (1a)

…… (2a)

The input of the system is u(K) and the output of the system is Y(K).

The transfer function is determined as follows:

The system transfer function is given by,

…… (3a)

The state equations are of the form of

…… (4a)

…… (5a)

Now comparing the equations (4a), (5a) with equations (1a), (2a), we obtain,

Consider, :

Where,

Hence,

Thus,

…… (6a)

We know that,

…… (7a)

where, is the transpose of the coefficient matrix

We have,

Now,

Therefore,

That implies,

…… (8a)

Now using the equations (6a) and (8a) in equation (7a), we get,

From equation (3a), we have:

Now substituting all the known terms in equation (3a), we obtain,

Therefore,

…… (9a)

The system transfer function is given by,

…… (10a)

Therefore, from equations (9a) and (10a), the system transfer function is,

PROBLEM B :

(b) A system is described by:

…… (1b)

…… (2b)

Using similarity transformation, we need to find the different state model for this system.

The state equations are in the form of:

…… (3b)

…… (4b)

Now comparing the equations (4b), (3b) with equations (1b), (2b), we get:

Now we can apply linear transformation,

…… (5b)

That is,

Hence, is a constant matrix and is the new state vector.

Note that it is necessary that , (the inverse of ), exists so that can be determined from .

Now by substituting equation (5b) into equation (3b), we obtain,

…… (6b)

Now multiplying on both sides of equation (6b), we obtain,

We know that,

Hence,

…… (7b)

Now substituting equation (5b) into equation (4b), we obtain,

…… (8b)

Now the equations (7b) and (8b) can be expressed as,

Where,

…… (9b)

…… (10b)

…… (11b)

…… (12b)

Thus for each different for which exists, a different state model of a given system can be found.

We arbitrarily choose a linear transformation matrix,

The inverse of is given by:

…… (13b)

Where, indicates the transpose, and denotes the matrix of cofactors of

Thus,

…… (14b)

Hence,

…… (15b)

Now using the equations (15b) and (14b) in equation (13b), we get,

Hence,

…… (16b)

From equation (9b) we have,

Now substituting the known values in equation (9b), we get:

From equation (10b), we have:

Now substituting the known values in equation (10b), we get:

From equation (11b), we have:

Now substituting the known values in equation (11b), we get:

Thus, the different state model is achieved and the new state equations are:

PROBLEM C :

(c) The transformed state equations are,

…… (1c)

…… (2c)

From equations (1c) and (2c), it is clear that,

The input of the system is and the output of the system is

We need to find the transfer function

The transfer function is given by,

…… (3c)

The state equations are in the form of,

…… (4c)

…… (5c)

Now by comparing equations (4c), (5c) with equations (1c), (2c), we obtain,

Consider, :

Where,

Hence,

Thus,

…… (6c)

We know that,

…… (7c)

where, is the transpose of the coefficient matrix

We have,

Now,

Therefore,

Thus we have,

…… (8c)

Now by using the equations (6c) and (8c) in equation (7c), we obtain,

From equation (3c), we have,

Now by substituting all the known terms in equation (3c), we obtain,

Therefore,

…… (9c)

The system transfer function is given by,

…… (10c)

Therefore from equations (9c) and (10c), the system transfer function is,

PROBLEM D :

(d) We need to verify that the given matrix and derived in part (b), satisfy the first three properties of similarity transformations.

We have:

Property 1:

The characteristic values of the matrix are unchanged under the transformation.

The characteristic values are the roots of the characteristic equation.

From equation (6c), we have the characteristic equation of as:

The roots of the above characteristic equation are

From equation (6a), the characteristic equation of is given by,

The roots of the above characteristic equation are

Hence, it is clear that the roots of the characteristic equations of are same.

Therefore, we can say that the characteristic values of the matrix are unchanged under the transformation.

Property 2:

The determinant of (which is the product of characteristic values) is equal

to the determinant of :

We have:

The determinant of :

The determinant of :

The product of the characteristic values is:

Hence, we can say that the determinant of (which is the product of characteristic values) is equal to the determinant of .

Property 3:

The trace of is equal to the trace of

Now,

Hence, we can say that the trace of is equal to the trace of

Property 4:

This property was verified in part (c)