1. Similarity transformations  Suppose that we are given a ss model as in (1).  Now define state vector v(t) that is the same order of x(t), such that the elements of v(t) is a linear combination of the elements of x(t), that is In matrix form (1) v(t) = Qx(t)= P  1 x(t) And the transfer function G(s) is given by G(s) = C(sI-A)  1 B = C  (s)B (2) There are many combination of matrices A, B, C, and D that will satisfy (1) for a given G(s). We will show that this combination is unbounded v 1 (t) = q 11 x 1 (t) + q 12 x 2 (t) +  + q 1n x n (t) v 2 (t) = q 21 x 1 (t) + q 22 x 2 (t) +  + q 2n x n (t) v n (t) = q n1 x 1 (t) + q n2 x 2 (t) +  + q nn x n (t) x(t) = Pv(t)  Matrix P is called the transformation matrix  this will transform one set of state vector to a different state vector  This transformation alter the internal model but not the input output relationship  This typo of transformation is called similarity transformation  Finding the S.S. model from Diff. Eq. or T.F. has been presented.  It has been shown that a unique state model does not exist.  Two general state model control canonical form and observer canonical form can always be found  The number of internal models (state model) is unbounded The state model of SISO system is

2. Similarity transformations We are going to develop the detail of similarity transformation. Assume the similarity transformation of the form The general output equation y(t) = Cx(t) + Du(t) (5) becomes, using (1) y(t) = CPv(t) + Du(t)(6) We have the state eq. expressed as x(t) and as transformed state eq. as v(t). x(t) = Pv(t) (1) Substituting (1) to ss model gives (2) (3) Solving this eq. result the state model for variable v(t) (4) The state eq. as a function of v(t) can be expressed for multivariable system as y(t) = C v v(t) + D v u(t) (8) where the subscription indicates the transformed matrices. Comparing (7) with (4) and (8) with (6) we se that the equations for the transformed matrices are A v = P  1 AP B v = P  1 B C v = CPD v = D v(t) = A v v(t) + B v u(t) (7)

3. Example Similarity Transformation Consider the system described by With the transformation The transformed system become The transformed state eq. are then Similarity transformation properties  The eigenvalues of A are equal to those of A v, or det (sI  A) = det (sI  A v )= (s  1 )(s  2 )...  The determinant of A is equal to determinant of A; det A =det A v = 1 2... n  The trace of A is equal the trace of A v trA=trA v = 1 + 2 +...+ n  The following transfer function are equal C (sI  A)  1 B= C v (sI  A v )  1 B v

4. 3.6 Digital Simulation  The practical procedure for finding time response is thru simulation rather than solving the differential equation.  L.T limits us to solving low order systems. For higher order systems, we need simulation.  Here we consider the numerical solution of D.E by integration algorithm.  Many algorithm are available one of them is Euler’s method.  The process of numerically integrating a time function will be illustrated. We wish to find x(t) where z(t)z(t) t 0 Suppose we know x[(k-1)H] and we want to calculate x(kH), H is called the numerical integration increment and is the step size of algorithm. Euler algorithm assume z(t) constant in this period. Then (k-1)H kH x(kH) in (2) is only approximation of (1). As an example suppose we wish to solve (1) (2) (3) Differentiating (1) and substituting to (3), we have Suppose we chose H=0.1solving (4) iteratively starting with k=1, we get For this case (2) becomes x(kH) = x[(k-1)H]+H{-x[(k-1)H]} (4)

5. 3.6 Digital Simulation x(0.1) = x(0) - Hx(0)=1.0-(0.1)(1.0)=0.9 x(0.2) = x(0.1) - Hx(0)=1.0-(0.1)(1.0)=0.9 As an example suppose we wish to solve (3) Differentiating (1) and substituting to (3), we have For this case (2) becomes z(t)z(t) t 0 (k-1)H kH x(kH) = x[(k-1)H]+H{-x[(k-1)H]} (4) Suppose we chose H=0.1solving (4) iteratively starting with k=1, we get x(1.0) = x(0.9) - Hx(0.9)=0.3487 If we solve (3) analytically we have (5) x(1.0)= e  1.0 = 0.3678 The error of numerical integration is error = 0.3678  0.3487 = 0.0191 If we chose H=0.01 the error will only be 0.0018

6. 3.6 Digital Simulation Next we consider system described by (1) To develop the Euler method of this equation consider the first element, x 1 (t) of (1) which can be expressed as (2) We may write (3) (4) The same procedures applies for any element (5) These equations may be written in terms of vectors Where from (1) we can write (6) The Numerical integration algorithm is then: 1.Let k = 1 3.Evaluate x(kH) in (5) 4.Let k=k+1 5.Go to step 2 This algorithm is easy to program requires smaller step compared to the other 2.Evaluate in (6)

7. Example of numerical integration of ss equation Consider the system described by Suppose we choose H=0.01s, and x(0)=0, let k = 1 then Recall that the algorithm is: 1.Let k = 1 3.Evaluate x(kH) 4.Let k=k+1 5.Go to step 2 2.Evaluate step 3 with unit step input, that is u(t) is unit step for the second iteration (k = 2) we have step 2 step 3 The iteration can be repeated again. For this system we also able find the solution analytically.

8. Table bellow shows the result of numerical algorithm computation versus the analytical one Numerical algorithm Analytical (exact) tX 1 (t)X 2 (t)X 1 (t)X 2 (t) 00000 0.010 0.000050.00999 0.020.000100.020.000200.01999 0.030.000300.030.000400.02999 It seem that the result is not accurate enough, making smaller H will increase its accuracy. Example of numerical integration of ss equation