Unit-2

State-Variable Analysis

Q1) Obtain the eigenvectors of matrix A =

Sol: For eigenvalues |iI-A|=0

|I-A|= - =0

|I-A| = =0

|I-A| = (+4)2-1=0

|I-A| =2+8+15=0

The eigenvalues are

1=-3, 2=-5

Q2) Obtain the eigenvalues of the system described as = + [u] and y= [1 0 0]

Sol: For eigenvalues |iI-A|=0

|I-A|= - =0

|I-A| = =0

3+62+11+6=0

(+1) (+2) (+3) = 0

Hence eigen values are 1=-1, 2=-2, 3= -3

Q3) Find eigen vectors for the following system = + [u] and y= [1 0 0]

Sol: refer above que 2. The eigenvalues are 1=-1, 2=-2, 3= -3

The eigenvectors will be

For 1=-1

|I-A|X = 0

= 0

-x1-x2 = 0

3x1 + x2 + 2x3 = 0

12x1+7x2+5x3=0

Solving above equations we get x1 = 1, x2= -1 as x3 = -1

Therefore, the eigen vectors are

For 2=-2

|I-A|X = 0

= 0

-2x1-x2 = 0

3x1 + 2x2 + 3x3 = 0

17x1+7x2+4x3=0

Solving above equations we get x1 = 2, x2= -4 as x3 = -1

Therefore, the eigen vectors are

For 3=-3

|I-A|X = 0

= 0

3x1+x2 = 0

3x1 + 3x2 + 2x3 = 0

12x1+7x2+3x3=0

Solving above equations we get x1 = 1, x2= -3 as x3 = 3

Therefore, the eigen vectors are

Q4) Find the eigen vectors of the matrix A=

Sol: For eigenvalues |iI-A|=0

|I-A|= - =0

|I-A| = =0

|I-A| = (+3)2-1=0

|I-A| =2+6+8=0

1=-2, 2=-4

For 1=-2 the eigenvectors are

- x Xi=0

- x = 0

= 0

-x1+x2=0

For x1=1 x2=1 the equation above is satisfied.

Xi=

For 2=-4

- x = 0

= 0

-x1-x2=0

For x1=1 x2= -1 the equation above is satisfied.

Xi=

Q5) For the given matrix A=

Sol: For eigenvalues

|I-A|= - =0

|I-A| = =0

|I-A| = (-3)( -2)-2=0

|I-A| =2-5+4=0

1=1, 2=4

For 1=1 the eigenvectors are

- x Xi=0

- x = 0

= 0

-2x1+2x2=0

For x1=1 x2=1 the equation above is satisfied.

X1=

For 2=4

- x = 0

= 0

x1+2x2=0

For x1=2 x2= -1 the equation above is satisfied.

X2=

Arrange eigenvectors in a matrix P =

P-1= - =

The transformation matrix =P-1AP

=

=

=

Q6) A system is represented by the state equation and output equation as

= + u(t) and Y=[1 2]

Find the poles of the system and comment on stability?

Sol: A=

The characteristic equation is given as

|I-A|= - =0

|I-A| = =0

|I-A| = (+3) (+2)-2=0

|I-A| =2+5+4=0

1=-1, 2=-4

The poles are -1 and -4. Both on the left half of the s-plane so the system is stable.

Q7) The state and output equation of the LTI system is = + [u] and y= [1 1 0]. Find the characteristic equation.

Sol: The characteristic equation is given as |I-A|=0

|I-A|= - =0

|I-A| = =0

The characteristic equation is

3-52-8+2=0

Hence poles are 1=-1, 2=-2, 3= -3

Q8) For matrix A= . Find the state transition matrix?

Sol: The state transition matrix is given by L-1[SI-A]-1=φ(t)

[SI-A]= -

=-

=

Taking inverse Laplace of above, we get

[SI-A]-1=/(S+5)(S+10)

=

Hence φ(t)=L-1[SI-A]-1=

Q9) Find state transition matrix if A =

Sol: The state transition matrix is given by L-1[SI-A]-1=φ(t)

[SI-A]=-

=

[SI-A]-1=

Hence φ(t)=L-1[SI-A]-1=

A= .

Q.10) Calculate characteristic equation and stability?

Sol: The characteristic equation is given as [SI-A]=0

S - =0

- = 0

=0

S(S+3)-(-1)*2=0

Hence, the characteristic equation is

S2+3S+2=0

(S+1)(S+2)=0

S=-1,-2

Both roots on left-half of s-plane, real and different, system stable.

Q11) A= Find the characteristic equation and comment on stability?

Sol: The characteristic equation is given by [SI-A] =0

-=0

=0

S(S+2)+2=0

S2+2S+2=0

S=-1±j

Roots on left-half of s-plane, complex conjugate, system stable.

Q12) A=. Find the state transition matrix?

Sol: The state transition matrix is given by L-1[SI-A]-1=φ(t)

[SI-A]= -

=-

=

Taking inverse Laplace of above, we get

[SI-A]-1=/(S+5)(S+10)

=

Hence φ(t)=L-1[SI-A]-1=

Q13) Find state transition matrix if A=

Sol: The state transition matrix is given by L-1[SI-A]-1=φ(t)

[SI-A]=-

=

[SI-A]-1=

Hence φ(t)=L-1[SI-A]-1=

Q14) Find the observable canonical realization of the system H(s)=

Sol: The above transfer function can be also written as

H(s)=

Comparing the above equation with the standard equation we conclude that

The gains of forwarding paths are

The feedback loop gains are

The SFG satisfying the above conditions will be

The observable canonical form can be obtained by converting the above SFG to a block diagram

Q15) Find the controllable canonical realization of the following systems

a) H(s)=

b) H(s)=

Sol: a) H(s)=

Let H(s)=

=

H1(s)=1/s+6

X(s)=sX1(s)+6X1(s)

sX1(s)= X(s) +6X1(s)

We can get X1(s) by passing sX1(s) through integrator. The above equation can be realised as

H2(s)=s+2

Similarly, Y(s)= sX1(s)+2X1(s)

H2(s)=s+2

The complete realization of the transfer function can be obtained by combining the above two realizations. The complete realization will be

b)H(s)=

Let H(s)= =

=s+3

Y(s)=sX1(s)+3X1(s)

The above transfer function can be realized as

Now, =

s2X1(s)=X(s)-2sX1(s)-5X1(s)

Assuming s2X1(s) is available the above transfer function can be realized as

The complete realization of the transfer function can be obtained by combining the above two realizations. The complete realization will be

Q16) Obtain the state space representation for the given electrical system

Sol: The state model shows that there are two energy storing elements L, C. As we already know that Number of state variables is equal to the number of energy storing elements.Hence we have two state variables[x1(t) and x2(t)]. We have one output V0(taken across the capacitor) and input u(t).

Here output is V0. But from above electrical circuit V0=ILR

V0=x2(t)R

y(t)= V0= [0 R] + [0] u(t)

The output equation is given as

Y(t)=CX(t)+DU(t)

C=[0 R] D=[0]

Now finding the state equation, we apply KCL in the given electrical circuit

I=IC+IL

=

But I-IL=IC

=

=

=

=x2(t)+ ……..(a)

=[0 1/C]+ [1/C 0] …….(b)

==

Applying KVL in the given electrical circuit we get

VC=VL+ILR

VC-ILR=VL=L

=

=[1/L -R/L] + [0] u(t) ………….(c)

From equation (b) and (c) we have

Now writing the state equation

=Ax(t)+Bu(t)

=+

Hence A= B=

Note: We should always take voltage across the inductor L, and current through capacitor C.

Q17) A system is represented by the following state model A= B=find whether it is controllable or not?

Sol: A= B=

So the order of the matrix is n=2

[S]=[B AB]

=

=0 hence, the system is not controllable.

Q18) A system is represented by the following state model A= B= , C=[1 1] test whether the system is observable or not?

Solution: AT= CT=

[Q]=[CT ATCT]

=

=

=

=0

=0 hence not observable.