Unit 5
Solution of State Equations
Q1) Obtain the eigenvectors of matrix A = 
A1) For eigenvalues |
iI-A|=0
|
I-A|=
- 
=0
|
I-A| = 
=0
|
I-A| = (
+4)2-1=0
|
I-A| =
2+8
+15=0
The eigen values are
1=-3, 
2=-5
Q2) Obtain the eigenvalues of system described as 
=
+ 
[u] and y= [1 0 0]
A2) For eigenvalues |
iI-A|=0
|
I-A|=
- 
=0
|
I-A| = 
=0
3+6
2+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]
.
A3) Refer above que 2. The eigen values are 
1=-1, 
2=-2, 
3= -3
The eigen vectors 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 eigenvectors of the matrix A= 
.
A4) 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=
A5) For eigen values
|
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.
A6) 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 left half of s-plane so system is stable.
Q7) The state and output equation of LTI system is 
= 
+
[u] and y= [1 1 0]
. Find the characteristic equation.
A7) The characteristic equation is given as |
I-A|=0
|
I-A|=
- 
=0
|
I-A| = 
=0
The characteristic equation is
3-5
2-8
+2=0
Hence poles are 
1=-1, 
2=-2, 
3= -3
Q8) For matrix A= 
. Find the state transition matrix.
A8) 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 = 
.
A9) 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=
Q10) A=
. Calculate characteristic equation and stability.
A10) 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 absolutely stable.
Q11) A= 
. Find the characteristic equation and comment on stability.
A11) 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 absolutely stable.
Q12) Find f(A)=eAt for A=
.
A12) The characteristic equation is
q(
)=|
I-A|=
=(
+1)2=0
The eigen values of A are 
1,
2=-1
Since, A is of second order, R(
) will be
=
0+
1
Then 
0 and 
1
f(A)=
0+
1
=te-t
=
1
0=(1+t)e-t
1=te-t
f(A)=eAt=
0I+
1
=
