Unit-2
State-Variable Analysis
The state of a system is a minimal set of variables known as state variables. The knowledge of these variables at any instance of time together with the knowledge of the inputs for the same instance of time determines the complete behaviour of the system. The fewer drawbacks in the transfer function method for representing any system led to the use of state variables in the analysis of the system. Few advantages are listed below:
Representation of state-space: The system shown below has ‘m’ inputs, ‘p, outputs, and ‘n’ number of state variables. The state equation gives us the relation between the state variables and the inputs.
Fig 1 State variable
So, the above system shown can be described through equations as
=f1 (x1, x2,…….xn , u1, u2……… ,um)
= f2 (x1, x2,…….xn , u1, u2……… ,um)
=f1 (x1, x2,…….xn , u1, u2……… ,um) (1)
The above set of equations can be represented as
=f(x(t),u(t)) (2)
As we are concerned for time invariant system, for which the term 
in (1) is a linear combination of state variables and input. So,
(t)=2tx1+x2+u1+u2 (3)
=a11x1(t)+a12x2(t)+…a1nxn(t)+b11u1(t)+b12u2(t)+…….+b1mum(t)
=a21x1(t)+a22x2(t)+…a2nxn(t)+b21u1(t)+b22u2(t)+…….+b2mum(t)
=an1x1(t)+an2x2(t)+…annxn(t)+bn1u1(t)+bn2u2(t)+…….+bnmum(t)
The above equation can be represented in matrix form as given below
(4)
The above coefficients aij and bji in equation (4) can be written in a vector-matrix form as
=Ax(t)+Bu(t) (5)
Key takeaway
The state of a system is a minimal set of variables known as state variables.
The vector-matrix form is given as 
=Ax(t)+Bu(t)
State Transition Matrix and its Properties:
y(t)=Cx(t)+Du(t)
x(t)=Ax(t)+Bu(t)
-Ax(t)=Bu(t)
SX(s)-X(0)-AX(s)=BU(s)
SX(s)-AX(s)=BU(s)+X(0)
[SI-A]X(s)=X(0)+BU(s)
X(s)=[SI-A]-1[X(0)+BU(s)]
X(s)=[SI-A]-1X(0)+[SI-A]-1BU(s) (a)
This is solution of state differential equation
L-1X(s)= L-1{[SI-A]-1X(0)+[SI-A]-1BU(s)}
x(t)= [SI-A]-1x(0)+ [SI-A]-1Bu(t) (b)
From above x(t) we can find output equation by replacing x(t) in output equation by its value from above equation(b)
For the given system below
Fig 2 Control system
TF=c(t)/r(t)
This c(t) is the output of the present system, which is not equal to above y(t), as their initial conditions are not considered.
If initial conditions are zero than both y(t) and c(t) will be equal to
L-1[SI-A]-1=φ(t)…….. state transition matrix
Φ(s)=[SI-A]-1
x(t)=φ(t)x(0)+L-1[φ(s)* BU(s)] (c)
State transition matrix satisfies the solution of state equation when input is zero. [u(t)=0]
=Ax(t)+Bu(t)
As u(t)=0
=Ax(t)
-Ax(t)=0
Solution to above equation is
y(t)=Ke-Pt+e-Pt∫ ePt Q d(t)
But 
-Ax(t)=0
Hence above equation becomes
X(t)=AeAt
Substitute t=0
x(0)=ke0
x(0)=k
x(t)=x(0)eAt (zero input response)
Properties of state transition matrix:
From equation (31) when u(t)=0
x(t)=φ(t) x(0)
and from zero input response we have
φ(t)=eAt
Property 1:
φ(0)= [I]
Property 2:
Φ-1(t)= [φ(t)]-1=e-At=eA(-t)
Φ-1(t)= Φ(-t)
Property 3:
ΦK(t)= [Φ(t)]K
ΦK(t)=[eAt]K=eA(tK)
ΦK(t)= Φ(Kt)
Property 4:
Φ(t1+t2)=eA(t1+t2)
=e(At1+At2)=eAt1 * eAt2
Φ(t1+t2)=Φ(t1)Φ(t2)
Property 5:
Φ(t2-t1) * φ(t1-t0)=eA(t2-t1) * eA(t1-t0)
=
=
=eA(t2-t0)
Φ(t2-t1) * φ(t1-t0)=Φ(t2-t0)
Key takeaway
x(t)= x (0) eAt (zero input response)
L-1[SI-A]-1=φ(t) state transition matrix
Φ(0) = [I]
Φ-1(t)= Φ(-t)
ΦK(t)= Φ(Kt)
Φ(t1+t2) =Φ(t1) Φ(t2)
Φ(t2-t1) * φ(t1-t0) =Φ(t2-t0)
Question: A=
. Find the state transition matrix?
Solution: 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=
Question: Find state transition matrix if A=
Solution: 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=
The nth order linear differential equation for given input y(t) and output u(t) is given as
y(n)+a1y(n-1)+………..+an-1
+any=bou(m)+……+bm-1
+ bm u. ….. (5)
The initial conditions are expressed as y(0),
,…..,y(n-1)(0).
The transfer function assuming all initial conditions zero is
T(s)=
=[b0sm+b1sm-1+….+bm-1s+bm/[sn+a1sn-1+……..+an-1s+an] ……(6)
Considering simple case when there are no zeros in the system, the transfer function becomes
T(s)=
=[b]/ [sn+a1sn-1+……..+an-1s+an] …….(7)
For above transfer function the differential equation is given as
y(n)+a1y(n-1)+………..+an-1
+any=bu …….(8)
Let x1=y
x2=
…….
xn=y(n-1)
From (8) the differential equation is given as
……….
=-anx1-an-1x2-…….-a1xn+bu
=Ax+Bu ……….(9)
Note: Matrix A is known as Bush form or companion form. Matrix B has all elements zero except the last element.
Key takeaway
We learned to write the state equation and its solution for LTI systems. The state equation can be written directly from the block diagram of the system but the above method helps to write the equation for any higher-order system.
Q1) Find the state equation from the given differential equation
Sol: Let y=x1
=
=x2
=
=x3
=
The above differential equation then becomes
+6
+11 x2+6 x1=u
(t) =u-6
(t)-11 x2-6 x1
Hence the state equation will be
= 
+
[u]
The state variables selected here are the physical quantities of the system, which can be measured. The selection of these variables can be directly related to the physical system because the solution of the state equation is related to the time variation of the system variables.
The number of energy storing elements in any system is equal to the number of state variables. Below shown are few electrical circuits, just to brush up on the concept of energy storing elements and state variable relation.
Fig 3 Electrical circuit with energy storing elements
Now calculating the state equation and output equation for state variable analysis.
State Equation and Output equation:
Number of output=Number of output equation
Fig 4 State Model
Considering multiple input and multiple output system, with two inputs u1(t)and u2(t), and two outputs y1(t) and y2(t) respectively.
y1(t)=c11x1(t)+c12x2(t)+d11u1(t)+d12u2(t)
y2(t)=c21x1(t)+c22x2(t)+d21u1(t)+d22u2(t)
=
+
The output equation is given as
Y(t)=CX(t)+DU(t)
Y(t)=
X(t)=
C=
D=
U(t)=
# Now finding State Equation
Number of energy-storing elements= Number of state variables
=
=a11x1(t)+a12x2(t)+b11u1(t)+b12u2(t)
=
=a21x1(t)+a22x2(t)+b21u1(t)+b22u2(t)
=
+
The state equation is then given as
=Ax(t)+Bu(t)
Key takeaway
The number of energy storing element= Number of state variables
We should always take voltage across the inductor L, and current through capacitor C.
Question: Obtain the state space representation for the given electrical system
Solution: The state model is given as
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).
The output equation is then given as
Y(t)=CX(t)+DU(t)
V0=Vc= x1(t) ….(a)
Hence output equation becomes
V0= x1(t)
y(t)=[1 0]
+[0]u(t)
so, C=[1 0] D=[0]
Now writing the state equation
=Ax(t)+Bu(t)
For that applying KVL in the above circuit
V=ILR+L
+Vc
State Equation is 
=Ax(t)+Bu(t)
=
x1(t)=Vc
=
IL=C
=
=
=
=(1/C) x2(t) ……….(b)
=
=
VL=L
=
=VL/L
From KVL
L
=VL=V-ILR-VC
=
=
=
…………….(c)
From equation (b) and (c)
= [0 1/c]
+ [0 0]u(t)
=[-1/L -R/L]
+ [1/L 0]
Now writing the state equation
=Ax(t)+Bu(t)
=
+ 
Hence A=
B= 
Question: Obtain the state space representation for the given electrical system
Solution: 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.
Decomposition is the construction of the state diagram from the transfer function between the input and the output of the system.
Types of decomposition are
Direct decomposition is applied to transfer functions that are in polynomial form.
i) Direct Decomposition to controllability canonical form
ii) Direct Decomposition to observability canonical form
Parallel Decomposition is applied to transfer function by partial fraction expansion form
i) Diagonal Canonical form
ii) Jordan Canonical form
Direct decomposition is applied to transfer functions that are in polynomial form.
i) Direct Decomposition to controllability canonical form
ii) Direct Decomposition to observability canonical form
The phase variables are defined as the state variables obtained from any one of the system variables and its derivatives. The nth order linear differential equation for given input y(t) and output u(t) is given as
y(n)+a1y(n-1)+………..+an-1
+any=bou(m)+……+bm-1
+ bm u. ….. (10)
The initial conditions are expressed as y(0),
,…..,y(n-1)(0).
The transfer function assuming all initial conditions zero is
T(s)=
=[b0sm+b1sm-1+….+bm-1s+bm/[sn+a1sn-1+……..+an-1s+an] ……(11)
Considering simple case when there are no zeros in the system, the transfer function becomes
T(s)=
=[b]/ [sn+a1sn-1+……..+an-1s+an] …….(12)
For above transfer function the differential equation is given as
y(n)+a1y(n-1)+………..+an-1
+any=bu …….(13)
Let x1=y
x2=
…….
xn=y(n-1)
From (13) the differential equation is given as
……….
=-anx1-an-1x2-…….-a1xn+bu
=Ax+Bu ……….(14)
Note: Matrix A is known as Bush form or companion form. Matrix B has all elements zero except the last element.
Fig 5 Phase Variable Form
The output equation is given as y=x1
y=Cx ……….(15)
where C =[1 0 0………..0]
The phase variable formation for the transfer function can be obtained by the below mentioned two canonical forms.
Controllable canonical form, Observable canonical form
Observable Canonical Form:
An Observable canonical form is used to analyze and design control systems because this form allows observability. A system is observable if all its states can be determined by the output. Consider a transfer function of third-order
=T(s)=(b0s3+b1s2+b2s+b3)/(s3+a1s2+a2s+a3)
T(s)=(b0+b1/s+b2s2+b3/s3)/[1-(-a1/s-a2/s2-a3/s3)] …………(16)
By Mason’s Gain Formula
T=
…………(17)
Pk
forward path transmittance of k+n path from a specified i/p node to n o/p nodes
While calculating ipnode to n o/p nodes.
it is the graphics determined which involve transmittances and multiple increases b/w non-touching loops.
= 1- [sum of all individual loop transmitting]
+[ sum of loop transmittance product of all possible non-touching loops]
-[sum of loop transmittance of all possible triples of non-touching loops]
path factor associated with concered path & involves all a in the graphic which is isolated from the forward path under consideration.
The path factor 
for kthis equal to graph determinant of SFG which effect after erasing the kth path from the graph.
The required SFG is shown below, we can conclude
y=x1+b0u
=-a1(x1+b0u)+x2+b1u
=-a1x1+x2+(b1-a1b0)u ……….(18)
=-a2x1+x3+(b2-a2b0)u ……….(19)
=-a3x1+(b3-a3b0)u …………(20)
y=[1 0 0]
+b0u ………….(21)
The resulting matrix is given below
The above equations in matrix form can be easily interpreted through the signal flow graph. From the equation (18) and (20) we can say that
a) Three feedback loops are touching each other and having to gain -a1/s, -a2/s2,and -a3/s3.
b) Four forward paths are touching the loop and having to gain b0, b1/s, b2/s2, b3/s3.
So, the signal flow graph is shown below
Fig 6 SFG for observable Form
Question: Find the observable canonical realization of the system H(s)=
Solution: 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
Fig 7 SGF for an observable canonical form
The observable canonical form can be obtained by converting the above SFG to a block diagram
Controllable Canonical Form: In this, the transfer function is divided into two parts as shown in the figure below.
Fig 8 A third-order system
Fig 9 Simplified representation of an above system
G(s)=
= 
=
………..(21)
=b0s3+b1s2+b2s+b3
The equation relating phase variable to input is given as
The equation relating output to the phase variable can be given as(refer eq.(21))
y=b0
+b1
+b1
+b3x1
=b0(-a3x1-a2x2-a1x3+u)+b1x3+b2x2+b3x1
=(b3-a3b0)x1+(b2-a2b0)x2+(b1-a1b0)x3+b0u
The above equation in vector matrix form can be written as
y=[((b3-a3b0) (b2-a2b0) (b1-a1b0)]
+b0u ……..(22)
The signal flow graph for the above vector-matrix equations is shown below
Fig 10 SFG for controllable form
Key takeaway
i) Matrix A is known as Bush form or companion form. Matrix B has all elements zero except the last element.
ii) A system is observable if all its states can be determined by the output
Question: Find the controllable canonical realization of the following systems
a) H(s)=
b) H(s)=
Solution: 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
Parallel Decomposition is applied to transfer function by partial fraction expansion form
i) Diagonal Canonical form: refer section 2.9
ii) Jordan Canonical form
Jordan’s Canonical form
This is a reduction technique used to find the state model with the help of the transfer function. The below given example makes the form clearer.
Q1) The closed-loop transfer function is given as T(s)=
. Calculate the state model?
Sol: The transfer function can be simplified using the partial fraction method as
=
S+2=As+10A+Bs+5B
Equating coefficients of s from both sides
A+B=1
Equating coefficients of s0 from both sides
10A+5B=2
Solving the above equations we get
A=-3/5
B=8/5
The transfer function will be
T(s)=
=-
+
Number of poles= Number of energy storing elements
The output equation is given as
y(t)=k1x1(t)+k2x2(t)
y(t)= [k1 k2]
+ [0] [u(t)]
y(t)= [-3/5 8/5]
+ [0] [u(t)]
C=[-3/5 8/5]
D=0
From the above block diagram
=u(t)-5 x1
=-5x1+u(t)
=-10x2+u
Therefore, the state equation is given as
=
+
A=
B=
Q2) The closed-loop transfer function is given as T(s)=
. Calculate the state model?
Sol: The transfer function can be simplified using the partial fraction method as
=
+ 
s+7=A[s2+7s+12] +B[s2+6s+8] +C[s2+5s+6]
Equating coefficients of s2 from both sides
A+B+C=0
Equating coefficients of s from both sides
7A+6B+5C=1
Equating coefficients of s0 from both sides
12A+8B+6C=7
Solving the above equations and finding values of A, B, and C
A=5/2
B=-4
C=3/2
The transfer function will be
T(s)=
+ 
=
+ 
The output equation will be
y=k1x1+k2x2+k3x3
y=[k1 k2 k3]
[5/2 -4 3/2]
C=[5/2 -4 3/2]
D=[0]
The state equation is given as
=u(t)-2x1
=-2x1+u(t)
=-3x2+u
= -4x3+u
Therefore
=
+
[u]
A=
B=
Key takeaway
Inferences from the above matrices (Jordan’s canonical form)
a) Matrix [A] is a diagonal matrix and the diagonals are closed-loop poles.
b) Matrix [B] is always of the form 
c) Matrix [C] contains residue of poles
d) Matrix [D] is always zero When the number of poles ≠ number of zeros
e) When many poles = number of zeros only change comes in the matrix [D]. It is the ratio of the coefficient of the highest power of s in the closed-loop transfer function.
Q3) The closed-loop transfer function is T(s)=
. Find the state equation?
Sol: Here the number of poles = number of zeros
T(s)=
=
=
=2+ 
=2+
Solving above by partial fraction method
88= As+2A+Bs+3B
A=-B
2A+3B=88
A=-88
B=88
The transfer function becomes
T(s)= 2-
The output equation will be
y(t)=2U+k1x1+k2x2
y=[-88 88]
+[2]u
The output equation will be
=u(t)-3 x1
=-3x1+u(t)
=-2x2+u(t)
Therefore, the state equation is given as
=
+
A=
B=
Controllability: A control system is said to be completely controllable if it’s possible to transfer the state of the system from the initial state to any other required state.
[S]= [B AB AB2 AB3 ------------------- An-1B]
Kalman’s Test: A system is said to be completely controllable if the determinant of a matrix is not equal to zero, or the rank of the following matrix should be to ‘n’.
n---- order of the matrix.
If 
≠0 then completely controllable
If 
=0 then not controllable
Question: A system is represented by the following state model A= 
B=
find whether it is controllable or not?
Solution: A= 
B=
Sothe order of the matrix is n=2
[S]=[B AB]
= 
=0 hence, the system is not controllable.
Observability: A system is said to be completely observable if it is possible to determine the initial state of the system by observing the output for a finite interval of time.
Calman’s Test: A system is said to be completely observable within the determinant of the following matrix is not equal to zero, or, the matrix rank should be equal to 1.
[Q]= [CT ATCT (AT)2CT ----------(AT)n-1CT]
If
≠0 then completely observable
=0 then not observable
Key takeaway
If
≠0 then completely observable
=0 then not observable
If 
≠0 then completely controllable
If 
=0 then not controllable
Question: 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.
Eigenvalues
Let the transfer function be
T(s)=
The characteristic equation is
=0
But we already know
sX(s)-X(0)=AX(s)+BU(s)
(sI-A)X(s)=X(0)+BU(s)
X(s)= (sI-A)-1X(0)+ (sI-A)-1BU(s)
Y(s)=C(sI-A)-1X(0)+C(sI-A)-1BU(s)+DU(s)
For zero initial conditions
Y(s)= C(sI-A)-1BU(s)+DU(s)
= C{[SI-A]-1B} + D
[SI-A]-1=
The denominator of the above equation is the characteristic equation
[SI-A]=0
As the eigenvalues of matrix A be 
1,
2,
3 ……then
|
iI-A|=0
Eigenvectors
Any non-zero vector Xi satisfies
|
iI-A| Xi =0
Xi=|
iI-A|-1. 0
=
. 0
But Xi is non-zero so, a solution exists if |
iI-A|=0 from which eigenvalues can be determined.
Let Xi=mi. The solution mi is called the eigenvector of A associated with Eigen valued 
i. Let m1,m2,m3…. Be the eigenvectors corresponding to the eigenvalues 
1,
2,
3…..respectively. Then the matrix formed by placing the eigenvectors together is known as model matrix P of A. This matrix is also known as the diagonalizing matrix.
=P-1AP
When A=
The model matrix is given as V= 
The above matrix is called a Vander Monde Matrix.
Key takeaway
Eigenvalues can be determined if |
iI-A|=0
The eigenvectors corresponding to the eigenvalues 
1,
2,
3….. are calculated.s
Q1) Find the eigenvectors 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=
Q2) 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
=
=
=
Diagonalization
A diagonal system matrix has the advantage that each state equation is the function of only one state variable. Hence, each state equation can be solved simultaneously. The technique of transforming a non-diagonal matrix into a diagonal matrix is known as diagonalization.
The state equation is given as
Y(t)=Cx(t)+Du(t)
Let the eigenvalues of matrix A be 
1,
2,
3 …… Then Let us transform X into a new state vector Z
X=PZ
P is n x m non-singular matrix
P
=APZ+BU
Y=CPZ+DU
=
Z+
U
Y=
Z+
U
=P-1AP
= P-1B
=CP
Diagonalization of system matrices with distinct and repeated eigenvalues
We can always diagonalize a matrix with distinct eigenvalues (whether these are real or complex). We can sometimes diagonalize a matrix with repeated eigenvalues. (The condition for this to be possible is that any eigenvalue of multiplicity m had to have associated with m linearly independent eigenvectors.) The situation with symmetric matrices is simpler. We can diagonalize any symmetric matrix. To take the discussions further we first need the concept of an orthogonal matrix. A square matrix A is said to be orthogonal if its inverse (if it exists) is equal to its transpose: A −1 = AT or, equivalently, AAT = ATA = I.
Let us understand using a few examples.
EX.1 Consider a matrix A=
The characteristic equation is given as |
I-A|=0
(1-
)(1-
)=0
1=
2=1, which means it has only one eigenvalue i.e 1 which is repeated. But has a multiplicity of 2.
Now, finding eigenvectors of A corresponding to 
=1
- 
x Xi=0
-
x 
= 0
x1=x2
-4x1+x2=x2
The solution is x1=0 and x2 is any arbitrary value. The possible eigenvectors can be 
, 
……….
Now, to form the eigenvector matrix P considering any two values of eigenvectors P=
But P has zero value of the determinant. Thus, P-1 does not exist, and finding the transformation matrix 
=P-1AP is also not possible. So, the diagonalization of the above matrix is not possible.
EX.2 Consider A=
The above matrix has non-repeated eigenvalues of 
1=-1, 
2=5 (distinct)
The eigenvectors corresponding to these values are X1=
X2=
Arrange eigenvectors in a matrix P = 
. The |P|≠0. Hence,P-1 exist, and finding the transformation matrix 
=P-1AP is also possible. So, diagonalization of the above matrix is possible.
Key takeaway
Reference:
1. I. J. Nagrath& M. Gopal, “Control System Engineering”, New Age International Publishers
2. A. Ambikapathy, Control Systems, Khanna Publishing House, Delhi.
2. Joseph J. Distefano III, Allen R. Stubberud, Ivan J. Williams, “Control Systems” Schaums Outlines Series, 3rdEdition, Tata McGraw Hill, Special Indian Edition 2010.
3. William A. Wolovich, “Automatic Control Systems”, Oxford University Press, 2010.
