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MODULE-I

State space analysis
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The conventional and old method for the design and analysis of control systems is the
transfer function method
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Advantages of state variable analysis
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▪ It can be applied to tile invariant systems
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▪ Its gives idea about the internal state of the system
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State vector: If n state variables are needed to completely describe the behaviour of a given system,
then these n state variables can be considered the n components of a vector X
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State space: The n-dimensional space whose co-ordinate axes consists of the x1 axis, x2 axis,
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xn are state variables: is called a state space
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𝑢𝑟 (𝑡)
m no of outputs 𝑦1 𝑡 , 𝑦2 𝑡 ,
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𝑥𝑛 (𝑡)
Then the state model is given by state & output equation
X t = AX t + BU t
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output equation
A is state matrix of size (n×n)
B is the input matrix of size (n×r)
C is the output matrix of size (m×n)

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D is the direct transmission matrix of size (m×r)
X(t) is the state vector of size (n×1)
Y(t) is the output vector of size (m×1)
U(t) is the input vector of size (r×1)

(Block diagram of the linear, continuous time control system represented in state space)
𝐗 𝐭 = 𝐀𝐗 𝐭 + 𝐁𝐮 𝐭
𝐘 𝐭 = 𝐂𝐗 𝐭 + 𝐃𝐮 𝐭

STATE SPACE REPRESENTATION OF NTH ORDER SYSTEMS OF LINEAR
DIFFERENTIAL EQUATION IN WHICH FORCING FUNCTION DOES NOT INVOLVE
DERIVATIVE TERM
Consider following nth order LTI system relating the output y(t) to the input u(t)
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Often the variables used is
the system output & the remaining state variables are then derivatives of the output
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0
0
⋮
1
−𝑎1

𝑛×𝑛

Output equation can be written as
Y t = CX t
𝐶= 1

0……
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The input is fa and the output is y
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6

Freebody Diagram

Equation

There are three energy storage elements, so we expect three state equations
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Therefore we choose
as our state variables x (the energy in spring k2 is ½k2x²), the velocity at x (the energy in
the mass m is ½mv², where v is the first derivative of x), and y (the energy in spring k1 is
½k1(y-x)² , so we could pick y-x as a state variable, but we'll just use y (since x is already a
state variable; recall that the choice of state variables is not unique)
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The equations of motion from the free body
diagrams yield

or

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with the input u=fa
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The input is ia and the output is e2
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Try
choosing i1, i2 and e1 as state variables
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The
voltage across the inductor L2 is e1 (which is one of our state variables)

so our first state variable equation is

If we sum currents into the node labeled n1 we get

This equation has our input (ia) and two state variable (iL2 and iL1) and the current
through the capacitor
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We start by solving the state equation for Q(s)

The matrix Φ(s) is called the state transition matrix
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e
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Example: State Space to Transfer Function
Find the transfer function of the system with state space representation

First find (sI-A) and the Φ=(sI-A)-1 (note: this calculation is not obvious
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Rules for inverting a 3x3 matrix are here
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>> % First define state space system
>> A=[0 1 0; 0 0 1; -3 -4 -2];
>> B=[0; 0; 1];
>> C=[5 1 0];
>> [n,d]=ss2tf(A,B,C,D)

n =
10

0

0

1
...
0000
3
...
0000

4
...
0000
d =

>> mySys_tf=tf(n,d)
Transfer function:
s + 5

s^3 + 2 s^2 + 4 s + 3

Transfer Function to State Space
Recall that state space models of systems are not unique; a system has many state space
representations
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Before we look at procedures for converting from a transfer function to a state space
model of a system, let's first examine going from a differential equation to state space
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First we start with an example demonstrating a simple way of converting from a single
differential equation to state space, followed by a conversion from transfer function to state
space
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We'll use
a third order equation, thought it generalizes to nth order in the obvious way
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We'll use a third order equation, thought it generalizes to
nth order in the obvious way
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Note: For an nth order system the matrices generalize in the obvious way (A has ones
above the main diagonal and the coefficients of the denominator polynomial for the last
row, B is all zeros with b0 (the numerator coefficient) in the bottom row, C is zero except
for the leftmost element which is one, and D is zero)
If we try this method on a slightly more complicated system, we find that it
initially fails (though we can succeed with a little cleverness)
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We can try the same method as before:

The method has failed because there is a derivative of the input on the right hand, and
that is not allowed in a state space model
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Now when we take the derivatives we get:

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The second and third equations are not correct, because ÿ is not one of the state
variables
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The process described in the previous example can be generalized to systems with
higher order input derivatives but unfortunately gets increasingly difficult as the order
of the derivative increases
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Clearly more straightforward techniques are necessary
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Controllable Canonical Form (CCF)
Probably the most straightforward method for converting from the transfer
function of a system to a state space model is to generate a model in "controllable
canonical form
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To see how this method of generating a state space model works,
consider the third order differential transfer function:

We start by multiplying by Z(s)/Z(s) and then solving for Y(s) and U(s) in terms of
Z(s)
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We can now choose z and its first two derivatives as our state variables

Now we just need to form the output

From these results we can easily form the state space model:

In this case, the order of the numerator of the transfer function was less than that of
the denominator
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A result
that works in all cases is given below; the details are here
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"
This term comes from Control Theory but its exact meaning is not important to us
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Now let's examine q2 and its derivative:

Again we note that y=q1 and we call the integral terms q3:

so

This is our second state variable equation
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Our state space model now becomes:

In this case, the order of the numerator of the transfer function was less than that of
the denominator
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A result
that works in all cases is given below; the details are here
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=

MATLab Code
Transfer Function to State Space(tf2ss)
Y(s)
s
= 3
2
U(s) s + 14s + 56s + 160
num=[1 0];
den=[1 14 56 160];
[A,B,C,D]=tf2ss(num,den)

A=

-14 -56 -160
1

0

0

0

1

0

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B=

1
0
0
C=
0

1

0

D=
0

Concept of Eigen Values and Eigen Vectors
The roots of characteristic equation that we have described above are known as eigen values
of matrix A
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Sum of eigen values of any matrix A is equal to the trace of the matrix A
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If we multiply a scalar quantity to matrix A then the eigen values are also get multiplied by
the same value of scalar
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If we inverse the given matrix A then its eigen values are also get inverses
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If all the elements of the matrix are real then the eigen values corresponding to that matrix are
1
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3
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either real or exists in complex conjugate pair
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Where𝜆𝑖, i = 1, 2, 3,
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This eigen vector may be obtained by taking cofactors of matrix 𝜆𝑖𝐼 − 𝐴 along any row &
transposing that row of cofactors
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𝑚𝑛 be the eigenvectors corresponding to the eigen value 𝜆1, 𝜆2 ,
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Then 𝑀 = 𝑚1 ⋮ 𝑚2 ⋮ ⋯ ⋮ 𝑚𝑛 is called diagonalizing or modal matrix of A
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Under this assumption original state model modifies to
V t = 𝐴V t + BU t
Y t = C V t + DU t
Where 𝑨 = 𝑴−𝟏𝑨𝑴 = 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑚𝑎𝑡𝑟𝑖𝑥,

𝑩 = 𝑴−𝟏𝑩 ,

𝑪 = 𝑪𝑴

The above transformed state model is in canonical state model
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If the system matrix A is in companion form & if all its n eigen values are distinct, then
modal matrix will be special matrix called the Vander Monde matrix
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3

1
𝜆𝑛
⋮
𝑛−2
𝜆𝑛
𝜆𝑛−1
𝑛

𝑛×𝑛

State Transition Matrix and Zero State Response
We are here interested in deriving the expressions for the state transition matrix and zero state
response
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L-1
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Now let us discuss some of the properties of the state transition matrix
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If we substitute t = 0 in the above equation then we will get 1
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2
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Mathematically we can write
θ(-t) = [θ(t)]-1
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We also another important property [θ(t)]n = θ(nt)
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This theorem provides a simple procedure for evaluating the functions of a matrix
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Its characteristic equation is given by
𝑞 𝜆 = 𝜆𝐼 − 𝐴 = 𝜆𝑛 + 𝑎1𝜆𝑛−1 + 𝑎2𝜆𝑛−2 + ⋯ + 𝑎𝑛−1𝜆 + 𝑎𝑛 = 0
If 𝑓 𝐴 is divided by the characteristic polynomial 𝑞 𝜆 , then
𝑓 𝜆
𝑅 𝜆
=𝑄 𝜆 +
𝑞 𝜆
𝑞 𝜆
𝑓 𝜆 = 𝑄 𝜆 𝑞 𝜆 +𝑅 𝜆
… … … … …
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𝜆𝑛 , then 𝑞 𝜆 = 0 and we have from
equation (1) 𝑓 𝜆 𝑖 = 𝑅 𝜆 𝑖 ; 𝑖 = 1,2, …
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𝑎𝑛−1 , can
be obtained by successfully substituting
𝜆1, 𝜆2 , … …
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Substituting A for the variable 𝜆 in equation (1), we get
𝑓 𝐴 =𝑄 𝐴 𝑞 𝐴 +𝑅 𝐴
As 𝑞 𝐴 𝑖𝑠 𝑧𝑒𝑟𝑜, 𝑠𝑜 𝑓 𝐴 = 𝑅 𝐴
⇒ 𝑓 𝐴 = 𝑎0𝐼 + 𝑎1𝐴 + 𝑎2𝐴2 + ⋯ + 𝑎𝑛−1𝐴𝑛−1
𝑤𝑕𝑖𝑐𝑕 𝑖𝑠 𝑡𝑕𝑒 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑟𝑒𝑠𝑢𝑙𝑡
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Kalman‟s test
Consider nth order multi input LTI system with m dimensional control vector
X t = AX t + BU t
is completely controllable if & only if the rank of the composite matrix Qc is n
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Kalman‟s test
Consider nth order multi input LTI system with m dimensional output vector
X t = AX t + BU t
Y t = CX t + DU t
is completely observable if & only if the rank of the observability matrix Qo is n
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➢ The Pair (AB) is controllable implies that the pair (ATBT) is observable
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Design of Control System in State Space
Pole placement at State Space
Assumptions:
• The system is completely state controllable
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ƒ
• Control input is unconstrained
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𝜇𝑛 which are their „desired locations‟
Necessary and sufficient condition: The system is completely state controllable
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𝑆 − 𝜇𝑛
Pole Placement Design Steps:Method 1 (low order systems, n ≤ 3):
• Check controllability
• Define 𝐾 = 𝑘1 𝑘2 𝑘3
• Substitute this gain in the desired characteristic polynomial equation
𝑆𝐼 − 𝐴 + 𝐵𝐾 = 𝑆 − 𝜇1 𝑆 − 𝜇2 …
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MATLab code place works for single- or multi-input system
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0000

55
...
0000

State Estimators or Observers
• One should note that although state feed back control is very attractive because of precise
computation of the gain matrix K, implementation of a state feedback controller is possible
only when all state variables are directly measurable with help of some kind of sensors
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• Only a subset of state variables or their combinations may be available for measurements
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• Hence the need for an estimator or observer is obvious which estimates all state variables
while observing input and output
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Reduced Order Observer : An observer that estimates fewer than ``n'' states of the
system is called reduced order observer
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Observer Block Diagram

Design of an Observer
The governing equation for a dynamic system (Plant) in statespace representation may be
written as:
X t = AX t + BU t
Y t = CX t
The governing equation for the Observer based on the block diagram is shown below
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𝑋 = 𝐴𝑋 + 𝐵𝑈 + 𝐾𝑒 (𝑌 − 𝑌)
𝑌 =𝐶𝑋
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Define the error in estimation of state vector as
𝑒 = (𝑋 − 𝑋)
The error dynamics could be derived now from the observer governing equation and state
space equations for the system as:
𝑒 = (𝐴 − 𝐾𝑒 𝐶)𝑒
𝑌 − 𝑌 = 𝐶𝑒
The corresponding characteristic equation may be written as:
|𝑆𝐼 − (𝐴 − 𝐾𝑒 𝐶) | = 0
You need to design the observer gains such that the desired error dynamics is obtained
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𝑆 − 𝜇𝑛
Solve for 𝑘1, 𝑘2 , 𝑘3by equating the like powers of S on both sides

Here 𝜇1, 𝜇2 , … …
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MODEL QUESTIONS
Module-1
Short Questions each carrying Two marks
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The System matrix of a continuous time system, described in the state variable form
is

x 0
A= 0 y
0 1

0
−1
−2

Determine the range of x & y so that the system is stable
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For a single input system
X = AX + BU
Y = CX
0
1
0
A=
; B=
; C= 1 1
−1 −2
1
Check the controllability & observability of the system
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Given the homogeneous state space equation X =
X;
−1 −2
Determine the steady state value 𝑋𝑠𝑠 = lim𝑡→∞ 𝑋(𝑡) given the initial state value
10
X(0) =

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State Kalman‟s test for observability
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5
...

[10]
6
...

[3
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Investigate the controllability and observability of the following system:
1 0
0
𝑋 =
𝑋+
𝑢; 𝑌 = 0 1 𝑋
[8]
0 2
1
8
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(b) state transition matrix
(c) MIMO systems
(d) hydraulic servomotor
(e) Principle of duality due to kalman
9
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Represent the system in
state space:
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𝑑3𝑋
𝑑𝑡3

𝑑2𝑋
+3

𝑑𝑡2

𝑑𝑋
+4

𝑑𝑡

+ 4𝑋 = 𝑢1 𝑡 + 3𝑢2 𝑡 + 4𝑢3(𝑡)

and outputs are
𝑦 =4

𝑑𝑋

1

𝑦 =
2

2
𝑑 𝑋

𝑑𝑡

+ 3𝑢
1

+ 4𝑢 + 𝑢
2

𝑑𝑡2

[8]

3

10
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(a) Obtain a state space representation of the system
C(s)
U(s)

=

10(s+2)
s3 +3s2+5s+15

[14]
[7]

(b) A linear system is represented by
−6 4
1
1 0
𝑋 =
𝑋+
𝑈;
𝑌=
𝑋
−2 0
1
1 1
(i) Find the complete solution for Y(t) when U(t)=1(t), 𝑋1 0 = 1, 𝑋2 0 = 0
(ii) Draw a block diagram representing the system
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Discuss the state controllability of the system
𝑋1
−3 1 𝑋1
1
+
u
=
−2 1
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13
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[3+4]

[5+4]

14
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15
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(a) Derive an expression for the transfer function of a pump controlled hydraulic system
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[8]
(b) Simulate a pneumatic PID controller and obtain its linearized transfer function
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Describe the constructional features of a rate gyro, explain its principle of operation and
obtain its transfer function
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(a) Explain how poles of a closed loop control system can be placed at the desired points
on the s plane
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[4]
19
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[7]
Y=

20
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What are the
advantages of state space analysis?
(b) A two input two output linear dynamic system is governed by
0
1
2 1
𝑋 (𝑡)=
X(t)+
R(t)
−2 −3
0 1
1 0
Y(t)=
X(t)
1 1
i)
Find out the transfer function matrix
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(a) A system is described by
−4 1
0
𝑋 (t)= 0 −3 1 X(t)
0
0 −2
Diagonalise the above system making use of suitable transformation X=PZ
(b) Show how can you compute 𝑒𝐴𝑡 using results of (a)
22
...

23
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[5×3]

24
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Find the matrix k=[𝑘1 𝑘2 ]
which is called the state feedback gain matrix for the closed loop poles to be located at -1
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j2
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6

0

X+

0

𝑈

[6]

1

25
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(b) Prove non uniqueness of state space model
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(a) Show the following system is always controllable
0
1
0
0
0
1 𝑥+ 0 𝑢
𝑥 𝑡 = 0
−𝑎3 −𝑎2 −𝑎 1
1
(b) Explain the design of state observer
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[12]
[4]

[4+4+4]



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