“State Space Form”

PowerPoint PresentationObjectives
Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
Engineering Technology, UTHM. @ Dr. HIKMA Shabani.
Upon completing this topic, students should be able to:
– find a mathematical model, called a state-space
representation for a linear, time-invariant (LTI) system
– model electrical and mechanical systems in state space
– convert a transfer function to state space
– convert a state-space representation to a transfer function
– linearize a state-space representation
Contents
1. Introduction
Function
6. Linearization
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CHAP_6. 1: Introduction
14/10/2021 Control System for Industrial Automation, Dept of Electrical Eng, Faculty of
5
So far, we did learn in class, the Classical Control Theory that is based on:
a simple input–output description of the plant usually expressed as a Transfer Function (TF);
it is a Frequency-Domain Technique. - Limitations:
Applicability: Limited to only LTI Systems limited to single input- single output (SISO) systems allows limited control of the closed-loop behavior
when feedback control is used.
Modern Control Theory solves many of the limitations of classical theory by using a much “richer” description of the plant dynamics;
The so-called “State-Space Approach.” It is a Time-Domain Technique. Applicability: to both LTI and Non-LTI Systems
• A general form of a State-Space System for Linear, Time Invariant (LTI) System with input and output is
presented as:
1 1
2 2

State: - It is the smallest set of variables, , , , , called:
State Variables or State Vectors.
- The number of state variables to completely define the dynamics of the system is equal to the number of integrators involved in the system (System Order)
A. Definition of State-Space System
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Inner State Variables/Vectors:
, , ,
• System Variable: - Any variable that responds to an input or initial conditions in a system
• State Variables: - The smallest set of linearly independent system variables such that the values of
the members of the set at time 0 along with known forcing functions (applied inputs) completely determine the value (behavior) of the set (System variables) at time ≥ 0.
• State Vector: - A vector whose elements are the state variables
• State-Space: - The − space whose axes are the state vectors/variables
• State-Space Equations: - A set of − differential equations with variables, where the
variables to be solved are the state variables
• Output Equation: - The algebraic equation that expresses the output variables of a system as linear
combination of the state variables and the inputs.
B. Terminologies
14/10/2021 7 • Linear Independence: A set of variables is linearly independent if none of the variables
can be written as a linear combination of the others.
CHAP_6. 2: State-Space Representation
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• A LTI System is represented in State-Space Model (Equations) by Two (2) vector-matrix differential equation (DE) as:
- State-Space Equation: = +
- Output Equation: = +
With: ≥ 0 and initial conditions 0
Where:
: Input Vector
: Output Vector
: Input Matrix
: Output Matrix
: Feedforward Matrix
6.2. State-Space Representation
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Dynamic Equations
Measurement Equations
• The Vectors and Matrices of State-Space Model (Equations) can be
presented in the following Formats:
=
1
2
0 ×1
= × , = × , = × , = ×
Vectors and Matrices of State-Space Model
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State Vectors Input Vectors Output Vectors State Vectors at initial condition
State/System Matrix Input Matrix Output Matrix Feedback Matrix
Time-Domain/State-Space Models: consider the internal behavior of a system can easily incorporate complicated output variables can represent multi-input/multi-output (MIMO) systems and
nonlinear, Time-Varying, Multivariable systems is a unified method for:
- modeling, analyzing, and designing a wide range of systems.
Classical Input/output or Frequency-Domain Models: are conceptually simple based on converting a system's differential equation to a
Transfer Function (frequency domain) that are more intuitive to practicing engineers
can be applied only to LTI systems or systems that can be approximated as such.
Time-Domain Vs. Frequency-Domain Models
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CHAP_6. 3: State-Space Model
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• Two (2) Steps to derive State-Space Equations (Model)
1.Select a particular subset of all possible system variables, and
call state variables
equations in terms of the state variables (state equations).
Note:
- If we know the initial condition of all of the state variables at 0
as well as the system input for ≥ 0, we can solve the equations
- The state-space model can be obtained from any one of the
following two mathematical models:
b) the Transfer Function Model.
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6.3. State-Space Model
• Consider the following -order system of Linear Differential
Equations (DE):

+ + −1 + =
• where:
- is the system output
- is the System input
The system is -order, then it has -integrators (State Variables) t
A.State-Space Model of Differential Equation
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=

1 = 2 2 = 3
−1 = = −01 − − −1 +
Or:
Where:
−0
0
−1
2) The Output Equation is: = +
- If, the output equation is: = 1 (see slide 14)
Then, = 1 0 0 and = 0
Therefore, the State-Space Equation is: = 1 0 0
1
2

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Ex_1. Let consider the mechanical system:
Solution
• The system linear differential equation is: + + =
⇒ = 1
• This system is of 2nd -Order:
This means 2 simultaneous 1st-Order differential equations are needed to solve for Two (2) State Variables.
To solve: - Firstly, let us define the state variables as:
1 = (2) and 2 = 1 = (3)
- Secondly, substitute the state variables into the differential equation (1):
= 2 = 1
−
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-Thirdly, we write the 2 resulting simultaneous 1st-Order differential equations:
• Combine the Eq. (3) and (4) to obtain:

2 = = 1
−
1) The State-Space Equation: = +
* From Eq. (2) and (5), the Vectors and Matrices are obtained as:
=
2
=

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2) Output equation: = +
* From the Eq. (2): = 1 , the Matrices are obtained as:
= 1 0 and = 0
Therefore, the Output Equation is: = 1 0 1
2
Ex_1: Solution (Cont’d)
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- Let consider the following R-L-C network:
Solution:
- In this system, the variables are:
, , , and
a) One of ‘State Variable’ choices:
* State Variables: and
⇒ +
+
1
Ex_2
- The state-variables must be linearly independent
Or: =
(2)
⇔ 2
⇔ 2
2 = − 1
2 = 1 =
(5)
2 (6)
2 = − 1
+ +
− −
Combine the Eq. (5) and (7) for the following 2 resulting simultaneous 1st-Order
differential equations:
2 = 2
1) The State-Space Equation: = +
* From Eq. (4) and (8), the Vectors and Matrices are obtained as:
=

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Ex_2: Solution (Cont’d)
= +
- For the output , the output equation is derived from Eq.(3) as:
= −
⇔ = −2 − 1
1 − 2 + (8)
Therefore, the Matrices are: = − Τ1 − and = 1
Hence, the Output Equation is: = − Τ1 −

+
b) Other choice of ‘State Variable’:
* State Variables: and
⇒
⇒
Or:
(2)
2
⇔ 2
2 = − 1
2 = 1 =
(5)
2 (6)
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2 = − 1
= +
- From Eq. (4), (5), (6) and (7), the
vectors and matrices are obtained as:
=
Equation can be obtained.
2) Output Equation: = +
- For the output , the equation is: -
y =
Hence, the Matrices are: = 1 0 and = 0
Therefore, the Output Equation is: = 1 0

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1) Find the State-Space Model representation and block diagram for a
system that is described by the following differential equation with the input signal and the output signal.
+ 9 + 26 + 24 = 24
2) Considering the voltage across capacitor as the output, find the
State-Space Representation of the following series of the RLC circuit.
Use and as the State Variables.
Try…
P.S: Compare your Solution with
those from Ex_2(a) and Ex_2(b)
Two (2) types of Transfer Function to consider:
1) Transfer Function having constant term in Numerator.
2) Transfer Function having polynomial function of ‘’ in Numerator.
Consider the following Transfer Function of a system;
=
Rearrange, the above equation as:
+ −1−1 + + 1 + 0 = 0
Applying the Inverse Laplace Transform on both sides, yields to:

−1 + + 1
+ 0 = 0 (1)
B. State-Space of Transfer Function Systems
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B.1. Transfer Function with constant term in Numerator
P.S: This Eq.1 becomes the -order system of Linear Differential Equation (Slide 14)
Find the state-space model for the system having the following transfer
function:
⇔ 2 + + 1 = (1)
Applying the Laplace Transform on both sides of Eq.1, yields to:
2
• Let:
2 = 1 =
(4)
2 (5)
Example
2 = −1 − 2 + (7)
• Combine the Eq. (4) and (7) for simultaneous
1st-Order differential equations as:
1 =
(8)
Hence:
= +
- The vectors and matrices are:
=
0
1
Therefore, the State-Space Equation (Model) is:
=
- For the output , the equation is: -
y = 1 from Eq. 3 (slide 26)
Hence, the Matrices are: = 1 0 and = 0
Therefore, the Output Equation is: = 1 0
1
2
Example: Solution (Cont’d)
Find the state-space model for the system having the following
transfer function:
−24 −26 −9
1
2
3
Try…
Consider the following Transfer Function of a system;

=
⇔
=
1
+−1−1++1+0 + −1−1 + 1 + 0 (1)
The above Eq. (1) is in form of product of Transfer Functions of two (2) cascaded
blocks;
Hence:
1
+−1−1++1+0 + −1−1 + 1 + 0 (2)
From the Eq. (2), two (2) cascaded Transfer Functions are defined as:
1)
2)
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B.2. TF with polynomial function of ‘’ in Numerator
Let solve the Eq. (3) and (4):
a) Solve:
- So, let rearrange the above equation to get:
+ −1−1 + + 1 + 0 =
- Apply the inverse Laplace Transform on both sides of the above equation to obtain:

−1 + + 1
+ 0 = (1)
Hence, solving this -order system of linear differential equation (as in Slide 14),
yields to:
= −01 − 12 − −1 + (I)
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b) Solve:
= + −1−1 + 1 + 0
=
+ −1 −1
−1 + + 1
+ 0
Again, solving this -order system of linear differential equation (as in Slide 14),
yields to:
y = + −1 + + 12 + 01 (II)
Next, substituting the Eq.(I) into Eq.(II), yields to:
y = −01 − 12 − −1 + + −1 + + 12 + 01
⇔ y = 0 − 0 1 + 1 − 1 2 + + −1 − −1 +
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(III)
Finally:
– From the Eq.(I): = −01 − 12 − −1 + , it is
obtained the Vectors and Matrices as:
=
1
2
−0
0
−1
0 0 0 1
– From the Eq. (III): y = 0 − 0 1 + 1 − 1 2 + + −1 − −1 + , it is obtained the Matrices as:
= 0 − 0 1 − 1 −1 − −1 and =
0
0
0

Note: if = 0, then: = 0 1 −1 and = 0
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transfer function:
From Eq. (2), the two (2) cascaded Transfer Functions are:
1)
2)
a) Solve:
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Example
a) Solve:
⇒ 3 + 92 + 26 + 24 =
3
⇔ 3
- Or , let:
3 = −93 − 262 − 241 + (I)
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- From Eq.(2) and (I), it is obtained:
=
−24 −26 −9
+ 0 0
24
b) Solve:
⇒ = 2 + 7 + 2
= 2
2 + 7
+ 2 (1)
= 3 + 72 + 21 (II)
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- From Eq.(II), it is obtained:
= 2 7 1
1
2
3
transfer function:
Try…
CHAP_6. 4: State-Space Analysis
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•Here, it is discussed:
2.The Controllability and Observability of Control Systems.
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6.4. State-Space Analysis
Let consider the following State-Space Model of a Linear Time-
Invariant (LTI) System:
- State-Space Equation: = + (1)
- Output Equation: = + (2)
Applying the Inverse Laplace Transform on both sides of the Eq.(1), yields
to:
⇒ = − −1 (3)
Applying the Inverse Laplace Transform on both sides of the Eq.(2), yields
to: = + (4)
Substituting the Eq.(3) into Eq.(4), yields to:
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A.Transfer Function from State-Space Model
= − −1 +
⇔ = − −1 +
⇒
= − −1
Example: - Let calculate the transfer function of the system represented in the state space
model as:
=
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- The matrices are derived from the state space model as:
= −1 −1
- Then, when = 0,
= − −1 (1)
- Substituting the values of the above matrices into Eq. (1), yields to:

1
0 =
Thus, the Transfer Function of the system is:
=
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- Find the transfer function of the system represented in the state space
model as:
−24 −26 −9
1
2
3
−1 −2 −3 +
= 1 0 0 Ans.: 10 2+3+2
3+32+2+1
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Controllability:
A control system is said to be controllable:
- if the initial states of the control system are transferred (changed) to
some other desired states by a controlled input in finite duration of
time.
The Kalman’s test is used to check the controllability of a control
system: Write the matrix in the following form:
= 2 −1
Find the determinant of matrix ; • if it is not equal to zero, then the Control System is controllable.
B.Controllability and Observability
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Observability:
A control system is said to be observable:
- if it is able to determine the initial states of the control system by
observing the output in finite duration of time.
The Kalman’s test is used to check the observability of a control
system: Write the matrix in the following form:
= 2 −1
Find the determinant of matrix ; • if it is not equal to zero, then the Control System is observable.
B.Controllability and Observability
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- Let us verify the controllability and observability of a control system which is
represented in the state space model as:
= 1
, = 1 0
1) For Controllability check.
For = 2, the matrix will be: = and = −1 −1 1 0
1 0
= −1 1
= 1 ≠ 0
Since the determinant of matrix is not equal to zero, the given control system is
controllable.
Example
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2) For Observability check.
and = −1 1 −1 0
0 1
= 1 0
= −1 ≠ 0
Since the determinant of matrix o is not equal to zero, the given control system is
observable.
Therefore, the given control system is both controllable and observable.
Solution (Cont’d)
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- The
Try…
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CHAP_6. 5: Linearization
10/14/2021 49Control System for Industrial Automation, Dept of Electrical Eng, Faculty of

“State Space Form”

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