Modern Control Systems (MCS)
Lecture-30-31
Design of Control Systems in Sate Space
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
Lecture Outline
• Introduction
• Pole Placement
– Topology of Pole Placement
– Pole Placement Design Techniques
• Using Transformation Matrix P
• Direct Substitution Method
• Ackermann’s Formula
Introduction
• One of the drawbacks of frequency domain methods
of design is that after designing the location of the
dominant second-order pair of poles, we keep our
fingers crossed, hoping that the higher-order poles
do not affect the second-order approximation.
• What we would like to be able to do is specify all
closed-loop poles of the higher-order system.
Introduction
• Frequency domain methods of design do not allow
us to specify all poles in systems of order higher
than 2 because they do not allow for a sufficient
number of unknown parameters to place all of the
closed-loop poles uniquely.
• One gain to adjust, or compensator pole and zero
to select, does not yield a sufficient number of
parameters to place all the closed-loop poles at
desired locations.
Introduction
• Remember, to place n unknown quantities, you need n
adjustable parameters.
• State-space methods solve
introducing into the system
this
problem
by
– Other adjustable parameters and
– The technique for finding these parameter values
• On the other hand, state-space methods do not allow
the specification of closed-loop zero locations, which
frequency domain methods do allow through
placement of the lead compensator zero.
Introduction
• Finally, there is a wide range of computational
support for state-space methods; many software
packages support the matrix algebra required by
the design process.
• However, as mentioned before, the advantages of
computer support are balanced by the loss of
graphic insight into a design problem that the
frequency domain methods yield.
Pole Placement
• In this lecture we will discuss a design method commonly called
the pole-placement or pole-assignment technique.
• We assume that all state variables are measurable and are
available for feedback.
• If the system considered is completely state controllable, then
poles of the closed-loop system may be placed at any desired
locations by means of state feedback through an appropriate
state feedback gain matrix.
Pole Placement
• The present design technique begins with a determination
of the desired closed-loop poles based on the transientresponse and/or frequency-response requirements, such
as speed, damping ratio, or bandwidth, as well as steadystate requirements.
• By choosing an appropriate gain matrix for state feedback,
it is possible to force the system to have closed-loop poles
at the desired locations, provided that the original system
is completely state controllable.
Topology of Pole Placement
• Consider a plant represented in state space by
𝒙 = 𝑨𝒙 + 𝑩𝑢
𝑦 = 𝑪𝒙
Topology of Pole Placement
• In a typical feedback control system, the output, y, is fed
back to the summing junction.
• It is now that the topology of the design changes. Instead
of feeding back y, we feed back all of the state variables.
• If each state variable is fed back to the control, u, through
a gain, ki, there would be n gains, ki, that could be adjusted
to yield the required closed-loop pole values.
Topology of Pole Placement
• The feedback through the gains, ki, is represented in
following figure by the feedback vector K.
𝒙 = 𝑨𝒙 + 𝑩(𝑟 − 𝑲𝒙)
𝒙 = 𝑨𝒙 + 𝑩𝑟 − 𝑩𝑲𝒙
𝒙 = (𝑨 − 𝑩𝑲)𝒙 + 𝑩𝑟
𝑦 = 𝑪𝒙
Topology of Pole Placement
• For example consider a plant signal-flow graph in phasevariable form
Topology of Pole Placement
• Each state variable is then fed back to the plant’s input, u,
through a gain, ki, as shown in Figure
Pole Placement
• We will limit our discussions to single-input, single-output
systems (i.e. we will assume that the control signal u(t) and
output signal y(t) to be scalars).
• We will also assume that the reference input r(t) is zero.
𝑦
𝒙 = (𝑨 − 𝑩𝑲)𝒙 + 𝑩𝑟
𝒙 = (𝑨 − 𝑩𝑲)𝒙
𝑢 = −𝑲𝒙
Pole Placement
𝒙 = (𝑨 − 𝑩𝑲)𝒙
• The stability and transient response characteristics are
determined by the eigenvalues of matrix A-BK.
• If matrix K is chosen properly Eigenvalues of the system
can be placed at desired location.
• And the problem of placing the regulator poles (closedloop poles) at the desired location is called a poleplacement problem.
Pole Placement
• There are three approaches that can be used to determine the
gain matrix K to place the poles at desired location.
• Using Transformation Matrix P
• Direct Substitution Method
• Ackermann’s formula
• All those method yields the same result.
Pole Placement (Using Transformation Matrix P)
• Following are the steps to be followed in this particular
method.
1. Check the state controllability of the system
𝐶𝑀 = 𝐵
𝐴𝐵
𝐴2 𝐵 ⋯ 𝐴𝑛−1 𝐵
Pole Placement (Using Transformation Matrix P)
• Following are the steps to be followed in this particular
method.
2. Transform the given system in CCF.
P  CM  W
 a n 1

a
 n2
W   

 a1
 1
an2

a1
a n3

1



1

0
0

0
A  P
1
1

0



0
0 
AP
𝑠𝐼 − 𝐴 = 𝑠 𝑛 + 𝑎1 𝑠 𝑛−1 + 𝑎2 𝑠 𝑛−2 + ⋯ + 𝑎𝑛−1 s+𝑎𝑛
1
B  P B
C  CP
Pole Placement (Using Transformation Matrix P)
• Following are the steps to be followed in this particular
method.
3. Obtain the desired characteristic equation from desired
Eigenvalues.
•
If the desired Eigenvalues are 𝜇1 , 𝜇2 , ⋯ , 𝜇𝑛
(𝑠 − 𝜇1 )( 𝑠 − 𝜇2 ) ⋯ 𝑠 − 𝜇𝑛 = 𝑠 𝑛 + 𝛼1 𝑠 𝑛−1 + 𝛼2 𝑠 𝑛−2 + ⋯ + 𝛼𝑛−1 𝑠+𝛼𝑛
Pole Placement (Using Transformation Matrix P)
• Following are the steps to be followed in this particular
method.
4. Compute the gain matrix K.
𝑲 = 𝛼𝑛 − 𝑎𝑛
𝛼𝑛−1 − 𝑎𝑛−1
⋯ 𝛼2 − 𝑎2
𝛼1 − 𝑎1
Pole Placement (Using Transformation Matrix P)
• Example-1: Consider the regulator system shown in following
figure. The plant is given by
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• The system uses the state feedback control u=-Kx. The desired
eigenvalues are 𝜇1 = −2 + 𝑗4, 𝜇2 = −2 − 𝑗4 ,𝜇3 = −1. Determine
the state feedback gain matrix K.
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-1
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• First, we need to check the controllability matrix of the system. Since
the controllability matrix CM is given by
𝐶𝑀 = 𝐵
𝐴𝐵
0 0
𝐴2 𝐵 = 0 1
1 −6
1
−6
31
• We find that rank(CM)=3. Thus, the system is completely state
controllable and arbitrary pole placement is possible.
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-2 (Transformation to CCF)
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• The given system is already in CCF
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-3
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• Determine the characteristic equation
𝑠𝐼 − 𝐴 = 𝑠 3 + 6𝑠 2 + 5𝑠 + 1 = 0
𝑠𝐼 − 𝐴 = 𝑠 3 + 𝑎1 𝑠 2 + 𝑎2 𝑠 + 𝑎3
• Hence
𝑎1 = 6,
𝑎2 = 5,
𝑎3 = 1
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-4
• The desired characteristics polynomial can be computed using
desired eigenvalues
𝜇1 = −2 + 𝑗4
𝜇2 = −2 − 𝑗4
𝜇3 = −1
(𝑠 − 𝜇1 )( 𝑠 − 𝜇2 ) ⋯ 𝑠 − 𝜇𝑛 = (𝑠 + 2 − 4𝑗)( 𝑠 + 2 + 4𝑗) 𝑠 + 10
𝑠 + 2 − 4𝑗
𝑠 + 2 + 4𝑗
𝑠 + 10 = 𝑠 3 + 14𝑠 2 + 60𝑠 + 200
= 𝑠 3 + 𝛼1 𝑠 2 + 𝛼2 𝑠 + 𝛼3
• Hence
𝛼1 = 14,
𝛼2 = 60,
𝛼3 = 200
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-4
• State feedback gain matric K is then calculated as
𝑎1 = 6,
𝛼1 = 14,
𝑎2 = 5,
𝛼2 = 60,
𝑲 = 𝛼3 − 𝑎3
𝛼2 − 𝑎2
𝑲 = 199 55 8
𝑎3 = 1
𝛼3 = 200
𝛼1 − 𝑎1
Pole Placement (Using Transformation Matrix P)
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• State diagram of the given system
𝑢(𝑡)
𝑥3
+
+
+
+
+
∫
𝑥3
𝑥2
∫
𝑥2
-6
-5
+
+
-5
𝑥1
∫
𝑥1
𝑥1
𝑢 = − 199 55 8 𝑥2
𝑥3
𝑲 = 199 55 8
𝑢 = −𝑲𝑥
199
+
+
+
55
+
8
+
-1
𝑥3
𝑢(𝑡) +
+
+
+
+
∫
𝑥3
𝑥2
∫
𝑥2
-6
-5
+
+
-5
𝑥1
∫
𝑥1
Pole Placement (Direct Substitution Method)
• Following are the steps to be followed in this particular
method.
1. Check the state controllability of the system
𝐶𝑀 = 𝐵
𝐴𝐵
𝐴2 𝐵 ⋯ 𝐴𝑛−1 𝐵
Pole Placement (Direct Substitution Method)
• Following are the steps to be followed in this particular
method.
2. Define the state feedback gain matrix as
𝑲 = 𝑘1
𝑘2
– And equating 𝑠𝐼 − 𝐴 + 𝐵𝐾
equation.
𝑘3 ⋯ 𝑘𝑛
with desired characteristic
(𝑠 − 𝜇1 )( 𝑠 − 𝜇2 ) ⋯ 𝑠 − 𝜇𝑛 = 𝑠 𝑛 + 𝛼1 𝑠 𝑛−1 + 𝛼2 𝑠 𝑛−2 + ⋯ + 𝛼𝑛−1 s+𝛼𝑛
Pole Placement (Using Direct Substitution)
• Example-1: Consider the regulator system shown in following
figure. The plant is given by
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• The system uses the state feedback control u=-Kx. The desired
eigenvalues are 𝜇1 = −2 + 𝑗4, 𝜇2 = −2 − 𝑗4 ,𝜇3 = −1. Determine
the state feedback gain matrix K.
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-1
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• First, we need to check the controllability matrix of the system. Since
the controllability matrix CM is given by
𝐶𝑀 = 𝐵
𝐴𝐵
0 0
𝐴2 𝐵 = 0 1
1 −6
1
−6
31
• We find that rank(CM)=3. Thus, the system is completely state
controllable and arbitrary pole placement is possible.
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-2
• Let K be
𝑲 = 𝑘1
𝑠𝐼 − 𝐴 + 𝐵𝐾 =
𝑠 0
0 𝑠
0 0
𝑘2
0
0
0 − 0
𝑠
−1
𝑘3
1
0
−5
0
0
1 + 0 𝑘1
−6
1
𝑘2
𝑘3
= 𝑠 3 + 6 + 𝑘3 𝑠 2 + 5 + 𝑘2 𝑠 + 1 + 𝑘1
• Desired characteristic polynomial is obtained as
𝑠 + 2 − 4𝑗
𝑠 + 2 + 4𝑗
𝑠 + 10 = 𝑠 3 + 14𝑠 2 + 60𝑠 + 200
• Comparing the coefficients of powers of s
14 = 6 + 𝑘3
𝑘3 = 8
60 = 5 + 𝑘2
𝑘2 = 55
200 = 1 + 𝑘1
𝑘1 = 199
Pole Placement (Ackermann’s Formula)
• Following are the steps to be followed in this particular
method.
1. Check the state controllability of the system
𝐶𝑀 = 𝐵
𝐴𝐵
𝐴2 𝐵 ⋯ 𝐴𝑛−1 𝐵
Pole Placement (Ackermann’s Formula)
• Following are the steps to be followed in this particular
method.
2. Use Ackermann’s formula to calculate K
𝐾 = 0 0 ⋯0 1 𝐵
𝐴𝐵
𝐴2 𝐵 ⋯ 𝐴𝑛−1 𝐵
−1 ∅(𝐴)
∅ 𝐴 = 𝐴𝑛 + 𝛼1 𝐴𝑛−1 + ⋯ + 𝛼𝑛−1 𝐴 + 𝛼𝑛 𝐼
Pole Placement (Ackermann’s Formula)
• Example-1: Consider the regulator system shown in following
figure. The plant is given by
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• The system uses the state feedback control u=-Kx. The desired
eigenvalues are 𝜇1 = −2 + 𝑗4, 𝜇2 = −2 − 𝑗4 ,𝜇3 = −1. Determine
the state feedback gain matrix K.
Pole Placement (Using Transformation Matrix P)
• Example-1: Step-1
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
• First, we need to check the controllability matrix of the system. Since
the controllability matrix CM is given by
𝐶𝑀 = 𝐵
𝐴𝐵
0 0
𝐴2 𝐵 = 0 1
1 −6
1
−6
31
• We find that rank(CM)=3. Thus, the system is completely state
controllable and arbitrary pole placement is possible.
Pole Placement (Ackermann’s Formula)
• Following are the steps to be followed in this particular
method.
2. Use Ackermann’s formula to calculate K
𝐾= 0
0 1 𝐵
𝐴𝐵
2 −1 ∅(𝐴)
𝐴
∅ 𝐴 = 𝐴3 + 𝛼1 𝐴2 + 𝛼2 𝐴 + 𝛼3 𝐼
• 𝛼𝑖 are the coefficients of the desired characteristic
polynomial.
𝑠 + 2 − 4𝑗
𝑠 + 2 + 4𝑗
𝛼1 = 14,
𝑠 + 10 = 𝑠 3 + 14𝑠 2 + 60𝑠 + 200
𝛼2 = 60,
𝛼3 = 200
Pole Placement (Ackermann’s Formula)
𝑥1
0
1
0 𝑥1
0
𝑥2 = 0
0
1 𝑥2 + 0 𝑢(𝑡)
𝑥3
−1 −5 −6 𝑥3
1
∅ 𝐴 = 𝐴3 + 14𝐴2 + 60𝐴 + 200𝐼
0
∅ 𝐴 = 0
−1
1
0
−5
0
1
−6
3
0
1
+ 14 0
0
−1 −5
0
1
−6
2
0
+ 60 0
−1
199 55
8
∅ 𝐴 = −8 159
7
−7 −34 117
1
0
−5
0
1
1 + 200 0
−6
0
0
1
0
0
0
1
Pole Placement (Ackermann’s Formula)
𝐵
𝐴𝐵
0 0
1
𝐴2 𝐵 = 0 1 −6
1 −6 31
𝐾= 0
𝐾= 0
199 55
8
∅ 𝐴 = −8 159
7
−7 −34 117
0 1 𝐵
0
0 1 0
1
𝐴𝐵
0
1
1 −6
−6 31
−1
2 −1 ∅(𝐴)
𝐴
199 55
8
−8 159
7
−7 −34 117
𝐾 = 199 55 8
Pole Placement
• Example-2: Consider the regulator system shown in following
figure. The plant is given by
𝑥1
1 2 1 𝑥1
1
𝑥2 = 0 1 3 𝑥2 + 0 𝑢(𝑡)
𝑥3
1 1 1 𝑥3
1
• Determine the state feedback gain for each state variable to place
the poles at -1+j, -1-j,-3. (Apply all methods)
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END OF LECTURES-30-31