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MODULE-IV
OPTIMAL CONTROL SYSTEMS
Introduction:
There are two approaches to the design of control systems
...
In other approach, for a given plant we find an overall
system that meets the given specifications & then compute the necessary compensators
...
Compensators are selected
that give as closely as possible, the desired system performance
...
Then, through a trial & error procedure, an
acceptable system performance is achieved
...
In
parameter optimization procedure, the performance specification consists of a single
performance index
...

Parameter Optimization: Servomechanisms
The analytical approach of parameter optimization consists of the following steps:(i)

(ii)

Compute the performance index J as a function of the free parameters
K1,K2,…
...
,Kn)
...
𝑛
… … … … … … … … …
...


Sufficient conditions
From the solution set of equation(2), find the subset that satisfies the sufficient conditions
which require that the Hessian matrix given below is positive definite
...


𝜕2𝐽

𝜕2𝐽 … …
𝜕𝑘1𝜕𝑘2
2

𝜕2𝐽
𝜕𝑘1𝜕𝑘𝑛
2

𝜕 𝐽
𝜕𝐾2 … …
...


………

2

𝜕2𝐽

𝜕𝑘𝑛𝜕𝑘1 𝜕𝑘𝑛𝜕𝑘2

……

𝜕2𝐽

2

𝜕𝐾𝑛
102

… … … … … (3)

𝜕2 𝐽

Since

𝜕𝑘 𝑖 𝜕𝑘 𝑗

(iii)

=

𝜕2 𝐽
𝜕𝑘 𝑗 𝜕𝑘 𝑖

, the matrix H is always symmetric
...

The set that has the smallest J gives the optimal parameters
...

The quadratic performance index, this can be done by using the Parseval‟s theorem which
allows us to write
∞

𝑥2

𝑡 𝑑𝑡 =

0

𝑗∞

1
2𝜋𝑗

𝑋 𝑠 𝑋 −𝑠 𝑑𝑠

… … … … … … … … (4)

−𝑗∞

The values of right hand integral in equation(4) can easily be found from the published tables,
provided that X(s) can be written in the form
𝑋 𝑠 =

𝐵(𝑠)
𝑏0 + 𝑏0𝑠 + ⋯ + 𝑏𝑛−1𝑠𝑛−1
=
𝐴(𝑠)
𝑎0 + 𝑎0𝑠 + ⋯ + 𝑎𝑛 𝑠𝑛

Where A(s) has zeros only in the left half of the complex plane
...

We may define error e(t)=c(t) - r(t)
The design objective in a servomechanism or tracking problem is to keep error e(t) small
...


103

EXAMPLE
Referring to the block diagram given below, consider 𝐺 𝑠 = 100 and 𝑠 = 1
...


100

=
100
s + 100K
Ks
s2
𝐸 𝑠
1
=
𝑅 𝑠
1+𝐻 𝑠
𝑅(𝑠)
𝑠 + 100𝑘
1 𝑠
⇒𝐸 𝑠 =
= 2
=
𝑠 + 100𝑘𝑠 + 100
1 + 𝐻 𝑠 1 + 100
𝑠 + 100𝐾
Here

b0=100K,

b1=1,

1 + G s Ks

a0=100,

=

1+

a1=100K,

a2=1

As E(s) is 2nd order
𝐽2 =

𝜕𝐽
𝜕𝐾

𝑏12𝑎0 + 𝑏02𝑎2

2𝑎0𝑎1𝑎2
= 0 gives K=0
...
1
...
(5)
0

may be unsatisfactory because it may lead to excessively large magnitudes of some control
signals
...
Therefore, a more realistic
PI should be to minimize
∞

𝐽=

𝑒2(𝑡) 𝑑𝑡 … … … …
...
(6𝑎)
The constant M is determined by the linear range of the system plant
...
e in order to implement the optimal design, nonlinear &/or time-varying
devices are required
...
(7)
0

Where λ, a positive constant, is called the weighting factor
...
As 𝜆 → 0 ,the
∞
contribution of u(t) becomes less significant & PI reduces to 𝐽 = 0 𝑒2(𝑡) 𝑑𝑡
...
If → ∞ , performance criterion given by equation(7) reduces to
∞

𝐽=

𝑢2(𝑡) 𝑑𝑡 … … … …
...

From these two extreme cases, we conclude that if λ is properly chosen, then the constraint of
(6a) will be satisfied
...
1

Solution
100
K 100 s
1
2
1
s
H s =
=
=
K1
s
+
K1K2100
1 + K1G(s)K2s
1 + 2 100K2s
s
𝐸 𝑠
1
1
s2 + K1K2s100
=
=
= 2
K 100
𝑅 𝑠
1+𝐻 𝑠
s + K1K2s100 + K1100
1+ 2 1
s
+
K
1K2100
1
𝑠 𝑠 + K1K2100
𝑠 + K K 100
𝑠
1 2
⇒𝐸 𝑠 =
=
2
2
s + K1K2s100 + K1100
s + K1K2s100 + K1100
K G(s)

Here

b0=K1K2100,

b1=1,

K1

a0=K1100, a1=K1K2100,
105

a2=1

As E(s) is of 2nd order, so PI
∞

1 + 100K 1𝐾 22
=
2𝑎0𝑎1𝑎2
200K1K2
0
100K1
100
𝐶 𝑠 =
=
𝑈(𝑠)
𝑠 s2 + K1K2s100 + K1100
𝑠2
𝑠K1
𝑈 𝑠 =
s2 + K1K2s100 + K1100
𝑒2(𝑡) 𝑑𝑡 =

𝐽𝑒2 =

Here

b0=0,

𝑏 12𝑎 0+ 𝑏 02𝑎

2

a0=K1100, a1=K1K2100,
∞
K1
2
𝐽𝑢2 = 𝑢 (𝑡) 𝑑𝑡 =
200K2

b1=K1,

a2=1

0

The energy constraint on the system is thus expressed by the equation
K1
= 0
...
(𝑎)
𝐽𝑢 =
200K2
2
1+100K1𝐾2

The PI for the system is 𝐽 = 𝐽𝑒 + 𝜆𝐽𝑢 =

200K1K2

K1

+ 𝜆 200K

2

𝜕𝐽
= 0 𝑓𝑜𝑟 𝑖 = 1,2 𝑔𝑖𝑣𝑒𝑠
𝜕𝐾𝑖
𝜆𝐾12 = 1 … … … …
...
(𝑐)
Solving equation (a),(b),(c), we get 𝜆 = 0
...
1

The Hessian matrix 𝐻 =

𝜕2 𝐽

𝜕2 𝐽

𝜕𝐾12
2
𝜕 𝐽
𝜕𝑘 2 𝜕𝑘 1

𝜕𝑘 1 𝜕𝑘 2
2

=

𝜕 𝐽
𝜕𝐾2
2

1

1−𝜆𝐾12

100K13K22
1−𝜆𝐾1
200𝐾 2𝐾 2

200𝐾12 𝐾22 2
1
100K1 𝐾2 −1
K − 100K 3K

1

2

2

2

1

For 𝐾1 = 2, 𝐾2 = 0
...
1 satisfy the necessary as well as sufficient conditions for J to
be minimum
...
For zero input, the output is zero if all
the initial conditions are zero
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The primary objective of the design is to damp out the
response due to initial conditions quickly without excessive overshoot & oscillations
...
The response(roll
angle Ө(t)) to this disturbance is highly oscillatory
...

If there is no constraint on “control effort”, the controller which minimizes the performance
index
...

𝜃𝑑 (𝑡) → desired roll angle which is clearly zero
...
If for
disturbance torque applied at t=to , the controller is required to regulate the roll motion which is
finite time (tf-to), a suitable performance criterion for design of optimum controller is to minimize
𝑡𝑓

𝐽=

𝜃2(𝑡) 𝑑𝑡 … … … …
...
(11)
A is constant matrix of size (n×n)
B is the constant matrix of size (n×m)
X(t) is the state vector of size (n×1)
U(t) is the control vector of size (m×1)
Find the control function u* which is optimal with respect to given performance criterion
...

The State Regulator Problem
When a system variable x1(t)(the output) is required to be near zero, the performance measure is

107

𝑡𝑓

𝐽=

𝑥12(𝑡) 𝑑𝑡
𝑡𝑜

A performance index written interms of two state variables of a system would be then
𝑡𝑓

𝐽=

(𝑥12 𝑡 + 𝑥22(𝑡))𝑑𝑡
𝑡𝑜

Therefore if the state x(t) of a system described by equation (11) is required to close to Xd=0,
a design criterion would be to determine a control function that minimizes
𝑡𝑓

𝐽=

(𝑋𝑇𝑋 ) 𝑑𝑡
𝑡𝑜

In practical, the control of all the states of the system is not equally important
...

The roll motion contributes much discomfort to passengers, in the design of passenger
ship,the value of 𝜆 will be less than one
...

The simplest form of Q is a diagonal matrix:
𝑞1
0
𝑄=
⋮
0

0…
...
0
⋮
⋮
0 …
...
The larger the value of qi relative to other values of q,the more control
effort is spent to regulate xi(t)
...

In the infinite time state regulator problem 𝑡𝑓 → ∞ , the final state should approach the
equilibrium state X=0; so the terminal constraint is no longer necessary
...

If PI is modified by adding a penalty term for physical constraints, then solution would be
more realistic
...

By giving sufficient weight to control terms, the amplitude of controls which minimize the
overall PI may be kept within practical bound, although at the expense of increased error in
X(t)
...
(12)
Find the optimal control law u*(t),𝑡 ∈ 𝑡0, 𝑡𝑓 , where 𝑡0 & 𝑡𝑓 are specified initial & final
times respectively, so that the optimal PI
𝑡𝑓

1
1
𝐽 = 𝑋 𝑇 𝑡𝑓 𝐹𝑋 𝑡𝑓 +
2
2

𝑋 𝑇 𝑡 𝑄𝑋 𝑡 + 𝑢 𝑇 𝑡 𝑅 𝑢 𝑡 𝑑𝑡 … … … … … … …
...
tf is fixed & given & X(tf) is free
...
We shall assume that, both
Q & F are not simultaneously zero matrices to avoid trival solution
...
(14)
With final condition P(t=tf)=F
Where P(t)=Riccati coefficient matrix,time varying matrix,symmetric positive definite
matrix
...
Numerical integration is
carried out backward in time; from t=tf to t=t0 with boundary condition P(tf)=F
...
(15)
Where K(t) = −R−1BTP t is called Kalman gain
...

It is desired to find the control Law that minimizes the PI

1
𝐽=

𝑡𝑓

2

1 2
2
3𝑥 + 𝑢 𝑑𝑡 , 𝑡𝑓 = 1 𝑠𝑒𝑐
4

𝑡𝑜

Solution
Comparing the state equation with equation(12), A=2, B=1
Comparing PI with equation(13), we get 𝐹 = 0, 𝑅 = 1 , 𝑄 = 3
4

As there is one state variable so P(t)=p(t), matrix reduces to scalar function
...
(16)
𝑡𝑜

Salient points of infinite time regulator problem:
(1) When 𝑡𝑓 → ∞ , 𝑋(∞) → 0 for the optimal system to be stable
...
e we set F=0 in
general quadratic PI
...

(4) Solve ARE to get P ,then the optimal control law is given by
u∗ t = −R−1BTP X t = −KX t
Where K(t) = −R−1BTP is called Kalman gain
...

(5) The optimal value of PI is
1 𝑇
𝑋 0 P𝑋 0
2

𝐽∗ =
EXAMPLE

Obtain the control Law which minimizes the performance index
∞

𝑥12 + 𝑢2 𝑑𝑡

𝐽=
0

For the system

𝑥1
0 1 𝑥1
0
=
+
𝑢
𝑥
𝑥
0 0 2
1
2

Solution
𝐴=

0

1

,𝐵 =

0 0

0

,𝑄 =

1

2

0

,𝑅 = 2

0 0

ARE
−

p11

p12 0 1

p12

p22 0 0

TP

−

−P A − A
0 0 p11 p12
1 0 p12

+

−1 TP − Q = 0
+p P BR
p12 B
p11
11
0 1

0

p12

p22

Simplifying
−

p22 1 2

1

p12

p12
p22

p12 2

+2=0
2
p p
p11 − 12 22 = 0
2
p22 2
−
+ 2p12 = 0
2
For P to be positive definite matrix we get the solution P = 2 2
2
The Optimal Control Law is given by

112

2 ,
2 2

−

2

0

0

0

=

0

0

0

0

1

∗

−1 T

u t = −R

B PX t =−

0

1

2 2

2

𝑥1(𝑡)

=−
𝑥

𝑡 − 2
𝑥

1
2
2
2 2 𝑥2(𝑡)
It can be easily verified that closed loop system is asymptotically stable
...
In the output regulator problem on the other hand, we ere concerned with
making the components of the output vector small
...
17
Y t = CX t
Find the optimal control law u*(t),𝑡 ∈ 𝑡0, 𝑡𝑓 , where 𝑡0 & 𝑡𝑓 are specified initial & final
times respectively, so that the optimal PI
𝑡𝑓

1
1
𝐽 = 𝑌 𝑇 𝑡𝑓 𝐹 𝑌 𝑡𝑓 +
2
2

𝑌𝑇 𝑡 𝑄𝑌 𝑡 + 𝑢 𝑇 𝑡 𝑅 𝑢 𝑡 𝑑𝑡 … … … … … … … (18)
𝑡𝑜

Is minimized,subject to initial state x(t0)=x0
...
Suppose that the
vector Z(t) is the desired output
...

Output Regulator as state regulator Problem
If the controlled process given by equation(17) is observable then, we can reduce the output
regulator problem to the state regulator problem
...
(19)
With boundary condition 𝑃 𝑡𝑓 = CTFC
The Tracking Problem
Here we shall study a class of tracking problems which are reducible to the form of the output
regulator problem
...

We define error vector e(t)=Y(t)-r(t)

Find the optimal control law u*(t),𝑡 ∈ 𝑡0, 𝑡𝑓 , where 𝑡0 & 𝑡𝑓 are specified initial & final
times respectively, so that the optimal PI
𝐽=

1
2

𝑒𝑇 𝑡 𝐹 𝑒 𝑡
𝑓

𝑓

+

1

𝑡𝑓

𝑒 𝑇 𝑡 𝑄𝑒 𝑡 + 𝑢𝑇 𝑡 𝑅 𝑢 𝑡 𝑑𝑡

2
𝑡𝑜

Is minimized
...

Z t = AZ t
r t = CZ t
The matrices A & C are same as those of the plant
...


Parameter Optimization: Regulators
Solution of control problem when some elements of feedback matrix K are constrained
Consider completely controllable process
X t = AX t + Bu t
The PI is 𝐽 = 1

∞

𝑋 𝑇 𝑡 𝑄𝑋 𝑡 + 𝑢 𝑇 𝑡 𝑅 𝑢 𝑡 𝑑𝑡

2 0

Optimal control is linear combination of the state variables u=KX(t)
With the above feedback law,closed loop system is described by
X t = AX t + BKX t = A + BK X(t)
Solution
(1) Determine elements of P as functions of the elements of the feedback matrix K from the
equations given below
A + BK T P + P A + BK + KTRK + Q = 0
...
(21)
2
If K1,K2,…
...
,Kn)
...
𝑛
(𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛)
𝜕𝐾𝑖
Hessian matrix is positive definite (sufficient condition)
115

Solution set Ki of equation(22) satisfies necessary & sufficient condition is obtained which
gives the suboptimal solution to the control problem
...
If all the parameters of P are
free, the procedure above will yield an optimal solution
...
(23)
EXAMPLE
Consider the second order system, where it is desired to find optimum 𝜁 which minimizes the
integral square error i
...
Since K2 = 2ξ,
⇒ ξ = 0
...

It can be easily verified that the suboptimal control derived above results in a closedloop
system which is asymptotically stable
...
For practical
2

reasons, it is desirable to have one control irrespective of what x(0) is
...

1
𝐸 𝑋 0 𝑋𝑇 0 = 𝐼
𝑛
New PI 𝐽 = 𝐸 𝐽 = 𝐸 1 𝑋 𝑇 0 𝑃 𝑋 0 = 1 𝑡𝑟 𝑝
2

2𝑛

𝑡𝑟 𝑝 = 𝑡𝑟𝑎𝑐𝑒 𝑜𝑓 𝑃 = 𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑃
The parameter K2 that optimizes 𝐽 is 2
...


INTRODUCTION TO ADAPTIVE CONTROL
To implement high performance control systems when the plant dynamic characteristics are
poorly known or when large & unpredictable variations occur, a new class of control systems
called nonlinear control systems have evolved which provide potential solutions
...
One of the goals of adaptive
control is to compensate for parameter variations, which may occur due to nonlinear
actuators, changes in the operating conditions of the process, & non-stationary
disturbances acting on the process
...

An adaptive control system may be thought of as having two loops
...
The other loop is a parameter
adjustment loop
...
The
parameter adjustment loop is often slower than the normal feedback loop
...
They are
(1) Model Reference Adaptive control Method
(2) Self-Tuning method
(1) Model Reference Adaptive control (MRAC)
The MRAC system is an adaptive system in which the desired performance is expressed
in terms of a reference model, which gives the desired response signal
...

(a) a plant containing unknown parameters
(b) A reference model for compactly specifying the desired output of the control system
...

(d) parameter adjustment loop is called as outer loop
...

118

Fig: Model Reference Adaptive controller
(2) Self-Tuning control
A general architecture for the self tuning control is shown below
...
They are
(1) Parameter estimation

(2) control law

(1) Parameter estimation: Parameter estimation is performed online
...
A number of recursive parameter estimation schemes are employed
for self tuning control
...

(2) Control law: The control law is derived based on control performance criterion
optimization
...
This approach of designing
controller using estimated parameters of the transfer function of the process is known
as indirect self-tuning method
...

1
...
Find the control law which minimizes the performance index

[2]

∞

𝐽=

(𝑋12 + 𝑈2) 𝑑𝑡
0

For the system
𝑋1
0 1 𝑋1
0
=
u
X +
0 0
1
𝑋2
2

[10]

3
...

(a) State regulator problem
(b) Pontryagin‟s minimum principle
4
...
5×2]

𝑋1 = 𝑋2
𝑋2 = −2𝑋1 − 3𝑋2 + 𝑢
Determine the optimal control law 𝑢𝑜𝑝𝑡 (𝑡) such that the following performance
index is minimised
1 ∞ 2
2
2
𝐽=
(𝑥1 + 𝑥2 + 𝑢 ) 𝑑𝑡
2
0

Derive the formula used
...
What are the different types of performance indices? Explain ISE & ITAE
...
(a) Explain the following error performance indices:
ISE, ISTE, IAE, ITAE
[6]
(b) Determine the optimal controller to minimize
∞

𝐽=

(𝑦2 + 𝑢2) 𝑑𝑡
0

for the process described by 𝑑𝑦 + 𝑦 = 𝑢
𝑑𝑡

7
...
A linear autonomous system is described in the state equation
𝑋 = −4𝐾 4𝐾 X
2𝐾 −6𝐾
Find restriction on the parameter k to guarantee stability of the system
...
A first order system is described by the differential equation
𝑋 𝑡 = 2𝑋 𝑡 + 𝑢(𝑡)
Find the control law that
minimises the performance index
1 𝑡𝑓
J=
(3𝑋2 + 1 𝑢2) dt
2 0

[15]

4

When 𝑡𝑓 =1 second
10

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