The controllers have been widely used in many industrial processes. The goal of accomplishing a practical control system design is to meet the functional requirements and achieve a satisfactory system performance. We will introduce the design method of the state feedback controller, the state observer and the servo controller with optimal control law for a linear system in this paper.

1. The Controller Design For Linear System:
A State Space Approach
HONG Yang
Matric. No. : HD 98-1284R
Department of Electrical Engineering
National University of Singapore
Singapore 119260
∗
Email: engp8799@nus.edu.sg
Abstract—The controllers have been widely used in many industrial processes. The goal
of accomplishing a practical control system design is to meet the functional requirements
and achieve a satisfactory system performance. We will introduce the design method of
the state feedback controller, the state observer and the servo controller with optimal
control law for a linear system in this paper.
Key Words— state feedback controller, state observer, optimal control, servo control.
1

2. 1 Introduction
The controllers have found wide applications in the industries. The goal of the control
system design is to obtain a desired system performance. Although most control system
are nonlinear, the nonlinearity is small enough to be neglected in many real-world system
design cases, where a linear analysis can describe the system dynamics effectively. We
will introduce the controller design methods of a typical plant for industrial process.
1.1 Description of the system
Figure 1 depicts a typical linear system in the industrial process,
)(sY+
+
)(sd
)(sU
2
2
+s s
1
Figure 1: A typical plant for industrial process
where u(t) is the input, y(t) is the output and d(t) is the disturbance. The transfer
function is
G(s) =
2
s(s + 2)
(1)
The objectives of our controller design are summarized as follows:
1. Zero steady-state output error when the reference input r(t) is a unit step.
2. Small response which goes to zero at steady state when the disturbance input d(t) is
2

3. a unit step.
3. Fast response when the reference input r(t) is a unit step.
1.2 Analysis of the system
For the convenience of the analysis of the system, the disturbance will be taken into
account later on. The control plant of linear system can be expressed as
Y (s)
U(s)
=
2
s(s + 2)
(2)
which corresponds to
Y (s)(s(s + 2)) = 2U(s) (3)
or in time domain,
d2
y(t)
dt
+ 2
dy(t)
dt
= 2u(t)
Without loss of generality, we define
x1(t) = y(t)
x2(t) = ˙y(t) =
dy(t)
dt
Then we can obtain
˙x1(t) = x2(t)
˙x2(t) = −2x2(t) + 2u(t)
If the disturbance is added, then
˙x1(t) = x2(t) + d(t)
˙x2(t) = −2x2(t) + 2u(t)
3

4. or in matrix form,
˙x =




0 1
0 −2



 x +




0
2



 u +




1
0



 d
y = 1 0 x (4)
See Figure 2 below.
2
+
+
d
y
-2
u
+
+
.
2x 2x
.
1x
∫∫
Figure 2: The state-space representation of the plant
From the characteristic equation of the system, one can see that the system places one
open-loop pole in the origin. Obviously the system is unstable.
When a unit step input is applied, the output y(t) diverges, as shown in Figure 3. When
the input u(t) is zero and a unit-step signal is applied as a disturbance, the output y(t)
also diverges, as shown in Figure 4. Thus this system cannot reject the disturbance d(t).
However, we can combine the integral control action with state feedback controller to
reject the disturbance. The concrete method will be introduced in Section 4.
2 The state feedback controller design using pole place-
ment
4

5. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time(sec)
Outputy(t)
Figure 3: The open-loop step response of the system
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time(sec)
Outputy(t)
Figure 4: The output of the system when the disturbance is unit-step input
First of all, let us study the state feedback controller by introducing the n-dimensional
5

6. linear system with m-inputs and p-outputs:
˙x = Ax + Bu
y = Cx (5)
To accomplish a stable feedback control system, a control law consisting of state feedback
Kx and feedforward Fr is applied, that is,
u = −Kx + Fr (6)
where r is a reference vector, K and F are constant matrices, and F is non-singular.
Substituting (6) into (5) yields
˙x = (A − BK)x + BFr
y = Cx (7)
Thus, by the control of (6), (A, B, C) is changed to (A − BK, BF, C). The structure of
the closed loop system is depicted in Figure 5. For later reference, (5) is known as the
r
F B C
A
K
u+
-
+
+
yx
.
x
∫
Figure 5: State feedback control system
open-loop plant, (6) as the state feedback controller, K as the state feedback gain. (5)
and (6), or equivalently (7), is known as the state feedback control system.
6

7. Next, we begin to design the state feedback controller of the system given in (4) using
pole placement method. The system given in (4) has the controllability matrix
B AB =




0 2
2 −4




with full rank, so the system is controllable.
Thirdly, we can tune the transient performance of the system by locating the proper
closed-loop poles. Consider a second order system whose closed-loop transfer function is
defined by
Y (s)
R(s)
=
ωn
2
s2 + 2ξωn + ωn
2
(8)
Its poles are located at −ξωn ± jωn 1 − ξ2 (with 0 < ξ < 1). It is well known that for a
step input, the closed-loop system possesses the following properties:
(a) Percentage overshoot Mp = e
− ξπ√
1−ξ2
∗ 100%;
(b) Settling time τs = 4
ξωn
.
Control engineering practice suggests the choice of the overshoot < 10% and the setting
time 2 second that can achieve a satisfactory transient response. This leads to the
following requirements for selecting ξ and ωn, that is, ξ > 0.6 and ξωn 2.
When the closed-loop poles are specified as −2±j2 (with ξ = 0.707 and ξωn = 2), the two
major specifications of step response of the system can be obtained as Mp = 4.3%, τs = 2s.
Finally, we design the state feedback controller for the system given in (4) according to
equation (6) and (7). The characteristic equation of closed-loop system becomes
φf (s) = (s + 2 − j2)(s + 2 + j2) = s2
+ 4s + 8
7

8. Hence Ackermann’s formula yields
K = [4 1]
To achieve a zero steady-state error for a unit step, the system needs to meet the constraint
C(−A + BK)−1
BF = 1
that is, F = 4.
Figure 6 depicts the closed-loop system with state feedback controller. The relative sim-
ulation model is shown in Figure 7.
2
+
+
d
y
-2
u
+
+
.
2x 2x
.
1x
∫∫
K1
K2
F
r +
-
-
Figure 6: The state-space representation of state feedback control system
2
s+2
T ransfer Fcn
t
T o Workspace1
y
T o Workspace
Step1
Step
Scope
s
1
Integrator
K2
Gain2
K1
Gain1
F
Gain
Clock
Figure 7: The simulation model of the closed-loop system
8

9. When a unit step is applied in the input r(t) and no disturbance is added, the output
y(t) is plotted in Figure 8. When a unit step is also applied as a disturbance at the same
time, the output y(t) is plotted in Figure 9.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time(sec)
Outputy(t)
Figure 8: The step response of the closed-loop system without disturbance
To prevent the overshoot in the step response of a closed-loop system, we can specify
the damping ratio to be ξ > 1 by placing two closed-loop poles at −5 and −6, and the
corresponding characteristic equation of closed-loop system is expressed as
φf (s) = (s + 5)(s + 6) = s2
+ 11s + 30
Hence Ackermann’s formula yields
K = [15 4.5]
Given that the steady-state error for a unit step is zero, we can obtain F = 15. When
a unit step is applied in the input r(t) and no disturbance is added, the output y(t) is
9

10. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time(sec)
Outputy(t)
Figure 9: The step response of the closed-loop system with disturbance
plotted in Figure 10. When a unit step is also applied as a disturbance at the same time,
the output y(t) is plotted in Figure 11.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time(sec)
Outputy(t)
Figure 10: The step response of the closed-loop system(2) without disturbance
10

11. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time(sec)
Outputy(t)
Figure 11: The step response of the closed-loop system(2) with disturbance
3 Full order observer design
The controller design method we developed in the previous section presuppose the avail-
ability of all state variables for feedback. However, in the real-world control system, the
only measured quantity are the output y of the system given in (5). This motivates us
to simulate a model with accessible state variables to estimate the states of the origi-
nal system - the model exhibits the same dynamics as the practical system. A famous
estimator, shown in Figure 12, is a closed-loop estimator which was first introduced by
Luenberger(1964) - it is commonly referred to as an asymptotic state estimator or simply
a full-order state observer. We make use of both input and output of the system given in
(5) to drive the estimator.
11

12. +
+
+
+
+
+ +-
A
A
B
B
C
C
L
u
y
xˆ
x
xˆ
x
yˆ
∫
∫
Figure 12: A closed-loop estimator (Luenberger Observer)
Consider an estimator of x(t) of the form:
˙ˆx = Aˆx + Bu(t) + L[y − Cˆx]
Here ˆx(t) denotes the estimate of x(t). Let the estimation error be denoted by ˜x so that
˜x = x − ˆx. It then readily leads to
˙˜x = (A − LC)˜x
with ˜x(0) = x(0) − ˆx(0).
If we choose L such that (A − LC) = A1 is stable, we have
˙˜x = (A − LC)˜x = A1 ˜x
˜x(t) = eA1t
˜x(0)
Clearly ˜x(t) −→ 0 as t −→ ∞. Thus the estimator output ˆx(t) will track x(t) asymp-
totically - this observer is called an asymptotic observer. In this paper, we use the pole
placement algorithms to adjust the rate of convergence of ˆx(t) to x(t). In practice, the
poles of the observer are usually chosen two or five times faster than the system response.
12

13. Suppose that we obtain the feedback gains of the controller with linear state feedback
under the assumption that all the state variables are available. In the real-world im-
plementation of the control policy, only the estimates ˜x(t) obtained using a Luenberger
observer is fed back. Figure 13 depicts a schematic of the observer/controller strategy,
where the overall system can be described as
˙x = Ax + Bu
˙ˆx = Aˆx + Bu + L[y − Cˆx]
y = Cx
u = r − Kˆx
It is convenient to write these equations in terms of x and ˜x, so that we get
+
+
+
+
+
++
-
A
A
B
B
C
C
L
x
∫
u
K
yˆ xˆ xˆ
∫
-
Figure 13: Schematic of the observer/controller structure




˙x
˙˜x



 =




A − BK BK
0 A − LC








x
˜x



 +




B
0



 r
y = Cx (9)
13

14. This allows us to design the state observer for the system given in (4) by use of Equation
(9). The system given in (4) has the observability matrix




C
CA



 =




1 0
0 1




with full rank, so the system is observable.
We use the control law specified in (6). From the above section, it is easily checked that
for F = 4 and K = [4 1], the system has the unit ’DC gain’ for the closed loop, while
the closed-loop poles are located at −2 ± j2.
If ˆx is generated via an observer, in order to place the observer poles at (−6, −6) (three
times as the closed-loop poles), the characteristic equation of observer becomes
det(sI − A + LC) = det(




s −1
0 s + 2



 +




l1 0
l2 0



) = det(




s + l1 −1
l2 s + 2



)
= s2
+ (l1 + 2)s + 2l1 + l2 = (s + 6)2
from which L = [10 16]T
can be derived.
Figure 14 illustrates the simulation model. Given that the input r(t) is zero and initial
condition is x1(0) = 1, x2(0) = 1, ˆx1(0) and ˆx2(0), we can obtain the output y(t) (dashed
curve) if true state feedback is used; similarly we can get the output yo(t) (solid curve) if
the observer is used. Figure 15 provides the performance comparison of both scenarios.
14

15. 1
s
T ra n sfer Fcn
(with in itia l o utpu ts)4
1
s
T ra n sfer Fcn
(with in itia l o utpu ts)3
1
s
T ra n sfer Fcn
(with in itia l o utpu ts)2
2
s+2
T ra n sfer Fcn
(with in itia l o utpu ts)1
t
T o Wo rksp a ce 1
y
T o Wo rksp a ce
Ste p
Sco p e
-2
G ain 5
L 2
G ain 4
L 1
G ain 3
K2
G ain 2
K1
G ain 1
Clo ck
2
F
Figure 14: The simulation model of full order observer/controller combination
0 0.5 1 1.5 2 2.5 3
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time(sec)
Output
y
without observer
yo
with observer
Figure 15: System performance with and without observer
4 Servo controller design
In the case of single-input/single-output (SISO) systems, a popular method for achieving
a zero steady-state error upon a step input is to employ integral control. As shown in
Figure 16, an integral control for the plant G(s) can maintain a zero steady-state error
15

16. even in the face of a step disturbance d
s
1
)(sK )(sG
y
d
+
+
+
-
r e
Figure 16: Classical integral control for a SISO system
To analyze control plant behavior, if given that r(t) = 0 and d(t) = 0, there holds
limt→∞ e(t) = limt→∞(y(t) − r(t)) = 0.
This is called the asymptotic tracking problem.
If given that r(t) = 0 and d(t) = 0, there holds
limt→∞ e(t) = limt→∞(y(t) − r(t)) = limt→∞ y(t) = 0.
This is called the asymptotic regulation problem.
If both asymptotic tracking and regulation are required, it is called the servo control
problem.
It is intuitively clear that an optimal controller for the plant cannot maintain a zero steady
state error (for a step input) and be robust at the same time. Usually integral controller
is combined with optimal controller to solve the servo problem.
16

17. The plant to be considered is
˙x = Ax + Bu + Ed
y = Cx (10)
where the input u is an m-vector, the state x is an n-vector, the controller vector y is a
m-vector and disturbance d is a q-vector. The objective of the servo control is to make y
follow the constant reference signal r in the presence of the constant disturbance d and
to stabilize the closed loop.
Let the models of the disturbance and reference signals be
˙xd = 0, d = xd,
˙xr = 0, r = xr.
and define the error e as e = y − r.
In presence of the constant reference signal and the constant disturbance (i.e., ˙d = 0 and
˙r = 0), the derivative of (10) gives
¨x = A ˙x + B ˙u
˙e = C ˙x (11)
which constitutes an augmented system with a matrix




¨x
˙e



 =




A 0
C 0








˙x
e



 +




B
0



 ˙u (12)
When e is taken as the output of this system, the output equation becomes
e = [0 1]




˙x
e




17

18. The control law to stabilize (12) is given by
˙u = −K1 ˙x − K2e ,
and u is given by
u(t) = −K1x − K2
t
0
e dτ + constant.
When the constant term is taken as zero, the control law is
u(t) = −K1x − K2
t
0
(y − r) dτ ,
One can see that integral control is applied in the control system, as shown in Figure 17,
The determination of K1 and K2 can be done by using optimal control for the criterion
function
J =
∞
0
( e 2
Q + ˙u 2
)dt (13)
which yields
(K1, K2) = −R−1
(BT
, 0T
)P (14)
where P is the positive definite solution of




A 0
C 0




T
P + P




A 0
C 0



 +




0
I



 Q[0 I] − P




B
0



 R−1
[BT
0T
]P = 0 (15)
Now we consider the system given in (4) and we specify the initial condition x(0) = 0.
With e = y − r and ˙e = C ˙x, the augmented system is




¨x
˙e



 =








0 1 0
0 −2 0
1 0 0












˙x
e



 +








0
2
0








˙u
18

19. d
++
-
r e
-
CBuAxx +=
s
K2
1K
yu
Figure 17: Closed loop system with integral control
Because
rank




C
CA



 = rank




1 0
0 1



 = 2, rank




A B
C 0



 = rank








0 1 0
0 −2 2
1 0 0








= 3
so the augmented system is observable and controllable.
we use optimal control for the criterion function given in (13), then R = 1 and
Q =








0 0 0
0 0 0
0 0 1








,
so we will derive P by the equation (15)








0 1 0
0 −2 0
1 0 0








T
P + P








0 1 0
0 −2 0
1 0 0








+








0 0 0
0 0 0
0 0 1








− P








0 0 0
0 4 0
0 0 0








P = 0
19

20. then
P =








2.5899 0.9175 1.6838
0.9175 0.3419 0.5000
1.6838 0.5000 1.8351








.
According to the equation (14), we will get
[K1, K2] = [1.8351 0.6838 1.0000]
which yields
˙u = −[1.8351 0.6838] ˙x − e
that is
u = −[1.8351 0.6838]x −
t
0
(y − r)dτ
The simulation model is depicted in Figure 18.
K2
s
T ransfer Fcn1
2
s+2
T ransfer Fcn
t
T o Workspa ce 1
y
T o Workspa ce
Step
Sco pe
K12
Gain2
K11
Gain1
Disturb ance
Clock
s
1
Figure 18: The simulation model of the system with integral control and optimal control
When a unit step input is applied and there is no disturbance, the output y(t) is shown in
Figure 19; When the input u(t) is zero and a unit-step signal is applied as a disturbance,
20

21. the output y(t) is shown in Figure 20; When a unit step input is applied and a unit-step
signal is also applied as a disturbance at the same time, the output y(t) is shown in
Figure 21.
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time(sec)
Outputy(t)
Figure 19: The step response of the system without disturbance
If we use the state observer to observe the state variables in the servo control above, then
we can use Figure 22 to depict the simulation model. We specify that initial condition of
the system is x1(0) = 1, x2(0) = 1, ˆx1(0) and ˆx2(0). When the input r(t) is zero and a
unit-step signal is applied as a disturbance, the output y(t) is shown in Figure 23; when
a unit step input is applied and a unit-step signal is also applied as a disturbance at the
same time, the output y(t) is shown in Figure 24.
21

22. 0 1 2 3 4 5 6 7 8 9 10
−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time(sec)
Outputy(t)
Figure 20: The output of the system when the disturbance is unit-step input
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time(sec)
Outputy(t)
Figure 21: The step response of the system with disturbance
22

23. disturbance
1
s
T ransfer Fcn
(with initial outputs)2
2
s+2
T ransfer Fcn
(with initial outputs)1
K2
s
T ransfer Fcn
t
T o Workspace1
y
T o Workspace
Step
Scope
-2
Gain5
L2
Gain4
L1
Gain3
K12
Gain2
K11
Gain1
Clock
1
s
T ransfer Fcn
2
1
s
Figure 22: The simulation model of the servo control system with observer
0 1 2 3 4 5 6 7 8 9 10
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time(sec)
Outputy(t)
Figure 23: The response of the system with observer when the input is zero and the
disturbance input is a unit step
5 Conclusions
We have presented the design method of the state feedback controller, the state observer
and the servo controller with optimal control law for a linear system. The major contri-
23

24. 0 1 2 3 4 5 6 7 8 9 10
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
Time(sec)
Outputy(t)
Figure 24: The step response of the system with observer when step disturbance is added
butions of this paper are:
(1) We have designed the state feedback controller by pole placement method to im-
prove the transient performance of the closed-loop system, but we can not cancel the
disturbance.
(2) We have designed the state observer to estimate the state variables based on the
knowledge of output and control variables.
(3) We have combined integral control with optimal control to solve the servo problem. By
suitable optimal control law, the system can achieve faster response and the zero steady-
state output error when the reference input r(t) is a unit step; by integral controller, the
response will approach zero at steady-state when the disturbance input d(t) is a unit step.
24

25. References
[1] K. Ogata, Modern Control Engineering, 3rd Edition, Prentice Hall, 1996.
[2] C.T. Chen, Linear System Theory and Design, 3rd Edition, Oxford University Press,
USA, 1998.
[3] T. Kailath, Linear Systems, Prentice-Hall, 1979.
[4] Q.G. Wang, Linear Systems, Lecture Notes, National University of Singapore, 1999.
Citation of this paper
Y. Hong, “The Controller Design For Linear System: A State Space Approach”, Technical
Report, National University of Singapore, November 1999.
Use case: Internet traffic control
Y. Hong and O.W.W. Yang, “Self-Tuning Optimal PI Rate Controller for End-to-End
Congestion With LQR Approach,” Proceedings of 20th International Teletraffic Congress
(ITC-20), Ottawa, Canada, June 2007, pp.829-840. Available on ResearchGate.
Discussions on control system design by ResearchGate members
“What are trends in control theory and its applications in physical systems (from a re-
search point of view)?”
https://www.researchgate.net/post/What are trends in control theory and its applications
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25