Numerical and Graphical Illustrations of Fundamental Concepts
in Automatic Control Systems
Robert J. Albright, Ph.D., P.E.
Donald P. Shiley School of Engineering
University of Portland
Portland, Oregon
Abstract
In my teaching of automatic control systems for many years, certain fundamental concepts
can be difficult for students to grasp. One concept is the nature and effects, both favorable and
unfavorable, of feedback; another concept is how steady-state errors can occur in systems even if
one could construct them with perfect components. This paper focuses on such concepts by
utilizing numerical and graphical methods, with analogies familiar to the students, to develop
conceptual understanding before applying detailed mathematical analyses. The basic equations
are presented and applied to enable the instructor to visually present the effects through
comparative graphs of the transient responses, such as the illustrative examples included in this
paper. If time and interest allows for the students to explore further, they can vary the
parameters of the systems and change the inputs to interactively experience the effects first hand
as formal assignments or informal exercises. An added benefit of the illustrations is that the
concepts of differentiation and integration first covered in calculus are visually reinforced and
enhanced for the students. Survey results indicate that the vast majority of my students find
these numerical and graphical illustrations with familiar analogies helpful to learning the
fundamentals of automatic control systems.
Keywords
control systems, control system theory, control system analysis, feedback, steady-state errors,
error analysis, error compensation, differentiation and integration
Introduction
Feedback and steady-state errors are important considerations of control systems for a
multitude of applications, such as the use of machine tools and robotics in manufacturing. As
such they form key aspects of the theory of control systems that students must understand. The
nature and effects of feedback and steady-state errors can be easily discovered by students fairly
early in their study of control systems. This preparation is helpful as background for the more
mathematically challenging aspects of corrective design measures [such as commonly used
proportional-integral-derivative (PID) compensation] which typically follow later in their study
of control systems.
Through multi-faceted, yet simple, graphical examples introduced by the instructor in
lectures, and investigated, if desired, in more depth by the students as formal assignments or
informal exercises, the instructor can guide his/her students to understand the nature and effects
of feedback and steady-state errors. The MATLAB® numeric computation software package,
such as utilized in this paper, can be readily used to perform the simple calculations and
construct plots of the results to demonstrate feedback and steady-state errors for various
scenarios.
The interactive graphical exercises presented in this paper are designed to match the
preferred learning style of most engineering students which is visual, sensing, inductive, and
active, and it provides balance to the traditional lecture presentation which is often auditory,
intuitive, deductive, and passive [1]. The results of the exercises are graphed to appeal to visual
students; the focus of the exercises is on the results of specific examples to appeal to sensing and
inductive learners; and the exercises provides a framework to allow students to easily experiment
with different scenarios to appeal to active learners.
I. Illustrations of the Nature and Effects of Feedback
To demonstrate the nature and effects of feedback, consider the linear control system shown
in Figure 1 where the output c(t) is the integral of an error signal e(t) which is the difference
between the input r(t) and the feedback b(t).
Figure 1. Block diagram of the system.
The input r(t) is set equal to the unit step function, u(t), in this example. The ideal integrator
has a gain (multiplier) K1 and the feedback has a constant (multiplier) K2. If K2 is a positive
number, the system operates with negative proportional feedback; if K2 is zero, the system has
no feedback and hence operates as an open-loop system; and if K2 is a negative number, the
system has positive proportional feedback.
Based on the use of rectangular integration to numerically approximate the ideal integrator,
the output of the system can be approximated by repeatedly solving the following discrete-time
equations:
c(n∆t) = c[(n – 1)∆t] + K1*e[(n – 1)∆t]∆t
b(n∆t) = K2*c(n∆t)
e(n∆t) = r(n∆t) – b(n∆t)
(1)
(2)
(3)
where n is an integer time index beginning at n = 0 and incremented by one after each iteration,
and ∆t is the computation time interval.
A. Numerical Results
With negative proportional feedback (for K2 = +1), and a computational time interval, ∆t, of
0.1 s, the output of the system increases linearly at first as shown in Figure 2, but then begins to
level off toward a stable value of one as the effect of the negative feedback becomes significant.
The error signal decreases toward zero, so the output of the integrator increases at a slower and
slower rate so the output eventually rises to the input value and stays there indicating a stable
response.
1
Signal Value
0.8
Input r(t)
Error e(t)
Output c(t) and Feedback b(t)
0.6
0.4
0.2
0
0
2
4
6
Time (sec)
8
10
Figure 2. Numerical signals for negative feedback.
With no feedback (that is, K2 = 0), meaning the system is operating open-loop, the error
signal is the same as the input unit step function and the output of the pure integrator is simply a
linear function of time as shown in Figure 3.
10
Input r(t) and Error e(t)
Output c(t)
Feedback b(t)
Signal Value
8
6
4
2
0
0
2
4
6
Time (sec)
8
10
Figure 3. Numerical signals for no feedback.
With positive proportional feedback (for K2 = -1), the output of the system increases
indefinitely toward an unstable value of infinity as shown in Figure 4 as the effect of the positive
feedback becomes more significant.
4
1.5
x 10
Input r(t)
Error e(t)
Output c(t)
Feedback b(t)
Signal Value
1
0.5
0
-0.5
-1
-1.5
0
2
4
6
Time (sec)
8
10
Figure 4. Numerical signals for positive feedback.
The numerical analysis cases of negative proportional feedback, no feedback, and positive
proportional feedback presented above demonstrate desirable and undesirable effect of feedback.
Additional results of changing the feedback constant (multiplier), as well as the integrator gain
and the computation time interval, are presented in reference [2].
B. Analytical Results
After the students perform the numerical analyses as homework assignment #1, they apply
Laplace transform theory [3] to solve for the signals analytically as the follow-up homework
assignment #2. The analytical (continuous-time) signal results for the case of negative feedback
are displayed in Figure 5.
1
Signal Value
0.8
Input r(t)
Error e(t)
Output c(t) and Feedback b(t)
0.6
0.4
0.2
0
0
2
4
6
Time (sec)
8
10
Figure 5. Analytical signals for negative feedback.
C. Comparison of the Numerical and Analytical Results
An important part of the learning experience for the students is to compare the accuracy of
their numerical results to the ideal analytical results. The differences between the numerical
results shown in Figure 2 and the analytical results displayed in Figure 5 are shown in Figure 6.
0.02
0.015
Difference in Error e(t)
Difference in Output c(t) and in Feedback b(t)
Signal Value
0.01
0.005
0
-0.005
-0.01
-0.015
-0.02
0
2
4
6
8
10
Time (sec)
Figure 6. Error between the numerical and analytical signals versus time.
The small difference between the numerical and analytical results in Figure 6 (maximum of
about 2% difference) depends on the numerical algorithm used and varies with the rate of change
of the signals. Students can also vary the computation time interval, ∆t, to see that the error
decreases as ∆t becomes smaller approaching the continuous-time analytical case.
II. Illustrations of the Nature and Effects of Steady-State Errors
Imagine that two cars, labeled as Car #1 and Car #2, are in a car performance race at an
automobile test track. Car #1, with its lighter mass M and more streamlined design yielding less
air drag as represented through a viscous damping coefficient B, will be considered as the
reference car. The following analyses will determine the differences in car performance of
displacements, velocities, and accelerations as a function of time, including the steady-state
differences which can be considered as analogous to steady-state errors in control systems. The
race scenario is represented in the block diagram of Figure 7.
x1(t)
Car #1
y1(t)
+
x2(t)
y3(t) = y1(t) – y2(t)
_
_
y2(t)
Car #2
Figure 7. Block diagram.
The motion of each car can be represented in the time domain by the following general
differential equation of Newton’s Second Law:
Md2y(t)/dt2 + Bdy(t)/dt + Ky(t) = x(t)
(4)
where M is the mass of a car, B is the viscous damping coefficient representing air drag, and K is
the “spring” constant taken as zero in this simple modeling of car motion, and where x(t) is the
input (force excitation resulting from pressing the gas pedal) and y(t) is the output (displacement
response of a car).
The Laplace transform of this differential equation is:
Ms2Y(s) + BsY(s) + KY(s) = X(s)
where X(s) and Y(s) are the Laplace Transforms of x(t) and y(t), respectively [3].
(5)
The displacement, velocity, and acceleration responses of the two cars for various car and
input parameters are illustrated and compared in the following sections. For simplicity, units for
the parameters and variables are not included in the equations of motion of the two cars.
A. Displacement, Velocity, and Acceleration Responses of Each Car to a Step Input
Select the following parameter values for the two cars, with Car #2 heavier (larger M value)
and less streamlined (greater B value) than Car #1:
Car #1:
Car #2:
M1 = 1; B1 = 1; K1 = 0
M2 = 1.25; B2 = 1.25; K2 = 0
1. Displacement for a Step Input
The transfer function for displacement is
1
TD(s) = Y(s)/X(s) = ------------------
(6)
Ms2 + Bs + K
where, for a step input, X(s) = A/s corresponding to x(t) = Au(t) where A is the amplitude of the
step with u(t) being the unit step function, giving the displacement, Y(s), as
1
DS(s) = ------------------ (A/s)
(7)
Ms2 + Bs + K
The displacements of the cars for identical step inputs with A1 = 1 and A2 = 1 are shown in
Figure 8.
10
9
8
Displacement
7
Car 1 Displacement
Car 2 Displacement
Displacement Difference
6
5
4
3
2
1
0
0
1
2
3
4
5
Time
6
7
8
9
10
Figure 8. Displacements of the cars for step inputs.
The difference of the car displacements for step inputs tends toward infinity as demonstrated
above in Figure 8.
2. Velocity for a Step Input
The transfer function for velocity is
s
TV(s) = [sY(s)]/X(s) = ------------------
(8)
Ms2 + Bs + K
where, for a step input of amplitude A, the velocity, sY(s), is
s
1
VS(s) = ----------------- (A/s) = ----------------- (A)
Ms2 + Bs + K
(9)
Ms2 + Bs + K
The velocities of the cars for identical step inputs with A1 = 1 and A2 = 1 are shown in Figure
9.
1
0.9
0.8
0.7
Car 1 Velocity
Car 2 Velocity
Velocity Difference
Velocity
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
Time
6
7
8
9
10
Figure 9. Velocities of the cars for step inputs.
The difference of the car velocities for step inputs tends toward a finite value as
demonstrated above in Figure 9. In addition, the velocity plots in Figure 9 correspond, as
expected, to the rate of change (slope or 1st derivative) of the displacement plots of Figure 8.
3. Acceleration for a Step Input
The transfer function for acceleration is
s2
TA(s) = [s2Y(s)]/X(s) = ------------------
(10)
Ms2 + Bs + K
where, for a step input of amplitude A, the acceleration, s2Y(s), is
s2
1
AS(s) = ------------------ (A/s) = ------------------ (As)
Ms2 + Bs + K
(11)
Ms2 + Bs + K
The accelerations of the cars for identical step inputs with A1 = 1 and A2 = 1 are shown in
Figure 10.
1
0.9
0.8
Acceleration
0.7
Car 1 Acceleration
Car 2 Acceleration
Acceleration Difference
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
Time
6
7
8
9
10
Figure 10. Accelerations of the cars for step inputs.
The difference of the car accelerations for step inputs tends toward zero as demonstrated
above in Figure 10. In addition, the acceleration plots in Figure 10 correspond, as expected, to
the rate of change (slope or 1st derivative) of the velocity plots of Figure 9.
B. Implementation of Changes to Impact Car Performance
The performance of a control system depends on the nature of both the system and its input,
as illustrated by varying the car and input parameters of this illustrative example, such as
follows.
1. Impact of Car Parameters on Performance
As observed in the above figures, the displacement, velocity, and acceleration of Car #2 with
its larger mass and greater damping (air drag) system characteristics lag those of Car #1 for
identical inputs. One way to enable Car #2 to catch up with Car #1 in performance is to redesign
it using lighter materials to reduce its mass and making it more streamlined to reduce its air drag.
To model this scenario, select the following identical parameter values for the two cars, with
the parameters of Car #2 equal to those of Car #1 as applied in Section II.A for Car #1:
Car #1:
Car #2:
M1 = 1; B1 = 1; K1 = 0
M2 = 1; B2 = 1; K2 = 0
To illustrate the car performances for this scenario, the drivers apply step inputs of equal
amplitudes to the gas pedals, specifically A1 = 1 for Car #1 and A2 = 1 for Car #2.
As expected with identical parameters values for both cars, the displacements, velocities, and
accelerations of the cars for step inputs of equal amplitudes are identical to those of Car 1 in
Figures 8, 9, and 10, respectively.
2. Impact of Input Parameters on Performance
a. Displacements of the Cars for Different Step Inputs
Another way to enable Car #2 to catch up with Car #1 in performance is for the driver of Car
#2 to apply more input to the gas pedal than by the driver of Car #1.
To model this scenario, select the same parameter values as applied in Section II.A for the
two cars:
Car #1:
Car #2:
M1 = 1; B1 = 1; K1 = 0
M2 = 1.25; B2 = 1.25; K2 = 0
To illustrate the car performances for this scenario, apply a step input of amplitude A1 = 1 for
Car #1 and a step input of larger amplitude A2 = 1.25 for Car #2.
With different parameters values for the cars, the displacements, velocities, and accelerations
of the cars for step inputs of the selected compensating amplitudes are identical to those of Car 1
in Figures 8, 9, and 10, respectively.
b. Displacements of the Cars for Step versus Ramp Inputs
Another way to enable Car #2 to catch up with Car #1 in performance is for the driver of Car
#2 to increase the input to the gas pedal over time, eventually exceeding the input by the driver
of Car #1.
To model this scenario, select the same parameter values for the two cars as used above in
Section II.B.2.a.
To illustrate the car performances for this scenario, apply a step input of amplitude A1 = 1 for
Car #1 and, for example, a ramp input of slope A2 = 0.5 for Car #2. The resulting displacements
of the two cars are shown in Figure 11.
20
Car 1 Displacement
Car 2 Displacement
Displacement Difference
15
Displacement
10
5
0
-5
-10
0
1
2
3
4
5
Time
6
7
8
9
10
Figure 11. Displacements of the cars for step versus ramp inputs.
As observed with different parameters values for the cars, the displacements of the cars for
step versus ramp inputs are different as shown above in Figure 11. Car #2 eventually catches,
and passes, Car #1. Similarly, the velocities and accelerations of the cars for step versus ramp or
parabolic inputs will be different (plots for these responses are not included in this paper).
C. Summary of the Steady-State Differences
The results of the steady-state differences (analogous for illustrative purposes to traditional
steady-state errors in control systems) of the displacement, velocity, and acceleration responses
for step, ramp, and parabolic inputs for this illustrative car example are summarized in Table 1.
Table 1. Summary of Steady-State Errors.
Displacement
Velocity
Acceleration
Step
Infinite
Finite
Zero
Ramp
Infinite
Infinite
Finite
Parabolic
Infinite
Infinite
Infinite
The results in Table 1 agree whether obtained by examining the responses in the time-domain
as t approaches infinity, or obtained in the s-domain by applying the Final-Value Theorem of
Laplace Transform Theory [3]. As expected, rapidly changing inputs make it difficult for Car #2
with its larger mass and greater air drag to match the performance of Car #1.
Assessment and Evaluation
Assessment and evaluation has been obtained informally for several years on the
effectiveness of the feedback illustrations used in EGR 404 Automatic Control Systems (3
semester credit hours). To assess and evaluate the effectiveness of the car race analogy to
explain steady-state errors, the students were asked to formally provide their opinions using a
survey form during the past two yearly offerings of the course.
The majority of the students had limited prior knowledge of feedback and even less, if any,
awareness of steady-state errors. The students indicated that the numerical and graphical
examples provided them valuable conceptual understanding prior to detailed mathematical
coverage of feedback and steady-state errors in the course.
Summary
Through multi-faceted, yet relatively simple, examples, the author has presented numerical
and graphical illustrations that allow students to discover the nature and effects of feedback and
steady-state errors in control systems. These flexible examples can also enable students to
explore various compensation measures. As an additional benefit, students can view the
derivative and integral relationships between the sets of graphs, such as those that relate
displacement, velocity, and acceleration, to complement and reinforce their learning in calculus,
physics, and engineering. These illustrative examples are designed to appeal to the preferred
learning style of most engineering students, namely visual, sensing, inductive, and active.
Acknowledgments
The author acknowledges the contributions of his students in automatic control systems and
the assistance of his colleague Dr. Joseph Hoffbeck during the past several years in helping to
refine the numerical and graphical examples illustrated in this paper.
References
[1] R. M. Felder and L. K. Silverman, “Learning and teaching styles in engineering education,” Eng. Educ., vol. 78,
no. 7, pp. 674-681, Apr. 1998.
[2] R. J. Albright and J. P. Hoffbeck, “Discovering the Effects of Feedback on Control Systems: Informative and
Interesting Numerical Exercises,” IEEE Trans. Educ., vol. 44, no. 2, pp. 104-108, May 2001.
[3] F. Golnaraghi and B. C. Kuo, Automat. Contr. Syst., 9th ed. Wiley, NY, 2010.
Biographical Sketch
Robert J. Albright received the B.S. and M.S. degrees in electrical engineering from Oregon
State University, Corvallis, in 1963 and 1965, respectively, and the Ph.D. degree in electrical
engineering from the University of Washington, Seattle, in 1971.
He is a Professor and Chair of Electrical Engineering at the University of Portland, Portland,
OR. A member of the Faculty of the University of Portland since 1970, he has served 33 years
as Chair of Electrical Engineering, 12 years as Chair of Computer Science, and one year as
Acting Dean of Engineering. He has been honored as a Tyson Distinguished Professor at the
University of Portland. His teaching, research, and consulting interests include energy
conversion, power systems, control systems, and engineering education.
Dr. Albright, a registered engineer in the State of Oregon, is a senior member of the IEEE
and a member of the ASEE.