Numerous analyses of the problem can be
found online; some running to 70 pages of fairly
dense mathematics. Is there a way to solve this
stabilization problem with just plain common
sense and a little intuition? That is what this
article hopes to do with nothing more than high
school level physics.
First, let’s get a little flavor of the classical
method that I applied in 1965. A system like the
motor driven inverted pendulum can be
represented by complex numbers plotted on a
Root Locus diagram like that shown in Figure 3.
In this diagram, the frequency dependent
characteristics of each component are represented by
“poles” (shown by Xs) and “zeros” (represented by circles).
For example, a charging capacitor with an RC time of 0.1s
would show as an X at - 10 on the real axis. Poles with an
imaginery component (off the real axis) always come in
matched upper and lower pairs, and represent oscillating
behavior such as a pendulum or LC circuit. A stable
oscillator with frequency w = 5 Hz would be represented by
poles at ± 5 on the imaginery axis. Any pole in the right half
plane indicates instability.
In Figure 3a, the poles/zeros of the inverted pendulum
are shown in red; for the motor drive, they are shown in
green. When the feedback loop is closed — that is, the
pendulum angle is amplified and applied to drive the robot
base (referred to as “cart” going forward) motor
— the poles move as a function of the loop gain
according to complex (but well-known) rules as
shown by the arrows. The path of the poles is
known as the Root Locus.
One of those rules states that a pole tends to
move toward a zero as gain increases, but can
never pass it. The right half plane pole in Figure
3a (representing the inverted pendulum
instability) can never escape the right half plane
because of the two zeros at the origin. It cannot
go up and over the origin because poles with an
imaginary component must come in pairs, and we
have only one pole in the right half plane. So,
there is no value of gain for which this closed-loop
system can be stable.
Figure 3a suggests two steps in the path to
stability: (1) We need a second pole in the right
half plane so that the two poles can come
together, break away from the real axis, and make
their way into the left half plane; and ( 2) We
need to add some zeros in the left half plane to attract the
two unstable poles.
Figure 3b shows in blue and purple how to add three
more poles and two more zeros to achive the desired
stability. The left half plane blue poles and zeros can be
implemented with a single operational amplifier (op-amp)
and a few resistors and capacitors. So, what about that
additional purple pole in the right half plane. That means
more instability. What is that all about?
If that control theory discussion leaves you confused,
you are not alone. In 1973, I was asked to give a talk on
the inverted pendulum problem for a group of engineers
and physicists. In 40 minutes, I went through the discussion
above with a lot more details and equations.
At the end of the talk, it was clear that I had not
conveyed any intuition about the problem, and few
By C.A. Hamilton
SERVO 12.2017 47
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Figure 3. Root Locus diagrams for the motor driven inverted
pendulum stability problem. The red poles/zeros represent
the inverted pendulum; the green pole is the motor drive
response; the purple pole is the instability from the
unstable position control loop; and the blue poles/zeros are
the stabilizing compensation function. The small 2s indicate
two zeros at the same point.
The 40:1 gearhead motor and drive wheels.
Rubber O-rings on the wheels provide the required traction.