Department of Mechanical Engineering

Massachusetts Institute of Technology

2.010 Modeling, Dynamics and Control III
Spring 2002

SOLUTIONS: Problem Set #1

Problem 1

This problem is aimed at helping you picture the different ways in which controls fit into
everyday life. This answers are not exhaustive of the possibilities. In particular for the
robot welder, there are many other possibilities on what to control in addition to position,
such as strength of weld, duration, speed, heat used.
Water closet Refrigerator Shower mixer Robot Iris of your
Welder eye
To flush toilet To let in the
Mixing of hot and Make the weld
Engineering/ each time with appropriate
Keep food cold cold water to in the correct
a customer appropriate
so it wont spoil maintain desired position each
amount of light
goal volume, refill to be able to see
temp. and volume time
tank properly
Being able to
Maintain
A measure of Height of water Proper amount of Position of weld adjust to see
temperature
b good in tank after each
inside relatively
flow at a constant from desired well in bright
performance flush temp. position light and in the
constant
dark.

Volume rate and

Output Temperature Size of opening
c variable
Water height
inside box
temperature of Position of weld
(pupil)
water

Position of shower Position of

knob. (this input is where weld
Reference Water height set Image seen by
d input by floater
Thermostat usually not in should be as
brain
degrees, but rather programmed by
relative position) the user

Eye ball, optic

Plant and Water tank and Icebox and Shower head, knob, Robot and piece
e actuator valve cooling system valve to weld
nerve, iris
muscles

Amount of light
The user feels the
Some type of in the eye as
f Measurement Floater Thermometer correct water temp,
position sensor sensed by the
volume rate
brain
Opening the Loss of hot water in
Floater gets stuck Vibration Sudden burst of
door water input.
g Disturbance Valve is broken
Inserting warm Drop in water
Position change light, dilated
Toilet is clogged of piece pupils
objects pressure
Problem 2

Functional Block Diagram:

Input Transducer: Pilot will input a desired angle through his controls which will be
converted to an input voltage.
Output Transducer: Gyroscope measures the actual angle and converts it to a voltage
Controller Aileron position
Plant Plane dynamics

Desired Roll
angle
input error aileron Actual Roll
voltage voltage Controller positon
Roll
angle
Input Trans. Plant rate
Ailerus integrator
Pilot controls Plane
position

Sensor
gyro

Problem 3

Functional Block Diagram

input
Q Controller position of rods
Input Trans. voltage Plant
Postion of Turbines
sa Reactor
rods
Q
Desired Power level
(heat)
voltage Actual heat
proportional
to power
Sensor
Neutron
Detector

It is important to understand and appreciate the difference between functional block

diagrams and mathematical block diagrams. The latter will be the focus of this
course, but Functional block diagrams are particularly useful in the early design
process of your control systems.
Problem 4

a) Time Constant
Note that this is a first order system which can be rewritten to fit the general form of:
d 1
X + X = input
dt

The equation becomes:

d 1 1
T + T = Qin
dt 60 18

Then it is obvious that = 60.

b) Steady State Gain (K)

d
You can find the steady state gain by letting all the transients go to zero. Namely = 0.
dt
Therefore:
T 10
K= =
Q 3
c) Transfer Function
Find the transfer function by applying the Laplace transform:
1 1
sT ( s ) + T ( s ) = Q( s )
60 18
Then solving for the transfer function
T ( s) 1 / 18
G ( s) = = .
Q( s ) s + 1 / 60

d) Plot
The equation for the curve is
t
10 10 60
T (t ) = e
3 3
Value of t/ t Value of function
1 60 2.0865
2 120 2.8746
3 180 3.1646
4 240 3.2713
5 300 3.3105
Problem 5
Note that this is a second order system.
a) Equation of motion

Equation of motion is found by doing a force balance on the system

Fm = Fx + Fk
m&x&2 = k ( x1 x 2 ) + b( x&1 x& 2 )
which is the same as:
mv&2 = k ( v v ) + b(v1 v 2 )
1 2
b) Transfer function

The transfer function can be obtained by taking the Laplace transform

Y ( s ) V2 ( s ) k + bs
G(s) = = =
U ( s ) V1 ( s ) ms + bs + k
2

c) System Parameters
Given the system parameters: k = 400 N/m, m = 1 Kg, b = 300 N/m/sec
We can calculate all the relevant information about the system.
First we change the denominator so that it looks like the general second order form,
namely:

s 2 + 2 n s + n
2

k +b s
V2 ( s )
G ( s) = = 2 m m
V1 ( s ) s + ( b ) s + ( k )
m m

Then we can easily figure out what each term is:

V2 ( s ) 300 s + 400
G ( s) = = 2
V1 ( s ) s + 300s + 400

k 400 N / m
= = = 20 Hz
m 1 Kg
1 b 1 300 N / m / sec
= = = 7.5 (VERY overdamped)
2 n m 40 Hz 1 Kg

Approximation for the 2% settling time for a second order system is given by:
4 4
Ts = = = 0.0267
n 7.5 20
d) Step Response
300 s + 400 1
T (t ) = L1 2
s + 300s + 400 s

f) Bode Plot
Notice that since the system is highly overdamped there is no peak for this second
order system.
g) Resonant Frequency
The resonant frequency of a second order system is at the peak of the magnitude of the
bode plot. Since the system is overdamped it will not resonate.