ANALOGUE COMPUTERS
INTRODUCTION
There are 2 main types of computer. What you probably think of when you say "'computer" is a
digital computer. Numbers are represented by sets of 1's and 0's where 1 and 0 are represented by
two different voltages, (typically +5V, and 0V). Operations are simple logical operations (i.e.
AND, OR, etc.) or arithmetic operations (i.e. addition or subtraction) .Calculus-type operations
are very complicated to do.
Another sort of computer is the analogue computer. In this sort of computer, numbers are
represented by continuously-varying quantities. Typically, voltage is the quantity used, though
other things might be used.
In the first part of this experiment, the fundamental building block of the electronic analogue
computer, the Operational Amplifier, is studied. The "OP-AMP" is placed in circuits where it can
be used to add, subtract, integrate (with respect to time), etc., various voltages.
In the second part of this experiment, these circuits are combined to solve various differential
equations.
It is left to the student to plan her/his own course of experimentation. The various exercises are
presented as possible 'Nays of investigating electronic analogue computers.
For further background and for ideas on pursuing the experiment, the student is referred to:
§
§
§
Jacob Millman, Microelectronics 1979; Chapters 15, 16. (Available in the Physics
Library).
J. Millman and C.C. Halkias, Integrated Electronics; Chapters 15, 16. (Available in the
Physics Library).
The Griffin Operational Amplifer Booklet. (Available at the wicket.)
To aid the student in investigating the analogue computer, the type 741 OP-AMP chips have been
conveniently mounted with power supply connections and extra resistors, capacitors and balance
controls in modular boxes. Other boxes are also available to provide variable attenuation and
voltages. Moreover, a control unit for setting initial voltage conditions is part of the apparatus. You
will use all these components as you proceed with the experiment.
PART 1: THE OPERATIONAL AMPLIFIER (OP-AMP)
Open Circuit Gain - Balancing the OP-AMP
EXERCISE: Connect the Op-Amp to the + and - 6 volt supply, observe the output with a SCOPE,
and with no feedback, vary the input voltage from positive to negative (See Fig. 1).
(Note: a D.C. variable input voltage may be obtained from the function generator's output by
setting the "Mode" button to trigger without supplying a "Trigger In" signal, and then by varying
the DC offset control. Then, if the "Mode" button is changed to "Continuous", the function
generator gives an appropriate A.C. signal superimposed on the D.C. voltage).
Notice that the unfed-back gain of the Op-Amp is very high (estimate how high) and that it is hard
not to have the Op-Amp's output voltage saturated. At what voltages does the output voltage
saturate?
EXERCISE: Now set the input. voltage in the above circuit to zero volts, and vary the Op-Amp
balance control, noting its effect. on the Op-Amp output. voltage. Set the balance control so that.
the Op-Amp provides zero output volts for zero volts input.
Note that. in your use of the Op-Amps, it. is important that. the balance control is always properly
set. Note also, that. it is important. that the Qp Amps are always used under-conditions that they are
functioning linearly, i.e. the output voltage should not swing to its saturation values.
Inverting and Non-Inverting Amplifier
An inverting amplifier is obtained by:
A non inverting amplifier is obtained by:
A unity gain (voltage follower) amplifier is obtained by:
EXERCISE: Observe the output voltages on the SCOPE for various input voltages obtained from
the function generator for the above circuits, for your own selected values of Rf' Ri. (Note that the
Op-Amp box has available in it Rf = 1 megohm and Ri = 1 megohm and 100 kilohm). Use A.C. and
D.C. input signals of various shapes and frequencies. Note the gain, frequency response, distortion
of wave-shape, and maximum output swing before saturation for these amplifier circuits. It is
instructive to obtain, for one circuit, a frequency response curve, log of (output/input) vs log
frequency, using sin wave input signals of frequencies from approximately 10 Hz to 2 MHz.
Summing Amplifiers
An inverting summing amplifier is obtained by:
A difference amplifier is obtained by:
EXERCISE: Observe the output voltage on the SCOPE for various combinations of input voltages
for the above circuits. A.C. or mixtures of A.C. and D.C voltages may be obtained from the
function generator. Also D.C. voltages may be obtained from the D.C. voltage source modules.
Fractional amounts of given input voltages may be obtained using resistors provided in potential
divider arrangements:
Integrator Circuits
Integration with respect to time is obtained by:
vo =
1
ò vi dt + v oi
RC
where voi is the value of vo at t = 0.
Provided that the times over which the integration is being considered are considerably less than
the (Op-Amp Gain) ´ (RC).
A summing integrator is obtained by:
Note that the integrator is a bit trickier to use than the previous circuits because its output not only
depends on the present input voltage, but also on the past history of input voltages. Thus, vo
depends on the initial voltage (at t = 0) voi and also the input voltage vi from t = 0 to the present.
The setting of voi may be done by setting the voltage across the capacitor C to this initial value. In
particular, if it is wished to set the integrator output voltage to an initial value of vo = voi = 0 volts,
Then the capacitor, C, should momentarily short circuited, imposing a 0 volt condition on it.
EXERCISE: Observe the output voltage on the SCOPE for a small constant D.C. input voltage.
Notice what happens to the output as you vary the input through positive and negative values. This
corresponds to integrating various constants with respect to time. Notice that the integrator output
voltage saturates at its maximum or minimum value if the integration goes on for sufficient time.
(Once saturated, the integrator stops integrating.) Thus it will be necessary, from time to time, to
reset the integrator by once more defining Voi, by setting the capacitor voltage.
Note that if the Op-Amp is not properly balanced, a zero input voltage will produce a continuously
drifting output voltage. This can be corrected using the balance control.
EXERCISE: With a short circuit (0 volt) input on the integrator, adjust the balance for minimum
drift.
EXERCISE: Observe the output voltage of the integrator on the SCOPE for various A.C. input
voltages from the function generator. Use a variety of wave shapes and frequencies in order to view
the output adequately, it will probably be necessary to choose frequencies lower than 20 Hz. It will
also be necessary to reset Voi from time to time.
Control Unit
This unit simplifies the use of up to two Op-Amps as integrators, when used in solving a
differential equation. It simultaneously permits the pre-setting of Voi to a selected value, while
facilitating the connecting or disconnecting of the input of the integrator. The control unit contains
two groups of sockets, switches and a 'Set' potentiometer to enable the setting up and operation of
two integrators and the simultaneous connection of the selected circuits. Fig. 9 shows the internal
circuitry for one group of sockets.
The integrator is connected to the control unit
as in Fig. 10. The yellow output socket and the
SJ socket on the module are connected to the
yellow OP socket and black SJ socket on the
control unit respectively. The blue input
sockets on the module are connected, as
necessary, to control unit blue sockets A or B.
Between A and the corresponding B (in the case
of the centre and right-hand set of sockets) ,
there is a 910 KW resistor which can be brought
into circuit by using socket A. This allows the
total input resistance of the operational
amplifier to be brought up to 1MW when using
one of the module's 100KW inputs, so making
CR = 15. Socket B should be used if connection
is made to the 1MW module input (for
CR = 15) or if it is to be connected to a
100 KW module input to give a constant
multiplier of 10(CR = 0.1s). Between A and B
in the left-hand set, the resistance is 10KW.
Outputs from other circuits are connected to
white sockets, C. These sockets are separated
from the corresponding B socket by an independent switch operated by the central control switch.
The latter has four positions labelled 'Set' , 'Hold' , 'Run' and 'Triggered'. The first three positions
are for manual operation and their use will be described first.
The 'Balance' and the 'Null' potentiometers are omitted for the sake of clarity.
When the integrator is connected to the control unit as in Fig. 10 the circuit diagram is as in Fig. 11.
With the control switch in the 'Set' position, each B socket switch is open, and the setting circuit is
connected across the operational amplifier so that it becomes an inverting amplifier with a gain of
one. The input to this is the potential at the sliding contact of the 'Set' potentiometer on the control
unit. Thus the output voltage of the integrator can be set to any initial value in the non-saturated
range. from which the integration will start. In the 'Hold' position the B socket switches remain
open and the setting circuit switch now opens. The feedback capacitor, C, has no route by which it
can discharge (except through the input resistance, r, of the operational amplifier for which the time
constant is (GAIN) ´ (RC) ~ 105s), and the 100 KW resistors of the setting circuit form a negligible
load on the amplifier. In the 'Run' position, each B socket is connected to the corresponding C
socket, with no change in the setting circuit switch so allowing integration to proceed.
EXERCISE: As a practise in using the control unit, set up an integrator connected to the control
unit as in Fig. 10. Provide an A.C. input voltage on v1 or v2 and observe vo. Familiarize yourself
with the various controls and functions of the control unit, and with the effects of switching from
.'Set.' to .'Hold" to "Run".
PART II - SOLVING DIFFERENTIAL EQUATIONS
The following experiments are exercises in problem-solving. It is instructive to think of each
experiment as consisting of four steps. The first step is the building of a 'model' of the problem.
The model is a mathematical description of the problem derived from known physical laws and/or
experimental data after making reasonable simplifications. It will include a set of algebraic and/or
differential equations, initial conditions, values of coefficients and range of variables of interest.
The second step is to rearrange the equations to get a set of 'computer equations' suitable for
interpretation. The third step is to interpret the equations term-by-term and from there construct the
computer. The fourth, and happiest, step is running the computer and recording the output. The
output, a plot of voltage against time, may be viewed on a SCOPE or may be plotted on an x-y
recorder, with the x-input connected to a .'Time Generator".
The following experiments involve the solution of three types of differential equations. Other
possible problems are also suggested in the Griffin Operational Amplifier Booklet, available at the
wicket.
Experiment 1: A Time Generator (The Integrator Revisited)
Recall that the integrator integrates with respect to time - real time, that is. Now it turns out that
some processes of interest vary with real time too quickly or too slowly to be computed
conveniently. Therefore, the passage of real time itself can be simulated by a voltage varying
linearly (with time) and the output of the computer can be expressed with respect to this voltage.
The voltage simulating time can also aid in displaying the output by providing the time base of an
oscilloscope or the x-drive of an xy plotter. For the circuit in Fig. 12, with vi = -1 Volt, D.C. and S
a switch which is closed at time, t = 0, and vio = 0, the output voltage T seconds after the closure is
T
vo (T) = - ò (-1)dt = T volts
0
Also, the voltage has increased at the rate of 1 volt/sec.
Clearly, the voltage is now simulating time -at a ratio of 1:1. The circuit of Fig. 1 can also be
thought of ~s a clock with displays the passage of time following the closure of S. The 1:1 ratio can
be changed by changing the forward gain of the integrator.
THE EXPERIMENT: Set up the circuit of Fig. 12 using an Op-Amp. module. Use a multimeter
or a SCOPE to display the output. To obtain vi use the module 'voltage source'. For 'S' use one of
the
switches (white-blue sockets) on the control unit. Balance the integrator. Momentarily short circuit
the capacitor, C. This procedure sets the initial condition of 0 volts. Switch the computer from
'SET' through 'HOLD' to 'RUN'. The output vo(t) should appear on the meter and should climb from
0 volts to about 6 volts in about 6 seconds.
Once the output reaches 6 volts, the voltage of the power supply, it will remain constant. The
amplifier is then saturated. To return the computer to zero, the starting condition, return the centre
switch to 'SET' and short circuit C momentarily. The output should return to zero. Once at zero,
you are ready for another run.
Now connect the output; to the input of the scope or the x-input of the chart recorder, or both, and
observe the use of the circuit to provide a time base.
With the construction of this real time computer, you now have a device which will prove useful in
the experiments to follow.
Experiment 2 - Solving
dv
= -ky
dt
This is an equation which may describe the charge on a capacitor, or a simple radioactive decay or
the cooling of a body under forced convection.
The equation may be computer modeled by noting that if the computer output voltage is y, its input
dv
must be
. Thus the following circuit (Fig. 13) may be used.
dt
Thus if the integrator has its initial conditions set so its output voltage is yo before S is closed, then
dv
when S is closed, is made equal to ky, and the circuit provides an integration of the equation
dt
subject to the initial conditions.
THE EXPERIMENT:
Set up the circuit in Fig. 14, using the control unit to provide initial conditions and to serve as
switch S. Connect the y voltage output to the y input of a SCOPE, and connect the x input to the
time generator circuit of Experiment 1.
Resetting the time generator circuit as in Experiment 1, observe the plots of y vs t obtained (as a
solution to this equation) by switching the control unit from "set" to "hold" to "run". Find solutions
for various values of k and various initial values of y.
Experiment 3: Radioactive Series (Two isotopes)
It is known that the number of radioactive nuclei of a given specie in a sample decreases
exponentially with time. Suppose now that the daughter nucleus is also radioactive. Then how do
the number of daughter nuclei vary with time? This experiment attempts to answer this question.
The computer requires three Op-Amps -an integrator, an inverter, and an integrator/adder.
This situation may be mathematically modelled as follows: A radioactive sample contains two
radioactive species of interest - a parent and daughter. At some instant of time let I be the number
of parent nuclei and X be the number of daughter nuclei. The rate of decay, or activity, of the
parent is
dI
= -a I
dt
or
-
dI
= aI
dt
where a is a constant, characteristic of the parent, called the decay constant. Also, a = 0.693/T
where T is the half-life. If the number of parent nuclei are decreasing at this rate, the number of
daughter nuclei must be increasing at the same rate. Concurrently, the number of daughter nuclei
must be decreasing at some rate determined by the decay constant, say b, of the daughter. Thus,
dX
= -bX + aI
dt
dX
= bX - a I
dt
These equations are computer modeled by two integrators and an inverting amplifier (gain of -1), as
in Figure 15.
Thus when S1, S2 and S3 are simultaneously closed,
-
dX
dI
= aI a n d = bX - a I
dt
dt
producing conditions. solutions to those equations subject to the initial conditions.
THE EXPERIMENT: Try solving the following problem.
A radioactive sample contains parent and daughter nuclei. At some time t = 0, the population of the
parent is estimated to be 10 times 106 atoms and the daughter 1/5 as many. The parent's half-life
is 0.693 sec. and the daughter's half-life twice as long. How does the daughter population vary with
time?
The job calls for the computer in Fig. 15. But what are the initial conditions?
Hints: An initial population of 10 times 106 atoms can be simulated by an initial voltage of 10 volts
on the I integrator. What then, is the initial condition on the X integrator? Calculate the values of a
and b and set the potentiometers accordingly.
What is the analytical solution of eq(2)? Make some quick checks on the accuracy of the computer.
Now proceed to plot curves for various values of a and b. Can you interpret them physically?
Experiment 4: Damped Free Oscillations
Many systems exhibit one dimensional oscillatory motion, as a result of obeying the second order
differential equation:
d2x
dt 2
+g
dx
+ w02 x = 0
dt
with g providing a damping term and wo being the natural frequency.
g
- wo )
2
is greater than, less than or equal to zero (over damping, under damping or critical damping.)
The nature of the solution of this equation depends on the extent of damping, or whether (
This equation may be computer modelled using two integrators (one of these a summing integrator)
and an inverter (gain of -1 amplifier), plus two attenuators.
THE EXPERIMENT: It is left to the student to computer mode this equation. This model should
include potentiometers to set g and wo2 and should include a way of setting the initial values of x
dx
and
. (Hint: The model is most easily set-up making one integrator the x integrator, and the
dt
dx
other the ( ) integrator.)
dt
Set-up the circuit and obtain solutions to the equation for various values of wo, g, and for various
values of the initial conditions. Explore values of g which produce over-, under-, and criticaldamped conditions.
Experiment 5 - Anything Else
Define a physical problem, and its corresponding equation. Model the equation, and obtain
analogue solutions. Consult with a demonstrator at each step of this experiment.