A Practical Approach to Analog Computers
APPLICATIONS REFERENCE LIBRARY
GENERAL PURPOSE ANALOG COMPUTATION
A PRACTICAL APPROACH
ANALOG COMPUTERS
ay Bee eo by
John D. Strong
George Hannauer
Electronic Associates, Inc.
®)
Abstract: The purpose of this paper is to introduce engineers to the basic principles
of analog computation. It explains briefly how this versatile problem-solving tech-
nique helps to increase engineering efficiency. Several types of computing modules
from presently available General Purpose Analog Computers are illustrated and de-
scribed, These modules represent typical details of hardware.
ELECTRONIC ASSOCIATES, INC., LONG BRANCH, NEW JERSEY
FIG. |. PACE 231R analog computer.
A PRACTICAL
APPROACH
TO ANALOG
COMPUTERS
JOHN D. STRONG
GEORGE HANNAUER
Electronic Associates, Inc.
The analog computer permits any engineer to
analyze and synthesize a system with speed and
efficiency. Here is a practical introduction to the
analog computer, and a guide to its use.
Reprinted from Instruments & Control Systems
Vol. 35, No. 8, August 1962
Shree &
tial record of achievement, especially in solving
problems of guided-missile trajectories and air-
frame design. Its use is spreading rapidly into other
fields, including process simulation, analysis of control
systems, power plant design, etc.
At first sight, the numerous dials, switches, and
indicators (Fig. 1) make the analog computer ap-
pear complex. However, the computer consists of a
large number of a few basic types of similar build-
ing blocks. This is why the average engineer can
be taught to operate it in as little as 8 hours; with
a week of practice and study he can successfully
set up and run worth-while problems. This article
nas been written for such an engineer—one who has
heard of what analog computers can do, and who
wants to know how the computer would help to
solve his engineering problems and thus increase his
engineering efficiency.
Although several types of analog computers are
available, the basic principles of all of them are
similar, so that a description of one type will pro-
vide an adequate introduction to the field. Refer-
ence will be made to the EAT Model 231R electronic
computer as a typical general-purpose analog com-
puter (Fig. 1) to illustrate details of hardware.
Gres ANALOG COMPUTER has a substan-
The Electric Analog of a Physical Variable
On an analog computer, physical variables such
as weight, temperature, or area are represented by
voltages—that is, voltage is the electrical analog of
cm
FIG. 2. PRE-PATCHED panel being inserted into com-
puter.
the variable being analyzed, which can be mechanical,
hydraulic, pneumatic or even electrical in nature. Arbi-
trary scale factors relate voltages in the computer
to the variables in the problem being solved.
The computer components are designed to operate
within an output voltage range of +100 volts, and
all computer variables are scaled to lie within this
range. Thus, a temperature T which can vary from
0 to 1000°C is represented on the computer by a
voltage that varies from 0 to +100 V (100 V=1000°C;
1°C= 0.1 V). The scale factor would be 1/10 volt
per degree Centigrade.
As another example, a displacement x which varies
from —5 to +10 inches could be represented by a
voltage that varies from —50 to +100 V, and the
scale factor would be 10 volts per inch.
The computer components are interconnected so
that the voltages in the computer are related by
the same mathematical equations as the original phys-
ical variables. Thus, if a voltage on the computer
represents temperature in a chemical reactor, and
is scaled for 1 volt per 5 degrees Centigrade, a graph
showing this voltage increasing from 10 to 100 volts
would mean that the reactor temperature varied from
50 to 500 degrees. It is from this analogy between
the problem variable and the computer voltage that
the analog computer derives its name.
Computer Elements
The computer consists of a few basic components,
or “building blocks,” which perform mathematical
operations such as addition, subtraction, multiplica-
tion, division, integration, etc. The basic components
are (1) amplifiers, (2) potentiometers, (3) multipliers
and (4) function generators. These components can
be connected to solve a variety of equations.
In practice, the computer elements are connected,
or “patched,” to set up (and solve) the equations
which are derived from the dynamics of the system
to be analyzed. The interconnections are made by
means of a patch panel (Fig. 2). Every computing
component has its input and output terminations
on this panel, and leads can be run between any
two holes on the panel, connecting the components
in any desired manner. The panel is removable, and
problems are usually patched external to the com-
puter. Hence the machine is not tied up while the
connections are being made; problems can be patched
on individual patch panels while the computer is
solving another problem.
The Amplifier
The basic computer component is the amplifier,
a direct-coupled amplifier with d-c gain of 100 million
or greater. This means that an input voltage of 10-8
volt (only 0.01 microvolt) could produce an output
of a volt or more. There is also-a sign (phase)
change associated with the amplifier—that is, a posi-
tive voltage input produces a negative voltage out-
put, and vice-versa. The symbol for an amplifier is
shown in Fig. 3.
OUTPUT
FIG. 3. HIGH-GAIN amplifier. The input voltage is mul-
tiplied by gain A (108 or greater); minus sign indicates
phase reversal in amplifier.
The Summing Amplifier
The amplifier can be used to add voltages as shown
in Fig. 4. R; and Ry, are called input resistors, and
> Z(-11 V)
FIG. 4. HIGH-GAIN amplifier used as a summer.
R; is the feedback resistor. Applying Kirchoff’s law
for current at the summing junction, we have:
im + ine + ine + ip = 0
where ig is the input current to the amplifier. If
the voltage at point S is e,, applying Ohm’s law:
Xa +: x + oe =~i
The amplifier itself is designed to draw as little
grid current (ig) as possible from the inputs X and
Y. As i, is negligible in comparison with the other
currents (ig = 10~-® amp), the right-hand side of
equation (1) can be set equal to zero without signifi-
cant error. Since the amplifier output Z = Ae,, where
A = the gain of the amplifier, e, = —Z/A. Sub-
stituting —Z/A for e, in equation (1) and solving for
the output voltage Z:
~ X 4 _Y_ a2, tfig~iad
2=—(a tae etal tate) ] ©
Now we see the reason for the extremely large
gain, A. If A is sufficiently large, we may ignore the
terms involving 1/A, and obtain the approximations:
Pee Rr
z=-[oco+ gem] @
If all three resistors are equal, then Z =—(X + Y).
Except for the sign reversal, we have succeeded in
adding two voltages. If all resistors are equal, the
error introduced by ignoring the 1/A terms is less
than 3 parts in 100 million.
This method is not limited to two inputs. With
additional input resistors, three or more voltages can
be added.
Multiplying By a Constant
When the input and feedback resistors are equal,
the amplifier is a simple summer. However, when the
input resistor is a fraction of the feedback resistor,
the output (Z) equals the input (X) multiplied by
the factor (R-/R), as shown in Equation 3.
Hence, if the feedback resistor Ry is 10 times the
input resistor, the output voltage is 10 times the
input voltage.
In actual computer diagrams the resistors are not
drawn, but a gain of 1 or 10 is indicated, as shown
in Fig. 5.
With a single input (gain of 1) the amplifier is
simply an inverter (Fig. 6). With summers and in-
verters, subtraction can be performed as shown in
Fig. 7. The amplifier at left is simply an inverter;
the amplifier at right is summing its two inputs.
The Potentiometer
We now can add and subtract voltages, change
sign (multiply by —1) and multiply by 10. To mul-
tiply a voltage by a factor other than 1 or 10, a
potentiometer (“pot” for short) is used. Fig. 8 shows
a pot schematic; Fig. 9 shows the symbol for a pot
(a circle); Fig. 10 shows the appearance of pots
on a computer; Fig. 11 shows use of pots.
In Fig. 8, the voltage on the arm of the potentiom-
eter is k times the input voltage, where the factor k
can be set to any value between zero and one.
Potentiometers on the computer can be equipped
with a calibrated dial, and the value of k set directly
on the dial—if high accuracy is not required.
However, this setting can be in error for several
reasons—electrical loading, mechanical misalignments
of the dial, backlash, etc. To avoid these errors and
to set a pot very accurately, the potentiometer is set
by monitoring its output with the actual load con-
nected to the pot wiper. Each pot is equipped with
a switch which, when depressed, disconnects the in-
put terminal of the pot from the circuit and connects
it to a fixed voltage (+100 volts). The arm of the
potentiometer is monitored on a voltmeter and the
dial is rotated until the correct reading is obtained.
For example, to set a pot to 0.5317, the operator
depresses the switch and then rotates the dial until
53.17 volts appears on the meter (Fig. 11). When
the switch is released the pot, now set correctly to
0.5317, is reconnected into the circuit.
The computer symbol for a potentiometer is simply
a circle, as shown in Fig. 9.
Reference Voltage
At various points in a problem, constant voltages
of different values are required. These are supplied
from highly stabilized circuits which provide d-c volt-
ages of +100 and —100 volts. These “reference” volt-
ages are made available on the patch panel. To
produce constant voltages other than 100 volts, the
pots are used in conjunction with the reference volt-
age. For example, if the pot in Fig. 11 is set to
0.5317, and connected to the —100-volt reference,
the output is —53.17 volts.
The Integrator
The basic equations of physics and engineering are
differential equations—that is, variables expressed in
terms of their derivatives, or rates of change. The
following is a simple ordinary differential equation:
:
If one has d?x/dt?, dx/dt can be obtained by per-
forming one integration. If one has dx/dt, one can
obtain x by performing an integration of dx/dt. Math-
ematically, this is written
d’x , _ dx . dx _
ik dt =at Ci; fie =x+Q
The device that performs integration in the analog
computer is called an integrator (Fig. 12).
The integrator itself is simply an amplifier with a
capacitor as its feedback*® (Fig. 13). Comparing Fig.
13 with Fig. 5, one can see that the integrator and
summer are identical except for the feedback; the
integrator uses a capacitor, the summer a resistor.
If several input resistors are used, the amplifier
integrates and adds simultaneously—that is, the out-
put is the negative of the integral of the sum of the
inputs. If the inputs are x;, x2... . X,, and the cor-
responding input resistors are R,, Re, .... Ra, then
the output is:
—-— ff) 2 4 _Xn_
z= iLakct & Fone ze |
*Making the same approximations as before (ignoring the
input current to the amplifier and dropping all terms in-
volving the factor 1/A), we obtain the expression for the
output of an amplifier with a feedback capacitor (Fig. 13).
1
Z=— RC Sxdt
i
sal
_—
IM 1
IM x | >
X Y 10
eee |
Y 0.1M FIG. 5. SUMMING amplifier (left) and its computer
symbol (top). Output Z = — (X ++ 10Y)
I -x 1
<> x — x-Y
Y
FIG. 6. WITH only one input as shown an amplifier FIG. 7. AMPLIFIERS used for subtraction. Here
is simply an inverter. two amplifiers are used to obtain X—Y.
INPUT
x K
R FIG. 8. A POTENTIOM- x (\
Kx ETER is simply an attenu- 4 Kx
K (OUTPUT) ator. FIG. 9. SYMBOL for a poten-
tiometer is a circle.
FIG. 10. POTENTIOMETERS on
the 231R. Note that each "pot"
is equipped with a switch for use
in setting to high accuracy.
100 0.5317 -~53.17 FIG. Il. PRODUCING a constant (— 53.17 volts) by using the ref-
{ \ erence voltage and a potentiometer.
LY
a a |
C
~dx ax 1t
d@x dt dt -X
=; R
dt2 X — wwe L__ 2
FIG. 12. INTEGRATOR symbol is a rectangle against — oo.
a triangle. FIG. 13. INTEGRATOR is simply an amplifier with
a feedback capacitor.
dx
"at +X FIG. 14. EACH INTEGRATION reduces the order of
the input derivative.
:
FIG. 15. SUMMING
the dx/dt and x terms
provides the right side
of equation (5); thus
equation (5) is satisfie 2
by joining the output ax dx
INS dx
* GK
and input as shown by dt dt 2
the dotted line. {
|
Lee |
“a dx
ad +x
=x FIG. 16. INTEGRATOR can sum as well as integrate.
ax/at Cat 7
Simple Programming
The output of an integrator is always one order
lower than the input. Conversely, the input of an
integrator is always one order higher than the output.
This fact is the key to setting up and using the analog
computer.
The basic steps in setting up the differential equa-
tion of the analog computer are:
1. Start with the highest-order derivative.
2. Integrate the highest-order derivative to obtain
the next lower-order derivative.
3. Repeat the process to obtain all derivatives of
desired order.
4, Multiply each term by the desired constants,
as given by the equation.
5. Add the terms as given by the equation.
6. Satisfy the equation by closing the loop with
the proper terms.
Equation (4) will be used as an example. The first
step is to rewrite the equation with the highest-order
derivative on the left side:
dx __ dx
“ae =~ a 7 * 6)
This equation says that the highest-order derivative
must equal the sum of the two terms on the right-
hand side.
If we have a second-order derivative, we can ob-
We wish to add dx/dt and x, but with the same
sign. Thus we invert the term (dx/dt) and add it to
x by using an adder, as shown in Fig. 15. The output
of the summer is now the desired sum on the right
side of equation (5), with the proper signs.
Equation (5) also says that this sum must equal
d’x/dt?. This is realized by connecting the output
of the summer (that is, the sum of the two terms
in equation 5) to the input of the first integrator
(which was assumed to equal d?x/dt? at the begin-
ning of the problem). The circuit in Fig. 15 now
satisfies equation (5), and all voltages inside the).
circuit must change only as demanded by equation ~~
(5).
Equation (5) might represent an actual physical
situation such as a wheel motion. The wheel system
is now simulated by the circuit in Fig. 15. The vari-
able x (which would be wheel displacement) can be
examined at the output of integrator 2; the term
dx/dt (which would be the velocity of the wheel)
can be examined at the output of integrator 1; the
term d?x/dt? (which would be the acceleration of
the wheel) can be examined at the output of the
summer or at the input of integrator 1. Note that
the displacement, velocity and acceleration of the
wheel all are simulated by voltages in the circuit of
Fig. 15.
The circuit shown in Fig. 16 uses the summation
provided by the integrator (at left) to eliminate the
separate summer shown at far right in Fig. 15. As
the sum of the inputs (—dx/dt—x) is equal to d2x/dt?,
tain thevfirst-order derivative by one integration, and this circuit also satisfies equation (5). Figs. 15 andagy, -
then obtain x by a second integration, as shown in 16 both solve the same problem, but Fig. 16 uses ™
Fig. 14.) one less element.
LLLLL LLL LL
FIG. 17. SIMPLE
pendulum.
FIG. 19. USING POTS to
d*e de
“dt? t +0
FIG. 18. IF d?6/dt? is available, one integration
gives —d6/dt; another integration gives +8.
multiply d@/dt by k/L,
and @ by g/L. Adding
these two terms gives
k/L (d0/dt) ++ o/L (8).
de de
FIG. 20. SINCE out-
put of summer-am-
plifier (4) is equel to
the input of inte-
grator (1), the cir-
cuit is completed as
#@ x 1:)
dt 4]> dt Ps +0
ae ‘shown. Pot | is con-
dt? -. nected to the Initial
4 | Conditiort (IC) ter-
minal of amplifier
2 to provide the
proper initial condi-
LM tion @.
Sample Problem
The three components described—summers, pots,
and integrators—are sufficient to solve many basic
differential equations that appear in physics and en-
gineering. The simple pendulum problem is an ex-
ample:
The pendulum bob shown in Fig. 17 is acted on by
two forces along its path—gravity and friction.
1. The tangential force due to gravity = Mg sin 6;
it depends on the mass M, gravity (g), and angular
_- displacement (6)
2. The air friction (damping) force = K(dé/dt).
Note that this force depends on velocity (d6/dt)
and a friction constant K.
The equation of rotational motion is obtained by
equating all torques about the pivot (including the
rotational inertia term) to zero:
Torque 1 due to gravity = MgL (sin 4)
Torque 2 due to friction K(d6/dt) = KL(d@/dt).
Rotational inertia of mass M = ML?
Rotational acceleration = d*0/d?
Torque due to rotational acceleration — (ML?) (d?6/dt)
Hence ;
d’6
ML? [=] +KL [2] + Mel (sin ¢) =0
The first step is to solve for the highest-order
derivative:
@o__ Kd) _ « ¢
ie =— ti | 4} L (sin 6) (6)
{7 To simplify matters, we can assume that 6 is al-
“~~ ways sufficiently small (less than 15°) so that we can
approximate sin @ by 6. This yields the equation:
de = KX ae | - £)
de = 7 on (4 8 (7)
If we have a voltage proportional to d*0/dt?, then
we can integrate once to find d6/dt and again to find
6, as in Fig. 18.
We can multiply (dé/dt) by (K/LM) and @ by
(g/L) as shown in Fig. 19. If we add these two
terms in a summer, we obtain the right-hand side of
equation (7), as shown in Fig. 19. This sum must be
d?6/dt? according to equation (7). (Note the sign
reversal every time an amplifier is used.)
Since amplifier 4 has the output d?6/dt?, we may
use it for the input to amplifier 1, as shown in Fig.
20. This completes the circuit and satisfies equation 7.
Let us consider the case where the bob is given
an initial displacement (6@,) of % radian and has an
initial velocity of zero.
Since 6 does not start at zero, the “initial condition”
is imposed on 6 by charging the feedback capacitor
of integrator 2, which develops @. This initial con-
dition 6, is fed into an IC terminal on amplifier 2
The voltage at the output of amplifier 2 (6) now
satisfies the differential equation and the initial con-
ditions 6=6, and dé/dt =0. The voltage represent-
ing @ therefore will vary in the same manner as the
angle 6 in the original problem.
Servomultipliers
Equation 7 is so simple that an analytic solution
is obtained easily and one normally would not use
a computer to solve such an equation. In solving
the pendulum problem, we made two assumptions—
(1) that the damping was proportional to velocity
and (2) that @ was so small that sin @ could be re-
placed by @ without serious error. These assump-
tions had the effect of making the differential equa-
tion linear. Suppose, however, that the damping term
is proportional to some power of the angular ve-
locity, say (d6/dt)?. Then equation 7 becomes
de _ K (de)?
@ 7 ila £ (8)
Note that we are still assuming that 6 is small.
Equation 8 is a nonlinear differential equation. An
analytic solution of (8) is difficult if not impossible.
To solve it on the computer, we use a device
that generates x* from x. This is done by using a
device that multiplies variables so that x can be
multiplied by x or y or x? or any desired variable.
As we have seen (Fig. 8), a variable voltage (x)
can be multiplied by a constant coefficient (K) by
using a potentiometer. To multiply one varying quan-
tity by another, we can use a pot whose slider is
automatically positioned by an electric motor to fol-
low a second variable. This type of multiplier is
known as a servomultiplier.
If we have two varying voltages (x and y in Fig. 21)
and wish to produce a voltage proportional to xy,
we use one potentiometer whose input is y and
whose sliding setting is proportional to x. The output
would by proportional to xy.
In practice, this is achieved by using two poten-
tiometers which are ganged— i. e., mounted on a
common shaft—and using a motor to turn this shaft
(Fig. 21). One of the pots is connected to a reference
voltage (100 volts). Its arm is used as one of the
inputs to a comparison network whose other input is
X. :
Pot 1 in Fig 21 has a wiper (arm) that is positioned
by the motor, which receives a signal only when
voltage on the wiper is not exactly equal to x. Thus
the wipers on both pots follow every change in x, and
the output from pot 2 is proportional to xy. As shown
in Fig. 21, the actual output is xy/100.*
There is no reason to restrict ourselves to just
two ganged pots; the servomultipliers in the 231R
use six, labeled A through F (Fig. 22). The “F”
potentiometer is connected to the reference voltage,
and plays the role of pot 1 in Fig. 21. If x is the
variable positioning the servo, then all six pots have
the arm position x/100, and pots A through E all are
available for multiplication by x. This arrangement
enables one to generate terms such as xy, xz, . . . etc,
*The output from pot 2 is xy/100, and not xy as desired,
because of the 100-volt supply used with pot 1. If a 1-volt
reference had been used for pot 1, the output from pot 2
would be exactly xy. When a 100-v supply is used on pot 1,
then x (output. from pot 1) = 100k, where k is the frac-
tion of pot 1. Hence k (fraction of the pot 1) is x/100.
Since pot 2 is ganged to pot 1, its setting is also x/100,
and its output must be xy/100.
The factor of 100 also can be appreciated by noting that
x and y each can vary from 0 to 100 volts. Hence the
maximum output of the product xy should be 100 x 100,
or 10,000 volts. However, the maximum voltage from pot
2 is 100 volts. Hence the output from pot 2 must be
xy/100. Note that if pot 1 had had a 1-volt supply, the
output from pot 2 would be exactly xy; if pot 1 had had
a 10-volt supply, the output from pot 2 would be xy/10,
etc.
with only one servomultiplier. This feature is es-
pecially useful in generating polynomials. In particu-
lar, it allows us to generate the cubic term in Equa-
tion 8 with just one servo, as shown in Fig. 23.
This scheme works only if x is a positive voltage
because only positive voltages are available along the
reference pot winding. If it is to be both positive and
negative, the arrangement in Fig. 24 is used. There
now is no restriction on the sign of the voltage x.
Note also that pot 2 now has both +-y and —y voltages
at its terminals. The variable —y is obtained simply
by using an inverter, as shown in Fig. 25, which also
shows the symbol for a servomultiplier. Note in the
symbol that one input goes to a box marked SM
(number ); the pot that is ganged to the motor-driven
pot is identified by having the same number fol-
lowed by a letter—(1A), (1B), etc. All pots that are
ganged to one servomultiplier have the same number.
Thus pot 4C means the third pot (C) ganged to
the number 4 servomultiplier.
Fig. 26 shows how nonlinear equation (8) is im-
plemented and solved. Note that the computer solu-
tion for equation (8) is scarcely more complicated
then that of equation (7)—yet the analytic solution
for equation (8) is much more difficult than that of
7).
The Quarter-Square Multiplier
Another-widely used multiplying device is the
quarter-square multiplier. This device uses the alge-
braic identity: ;
Jee W= ML(x+y)*— (x= y)*]*
To generate xy we need only perform addition
(x + y), subtraction (x — y) and squaring of the two
terms. Squaring turns out to be much simpler than
multiplication because it involves only one variable.
(The means of generating the square and other
functions of a single variable will be discussed later.)
As the quarter-square multiplier is all electronic, it
has a much higher frequency response than the servo-
multiplier.
Function Generators
Equation (8) was simplified by the replacement
of sin 6 by 6. If we are to remove this restriction,
we need a device which will accept an input voltage
proportional to @ and produce a voltage proportional
to sin 6. This can be done by a function generator.
Many types of function generators are available,
but the variable diode function generator (DFG) is
typical. It employs a network of resistors and diodes
to approximate the given function by use of straight-
line segments, as shown in Fig. 27. The slope and
breakpoint of each segment can be individually ad-
justed to allow a best fit to the curve (function).
A fixed DFG consists of a printed-circuit card
with specific components chosen to produce the de-
sired function. Commonly used functions such as log
x, x’, x#, tan x, sin x, etc. are available. The symbol
for either the fixed or variable type of DFG is given
in/Fig. 28.
*xy = % [(x+y)? — (x-— y)?]
== 4 (x? + 2xy + y*—x? + 2xy — y’)
= %4(4xy) = xy
~a™
cy
ae
~)
DIFFERENCE
AMPLIFIER
100 k
—_—— MOTOR po Bence Qoovery
x | ERROR _ xe
SIGNAL
AiG. 21. PRINCIPLE of multiplica-
tion of two variables. If top of
pot is a variable Y, and setting is
a variable X, then output is pro-
portional to XY.
FIG. 22. SERVOMULTIPLIFER po-
sitions the taps on six ganged pots
—one follower pot (F] and five
multiplying pots (A-E}.
FIG. 23. GENERATING the cube
of a variable with only one servo-
multiplier.
|
+10oo0.=—s«&F'Y
2 ___2._, xy
/ ’ 160
1400 -¥
+X
SMI
POT#i POT#2
FIG. 24, SERVOMULTIPLIER in
which variable X can have any .
value between —100 and -+-100
volts. Note that both references
(— 100 and + 100 v) are needed,
and also +X and —Y.
FIG. 25. SYMBOLS for a servo-
multiplier. Five pots (A-E) are
ganged to the motor-driven pot
(F) to permit X to be multiplied by
five variables simultaneously.
FIG. 26. CIRCUIT for solving the
pendulum problem with cubic
damping term. Note that we can
perform two multiplications by the
factor d0/dt to produce (d6/dt}®
with only one servomultiplier.
Jo ~~ FIG. 27. AP-
PROXIMATING
a smooth curve
by straight-line
segments.
X \ fbx)
|
FIG. 28. SYMBOL for a diode
function generator.
FIG. 30. SOLUTION to d?@/d#=— 2
2(d6/dt)®— 126, with initial conditions: a
6=-0.25 radian and dé/dt=0. @
FIG. 29, CIRCUIT for solving the equation
d?6/dt2=—{K/LM)(d6/d#)8—g/L(sin 6}.
Note that the DFG is the only difference
between this circuit and that of Fig. 26.
Me it oot g fae Lae f
P\ NAA AALA
PUY UV N VV
725 —— gcouba a:
WU VIVA)
VY AV Waa
The accuracy (closeness of fit to original function)
of the DFG depends primarily on the number of
segments used. Ten- and twenty-segment DFG’s are
available on the 231-R; these are sufficient for most
practical problems.
The pendulum problem. can now be solved with-
out any simplifying assumptions because we can use
a DFG set up to give an output sin @ for an input 6.
The computer program for. the nonlinear pendulum
is given in Fig. 29. (Other techniques can be em-
‘ployed to give a better approximation to sin 6, but
this simple application illustrates the basic principles. )
Recording Devices
We have seen: how to use a computer to obtain
time-varying voltages that represent the solution to
a mechanical problem. To observe and interpret these
results we need measuring and recording devices.
One such unit, the strip-chart recorder employs
a roll of paper drawn at a constant speed past a
pen which is deflected proportional to the input volt-
age. The pen thus draws a graph of the computer
voltage as a function of time. Typical models have
parallel channels on a single strip of paper, allowing
the operator to record several variables simultaneously.
Fig. 30 shows the solution of equation (8) recorded
on a strip-chart recorder. The three channels shown
record the displacement (6), velocity (d6/dt) and the
velocity cubed term (d6/dt)®. In a more complex
problem, more channels might be needed.
Another recording device, the X-Y plotter, uses a
moving pen and stationary paper. This device has »)
two voltage inputs, enabling one to plot one varying
voltage against another. The pen moves back and
forth along a straight arm, its deflection being pro-
portional to one voltage. The arm itself moves in
a perpendicular direction positioned by the second
voltage. On 10” x 15” paper ruled in 1/10” squares;
the resulting graph can be read with an accuracy of
about 1 part in 1000. Plotters for 11” x 17” and 30”
x 30” graphs are popular; even larger sizes are avail-
able.
Problem Checking
Many tests have been devised to assure the opera-
tor that the problem has been programmed correctly
and that all components are functioning properly.
One of these, the static test, is a check on both the
program and the equipment.
The standard procedure consists of choosing an
arbitrary set of initial conditions—that is, a set o
>
values for the unknown variable and all its derivatives
\ppearing in the problem, except the highest deriv-
~ative. In the pendulum problem, this means chosing
initial values for 6 and dé@/dt. The highest derivative
(in this case d?@/dt?) then can be computed, and
from this the voltage representing d?6/dt? can be
calculated. For example, suppose we are solving equa-
tion (7) for the case, k/LM = 0.5 and g/L= 08. If
we assume 6=0.25 radian (about 15°) and dé6/dt =
1 rad/see as initial conditions, we may solve equation
<7) for the initial value of d?6/dt?:
“d4_
ja = — 05(1) —,0.8(0.25) = — 07 rad/sec’
The assigned initial values for d@/dt and 6 corre-
-spond on the computey-Giagram (Fig. 29) to scaled
initial voltages of theoutputs of integrators 1 and 2.
Similly, the calculated value for d?6/dt? corre-
sponds to 2-velfage output of the summing amplifier
No. y-Since we know the initial outputs of the
Ertegrators, we can calculate the output of amplifier
from the circuit diagram. Comparing this scaled
oltage to the calculated value for d?6/dt?, we have
a check on the correctness of the diagram.
The final step in the static-check procedure is taken
after the problem is put on the computer, as follows:
Voltages proportional to the initial conditions (in
this case dé/dt-==-1 and 6=—0.25) are established
at the “Initial Condition” terminals of amplifiers 1
and 2, and the output of amplifier 4 is measured
and compared with the calculated value. This check
verifies the patching and the functioning of the com-
puter components. Most errors, human or mechanical,
are discovered and corrected during the static-test
procedure.
Automatic Setup
As computers increase in size to allow solution
of more complex problems, there is a trend toward
automatic devices to reduce set-up time and minimize
the possibility of human error. The ADIOS (Auto-
atic Digital Input-Output System) together with
AS (Digital Attenuator System) available on the
~ 231-R is an example. Consisting of an electric type-
writer, keyboard, paper-tape punch and tape reader,
together with the necessary relays and switches, the
ADIOS enables the operator to set up potentiometers
and check programming from either the keyboard or
tape. To set a potentiometer for 0.2317, the operator
simply punches out on the keyboard an address code
for the potentiometer desired and the digits 2317.
A servomechanism in the computer automatically sets
the pot to the desired value. Both potentiometers
and function generators can be set up in this fashion,
resulting in a much faster computer set-up and
problem check. The combination of a paper tape and
patch panel allows for complete problem storage.
Repetitive Operation
When the variation of many parameters is required
in a design study, full advantage may be taken of
the analog’s high speed by the addition of repetitive
Pperation (rep-op), a feature available on many com-
puters. In repetitive operation, the speed of solution
Printed in U, S. A,
is increased, generally, by changing the value of the
feedback capacitors on all integrators, and using high-
speed relays to repeatedly reset the initial conditions
on these. integrators, allow the solution to run, and
then reset again. The resultant voltage output (viewed
on an oscilloscope) appears as a continuous curve—
a graph of the solution versus time. When the pots
(representing parameters) are varied, the result is
an almost instantaneous change in the output graph.
This visual display enables the operator to find the
optimum parameters quickly.
Analog Versus Digital
Invariably the question arises—Which is better, an
analog or a digital computer? The answer is that
neither is always better; the best choice depends on
the particular problem to be solved. While there
are many considerations, in general, the principal
advantage of the analog computer is speed, while
the advantage of the digital computer is accuracy.
The speed advantage of the analog computer arises
from the fact that while the digital computer per-
forms its calculations with discrete numbers in se-
quence., the analog computer has all components
(summers, integrators, etc.) performing their opera-
tions simultaneously and continuously. Thus, the digi-
~ tal computer must wait until each calculation is com-
pleted before moving on to the next. The more com-
plex the problem, the more calculations are required
and the longer the time required for solutions. The
time required for solutions on the analog is virtually
independent of the complexity of the problem. A
typical analog computer solution takes from 10 to 60
seconds, with 20 seconds a good average figure. One
large analog computing laboratory which has acquired
a digital computer for checking purposes reports an
analog speed advantage of over 100 to. 1 in some
problems.
This speed advantage is especially important if a
problem is to be solved for many parameter varia-
tions. The stiffness of a spring, the wing span of an
airplane, and the gain of a pneumatic controller are
all examples of parameters that can be set on pots
and varied between runs, enabling the operator to
judge quickly the effect of any change in parameters.
In a design problem with many adjustable para-
meters, this speed advantage usually enables the
optimum design to be reached more quickly than by
digital methods.
The accuracy of an analog computer is limited by
the accuracy of the electrical components, and three
significant figures is the limit of accuracy presently
attainable in medium-sized problems. On a digital
computer, the accuracy is determined chiefly by the
number of decimal places (actually binary places)
retained in calculation and, with sufficient equipment
and time, high accuracy can be achieved. If ac-
curacies higher than about 1% are necessary, a digital
solution is indicated; however, most engineering prob-
lems do not require this high accuracy because in
many cases the input data are not this accurate. In
these cases the high speed of an analog, and the fact
that it produces easily interpreted continuous curves
rather than discrete points, make an analog solution
preferable.
BULLETIN NUMBER ALAC 916-1
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