Rensselaer Polytechnic Institute
ECSE-4760
Real-Time Applications in Control & Communications
EXPERIMENTS IN OPTIMAL CONTROL
Number of Sessions – 4
INTRODUCTION
In modern control systems design it is sometimes necessary to design controllers that not only
effectively control the behavior of a system, but also minimize or maximize some user defined criteria
such as energy or time conservation, or obey other physical or time constraints imposed by the
environment. Optimal Control Theory provides the mathematical tools for solving problems like
these, either analytically or through computer iterative methods, by formulating the user criteria
into a cost function and using the state equation representation for the system dynamics.
The optimal control experiments in this lab involve the use of a PC to illustrate the efficacy of
microprocessors as system controllers. Two types of control will be considered:
1) Continuous (analog)
2) Discrete (digital)
The general formulation of the optimal control problem is as follows:
Let the system states (i.e. the state variables) be represented by an n-dimensional vector x, where n
is the order of the system. Let the control variables (input) be represented by an m-dimensional
vector u, and the system output by an r-dimensional vector y. The system can, in general, be modeled
by a set of differential equations of the form:
.
x =  (x, u) ,
y =  (x, u)
x(t0) = c
(1)
where  and  are general (possibly nonlinear and time varying) expressions of x and u, t0 is the
initial time and c is an n-dimensional set of initial conditions. The objective is to find a control input
u(t) that will drive the system from the initial point c ( = x(t0) ) in the state space to the final point
x(tf), and at the same time minimize a cost functional J (formally called the performance
index/criterion) given by equation (2):
tf
J = h(x(tf), tf) +

 g(x(t), u(t), t) dt
t0
1
(2)
where:
tf represents the end of the control interval,
h and g are user defined penalty expressions.
PART I – CONTINUOUS CONTROL
MATHEMATICAL FORMULATION – RESULTS
Even though the general problem is very hard to solve, there are very useful simplifications that
lead to closed form (analytic) solutions. The most common simplification (and the one used for this
experiment) is the LQR problem, where the system is Linear and the controller (Regulator) must
satisfy a Quadratic cost functional. Assuming that the system is time invariant as well, then state
and output equations (1) become:
.
x(t) = Ax(t) + Bu(t) ,
y(t) = Cx(t) + Du(t)
x(0) = x0
(3)
where:
x(t) is the (n x 1) state vector,
x0 is the initial state vector,
u(t) is the (m x 1) control input vector,
y(t) is the (r x 1) output vector,
A is the (n x n) state dynamics matrix,
B is the (n x m) control dynamics matrix,
C is the (r x n) state-output matrix,
D is the (r x m) input-output matrix (for all practical purposes assumed 0 thereafter).
Corresponding to the system is a performance index represented in a quadratic form as:
tf
1
1
J(u(t), x(t), t) =
x'(t)Hx(t) + 
{ x'(t)Qx(t) + u'(t)Ru(t) } dt
2
2
(4)
0
where:
H is the (n x n) terminal state penalty matrix,
Q is the (n x n) state penalty matrix,
R is the (m x m) control penalty matrix.
It is important to emphasize that the H, Q, and R matrices are user selectable, and it is through
the proper selection of these that the various environment constraints are to be satisfied. If H, Q,
and R are positive definite and symmetric, then it can be shown that a closed loop optimal control
function u(t) exists (called u*(t)), and is uniquely given by:
u*(t) = -R-1B'P(t)x(t) = G(t)x(t)
.
G(t) = -R-1B'P(t)
2
(5)
where:
G(t) is the (m x n) optimal feedback gain matrix,
P(t) is an (n x n) symmetric and positive definite matrix that satisfies the continuous matrix
differential Riccati equation given by:
.
P (t) = -P(t)A - A'P(t) + P(t)BR-1B'P(t) - Q ,
P(tf) = H
(6)
From the above it is obvious that, even for the LQR problem, the set of coupled differential
equations (6) must be solved before the controller gains can be found. It should be noted that these
gains are functions of time. The significance of the result of equation (5) is that the gains can be
computed ahead of time, and the optimal controller is easily implemented on any digital computer.
In all but the simplest cases of the real life applications, computer iterative methods are utilized to
pre-calculate and store the P matrix and the gains, resulting in very efficient and robust open and
closed loop controllers.
A further simplification to the above solution takes place when tf approaches ∞, or in more
realistic terms the control is applied for a very long period of time. In this case it is shown that: if (1)
the system is controllable, (2) H is zero, (3) the A, B, Q, and R, are all constant matrices, then the
P(t) matrix converges to a constant symmetric, real, positive definite matrix Pss. Obviously then
.
P (t) is zero, and the matrix differential Riccati (6) is transformed into a matrix algebraic equation
(called the Steady State Riccati Equation) given by:
0 = -PA - A'P + PBR-1B'P - Q
(7)
The steady state solution Pss need not be unique. However, only one real positive definite
solution can exist. Since P is of dimensions (n x n), we obtain n2 equations to solve for the n2
components of P. Yet due to the fact that it is symmetric, the number of computations is greatly
reduced. Note that the steady state Riccati equation for the continuous case is easier to solve without
the aid of a computer than is the discrete equivalent to be computed later, because no inverse of the
unknown P appears in the equation. Once the P matrix is determined, the optimal control gain
matrix G can be found from equation (5). Of course now the gains are constants as expected. Solving
the general (non steady state) LQR problem requires substantial storage for the gains that depends
on the tf and the sampling interval, and exceeds the scope of this experiment that will hereafter focus
on the steady state case.
Another obvious but never the less important result of the steady state case is that the system
states must converge to zero regardless of their initial condition vector x0. An intuitive proof by
contradiction of the above result can be derived using the following notion: The performance index is
always positive as the sum of positive factors since all matrices are positive definite. If it is to be
minimized over an infinitely large time, then this minimum better be constant. Yet if the states are
not approaching zero, then from equation (5) the control isn't either, and we end up summing two
positive quantities, inside the integral, for an infinite time. This of course can never give a constant
(minimum) value; therefore the states must go to zero.
Applications exploiting this result are controllers designed to keep the states always near zero,
counteracting random abrupt changes in the state values (or any other causes that can be modeled as
impulses). The following example sets up a typical infinite time LQR problem and analytically solves
the steady state Riccati equation, giving the student sufficient experience to deal with the actual
experiment.
3
EXAMPLE
Consider the following second order system:
x. (t)
 1  =  0 1 x1(t) +
. 
 -1 0 x2(t)
x2(t)
x1(t)
y(t) = [ 1 0 ] x (t)
 2
 0  u(t)
1



with the corresponding performance index given by
∞

 1 0 x1(t) + u2(t) } dt
J = { [x1(t) x2(t)]
 0 1 x2(t)

0
The continuous steady state Riccati equation (7) can be written as:
PBR-1B'P = PA + A'P + Q
(8)
and using this example's data it becomes:
1 0
0 -1  P11 P12 
 P11 P12 0
 P11 P12 
 P11 P12  0 1
 P P 1 [1][ 0 1 ]  P P  =  0 1  +  1 0  P P  +  P P  -1 0 
 21 22 
 21 22 
 21 22 
 21 22 
The matrix equation can be rewritten as three simultaneous algebraic equations (remember
P12 = P21):
2
P12 = 1 - 2P12
P12P22 = P11 - P22
2
P22 = 1 + 2P12
Solving the three equations, two real solutions for P are obtained:
 1.912
 0.4142
0.4142
1.352
 -1.912
 0.4142
 ,

0.4142
-1.352


By checking the principal minors of each of the two matrices, it is found that only the first of these is
positive definite*, therefore it's the accepted solution.
The feedback gain matrix is then determined from equation (5) to be:
G = [ -0.4142
*
-1.352 ]
A matrix is positive definite if all principal minors are positive. The principal minors are the determinants
of the submatrices along the principal diagonal of the matrix which include the topmost left element of the
diagonal. Thus the first principal minor is the element (1,1), the second is the determinant of the 2x2
submatrix in the upper left corner, and so on.
4
Thus, the optimal control input u(t) is given by:
u(t) = -0.4142x1(t) - 1.352x2(t)
(9)
The student is expected to work through this example prior to attempting the problem for the
experiment, to ensure a thorough understanding of the procedure.
FIGURE 1. Block diagram of the open loop system.
PROBLEM FORMULATION
This experiment is based on the Steady State case of the LQR problem. Here the plant to be
controlled is a 2nd order linear and time invariant system implemented on the Comdyna analog
computer, the penalty matrices Q, R are constants and H is zero. The optimal control input u(t) will
drive the system states x1 and x2 from the initial values of -3 volts and 0 volts respectively to zero,
and it will be generated either using the Comdyna analog computer, or a PC running the appropriate
program. The second order system to be controlled is modeled by the transfer function:
Output
Y(s)
H(s) = Input =
U(s)
=
0.5
s(s + 0.5)
(10)
which is to be implemented on the Comdyna analog computer following a suitably designed
simulation diagram.
The state variable representation of the system transfer function (10) is of the same general form
as equations (3) with D equal to zero, and is given by:
x. (t)
 1  =  0 1 x1(t) +  0  u(t)
. 
 0 -0.5 x2(t)
 0.5 
x2(t)
x1(t)
y(t) = [ 1 0 ] x (t)
 2
,



-3
x1(0)
x (0) =  0 
 2 
(11)
Given the state variable equations, the open loop system block diagram can be obtained as in
FIG. 1 and the actual analog computer simulation as in FIG 2. The student is urged to verify both
results, paying particular attention to the initial conditions in FIG. 2. Note that the integrators
invert the voltage on the IC input so a +3 volts will give the desired initial output voltage of -3 volts.
5
-1
AO 0
-1
AI 0
AI 1
FIGURE 2. Analog computer simulation of the open loop system.
The performance index for this system is given in accordance with equation (4) by:
∞

 2 0 x1(t) + 0.4u2(t) } dt
J = { [x1(t) x2(t)]
 0 1 x2(t)

(12)
0
EXPERIMENTAL PROCEDURE
The aim of the experiment is to design a closed loop optimal controller that will drive the states
to zero using the feedback gains obtained via the equations (5). For this purpose the following steps
must be sequentially implemented:
1) The open loop system is to be built on the Comdyna analog computer in accordance with the
simulation diagram of FIG. 2. Special care must be taken when implementing the initial conditions
and the gains because of the sign inversions at the output of the amplifiers. The DSO (oscilloscope)
must be thoroughly calibrated and its various scales mastered, before any useful work can be done.
Remember, the Comdyna dial must be on the Pot Set position and the pushbutton labeled IC be
pressed in during setup; during operation the dial must always be on the Oper position, and the
pushbutton labeled OP must be pressed in just before each run starts.
2) The impulse response of the system is to be simulated using the Comdyna and the PC, and the
natural (uncontrolled) modes of the states x1, x2 are to be saved from the DSO screen. The concept
behind this is that when an Asymptotically Stable system is excited with an impulse function (t),
then the states converge to zero regardless of their initial values, producing the same final result as
the optimal controller does. Hence the plots of the state trajectories obtained in this part, are to be
compared with the ones produced by applying the closed loop optimal control. Before running this
part it is necessary to verify that both states are indeed stable by solving the system differential
equations. However, it should be noted that this system is not strictly Asymptotically Stable due to
the pole at the origin.
To implement the impulse function, connect the AO 0 port of the National Instruments BNC2110 board to the input of the system (details on running the PC program are provided later). The
states x1(t) and x2(t) as well as the control signal u(t) should be observed on the scope. Follow the
steps to load the LabVIEW program in 5) below. Select the Impulse Response tab at the top of the
VI front panel. After resetting the analog computer system’s I.C.s, (IC pushbutton) begin the scope
6
sweep, go to operate mode (OP pushbutton) on the Comdyna, and click the Run arrow icon in the VI
menu bar. Be sure to use Run, not Run Continuously. The weighted impulse (created by a 10 volt
square pulse held for 0.3 s) will drive x1(t) and x2(t) to zero. The controller stops automatically at the
end of the cycle. It is important to understand that since the (t) function is a theoretical pulse of
infinite amplitude and infinitesimal duration, no human realizable input can reproduce it. Hence a
+10 volt step function (the largest voltage the analog card is capable of) is applied for a period of 0.3 s
to produce the same net result. The duration is valid only for the -3 volt initial condition of state x1
(10 x 0.3 = 3, the negative value of the I.C.). Also you may observe that state x1 reaches the desired
zero as theoretically expected, yet it then very slowly continues to change linearly with time. This
should be attributed to leaks and offsets within the Comdyna integrators rather than to a theoretical
inconsistency.
FIGURE 3. Block diagram of the closed loop system.
3) The algebraic Riccati equation for the above system is to be solved and the P matrix elements
and the optimal feedback gains g1, g2 are to be computed using equation (5) and the example. Even
though MATLAB or any other equivalent math package can be used for verification purposes, a
detailed analytical solution of the P matrix elements and the gains is mandatory. Having found the
gains, the optimal controller is given by equation (13) as:
u*(t) = g1x1(t) + g2x2(t)
(13)
The block diagram of the complete closed loop system is given in FIG. 3, and its simulation
diagram in FIG. 4. The control calculation (equation (13)) of FIG. 4 is to be implemented on the
Comdyna analog computer and PC through the Optimal.vi LabVIEW program.
NOTE: The gain calculated from the Riccati solution will be negative numbers. These
values will be entered as positive attenuations on the analog computer pots with the
summing amp providing the sign inversion. Diagrams showing the negative gain of the
summing amp will also show negated feedback gains (which end up as positive numbers)
for consistency. This convention also applies to the LabVIEW implementation where the
gains will still be entered as positive values.
4) The complete closed loop system, both the plant and the feedback controller, is to be built on
the Comdyna analog computer using the FIG. 4, and a DSO screen recording of the state trajectories
be obtained for comparison with the results from the other runs. This controller should produce the
best results for a given set of dynamics and penalties. (Why?) The implementation on the Comdyna
7
can be done either using the multipliers, or passing the states individually through the .1 inputs of
adders to effectively increase them by a factor of 10, and then through potentiometers with values .1
times the gains.
-1
AO 0
-1
AI 0
AI 0
-1
Figure 4. Analog computer simulation of the closed loop system.
5) The experiment is to also be run using a LabVIEW program to implement the closed loop
controller. For this part, the system built on the Comdyna is to be connected to the PC as in FIG. 4.
State x1 must be fed to the BNC-2110 input port AI 0, state x2 to input port AI 1, and the system
input u to output port AO 0. All other ports are to be left free. The control algorithm is to be executed
using different sampling periods (at least ten runs ranging from 50 to 1000 ms), and the plots of the
state trajectories be compared with the previous results. To access the LabVIEW program do the
following:
•
•
•
•
•
•
Turn the PC on (if off) and go to the OPTIMAL subdirectory (My Computer\Local Disk):
C:\CStudio\RTA_lab\Optimal.
Double click Optimal Controller DAQmx+.vi to load the program. A LabVIEW program will
display with two different tabs at the top of the screen. When moving through the parameter
fields, the tab button will not work. You must use the mouse to manually select them.
Be sure to push the operate button (OP) on the analog computer before starting the VI program,
otherwise the controller will not behave properly. Push IC when done to reset the plant for the
next run.
Press the right arrow button at the top left-hand corner of the screen in order to start execution.
In order to pause; press the STOP button on the screen, not the stop sign at the top-left of
the page (this will help to calculate actual sampling times, and reset the output to 0
Volts). It may take a few seconds for the program to respond.
Although some values can be changed during execution by user input, it is important, in order to
properly ensure correct measurement, to stop the controller (by using the STOP button) before
altering parameters values as necessary.
If you notice the controller isn’t working properly:
o Press the stop button then run the LabVIEW program again. This should reset the program
and enable you to start from scratch.
o Wiggle the T-connectors; make sure the connection is good. If when wiggling you notice a
difference in the response, change connectors. Occasionally the PC may need to be restarted.
In almost all the cases data entry consists of the sampling period T in ms. For the Continuous
Case the pre-calculated gains in step 2 must be entered into the relevant fields once since they are
constant, and only the sampling period is to be changed before each new run. Since the computer
program does invert the controller sign in the output, the negated calculated gains must be entered
8
in the relevant fields (as with the analog implementation of step 3). Record your observations and
make suitable DSO screen recordings of the states.
WRITE-UP AND ANALYSIS
Your report should contain:
1) Solution of the continuous steady state Riccati equation and calculation of the feedback gain.
(Show that the solution to the steady state Riccati equation is positive definite.)
2) Comments on the asymptotic stability of the system.
3) Various DSO screen recordings for procedure parts 1), 3), and 4). These should be
appropriately labeled.
4) Discussion of results, including a comparison of the recordings and the significance of the
sampling times (especially when it increases). When discussing the sampling times, reference
must be made to the DSO screen recordings.
5) Conclusion.
PART II – DISCRETE CONTROL
MATHEMATICAL FORMULATION – RESULTS
In the case of digital control, the optimization must be treated as a discrete time problem. Thus,
the sampled system is modeled by the difference equations:
x(k + 1) = Ax(k) + Bu(k) ,
y(k) = Cx(k) + Du(k)
x(0) = c
(14)
and the discrete performance index in a quadratic form is given by
N
1
1
J =
x'(N)Hx(N) + { x'(k)Qx(k) + u'(k)Ru(k) }
2
2
(15)
k=0
where A, B, C, D, H, Q and R are similar to those in the continuous case and N is a fixed number of
time intervals. By applying the conditions of optimality to the problem, the optimal control is found
to be:
u*(k) = G(k)x(k)
(16)
where G is again the (m x n) feedback gain matrix given by:
G(k) = -R-1B'(A')-1[P(k) - Q ]
(17)
and P is the (n x n) real positive definite solution of the discrete matrix difference Riccati equation
given by:
P(k) = Q + A'[P-1(k + 1) + BR-1B']-1A
9
(18)
When N approaches ∞ and the same conditions mentioned in the continuous case apply, then the
P(k) matrix converges to a constant symmetric, real, positive definite matrix P. Hence:
P(k) = P(k + 1) = P
(19)
and the matrix difference Riccati becomes a matrix algebraic equation as before. Yet even in the
steady state case equation (18) is very difficult to solve analytically, hence iterative methods
implemented on computers handle this task. Once the P matrix is found, the gains and the controller
are easily computed using equations (17) and (16).
The following example should provide some reference as to the procedure to follow to obtain the
feedback gain matrix for a simple system.
EXAMPLE
Given a discrete system described by the state and output equations:
x1(k + 1)  cosT sinT x1(k)
x (k + 1) =  -sinT cosT x (k)
 2

 2 
x1(k)
y(k) = [ 1 0 ] x (k)
 2
+
 1 - cosT  u(k)
 sinT 



where T is the sampling interval in seconds. The discrete performance index is given by:
∞
J =
{
x'(k)
 1 0  x(k) + u2(k) }
0 1
k=0
If we let the sampling interval T be equal to /40 seconds, the discrete equations become:
0.997 0.0785 x1(k)
0.00308
x1(k + 1)
x (k + 1) =  -0.0785 0.997 x (k) +  0.0785  u(k)
 2

 2 
The discrete Riccati equation is not easily solved by hand due to the fact that a complicated cubic
equation has to be solved. Using the class web page MATLAB m-file program, the solution is found:
P =
 24.6 5.15 
 5.15 17.5 
Thus, from equation (17) the feedback gain matrix is:
G(k) = [ -0.331 -1.28 ]
and from (12) the discrete optimal control is becomes:
u(k) = -0.331x1(k) - 1.28x2(k)
PROBLEM FORMULATION
10
This part of the experiment implements the steady state case of the Discrete Linear Quadratic
Regulator problem. The same linear and time invariant system as before is used, but now for all
calculations it is treated as a sampled data system modeled by the following state and output
equations:
x1(k + 1)
 1 2 - 2e-.5T x1(k) +  T - 2 + 2e-.5T  u(k) ,



x (k + 1) = 

 0 e-.5T x2(k)
 1 - e-.5T 
 2

x1(k) 
y(k) = [ 1 0 ] x (k) 
 2 
-3
x1(0)
x (0) =  0 
 2 
(20)
where T is the sampling interval. The discrete performance index for this system is given by:
∞
J =

2 0 x1(k)
{ [x1(k) x2(k)] 0 1 x (k) + 0.4u2(k) }
 2 
(21)
k=0
It should be noted here that the discrete A state dynamics and B control dynamics matrices have
been chosen (out of the infinitely many possibilities representing the given system) so that the
discrete states x(k) are the continuous states x(t). Given a continuous system to be modeled by a
discrete system with zero order holds on the outputs, the discrete matrices Ad and Bd may be
obtained from the continuous matrices Ac and Bc by using the MATLAB function c2d or by:
A T
Ad = e c
-1
Bd = [Ad - I]Ac Bc
where T is the sample period. Note that Cc = Cd and Dc = Dd. (Why?) The section "Discrete State
Equations" in the chapter "Open-Loop Discrete-Time Systems" of Digital Control System Analysis
and Design by C. L. Phillips and H. T. Nagle provides more details on this continuous to discrete
transformation.
EXPERIMENTAL PROCEDURE
The procedure for this section is similar to that of the PC based Continuous Control section, yet
now the system and input dynamics (A, B) vary with sampling, hence they must be calculated in
advance. Use any tool with which you feel comfortable (e.g. Excel or MATLAB) to compute them. A
MATLAB m-file is available for download from the course web page that will solve the discrete
Riccati equation for the P matrix and the feedback gain matrix. The system is implemented on the
Comdyna analog computer and connected to the BNC-2110 according to FIG. 4 as before. Load the
LabVIEW program, enter/edit the data in the various fields and select the arrow at the top left
corner of the window to begin execution. Care must be taken to avoid typos when editing the m-file
values, especially R, because divisions by zero will crash the program. It is recommended that all
control values be saved and tabulated before starting the actual run. Record your observations, and
make appropriate DSO screen recordings.
Note that the LabVIEW program assumes negative feedback and negates the output, which is
the sum of the products of the system states and the feedback gains. This means that positive values
11
of the gains g1 and g2 entered into the program will produce the desired negative feedback control
signal.
WRITE-UP AND ANALYSIS
Your report should contain:
1) Parameters that are input for each of ten sampling times, and the P matrix for each of the
ten sampling times (from 100 ms to 1000 ms in steps of 100 ms).
2) Hardcopy recordings of x1 and x2 for each of the ten sampling times.
3) Discussion of results, which should include a comparison of the discrete control to the
continuous control problem, with regard to sampling times, etc. References must be made to
the recordings.
4) Conclusions.
PART III – DISCRETE CONTROL SUPPLEMENT
This is a variation of PART II, where the Q matrix and the value of R in the discrete
performance index are changed. As mentioned at the beginning, these matrices are "freely" selected
by the designer. Thus it becomes of great importance to observe and analyze how their values affect
the states and the control input. Given are four sets of Q and R, selected in such a way that each
time a different state or control will be heavily penalized. The procedure for PART II with the
exception of the sampling times (only 100 ms is used for this part), must be followed when using each
of the sets below. The write-up format should also be adhered to.
Q =
2 0,
0 1
R = 0.1
Q =
2 0,
0 1
R = 1.0
Q =
9 0,
0 1
R = 0.4
Q =
2 0,
0 9
R = 0.4
The effects of changing Q and R are to be noted in the write-up. The behavior of the responses for the
different cases must be explained (for example, the difference in response times).
12
REFERENCES
For the state variables and feedback control only introductory material is necessary. Suggested
reading:
Frederick, D. K. and Carlson, A. B., Linear Systems in Communication and Control, Section 3.5,
pp. 79-88.
DeRusso, P. M., Roy, R. J., and Close, C. M., State Variables for Engineers, Sections 8.1 - 8.5.
For the continuous and discrete optimal control two independent approaches exist: the Linear
Programming methodology and the Calculus of Variations approach. Kirk covers both approaches
from a tutorial point of view, yet the material is still extremely difficult to digest. The perspective
reader should concentrate on the notions and the examples rather than trying to verify the
mathematical derivations. Sage is more application-oriented and extends his results to the
Estimation Problem. Linear Algebra and Advanced Calculus are prerequisites for anything other
than browsing. Suggested reading:
Kirk,D., Optimal Control Theory, Prentice-Hall, NJ 1970
Chapter 1 pp. 3-21: general introduction to systems and terminology.
Chapter 2 pp. 34-42: Typical 2nd Order LQR example, very useful to read.
Chapter 3 pp. 84-86: Example of the Discrete LQR detailed in section 3.10.
Chapter 4: Complete introduction to Calculus of Variations, contains proofs of all subsequent
results.
Chapter 5: pp. 209-218: The LQR problem with examples, state transition matrix and Kalman
solutions.
Sage, A. P. and White, C. C.,Optimum Systems Control, Prentice -Hall, NJ 1977
Chapter 5 pp. 88-100: The LQR problem with examples, only the Kalman solution.
Chapter 6 pp 128-132 & pp. 135-136: The Discrete LQR derivation and examples.
13
APPENDIX – ANALOG COMPUTER WIRING DIAGRAMS
POT 3
AO 0
POT 1
AMP 1
-1
AMP 2
-1
AMP 7
POT 2
14
AI 0
AI 1
INTEGRATOR 1
3
SUMMER 3
INTEGRATOR 2
POT 3
(B)
AO 0
AMP 1
-1
POT 1
AMP 7
POT 4
POT 2
(B)
AMP 2
-1
____
10
AMP 3
-10
____
10
POT 5
15
AI 0
AI 1