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Case Study: Space Flight and
Control Systems
In this case study, the design of engineering control systems (such as the one in Figure 1
on page 216 of your text) is studied. Special attention is paid to how concepts from
Chapter 4 may be used in this analysis.
In Figure 1 on page 216, each box represents some process (which could be a piece of
equipment, a computer program, a measuring device, etc.) which takes in an input signal
and produces an output signal. These signals are represented by arrows in the diagram.
The crossed circles are called summing junctions; signals are combined at these points
for delivery to some process as input. For example, the commanded pitch is summed with
input from the inertial measuring unit, which produces a pitch rate that is used by process
K1 as input. Notice that this system uses feedback; that is, the final result (pitch) is used
as input to the system.
In order to build a mathematical model for systems such as in Figure 1, first assume that
the output of the system is some vector x. In Figure 1, x would be a vector representing
the pitch angle of the shuttle's nose cone. Also assume (for the time being) that the
system has no external inputs. Thus if x0 is the initial pitch of the nose cone, the system
will generate a new pitch x1 from the input x0, and then a new pitch x2 will be generated
from x1, and so on, with each xk giving rise to a new xk+1. The final assumption is that
the process by which xk becomes xk+1 is a linear transformation and that it is the same for
all times k. Thus the system may be modelled by the equation xk+1 =Axk for k=0,1,2,. . .
given x0 as the initial state of the system.
 0 0
 1
 and x0 =   ,
 1 2
 1
Example 1: If A = 
 0
 0
0
 
 

then x1 = Ax0 =   , x 2 = Ax1 =   , x3 = Ax 2 =   ,
3
6
12

etc.
5
1 1 2 
1
 22 
 


 
Example 2: If A =  1 −1 −2  and x0 =  0  , then x1 = Ax0 =  −3  , x 2 = Ax1 =  −12  ,
 10 
2 2 3 
 2
 44 
 


 


 98 


x 2 = Ax3 =  −54  , etc.
 196 


Notice that in each example the outputs from the system seem to be careening out of
control. Of course, this type of behavior should be expected since the system is being left
alone: no external inputs have been allowed. In practice, external inputs are certainly
allowed. Consider the situation in Figure 1: all of the internal inputs may help, but the
pilot (whether human or computer) must input what ultimate pitch is desired. Thus the
Case Study Chapter 4: Control Systems
Page 2 of 7
pilot is attempting to control this system; the pilot is trying to get the system to deliver
the desired output vector.
Mathematically, assume that there is a sequence of external input vectors, u0 , u1 , u2 , …
and a matrix B so that xk+1 =Axk +Buk for k=0,1,2, . . . . Assume further that A is an n×n
matrix, B is a n×m matrix, that xk is in Rn for k=0,1,2,. . . and that uk is in Rm for
k=0,1,2,. . . . Also assume that the initial vector is x0 = 0. The sequence u0 , u1 , u2 , … is
called a sequence of controls for the system.
 1
Example 1 (cont.) : Suppose that the matrix B =   is given. Notice that uk is in R1
 3
since B has only one column, so the sequence of controls is really a sequence of numbers
1
in this case. For example, if u0=u1=1, then x1=Ax0+Bu0 =B =   , x2=Ax1+Bu1 =
3
 
 1  1  1 
A  +   =   .
 3   3  10 
1
Example 2 (cont.): Suppose that the matrix B =  1  is given. Again the matrix B has
 2
 
only one column, so the sequence of controls is a sequence of numbers. For example, if
1
1 1  5 
 
u0=1 and u1=-1, then x1=Ax0+Bu0 =B =  1  , x2=Ax1+Bu1 = A  1  +  1  =  −5  .
 2
 2   2   10 
 
     
The sequence of controls and the matrix B are used to force the system to deliver the
desired output within a short number of steps. In the shuttle example, this means that a
combination of external control and internal feedback ought to produce a desired pitch
within as few iterations of the system as possible. In linear algebra terms, a target vector
y in Rn is given for the output. A sequence of controls u0 , u1 , u2 , … is imposed so that
xk =y for some time k. Ideally y should be any vector in Rn. Thus the following
definition is given.
Definition: If a system has the form xk+1 =Axk +Buk , x0 = 0 where A is an n×n matrix
and B is a n×m matrix, and y is any vector in Rn, then the pair of matrices (A,B) is
controllable if there exists a sequence of controls u0 , u1 , u2 , … so that xm=y at some
time m.
In order to determine whether a pair is controllable, closely examine the sequence x0 , x1 ,
x2 , etc. Before beginning, here is a useful fact about the number of steps needed to
produce the target vector y.
Fact: Let xk+1 =Axk +Buk , x0 = 0 where A is an n×n matrix and B is an n×m matrix,
and y is any vector in Rn. Then the pair (A,B) is controllable if and only if there exists a
Case Study Chapter 4: Control Systems
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sequence of controls, u0 , u1 , u2 , … so that xm=y at some time m≤n. That is, the pair is
controllable if and only if the system can forced to any output vector in at most n steps.
The system of equations xk+1 =Axk +Buk , x0 = 0 may be solved iteratively:
x0 = 0
x1 = Ax0 + Bu 0 = Bu 0
x 2 = Ax1 + Bu1 = ABu 0 + Bu1
x3 = Ax 2 + Bu 2 = A( ABu 0 + Bu1 ) + Bu 2
= A2 Bu 0 + ABu1 + Bu 2
x k = Ak −1 Bu 0 + Ak − 2 Bu1 + Ak −3 Bu 2 + … + ABu k − 2 + Bu k −1
x n = An −1 Bu 0 + An − 2 Bu1 + An −3 Bu 2 + … + ABu n − 2 + Bu n −1
By the Fact above, the pair (A,B) is controllable if and only if each target vector y in Rn is
one of the vectors x0 , x1 , x2 , … xn for a suitable control sequence u0 , u1 , u2 , … un.
Notice that each of the vectors xk is a linear combination of the columns of the matrices
B, AB, A2B,. . . , An-1B because each term in the sum for xk is one of these matrices times a
vector uj of weights.
All of the columns of B, AB, A2B,. . . , An-1B may be placed into one large matrix by
constructing the partitioned matrix M=[B AB A2B . . . An-1B] . Then each target vector
y must be a linear combination of the columns of M. Since this should be true for all y in
Rn, the columns of M must span Rn. This explains the following result.
Result 1: A pair (A,B) (with x0 = 0) is controllable if and only if each vector y in Rn is a
linear combination of the columns of the matrix M=[B
AB A2B . . . An-1B] .
Equivalently, the pair is controllable if and only if the columns of M span Rn .
The matrix M is clearly fundamental to the analysis of the controllability of the system; it
is called the controllability matrix for the system.
Definition: If A is an n×n matrix and B is an n×m matrix, then the n×nm matrix M=[B
AB A2B . . . An-1B] is called the controllability matrix of the system xk+1 =Axk +Buk for
k=0,1,2, . . . .
Example 1 (cont.): Since A is 2×2 and B is 2×1, the controllability matrix M for this
1 0
.
3 7
system is M = [ B AB ] = 
Case Study Chapter 4: Control Systems
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Example 2 (cont.): Since A is 3×3 and B is 3×1, the controllability matrix M for this
system is M =  B
AB
 1 6 26 



A B =  1 −4 −14  .

 2 12 52 


2
By Theorem 4 in Section 1.4, the pair (A,B) (with x0 = 0) will be controllable if the
controllability matrix M has a pivot in each row; that is, if the number of pivots in M is n.
In Section 4.6, it is shown that the rank of a matrix is the number of pivots it possesses;
thus the following result is proven:
Result 2: A pair (A,B) (where A is an n×n matrix and B is an n×m matrix ) is controllable
if and only if the rank of the controllability matrix M is n.
 1 0  1 0 
∼
 , the rank of M is 2 and the system is
 3 7  1 0 
Example 1 (cont.): Since M = 
controllable.
1
6
26 
 1 0 2
 2 12

52 
 0 0 0


Example 2 (cont.): Since M =  1 −4 −14  ∼  0 1 4  , the rank of M is 2<3 and thus the
system is not controllable.
1 1 
Example 3: Consider the same matrix A as in Example 2, but now let B =  1 2  . Since
1 3 


6 9
 26 38 




2
AB =  −4 −7  and A B =  −14 −20  , the controllability matrix M for this system is
 12 18 
 52
76 



26 38 
1 1 6 9


M =  B AB A2 B  =  1 2 −4 −7 −14 −20 


 2 3 12 18 52 76 


Since M is row equivalent to
 1 0 0 −3 / 5 2 16 / 5 


0
0
0 
0 1 0
 0 0 1 8 / 5 4 29 / 5 


the rank of M is 3 and the system xk+1 =Axk +Buk , x0 = 0 is controllable.
Once it is determined that a pair is controllable, the exact sequence of controls which will
be needed to produce a desired target vector y may be computed.
Case Study Chapter 4: Control Systems
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 3
Example 1 (cont.): Let the target vector be y=   . This system is controllable, so it
 5
should be possible to specify u0 and u1 with y = ABu0 +Bu1. This system of equations has
3 
1 0
 1 0 3
 . Thus
 , which reduces to 
0 1 −4 7 
 3 7 5


 3
the controls u1 =3 and u0 = -4/7 will drive the system to the vector y =   .
 5
augmented matrix [ B AB y ] = [ M y ] = 
Example 4: Consider the system with matrices
1 3 1
1 0 




A =  2 6 2  and B =  2 1  .
 3 9 3
 1 −1




In Question
4 you will be asked to confirm that the controllability matrix for this system is
 1 0 8 2 80 20 


M =  2 1 16 4 160 40 
 1 −1 24 6 240 60 


and that this system is controllable. Given the output vector y
 2
=  3  ,
 0
 
the sequence of
controls which will drive the system to y are needed. To compute the controls, form the
 u 01 
 u11 
 , u1 = 
,
u
 u12 
 02 
system of equations y = A2 Bu0+ABu1+Bu2. Note that the vectors u0 = 
 u 21 
 are
 u 22 
and u 2 = 
in R2, thus the 6 entries u01, u02, u11, u12, u21, and u22 must be found to
solve this problem. The system of equations has augmented matrix
B

AB
A B y  = [M

2
 1 0 8 2 80 20 2 


y ] =  2 1 16 4 160 40 3 
 1 −1 24 6 240 60 0 


To emphasize the fact that the unknowns in this system are u01, u02, u11, u12, u21, and u22,
this augmented matrix may be expressed as a system of equations:
u21 + 0u22 + 8u11 + 2u12 + 80u01 + 20u02 = 2
2u21 + u22 + 16u11 + 4u12 + 160u01 + 40u02 = 3
u21 − u22 + 24u11 + 6u12 + 240u01 + 60u02 = 0
Since the augmented matrix reduces to
1 0 0

0 1 0

0 0 1

7 
0
2 
−1 
0

−3 
1 10 5
4
2
16 
0
0
0
0
the reduced system of equations u21=7/2, u22= -1, and u11 + (1/4) u12 + 10 u01+ (5/2) u02= 3/16 is produced. So u12, u01, and u02 are the free variables; for simplicity's sake, let them
Case Study Chapter 4: Control Systems
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 −3

7 
 0 
 −1 
 0
all equal 0. Thus it has been found that u0 =   , u1 =  16  , and u 2 =  2  is a set of
 


0
 
 2
controls which will drive the system to the point  3  .
 0
 
Questions:
Note: If the m-file case4.m accompanying this case study has been downloaded, you can
make it accessible to MATLAB by using the Path Browser feature in the MATLAB
command window. To get the data for questions 2, 4, 5, and 5, type case4 at the
MATLAB prompt and then give the appropriate question number. The m-file is not
necessary to complete the following questions.
1. Confirm that the controllability matrices for Examples 1 and 2 are as given above.
 1 1 2
 1


2. Consider the system with matrices A =  2 3 4  and B =  0  . Determine whether the
 1 2 2
 1


 
pair (A,B) is controllable.
3. If B is invertible, what can be said about the controllability of the pair (A,B)?
4. Confirm that the controllability matrix in Example 4 is as given above, and that the
system is controllable.
 1
 1 2 −1


5. Consider the system with matrices A =  0 −3 4  and B = 1 .
 1
0 0 2 
 


a. Find the controllability matrix for this system.
b. Show that system xk+1 =Axk +Buk , x0 = 0 is controllable.
 −7 
c. Find controls u0 , u1 , and u2 which will drive this system to the point  3  .
1
 
1 1 
 1 3 −2 


6. Consider the system with matrices A =  −2 −6 4  and B =  2 −1 .
1 0 
 5 15 −10 




.
a. Find the controllability matrix for this system.
b. Show that system xk+1 =Axk +Buk , x0 = 0 is controllable.
Case Study Chapter 4: Control Systems
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4
c. Find controls u0 , u1 , and u2 which will drive this system to the point  −1 .
3
 
4
d. Suppose that u0=(1,1). Can the system still be driven to  −1 ? If so, find
3
 
controls u1 and u2 to make this happen.