ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 9:
y Then, clearly,
which may be re‐written as ……………… 74
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 9:
y Noting that , equation (72) can be rewritten as:
or ……………… 75
Also from equation (71) and (72) we obtain:
Also from equation (71) and (72), we obtain:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 9:
y Taking inverse Laplace transform of this last equation becomes:
……………… 76
y Combining equations (74) and (75) into one vector matrix differential C bi i i
( ) d ( ) i i diff
i l equation, we obtain equation (69). y Equation (76) can be rewritten as given by equation (70).
q
(7 )
g
y q
(7 )
y Equations (69) and (70) are said to be in the controllable canonical forms. y Figure on the next slide shows the block diagram representation of the system defined by equation (69) and (70).
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 9:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 10: Consider the following transfer function system:
……………… 77
Drive the following observable canonical form of the state space representation for this transfer function:
……………… 78
78
79
……… 79
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 10: Equation (77) can be modified into the following form:
By dividing the entire equation by sn and rearranging:
……………… 80
80
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 10: Now define state variables as follows:
……………… 81
y The equation (80) can be written as:
……………… 82
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 10: Next substitute equation (82) into equation (81) and multiplying both sides of equation by s, we obtain:
y Taking the inverse Laplace transforms of the preceding n equations and writing them in reverse order we get:
writing them in reverse order, we get:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 10:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 10: Also, the inverse Laplace transform of equation (82) Also the inverse Laplace transform of equation (82) gives:
y Rewriting the state and output equations in the standard vector – matrix form gives equations (78) and (79) form gives equations (78) and (79). y Figure on the next slide shows a block diagram representation of the system defined by equations (78) and (79).
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 10:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 11: Consider the transfer function system defined as:
……………… 83
where Drive the state space representation of the system in the following diagonal canonical form. g
g
……………… 84
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 11:
……………… 85
85
y Solution Solution –– 11: Equation (83) may be written as:
……………… 86
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 11: Define the state variables as follows:
which may be rewritten as: ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 11: The inverse Laplace transform of these equations give:
……………… 87
The n equations make up a state equation. y In terms of the state variables equation (86) can be written as: ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 11: The inverse Laplace transform of this last equation is:
……………… 88
which is the output equation.
y Equation (87) can be put in the vector – matrix equation as given by equation (84).
equation (84)
y Equation (88) can be put into the form of equation (85).
y Figure on the next slide shows a block diagram representation of the g
g
p
system defined by equation (84) and (85).
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 11:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 12: Consider the system defined by
……………… 91
where the system involves a triple pole at s = ‐pi. (Assuming that, except for the first three pi’ss being equal, the p
being equal the pi’ss are different from one another). are different from one another) Obtain the Jordan canonical form of the state‐representation for this system.
Solution –– 12: The partial fraction expression of equation (91) becomes
Solution which may be written as:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 12:
……………… 92
y Define
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 12: Notice that the following relationship exist among X1(s), (s) X2(s) and X3(s):
y Then from preceding definition of state variables and the relationships:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 12: The inverse Laplace transform of the preceding n
equations give:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 12: The output equation (92) can be rewritten as:
The inverse Laplace transform of this output equation is:
Thus, the state space representation of the system for the case when the denominator polynomial involves a triple root pi can be given as follows: denominator polynomial involves a triple root –
……………… 93
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 12:
……………… 94
y The state space representation in the form given by equation (93) and (94) is said to be in the Jordan canonical form.
y Figure on the next slide shows a block diagram representation of the system given by equation (93) and (94)
system given by equation (93) and (94).
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 12:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 13: Consider the system defined by
Obtain the response of the system to each of the following inputs:
The r components of u are impulse functions of various magnitudes.
b. The r components of u are step functions of various magnitudes.
c. The r components of u are ramp functions of various magnitudes.
a.
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 13:
y Impulse Response: Referring to equation (43), the solution to the given state equation is:
g
Substituting t0 = 0‐ into this solution:
Let us write impulse input u(t) as
where w is a vector whose components are the magnitudes of r
impulse functions applied at t = 0 impulse functions applied at t = 0. ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 13:
y The solution of the state equation when the impulse input
s g ve at t 0 s:
is given at t = 0 is: ……………… 95
y Step Response: Let us write the step input u(t) as:
where k is a vector whose components are the magnitudes of r step f
functions applied at t = 0. The solution to the step input at t = 0 is i
li d Th l i h i
i given by: ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 13:
……………… 95
y If A is nonsingular, then this last equation can be simplified to give:
……………… 96
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 13:
y Ramp Response: Let us write the ramp input u(t) as
where v is a vector whose components are magnitudes of ramp p
g
p
functions applied at t = 0 is: ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 13:
y If A is nonsingular, then this last equation can be simplified to g
give:
……………… 97
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 14: If an n x n matrix A has n distinct eigenvalues, then the has n distinct eigenvalues then the minimal polynomial of A is identical to the characteristic polynomial. Also, if the multiple eigenvalues of A are linked in a Jordan chain, the minimal pol nomial and the characteristic pol nomial are identical If minimal polynomial and the characteristic polynomial are identical. If, however, the multiple eigenvalues of A are not linked in a Jordan chain, the minimal polynomial is of lower degree than the characteristic polynomial. l
i l y Using the following matrices A and B as examples, verify the above statement about the minimal polynomial when multiple eigenvalues
p y
p
g
are involved.
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 14: First, consider the matrix A. The characteristic First consider the matrix A The characteristic polynomial is given by as:
Thus the eigenvalues
g
of A are 2, 2, and 1. It can be shown that the Jordan , ,
J
canonical form of A is:
are linked in Jordan chain as shown
and the multiple eigenvalues are linked in Jordan chain as shown.
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 14:
y T0 determine the minimal polynomial, let us first obtain It is given by as: y Notice there is no common divisor of all the elements of Hence, Thus, the minimum polynomial is identical to the characteristic polynomial, or ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 14:
y A simple calculation proves that
but
y Thus, we have shown that the minimal polynomial and the characteristic p y
polynomial of this matrix A
are the same.
y Next, consider the matrix B. The characteristic polynomial is given by as:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 14:
y A simple calculation reveals that the matrix B has three eigenvectors, and the Jordan canonical form of B is given by as:
y Thus, the multiple eigenvalues are not linked. To obtain the minimal polynomial, first compute ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 14:
y From last expression it is evident that:
y Hence,
y As a check, let us compute ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 14:
y For the given matrix B, the degree of the minimum polynomial is lower by g
,
g
p y
y
1 than that of the characteristic polynomial.
y As shown here, if the multiple eigenvalues
A h
h
if h l i l i
l
of the n x n matrix are not f h i linked in a Jordan chain, the minimum polynomial is of lower degree than the characteristic polynomial.
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 15: Discuss the state controllability of the following system:
y Solution Solution –– 15: For this system,
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 15:
Since
we see that vectors B and AB are not linearly independent and the rank of the matrix is 1 the matrix is 1. Therefore, the system is not completely controllable. y In fact the elimination of x2 from the following two simultaneous g
equations:
yields:
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 15:
or, in the form of a transfer function,
y Notice the cancellation of the factor (s + 2.5) occurs in the numerator and N i h ll i f h f
( ) i h d denominator of the transfer function.
y Because of this cancellation, this system is not completely state ,
y
p
y
controllable.
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 16: A state space representation of a system in the controllable canonical form is given by as:
………………122
122
………………123
123
The same system may be represented by the following state equation y
y
p
y
g
q
which is in the observable canonical form: ………………124
124
125
………………125
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Example Example –– 16:
y Show that the state space representation given by equation (122) and (123) gives a system that is state controllable, but not observable.
y
y Show on the other hand that the state space representation defined by equation (124) and (125) gives a system that is not completely state controllable, but is observsable. y Explain what causes the apparent difference in the controllability and observability of the same system. ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 16:
Consider the system defined by equation (122) and (123). The rank of the controllability matrix is 2. Hence, the system is completely state controllable. The rank of the observabilityy matrix
is 1. Hence the system is not observable. ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 16:
Next consider the system defined by equation (124) and (125). The rank of the controllability matrix is:
is 1. Hence the system is not completely state controllable. The rank of the observabilityy matrix
is 2. Hence the system is observable. ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 16:
The apparent difference in controllability and observability of the same system is caused by the fact that the original system have a pole‐zero cancellation in the transfer function. ll
h
f f
From the equation below for D = 0 , we have: If we use equations (122) and (123), then
ct
Ele
om
y.c
ud
st
al4
ric
Additional
Additional Solved Examples
Solved Examples
y Solution Solution –– 16:
y If the pole‐zero cancellation occurs in the transfer function, then the p
,
controllability and observability vary, depending on how the state variables are chosen.
y To be completely state controllable and state observable, the transfer function must not have any pole zero cancellations.
om
y.c
ud
st
al4
ric
ct
Ele
Assignment Problems
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 1: Consider the following transfer function system
Obtain the state space representation on this system in
p
p
y
y Controllable canonical form
y Observable canonical form.
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 2: Consider the following system
Obtain the state space representation on this system in a diagonal p
p
y
g
canonical form.
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 3: Consider the system defined by
Transform the system equations into the controllable canonical form.
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 4: Consider the system defined by
Obtain the transfer function Y (s) / U (s).
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 5: Consider the following matrix A.
Consider the following matrix A
Obtain the eigenvalues
of the matrix A. Then obtain a transformation matrix P such that ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 6: Consider the following matrix A
Compute by three methods.
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 7: Given the system equation:
Find the solution in terms of the initial conditions x1(0), x2(0) and x3(0).
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 8: Find the of the system described by
where the initial conditions are
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 9: Consider the following state equation and output equation:
Show that the system equations can be transformed into the following form by the use of a proper transformation matrix.
z2 and z3.
Obtain the output y in terms of z1, z
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 10: Consider the system defined by:
Is the system completely state controllable and completely observable?
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 11: Consider the system defined by:
Is the system completely state controllable and completely observable?
b
bl ? Is the system completely output controllable?
I h l l ll bl ?
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 12: Is the following system completely state controllable and control observable?
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 13: Consider the system defined by
Except for an obvious choice of c1 = c2 = c3 = 0, find an example of set of c1, c2, c3 that will make the system unobservable.
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 14: Consider the system
The output is given by as:
Show that the system is not completely observable.
ct
Ele
om
y.c
ud
st
al4
ric
Assignment Problems
Assignment Problems
y Problem Problem –– 14:
Show that the system is completely observable if the output is given by as: