12
Lipschitz Continuity
Calculus required continuity, and continuity was supposed to require
the innitely little, but nobody could discover what the innitely
little might be. (Russell)
12.1 Introduction
When we graph a function f (x) of a rational variabl
Homework 4 Solutions ECE6550, Fall 2014
1
The transfer is possible if and only if
x(T ) eAT x(0) R().
Moreover,
=
1
1
0
0
1
0
R() = span
.
And, as A is already on Jordan form, we know that
eT
0
eAT =
T eT
eT
T eT
eT
eAT x(0) =
.
Hence, transfer between
Homework 3
ECE6550 Linear Control Systems
Magnus Egerstedt
Due: October 10, 2014 (Oct. 17 for DL students)
1.
Show that the controllable canonical realization does in fact result in a completely controllable system.
2
Show that if A is a stability matrix,
Homework 2
ECE6550 Linear Control Systems
Magnus Egerstedt
Due: September 19, 2014 (Sept. 26 for DL students)
1.
The way a continuous time control system
x = Ax + Bu
is turned into a discrete-time, sampled system is to assume that the control signal u is
Homework 2 Solutions ECE6550, Fall 2014
1
(k+1)
xk+1
=
eA(k+1) ) Bu( )d
eA(k+1)k) xk +
k
= eA xk +
eA(r) Bdruk
0
= eA xk +
eAs Bdsuk .
0
As such, we have
A = eA , B =
eAs Bds.
0
2
a
xk+1 = xk + (Axk + Buk ) = (I + A)xk + Buk ,
i.e.,
A = I + A, B = B.
b
Ex
Homework 1 Solutions ECE6550, Fall 2014
1
a
We have
x = Ax + Bu L (sI A)X(s) = BU (s) X(s) = (sI A)1 BU (s).
Moreover
y = Cx + Du L Y (s) = CX(s) + DU (s) Y (s) = C(sI A)1 BU (s) + DU (s),
and hence
G(s) = C(sI A)1 B + D.
b
Using the change of coordinates
Homework 5
ECE6550 Linear Control Systems
Magnus Egerstedt
Due: November 24, 2014 (Dec. 1 for DL students)
1.
Let
x = Ax + Bu,
be completely controllable, with x R and u Rm . Assume someone has designed a stabilizing state
feedback controller
u = K1 x,
n
Homework 1
ECE6550 Linear Control Systems
Magnus Egerstedt
Due: September 5, 2014 (Sept. 12 for DL students)
1.
Given
x = Ax + Bu
y = Cx + Du.
a
What is the transfer function G(s) for this system, where Y (s) = G(s)U (s)?
b
Introduce a change in coordinat
GEORGIA INSTITUTE OF TECHNOLOGY
ECE 6550
Solutions Test #2
Solution Problem 1:
i) Adapt the proof given for the time-invariant case as follows: If |u(t)| < B for all t > 0,
then
Z t
|y(t)| = h(t, )u( ) d
0
Z t
|h(t, )|u( )| d
0
Z t
B
|h(t, )| d.
0
Hence
GEORGIA INSTITUTE OF TECHNOLOGY
ECE 6550
Solution Take Home Re-test.
Solution Problem TH1.1
i) In the absence of predators,
N1
.
N1 = rN1 1
K
The equilibria are N1 = 0 and N1 = K. In the first case there is nothing, and thus the
only consistent reproduct
GEORGIA INSTITUTE OF TECHNOLOGY
ECE 6550
Take home re-test (20 points to add to test 1)
Date Assigned: October 31, 2016.
Due (for section A) : Wed., November 2, 2016
Distance learning students return the exam 48 hours after receiving it. All exams must be
Solution T1.1
From the mod-4 multiplication table we have that 2 2 = 0. Hence 2 cannot have a multiplicative inverse, and therefore the set cfw_0, 1, 2, 3 with addition and multiplication mod 4
cannot be a field.
Comments T1.1:
- Most of you correctly pro
Homework 4
ECE6550 Linear Control Systems
Magnus Egerstedt
Due: October 31, 2014 (Nov. 7 for DL students)
1.
Let
1
0
x=
1
1
x+
1
0
u.
For what values of T > 0 (if any) is it possible to drive this system from
0
1
x(0) =
to x(T ) =
2
2
?
2
Let
A=
0
1
1
0
,
Homework 5 Solutions ECE6550, Fall 2014
1
Since eigenvalue placement involves turning n coecients in the characteristic polynomial into the desired
values, we need (in general) n variables (n equations and n unknowns). But, K is m n, i.e., it has
nm compo
q
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yrcfw_
krq
yrcfw_
ircfw_
yrcfw_
ircfw_
cfw_
iqwiF)jkdHel"
z u n
xw% i n y
pkXyeefFkvwesHkvF)elenSntedn7xhH"RkueuXelen l ~ 6 kd Fuv
n s v nu n x v
k|9yl euhnvy Fv Eekut egRv el8hp
xvxF9HV8yxv9ywFmhd9j%e7euhnvyFEk~||8hueptyrcfw_27yxxz 8wu xnEw
1
Section 5.3
Open and Closed Sets
Section 5.3 Open and Closed Sets
Purpose of Section To introduce metrical concepts of the real number
system, such as open and closed sets, accumulation points, interior and
boundary points of a set. These concepts will
ECE 6550
Homework 2
Due on Wednesday, September 2
1. Let f : Rk be a function, where Rn . Suppose that is compact, and f is
locally Lipschitz continuous on . Prove that f is globally Lipschitz continuous on .
2. Let fn : Rk be a sequence of continuous fun
ECE 6550
Homework 1
Due on Wednesday, August 26
1. This problem concerns open sets and closed sets. Recall that a set Rn is open if for
every x there exists > 0 such that B(, ) .
x
Let An , n = 1, 2, . . ., be a sequence of open sets in Rn .
(a) Prove tha
Final Solutions ECE6550, Fall 2011
1
The closed-loop dynamics is
x = (A BLC)x =
1L
1 L
2
x,
with characteristic polynomial
ABLC () = 2 + (1 + L ) (1 + L) + 2(L 1),
which must be compared to ( + 1)2 = 2 + 2 + 1.
Identication of the coecients yields
1+L=2 =
Final: ECE6550
Magnus Egerstedt
Georgia Institute of Technology
Dec. 13, 2011, 11:30-2:20
Closed books, closed notes, closed calculator exam.
This exam gives a maximum of 60 points. (10 points/question.) For maximum credit, be explicit
about the dierent s
Final: ECE6550
Magnus Egerstedt
Georgia Institute of Technology
Dec. 11, 2008. 11:30-2:20
Closed books, closed notes, closed calculator exam.
This exam gives a maximum of 50 points. (10 points/question.) For maximum credit, be explicit
about the dierent s
Final: ECE6550
Magnus Egerstedt
Georgia Institute of Technology
Dec. 11, 2008. 11:30-2:20
Closed books, closed notes, closed calculator exam.
This exam gives a maximum of 50 points. (10 points/question.) For maximum credit, be explicit
about the dierent s
FINAL AND SOLUTIONS 2005
1
[
Let
x=
1 0
0 1
]
[
x+
1
1
]
u.
From x(t0 ) = 0, to what x1 is it possible to drive to?
1 SOLUTION
We know that we can drive from x0 to x1 in time t1 if and only if
x1 eAt1 x0 R().
In this case x0 = 0 so all we need to check is
Solutions to Midterm: ECE6550
Magnus Egerstedt
Georgia Institute of Technology
1
Fall 2008
1A
a
=
A BA
=
1 1
0 0
.
Hence rank() = 1 < 2 and the system is not completely controllable!
b
Since eAt = L1 [(sI A)1 ]. we can compute
sI A =
s1
1
0
s+1
(sI A)1 =
Solutions to Midterm: ECE6550
Magnus Egerstedt
Georgia Institute of Technology
1
The characteristic polynomial is
A () = det(I A) = ( + 2) 1 = 0
i.e.
2 + 2 1 = 0
= 2 + 1.
A is a stability matrix i Re(A) < 0 And, since 2 + 1 > 0, , the imaginary part of =
Midterm: ECE6550
Magnus Egerstedt
Georgia Institute of Technology
Oct. 11, 2011. 3:05-4:25
Closed books, closed notes, closed calculator exam.
This exam gives a maximum of 60 points. (10 points/question.) For maximum credit, be explicit
about the dierent
Midterm: ECE6550
Magnus Egerstedt
Georgia Institute of Technology
Oct. 7, 2008. 3:05-4:25
Closed books, closed notes, closed calculator exam.
This exam gives a maximum of 50 points. (10 points/question.) For maximum credit, be explicit
about the dierent s
Chapter 2
Mathematical Structures
2.1
Semigroups, Groups, Rings, Fields
I hope that everyone will agree on the premises that linear algebra is central to linear system
theory (ample motivation for this will follow later), and that adding, multiplying or s
Chapter 4
State Space Realizations
This chapter discusses first the notion of a state space representation of a linear time invariant
system. Although this is a bit premature, as we will not precisely define the notion of state
until a later section, it p
GEORGIA INSTITUTE OF TECHNOLOGY
ECE 6550
SOLUTIONS Problem Set No. 1
Problem 1.1
See the gure for the internal description. The state equations are
u
+
R
+
y
x1
R
x2
+
-2
-1
Figure 1: Internal description
[
x 1
x 2
]
[
=
1 2
0
1
][
x1
x2
]
[
+
0
1
]
[
u,
GEORGIA INSTITUTE OF TECHNOLOGY
ECE 6550
Problem Set No. 4
Date assigned: September 19, 2016
Solution Problem 4.1
N (D2 + 1) = cfw_y|
y + y = 0 = cfw_ cos t + sin t|, R.
Obviously, since there are 2 degrees of freedom, dimN (D2 + 1) = 2.
A nice basis is c
GEORGIA INSTITUTE OF TECHNOLOGY
EE 6550
Solutions Problem Set No. 2
Solution Problem 2.1:
This can be solved either over R or C, as long as it is done consistently (same field throughout).
One knows from the theory of ODEs that the solution to a homogeneo
GEORGIA INSTITUTE OF TECHNOLOGY
EE 6550
Solutions Problem Set No. 3
Solution Problem 3.1
Let iC (t) denote the current through the capacitor (from + to )
diL
+ R(iL + iC )
u1 = L
dt
u2 = R(iL iC ) vC
But
iC =
d
d
qC = C vC
dt
dt
Thus
diL
d
+ R(iL C vC )
d