ECE 553, Spring 2008
Problem Set #1 Solution
Posted: February 16th, 2008
Author: Yi Ma
Solutions:
t(1 t)
C1
= | 1/2.
(b) t(1 t)
= 1 | 1/2 | 2.
4
1. (a)
(c) The weak neighborhood is larger than the strong neighborhood.
2. (a) We derive the rst order condit
ECE 553, Spring 2008
Problem Set #2 Solution
Posted: February 28th, 2008
Solutions:
1
1. The objective functional is J (x) = 0 x2 (t) dt. From the previous homework, the associated
d2
E-L equation dt2 x = 0 admits a solution of the general form:
xo (t) =
ECE 553, Spring 2008
Problem Set #3 Solution
Posted: March 2nd, 2008
Solutions:
1. The problem is quite simple once you realize that the terminal time condition t = p(x, t) is
x(tf ) = c(tf ) and then t = p(c(t), t). Therefore, 1 = p c + p yields
c
t
1
c
ECE 553, Spring 2008
Problem Set #4 Solution
Posted: April 1, 2008
Solutions:
1. (a) Since P (t) = (t)X 1 (t), dierentiate it to obtain
P (t) = (t)X 1 + (t)X 1 (t)X (t)X 1 (t)
= Q AT Y (t)X 1 (t) Y (t)X 1 (t)A + Y X 1 (t)BR1 B T Y (t)X 1 (t)
= Q AT P P A
ECE 553, Spring 2008
Problem Set #5 Solution
Posted: April 15th, 2008
Solutions:
1. The eigenvalues of the system matrix A are 1 and 2 which are real. There can be at most
1 switch. The switching curve is a curve in 2-D space, which will be followed by an
ECE 553, Spring 2008
Problem Set #6 Solution
Posted: April 23th, 2008
Solutions:
1. Introduce as the Lagrange multiplier to the terminal state constraint x2 (tf ) + x2 (tf ) = 1.
1
2
1
The Hamiltonian is H = 2 2 x2 + 1 x2 + p1 (x2 ) p2 (x2 + x1 u) and it
ECE 553, Spring 2008
Problem Set #7 Solution
Posted: May 2nd, 2008
Solutions:
1. The optimal controller is still the one given in the solution to the Problem 6 in Homework
#5:
u (x, t) = p(t)x k (t), t 0.
The minimum expected cost is:
1
min E cfw_J = m(0