Unformatted text preview: n be obtained using the augmented A, A, as
ˆ
�
��
�
A�
1
�
˜� (AA� )−1 y =
˜˜
u=A
ˆ
AA +
y.
1
√
r
r
By noting that n 1 �2 1
AA + =
hi + ,
r
r
� i=1 we can obtain uj as follows
ˆ
hj y
2
i=1 hi + uj = �n
ˆ 1
r for j = 0, · · · , n − 1. ii) b) When r = 0, it can be interpreted that the error can be anything, but we would like to
minimize the input energy. Thus we expect that the solution will have all the ui � s to be zero. In
fact, the expression obtained in ii) a) will be zero as r → 0. On the other hand, the other situation
is an interesting case. We put a weight of ∞ to the ﬁnal state error, then the expression from ii)
a) gives the same expression as in i) as r → ∞. Exercise 3.3 This problem is similar to Example 3.4, except now we require that p(T ) = 0. We
˙
�t
can derive, from x(t) = p(t), that p(t) = x(t) ∗ tu(t) = 0 (t − τ )x(τ )dτ where ∗ denotes convolution
¨
and u(t) is the unit step, deﬁned as 1 when t > 0 and 0 when t < 0. (One way to derive this
is to take x(t) = p(t) to the Laplace domain, taking into account initial conditions, to ﬁnd the
¨
transfer function H (s) = P (s)/X (s), hence the impulse response, h(t) such that p(t) = x(t) ∗ h(t)).
�t
�T
Similarly, p t) = x(t) ∗ u(t) = 0 x(τ )dτ . So, y = p(T ) = 0 (T − τ )x(τ )dτ and 0 = p T ) =
˙(
˙(
�T
�T
x(t) ∗ u(t) = 0 x(τ )dτ . You can check that < g (t), f (t) >= 0 g (t)f (t)dτ is an inner product on
the space of continuous functions on [0, T ], denoted by C [0, T ], which we are searching for x(t). So,
˙(
we have that y = p(T ) =< (T − t), x(t) > and 0 = p T ) =< 1, x(t) >. In matrix form,
���
�
�
�
y
< T − t, x(t) >
=
=� T − t 1 , x(t) �
0
< 1, x(t) >
where � ., . � denotes the Grammian, as deﬁned in chapter 2. Now, in chapter 3, it was s...
View
Full
Document
This document was uploaded on 03/19/2014 for the course EECS 6.241J at MIT.
 Spring '11
 EmilioFrazzoli
 Computer Science, Electrical Engineering

Click to edit the document details