assignment3

# Ii b when r 0 it can be interpreted that the error can

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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...
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## This document was uploaded on 03/19/2014 for the course EECS 6.241J at MIT.

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