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assignment3 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.241: Dynamic Systems—Fall 2007 Homework 3 Solutions Exercise 3.2 i) We would like to minimize the 2-norm of u , i.e., u 2 . Since y n is given as 2 n y n = h i u n 1 we can rewrite this equality as i =1 y n = h 1 h 2 h n · · · u n 1 u n 2 . . . u 0 We want to find the u with the smallest 2-norm such that y ¯ = Au. where we assume that A has a full rank (i.e. h i = 0 for some i , 1 i n ). Then, the solution reduces to the familiar form: u ˆ = A ( AA ) 1 ¯ y. n h 2 i By noting that AA = , we can obtain ˆ u j as follows; i =1 u ˆ j = h j y ¯ n h 2 , i =1 i for j = 0 , 1 , · · · , n 1 . ii) a) Let’s introduce e as an error such that y n = y e . It can also be written as y y n = e . Then now the quantity we would like to minimize can be written as r ( y y n ) 2 + u 0 2 + · · · + u n 2 1 where r is a positive weighting parameter. The problem becomes to solve the following minimization problem : n u ˆ = arg min u i 2 + re 2 = arg min( u 2 2 + r e 2 2 ) , u u i =1 from which we see that r is a weight that characterizes the tradeoff between the size of the final error, ¯ y y n , and energy of the input signal, u . In order to reduce the problem into the familiar form, i.e, y Ax , let’s augment re at the bottom of u so that a new augmented vector, ˜ u is u u ˜ = , · · · re 1
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This choice of ˜ u follows from the observation that this is the ˜ u that would have u ˜ 2 2 = u 2 2 + re 2 , the quantity we aim to minimize . Now we can write y as follows u y = A . . . 1 = A ˜ u ˜ = Au + e = y n + e. r · · · re Now, ˆ u can be obtained using the augmented A , A ˜ , as u ˆ = A ˜ ( A ˜ A ˜ ) 1 y = A 1 AA + 1 y. r r By noting that n 1 1 AA + = h i 2 + , r r i =1 we can obtain ˆ u j as follows u ˆ j = n h j h y 2 1 for j = 0 , · · · , n 1 . i =1 i + r 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 u i 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 ) = ( t τ ) x ( τ )d τ where denotes convolution 0 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 )).
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