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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Aeronautics and Astronautics 16.060: Principles of Automatic Control Fall 2003 PROBLEM SET 4 Solutions Problem 1 Step responses A and D clearly exhibit a “long tail” due to a combination of a slow, real-valued p ole and a nearby zero, so they must correspond to transfer functions (3) and (4). To tell which is which, we need to find the sign of the residue for the real-valued p ole in the step responses for (3) and (4). In (3), there will b e a p ositive vector from the zero combined with a negative vector from the step input p ole at s = 0, so the residue will b e negative. That implies that the mode due to the slow p ole will b e subtracted from the steady-state value, and so the output will approach its steady-state value from b elow. Therefore, step response D corresponds to transfer function (3). Similarly, the residue for the real-valued p ole in the step response for (4) will b e p ositive, so the output will approach steady-state from above, so step response A must correspond to transfer function (4). The other two step responses are easy: as the zero moves towards the right, p eak time decreases and P.O. increases. Therefore B corresponds to (2), and C corresponds to (1). Problem 2 1. ⎤ ⎥⎤ ⎥ 1 −1 0 10 = |A| = −1 � A = 2 −1 −1 −2 1 ⎤ ⎥⎤ ⎥ 1 8 −3 4 −1.5 −1 |B | = 2 � B = = −1 0.5 2 −2 1 −1 |C | = 0 � C is singular, so C −1 does not exist. 2. � � � �2 1 � � � − (4) � 2 1 |A| = (1) � � 0 −1 � � 1 −1 = (1)(−2) − (4)(−3) = 10 � � � � � + (0) � 2 2 � �1 0 � � � � Since |A| �= 0, the matrix must have full rank, so A has rank = 3. � � � � � �4 � � −1 0 � � 0� � −1 4 � � |B | = (1) � � −8 18 � − (2) � 5 18 � + (9) � 5 −8 = (1)(72) − (2)(−18) + (9)(−12) =0 1 � � � � Since |B | = 0, the matrix must not have full rank. By inspection, the first two rows of B are linearly independent, so the third row must b e a linear combination of the first two rows (in fact, the third row is equal to 2 times the first row minus 3 times the second row). So B has rank = 2. A is non-singular, but B is singular b ecause |B | = 0. A would b e a valid transformation matrix for a 3-element state vector, but B would not, b ecause of its singularity. 3. (a) � � � � − 1 −3 � �=0 det(�I − A) = 0 � � �0 �−2 � � (� − 1)(� − 2) − (−3)(0) = 0 � �2 − 3� + 2 = 0 � �1 = 1, �2 = 2 � = �1 = 1 � (�1 I − A)v1 = 0 � ⎤ ⎥ ⎤⎥ 0 −3 0 � v1 = � 0 −1 0 ⎤⎥ 1 � v1 = � 0 � = �2 = 2 � (�2 I − A)v2 = 0 � ⎤ ⎥ ⎤⎥ 1 −3 0 � v2 = � 00 0 ⎤⎥ 3 � v2 = � 1 Similarly for B : det(�I − B ) = 0 � �2 − 7� + 12 = 0 � �1 = 3, �2 = 4 ⎤ ⎥ ⎤⎥ ⎤⎥ −2 2 0 1 � = �1 = 3 � � v1 = v= � � −1 1 1 0 1 ⎤ ⎥ ⎤⎥ ⎤⎥ 0 2 −1 2 � v2 = � v= � � = �2 = 4 � 1 −1 2 2 0 ⎤⎥ 1 (b) i. B � = x so the vector � has b een rotated from an angle of 63 � to an angle of 79� , and its x 5 � � length has b een scaled from 5 to 26. ⎤⎥ ⎤⎥ 1 3 ii. If � = v1 = v � , then B� = v = 3� . So the vector � has not b een rotated, but its v v 1 3 length has b een scaled by a factor of 3, which is equal to the eigenvalue associated with that eigenvector. Problem 3 1. Define x1 = y , x2 = y and x3 = y . So we have the state equations: ˙, ¨ x1 = x2 ˙ x2 = x3 ˙ x3 = −2x1 − 3x2 + 2x3 + u ˙ 2 And the output equation: y = x1 The state-space model is then: � 0 ˙ � = �0 x −2 � y= 10 ⎡ �⎡ 10 0 0 1 ⎣ � + �0⎣ u x −3 2 1 � �� 0 �+ 0 u x 2. Split up the system into p oles and zeros as follows: U (s) � X1 (s� ) 1 s2 +4s−1 3s − 2 Y (s) � From the first block, we have: x1 + 4x1 − x1 = u ¨ ˙ Let x1 b e the first state variable, and x2 = x1 b e the second state variable. Then the state equation ˙ is: ⎥ ⎤⎥ ˙= 0 1 �+ 0 u � x x 1 −4 1 Looking at the second block, we have: y = 3x˙1 − 2x1 = 3x2 − 2x1 And so the output equation is: Problem 4 � � �� y = −2 3 � + 0 u x Use the formula: G(s) = C (sI − A)−1 B + D (a) ⎤ ⎥ s 1 sI − A = 0 s+2 � ⎥ ⎦1 ⎤ −1 1 s + 2 −1 +2) = s s(s1 � (sI − A)−1 = 0 s 0 s(s + 2) s+2 �⎤ ⎥ ⎦ � ⎤ ⎥ ⎦1 −1 − s(s1 1 0 s s(s+2) 0 +2) = � G(s) = 1 1 01 0 1 s+2 s+2 3 (b) sI − A = ⎤ s −1 2 s+3 ⎥ ⎤ ⎥ ⎤ ⎥ 1 1 s+3 1 s+3 1 � (sI − A) = = (s + 1)(s + 2) −2 s s(s + 3) + 2 −2 s ⎤ ⎥⎤ ⎥⎤ ⎥ 1 10 s+3 1 0 1 � G(s) = −2 s 1 0 (s + 1)(s + 2) 1 −1 ⎤ ⎥⎤ ⎥ 1 10 1 s+3 = (s + 1)(s + 2) 1 −1 s −2 ⎦ � −1 = 1 (s+1)(s+2) s−1 − (s+1)(s+2) s+3 (s+1)(s+2) s+5 (s+1)(s+2) Problem 5 Define d1 as the p osition of m1 , d2 as the p osition of m2 , and d3 as the p osition of m3 , where d1 , d2 , and d3 are measured from the left of the diagram (the same way as y is measured). Then write down the mass-spring-damper equation for each mass: ¨ m1 d1 = F − c1 (d˙1 − d˙2 ) − k1 (d1 − d2 ) ¨ m2 d2 = −c2 (d˙2 − d˙3 ) − k2 (d2 − d3 ) + c1 (d˙1 − d˙2 ) + k1 (d1 − d2 ) ¨ m3 d3 = c2 (d˙2 − d˙3 ) + k2 (d2 − d3 ) Now define our states as: x1 = d1 , x2 = d2 , x3 = d3 , x4 = d˙1 , x5 = d˙2 , x6 = d˙3 . Also define the input as u = F . So the above equations can b e rewritten as: x˙4 = x˙5 = x˙6 = 1 (−k1 x1 + k1 x2 − c1 x4 + c1 x5 + u) m1 1 (k1 x1 − (k1 + k2 )x2 + k2 x3 + c1 x4 − (c1 + c2 )x5 + c2 x6 ) m2 1 (k2 x2 − k2 x3 + c2 x5 − c2 x6 ) m3 In addition, from the way we defined our states, we have: x1 = x4 ˙ x2 = x5 ˙ x3 = x6 ˙ And finally, the output equation is: y = x3 So now we can write the state-space model: 4 � 0 0 0 0 0 0 0 0 0 0 1 0 0 c − m11 � � � � ˙ � = � − k1 x k1 � m1 m1 � k1 k2 c1 k1 +k2 � m2 − m2 m2 m2 k k2 − m23 0 0 m3 � � �� y = 0 0 1 0 0 0 �+ 0 u x Problem 6 + R1 0 1 0 c1 m1 c1 +c2 − m2 c2 m3 0 0 1 0 c2 m2 c − m23 ⎡ �⎡ 0 ⎢ �0⎢ ⎢ �⎢ ⎢ �0⎢ ⎢ x �⎢ � + � ⎢u ⎢ ⎢ �1⎢ ⎢ �0⎣ ⎣ 0 + R2 + e_i C v i e_o L - - Define our states to b e x1 = v , the voltage across the capacitor, and x 2 = i, the current through the inductor. The input is u = ei and the output is y = eo . Applying KCL at the node b etween R1 and R2 gives: dv v − ei +C +i=0 R1 dt � dv 1 1 1 =− ei v− i+ dt R1 C C R 1 C Applying KVL around the loop containing C , R 2 , and L, gives: di =0 dt Finally, the ouptut voltage eo is given by: v − R2 i − L � di 1 R2 = v− i dt L L eo = v − R2 i So the state-space model is: ⎤1⎥ 1⎥ −C � + R1 C u x 1 0 − R2 L L � � �� x y = 1 −R2 � + 0 u ˙ �= x ⎤ 1 − R1 C 5 ...
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This note was uploaded on 11/06/2011 for the course AERO 100 taught by Professor Willcox during the Fall '03 term at MIT.

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