15 07 2023 E TS Mida II

Pagina 1

1 MODEL IDENTIFICATION AND DATA ANALYSIS 2nd module (MIDA2) Academic Year 2022/2023 – 15/7/2023 – Prof. Sergio M. Savaresi, Ing. Giacomo Delcaro, Ing. Gabriele Marini Surname Name University ID number Signature ................................ ............................... .....................……………….…. .....................………. It is not allowed to consult books, notes, lecture notes etc. =============================================================================================================== 1. Given the following impulse response coefficients: 𝜔𝜔(0) = 1 𝜔𝜔(1) = 0 𝜔𝜔(2) = − 1 2 𝜔𝜔(3) = 0 𝜔𝜔(4) = 1 4 𝜔𝜔(5) = 0 𝜔𝜔(6) = − 1 8 ω(7) = 0 answer the following questions: a. Identify the system matrices using the 4SID method. What is the system order? b. Compute the transfer function of the identified system. c. Identify the system matrices from the transfer function. Did you retrieve the same matrices of question (a)? d. Provide the analytical expression of the impulse response 𝜔𝜔(𝑡𝑡) for the identified system. SOLUTION a) 4SID method: 1) STEP 1: identify the system order. 𝐻𝐻1 = [0] 𝐻𝐻2 = � 0 −1/2 −1/2 0 � 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟(𝐻𝐻2) = 2 𝐻𝐻3 = � 0 −1/2 0 −1/2 0 1/4 0 1/4 0 � 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟(𝐻𝐻3) = 2 ⇒ system order 𝑟𝑟= 2 2) S TEP 2: factorize 𝐻𝐻3 𝑅𝑅3 = � 0 −1/2 0 −1/2 0 1/4 � 𝑟𝑟𝑟𝑟𝑎𝑎 𝑂𝑂3 = � 1 0 0 1 −1/2 0 � 3) S TEP 3: estimate the system matrices 𝐻𝐻� = 𝑂𝑂3(1, ∶) = [1 0 ] 𝐺𝐺� = 𝑅𝑅3(: , 1) = � 0 −1/2� 𝐹𝐹� = 𝑂𝑂3(1: 2, :)−1𝑂𝑂3(2: 3,∶) = �1 0 0 1 � −1 � 0 1 −1/2 0 � = � 0 1 −1/2 0 � 𝐷𝐷� = 𝜔𝜔(0) = 1 b) The transfer function is defined as 𝑊𝑊(𝑧𝑧) = 𝐻𝐻� �𝑧𝑧𝑧𝑧− 𝐹𝐹� � −1 𝐺𝐺� + 𝐷𝐷� 𝑊𝑊(𝑧𝑧) = [1 0 ] � 𝑧𝑧 −1 1/2 𝑧𝑧� −1 � 0 −1/2� + 1 = 1 𝑧𝑧2 + 1/2 [𝑧𝑧1] � 0 −1/2� + 1 = 𝑊𝑊(𝑧𝑧) = −1/2 𝑧𝑧2 + 1/2 + 1 = 𝑧𝑧2 𝑧𝑧2 + 1/2 c) Using the controllable canonical form: 𝑊𝑊(𝑧𝑧) = −1/2 𝑧𝑧2 + 1/2 + 1 = 0𝑧𝑧 −1/2 𝑧𝑧2 + 0𝑧𝑧+…