Second midterm exam for Model Identification and Data Analysis in the Computer Engineering degree programme at Politecnico di Milano. The document covers: 1 MODEL IDENTIFICATION AND DATA ANALYSIS 2nd module (MIDA2) Academic Year 2022/2023 – 8/1/2024 Surname Name Matr. Number Signature ................................ ............................... .....................………. .....................………. It is not allowed to consult
Second midterm exam for Model Identification and Data Analysis in the Computer Engineering degree programme at Politecnico di Milano. The document covers: 1 MODEL IDENTIFICATION AND DATA ANALYSIS 2nd module (MIDA2) Academic Year 2022/2023 – 8/1/2024 Surname Name Matr. Number Signature ................................ ............................... .....................………. .....................………. It is not allowed to consult
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1 MODEL IDENTIFICATION AND DATA ANALYSIS 2nd module (MIDA2) Academic Year 2022/2023 – 8/1/2024 Surname Name Matr. Number Signature ................................ ............................... .....................………. .....................………. It is not allowed to consult books, notes, lecture notes etc. =============================================================================================================== 1. Given the following impulse response: 𝜔𝜔(𝑡𝑡) = � 0 𝑡𝑡< 𝑡𝑡0 1 3 (−2)2−𝑡𝑡 𝑡𝑡 ≥ 𝑡𝑡0 where 𝑡𝑡0 ≥ 0 answer the following questions: a. Is the system asymptotically stable? For which values of 𝑡𝑡0 is it strictly proper? From now on, set 𝑡𝑡0 = 0. b. Compute the transfer function associated with the system. What is the order of the system? c. Identify the system matrices from the transfer function. d. Identify the system matrices using the 4SID method. What is the order of the retrieved system? Did you retrieve the same matrices of question (c)? SOLUTION a) Since 𝜔𝜔(𝑡𝑡) → 0 for 𝑡𝑡 → ∞, the system is asymptotically stable. If 𝑡𝑡0 = 0, then 𝜔𝜔(0) = 4 3 ≠ 0 so the system is NOT strictly proper. If 𝑡𝑡0 > 0 then 𝜔𝜔(0) =0 and the system IS strictly proper. b) The transfer function is: 𝑊𝑊(𝑧𝑧) = � 𝜔𝜔(𝑡𝑡)𝑧𝑧−𝑡𝑡 ∞ 𝑡𝑡=0 = � 1 3 (−2)2−𝑡𝑡𝑧𝑧−𝑡𝑡 ∞ 𝑡𝑡=0 = (−2)2 3 � (−2)−𝑡𝑡𝑧𝑧−𝑡𝑡 ∞ 𝑡𝑡=0 = 4 3 � �− 1 2 𝑧𝑧−1� 𝑡𝑡∞ 𝑡𝑡=0 = = 4 3 1 1 + 1 2 𝑧𝑧−1 = 4 3 𝑧𝑧 𝑧𝑧+ 1/2 The system is of order 𝑛𝑛= 1. c) We can use the state-space realization in controllable canonical form: 𝑊𝑊(𝑧𝑧) = 4 3 𝑧𝑧 𝑧𝑧+ 1/2 = 4 3 𝑧𝑧+ 1/2 − 1/2 𝑧𝑧+ 1/2 = 4 3 −1/2 𝑧𝑧+ 1/2 + 4 3 𝐹𝐹= − 1 2 𝐺𝐺= 1 𝐻𝐻= − 2 3 𝐷𝐷= 4 3 d) The first 4 samples of the IR are: 𝜔𝜔(0) = 4 3 , 𝜔𝜔(1) = − 2 3 , 𝜔𝜔(2) = 1 3 , 𝜔𝜔(3) = − 1 6 4SID method: 1) STEP 1: identify the system order. 𝐻𝐻1 = [−2/3] full rank 𝐻𝐻2 = �−2/3 1/3 1/3 −1/6� 𝑟𝑟𝑟𝑟𝑛𝑛𝑟𝑟(𝐻𝐻2) = 1 ⇒…
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