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1 MODEL IDENTIFICATION AND DATA ANALYSIS (INF-MAT-ELT), A.Y. 2018/2019 Prof. Sergio M. Savaresi – Prof. Sergio Bittanti - July 18th, 2019 Surname Name University ID Number Signature ................................ ............................... .....................………………….. ....................…………………….. ========================================================================================= ================== - Write the solutions (including procedures and intermediate steps) in the blank areas (use the back of the page, if needed) - Additional papers will not be considered. - Clarity, order and precision are strongly considered for the final evaluation. ========================================================================================= ================== 1. Consider the process 𝑦(𝑡) at the output of the system in the figure below: where 𝜂1(𝑡)~𝑊𝑁(1,1), 𝜂2(𝑡)~𝑊𝑁(0,1) are uncorrelated, namely Ε[(𝜂1(𝑡1)−𝐸[𝜂1(𝑡1)])(𝜂2(𝑡2)−𝐸[𝜂2(𝑡2)])]=0,∀𝑡1,𝑡2 Is y(𝑡) a stationary process? Why? Compute the mean value of 𝑦(𝑡). Find the expression of the spectrum Γ𝑦(𝜔) of 𝑦(𝑡). Draw the qualitative plot of the spectrum. The process is stationary because it is the sum of two stationary processes, i.e., 𝑦1(𝑡) and 𝑦2(𝑡), where 𝑦1(𝑡)=𝑧+5 𝑧−1 5 𝜂1(𝑡),𝑦2(𝑡)=𝑧+5 𝑧+1 5 𝜂2(𝑡) The processes 𝑦1(𝑡) and 𝑦2(𝑡) are stationary, since they are generated as filtered white noises, with stable filters (with poles that lie strictly inside the unit circle). We compute that 𝑦̅=𝐸[𝑦(𝑡)]=𝐸[𝑦1(𝑡)]+𝐸[𝑦2(𝑡)]=15 2 +0 Since 𝜂1(𝑡) and 𝜂2(𝑡) are uncorrelated, also 𝑦1(𝑡) and 𝑦2(𝑡) are uncorrelated with each other. In view of this , Γ𝑦(𝜔)=Γ𝑦1(𝜔)+ Γ𝑦2(𝜔), where Γ𝑦1(𝜔)= 26+10cos(𝜔) 26 25−2 5cos(𝜔) and Γ𝑦2(𝜔)=25, in view of the fact that 𝑦2(𝑡) is indeed a white noise process. We obtain that Γ𝑦(𝜔)= 52 26 25−2 5cos(𝜔) .…