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- Politecnico di Milano
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- Computer Engineering
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- Model Identification and Data Analysis
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Topic-based study materials for Model Identification and Data Analysis in the Computer Engineering degree programme at Politecnico di Milano. The document covers: Model identification February 6, 2017 1 Black box approach in identification We have a set of data y(1), y(2), ... (and u(1), u(2), ...) and we want to find the best model that approximate these data. We operate like this: 1. Select a family of models 2. We get an instance of that
Topic-based study materials for Model Identification and Data Analysis in the Computer Engineering degree programme at Politecnico di Milano. The document covers: Model identification February 6, 2017 1 Black box approach in identification We have a set of data y(1), y(2), ... (and u(1), u(2), ...) and we want to find the best model that approximate these data. We operate like this: 1. Select a family of models 2. We get an instance of that
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Model identification February 6, 2017 1 Black box approach in identification We have a set of data y(1), y(2), ... (and u(1), u(2), ...) and we want to find the best model that approximate these data. We operate like this: 1. Select a family of models 2. We get an instance of that family that best fit the provided data 3. We validate the model 2 Least squares method We use this method to estimate AR(X) models, which can be stated like A(z)y(t) = B(z)u(t) + η(t) Let’s call ϑ = a1 ... ana b1 ... bnb ϕ(t) = y(t− 1) ... y(t− na) u(t− 1) ... u(t− nb) We want to minimize the following cost function J(ϑ) = 1 N N∑ t=1 εϑ(t)2 with εϑ(t) = y(t)− ˆy(t|t− 1) = y(t)− ϑT ϕ(t) = y(t)− ϕ(t)T ϑ Therefore, we have J(ϑ) = 1 N N∑ t=1 ( y(t)− ϕ(t)T ϑ )2 1 The minimization of J(ϑ) happens when ∂J (ϑ) ∂ϑ = 0 = 2 N ∑N t=1 ε(t)∂ε(t) ∂ϑ which leads to ˆϑ = [ N∑ t=1 ϕ(t)ϕ(t)T ]−1[ N∑ t=1 ϕ(t)y(t) ] 2.1 Identifiability We say that the system is identifiable when the above equation admits only one solution. This happens when the matrix R(N) = [ 1 N N∑ t=1 ϕ(t)ϕ(t)T ] = [ Ryy(N) Ryu(N) Ruy(N) Ruu(N) ] is invertible. If we want to estimate the covariance function of the system directly from the data we can do ˆγv(τ) = 1 N N−τ∑ t=1 v(t)v(t + τ) which, for N→∞ leads to the correct γv(τ). Considering Ruu(N) = ˆγu(0) ˆ γu(1) ˆ γu(2) . . . ˆγu(1) ˆ γu(0) ˆ γu(1) . . . ˆγu(2) ˆ γu(1) ... . . . ... ... ... ... →N→∞ ¯Ruu = γu(0) γu(1) γu(2) . . . γu(1) γu(0) γu(1) . . . γu(2) γu(1) ... . . . ... ... ... ... which is called Toeplitz matrix (has the same elements per diagonal), we have that the matrix ¯R = [ ¯Ryy ¯Ryu ¯Ruy ¯Ruu ] is invertible if and only if matrix ¯Ruu is invertible. An other way of saying the same thing is to say that…
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