Solve a parameter estimation problem for a single-input single-output system using the non-recursive least squares method by defining parameter and data vectors | Step-by-Step Solution

Hey there! This is a great system identification problem that connects differential equations with practical parameter estimation. Let's break it down together! 🎯

What We're Solving:

We have a system with transfer function G(s) = 1/(c₁s² + c₂s + c₃), and we want to estimate the unknown parameters c₁, c₂, c₃ using experimental data. We need to set up the least squares framework by defining the parameter vector and data structures.

The Approach:

The key insight here is that we're transforming our continuous system into a form where we can use discrete data points to estimate parameters. We do this by:

• 1. Rearranging our system equation to separate known data from unknown parameters
• 2. Setting up vectors and matrices so we can use linear algebra to solve for the parameters

Think of it like solving y = mx + b for slope m and intercept b, but with more parameters and a more complex relationship!

Step-by-Step Solution:

Part (a): Defining θ̂ and ψ(k)

Let's start with our transfer function: G(s) = Y(s)/U(s) = 1/(c₁s² + c₂s + c₃)

Cross-multiplying: Y(s)(c₁s² + c₂s + c₃) = U(s)

This gives us: c₁s²Y(s) + c₂sY(s) + c₃Y(s) = U(s)

Converting back to time domain (where s becomes d/dt): c₁ÿ(t) + c₂ẏ(t) + c₃y(t) = u(t)

Rearranging to isolate the output: y(t) = u(t) - c₁ÿ(t) - c₂ẏ(t)

Now, at discrete time k, this becomes: y(k) = u(k) - c₁ÿ(k) - c₂ẏ(k)

We can rewrite this as: y(k) = [u(k) -ÿ(k) -ẏ(k)] [1 c₁ c₂]ᵀ

Wait! Let me reconsider this more carefully. Since we want to estimate ĉ₁, ĉ₂, ĉ₃, let's rearrange:

From c₁s²Y(s) + c₂sY(s) + c₃Y(s) = U(s), we get: y(k) = -c₁ÿ(k) - c₂ẏ(k) - c₃y(k-1) + u(k)

Actually, let me approach this correctly: y(k) = u(k) - c₁ÿ(k) - c₂ẏ(k) - c₃∫y(k)

For discrete implementation: y(k) ≈ [-ÿ(k) -ẏ(k) -y(k)] [ĉ₁ ĉ₂ ĉ₃]ᵀ + u(k)

Therefore:

• θ̂ = [ĉ₁ ĉ₂ ĉ₃]ᵀ (the parameter vector we want to estimate)
• ψ(k) = [-ÿ(k) -ẏ(k) -y(k)]ᵀ (the data vector at time k)

Part (b): Data Matrix Ψ for N = 2 (three data points)

For three data points (k = 0, 1, 2), we stack all the data vectors ψ(k) as rows:

Ψ = [ψ(0)ᵀ] = [-ÿ(0) -ẏ(0) -y(0)] [ψ(1)ᵀ] [-ÿ(1) -ẏ(1) -y(1)] [ψ(2)ᵀ] [-ÿ(2) -ẏ(2) -y(2)]

The Answer:

Part (a):

• θ̂ = [ĉ₁ ĉ₂ ĉ₃]ᵀ
• ψ(k) = [-ÿ(k) -ẏ(k) -y(k)]ᵀ

Part (b): ``` Ψ = [-ÿ(0) -ẏ(0) -y(0)] [-ÿ(1) -ẏ(1) -y(1)] [-ÿ(2) -ẏ(2) -y(2)] ```

Memory Tip:

Remember "Data in rows, parameters in columns!" The data matrix Ψ has one row for each time point and one column for each parameter you're estimating. The negative signs come from moving terms to the right side of the equation. Think of it as: "What we measured" = "Data matrix" × "What we want to find"! 🧠✨

Great job working through this system identification problem! The beauty of least squares is that once you set up these matrices correctly, linear algebra does the heavy lifting to find your parameter estimates.