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Unformatted text preview: ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 3 – Developing Models From Data Monday March 2, 2009 1 Today’s Objectives • Linearization examples • Function identification and parameter estimation • Fitting models to scattered data 2 Linearization Examples 2.1 Classic Pendulum Example ¨ θ + g L sin( θ ) = 0 = ⇒ f ( θ, ¨ θ ) = ¨ θ + g L sin( θ ) = 0 We want to linearize about the reference angle, θ r . ¨ θ r = 0 (since θ r is a constant) Now, write out Taylor series expansion to the first derivative. Note that this is a multivariable expansion in θ and ¨ θ r : f ( θ, ¨ θ ) ≈ f ( θ r , ¨ θ r ) + ∂f ∂θ θ r , ¨ θ r ( θ- θ r ) + ∂f ∂ ¨ θ θ r , ¨ θ r ( ¨ θ- ¨ θ r ) f ( θ, ¨ θ ) ≈ f ( θ r , 0) + ∂f ∂θ θ r , ( θ- θ r ) + ∂f ∂ ¨ θ θ r , ( ¨ θ- ¨ θ r ) 1 Now we evaluate terms on right hand side: f ( θ, ¨ θ ) = ¨ θ + g L sin( θ ) f ( θ r , 0) = g L sin( θ r ) ∂f ∂θ θ r , = g L cos( θ r ) ∂f ∂ ¨ θ θ r , = 1 Thus, f ( θ, ¨ θ ) ≈ g L sin( θ r ) + g L cos( θ r )( θ- θ r ) + 1( ¨ θ- ¨ θ r ) To have a valid solution, the first term must equal zero (recall that f ( θ r , ¨ θ r ) = 0) g L sin( θ r ) = 0 → θ r = nπ , n ∈ Z (all multiples of π ). Hence f ( θ, ¨ θ ) ≈ g L cos( θ r )( θ- θ r ) + ( ¨ θ- ¨ θ r ) where cos θ r = +1 pendulum is “down” ( n even) cos θ r =- 1 pendulum is “up” ( n odd) Now we will consider stable and unstable solutions to the pendulum problem 2.2 Stable solutions When θ r = 0 , ± 2 π, ± 4 π, . . . . We assume that θ r = 0, since these are all equivalent physically.= 0, since these are all equivalent physically....
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This note was uploaded on 03/02/2010 for the course MECH 530.343 taught by Professor Sun during the Spring '08 term at Johns Hopkins.
- Spring '08