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By Andrew D. Lewis

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Of course, once we have the solution for the state variable x, it is a simple matter to determine the output y: t A(t−t0 ) y(t) = Ce CeA(t−τ ) Bu(τ ) dτ + Du(t). x0 + t0 Our aim in this book is to study the response of the output y(t) to various inputs u(t), and to devise systematic ways to make the output do things we like. Let’s look at an example. 8 Example We return to our mass-spring-damper example. Let us be concrete for simplicity, and choose m = 1, k = 4, and d = 0. The system equations are then x˙ 1 0 1 = x˙ 2 −4 0 x1 0 + u(t).

Let x1 , x2 ∈ V and define r : R → V by −x1 +tx2 . Then we have r(0) = −x1 and r(0) = x2 . Therefore there must exist a continuous u : R → R so that x2 = −Ax1 + bu(0). Since this construction can be made for any x1 , x2 ∈ V , we must have Ax1 + x2 ∈ span {b} for every x1 , x2 ∈ V . 24 this means that V = T (A, b). Thus we see for single-input systems, the state trajectories we may exactly follow are actually quite limited. Nevertheless, even though one cannot follow all state trajectories, it is possible for a system to be controllable.

You may assume that there is no friction in the system. Linearise ˙ = (x0 , 0, 0, 0) and (x, θ, x, ˙ = (x0 , π, 0, 0), the equations about the points (x, θ, x, ˙ θ) ˙ θ) where x0 is arbitrary. 4. 4 Double pendulum m1 , and the second link has length 2 and mass m2 . The links have a uniform mass density, so their centres of mass are located at their midpoint. You may assume that Exercises for Chapter 1 17 there is no friction in the system. What are the equilibrium points for the double pendulum (there are four)?

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